IB Applications and Interpretation SL · Basic algebra and arithmetic · End of topic

IB exam questions for topic 1

114 exercises with solutions · By Álvaro López

Past IB exam questions grouped by type and, within each type, from easiest to hardest. Try each one before opening the solution.

Level

Paper

Solutions

Sets and Venn diagrams

Exercise 1 BasicIB · P1 TZ0 November 2005

Write down the expression for the shaded area in each of the following Venn diagrams: Four Venn diagrams with U and a hatched region. (a) A and B overlapping, the intersection A ∩ B hatched. (b) A and B, the outside of the union hatched. (c) A and B, the part of B outside A hatched. (d) A, B and C overlapping, the regions A ∩ C and B ∩ C hatched

View solution

We identify each hatched region.

(a) Only the area common to the two circles: $A \cap B$ .
(b) Everything outside the two circles: the complement of the union, $(A \cup B)^{'}$ .
(c) The part of $B$ that is not in $A$ : $B \cap A^{'}$ .
(d) $A \cap C$ and $B \cap C$ are hatched, including the area where all three overlap. It is the union of the two intersections: $(A \cap C) \cup (B \cap C)$ , which is equivalent to $(A \cup B) \cap C$ .

Answer: (a) $A \cap B$ ; (b) $(A \cup B)^{'} = A^{'} \cap B^{'}$ ; (c) $A^{'} \cap B$ ; (d) $(A \cap C) \cup (B \cap C)$ .

Exercise 2 BasicIB · P1 TZ0 November 2003

The following Venn diagram shows the sets $U$ , $A$ , $B$ and $C$ : Venn diagram with U, overlapping circles A and B, and a small circle C inside B on its right-hand side, disjoint from A; nothing shaded State whether the following statements are true or false according to the information given in the Venn diagram.

a) $A \cap C = \emptyset$

b) $C \cup B = C$

c) $C \subset A \cup B$

d) $A \subset C^{'}$

Show answer

Answer: (a) True; (b) False; (c) True; (d) True.

Show working and answer

In the diagram $C$ is inside $B$ and does not touch $A$ .

(a) $A \cap C = \emptyset$ : $A$ and $C$ do not overlap, they have no common elements. True.
(b) $C \cup B = C$ : since $C \subset B$ , the union is the larger set, $B$ , and $B \neq = C$ (there are elements of $B$ outside $C$ ). False.
(c) $C \subset A \cup B$ : all of $C$ is in $B$ , and $B$ is in $A \cup B$ . True.
(d) $A \subset C^{'}$ : $C^{'}$ is everything not in $C$ ; since $A$ has nothing in common with $C$ , all of $A$ is in $C^{'}$ . True.

Answer: (a) True; (b) False; (c) True; (d) True.

Exercise 3 BasicIB · P1 TZ0 November 2008

The following Venn diagram shows the number of students at a school who study science subjects: Physics (P), Chemistry (C) and Biology (B). Venn diagram with P, B and C inside U: only P 7; P and B without C 4; only B 8; P and C without B 5; all three 2; B and C without P 3; only C 9; outside the three circles 6

a) Write down the number of students who study only Chemistry.

b) Write down the number of students who study exactly two science subjects.

c) Write down the number of students who do not study Physics.

d) Find $n [(P \cup B) \cap C]$ .

Show answer

Answer: a) $9$ . b) $12$ . c) $26$ . d) $10$ .

Show working and answer

a) Those who study only Chemistry is the region of $C$ that does not touch $P$ or $B$ : $9$ .
b) Exactly two subjects are the three regions shared by two sets but not the third: $4 + 5 + 3 = 12$ (the 2 in the centre, who study all three, are not counted).
c) Those who do not study Physics are everyone outside $P$ , that is, only $B$ ( $8$ ), $B \cap C$ ( $3$ ), only $C$ ( $9$ ) and those outside all circles ( $6$ ): $8 + 3 + 9 + 6 = 26$ .
d) $(P \cup B) \cap C$ : the Chemistry students who also study Physics or Biology. These are the regions of $C$ that touch $P$ or $B$ : $P \cap C$ without $B$ ( $5$ ), all three ( $2$ ) and $B \cap C$ without $P$ ( $3$ ): $5 + 2 + 3 = 10$ .

Answer: a) $9$ . b) $12$ . c) $26$ . d) $10$ .

Exercise 4 BasicIB · P1 TZ2 May 2012

The school offers students music lessons in Piano (P), Violin (V) and Flute (F). The Venn diagram shows the number of students who are learning to play each instrument. Venn diagram with P, V and F inside U: only P 21; P and V without F 8; only V 56; P and F without V 5; all three 4; V and F without P 10; only F 37; outside the three circles 4

a) Write down the total number of students in the school.

b) Write down the number of students who

(i) are learning to play only the violin;

(ii) are learning to play the piano, the flute or both;

(iii) are not learning to play the flute.

c) Explain, in words, the meaning of the part of the diagram that represents the set $P \cap F^{'}$ .

Show answer

Answer: a) $145$ . b) (i) $56$ ; (ii) $85$ ; (iii) $89$ . c) Students learning piano but not flute (29 students).

Show working and answer

a) The total is the sum of all regions, including those outside: $21 + 8 + 56 + 5 + 4 + 10 + 37 + 4 = 145$ .
b) (i) Violin only: $56$ . (ii) Piano, flute or both is $P \cup F$ : all regions inside $P$ or $F$ , that is $21 + 8 + 5 + 4 + 10 + 37 = 85$ . (iii) Not flute is everything except $F$ : only $P$ ( $21$ ), $P \cap V$ ( $8$ ), only $V$ ( $56$ ) and outside ( $4$ ): $21 + 8 + 56 + 4 = 89$ .
c) $P \cap F^{'}$ are the students who are in $P$ but not in $F$ : those learning piano but not flute (piano only, or piano and violin). There are $21 + 8 = 29$ such students.

Answer: a) $145$ . b) (i) $56$ ; (ii) $85$ ; (iii) $89$ . c) Students learning piano but not flute (29 students).

Exercise 5 BasicIB · P2 TZ0 November 2006

Given a universal set $U = {cars}$ , $S = {sports cars}$ , $G = {green sports cars}$ .

a) Draw a Venn diagram to illustrate this information.

b) Shade the set $S \cap G^{'}$ on your diagram.

c) Write in words the meaning of $S \cap G^{'}$ .

View solution

a) 1. Every green sports car is a sports car, so $G$ is completely contained in $S$ ( $G \subset S$ ). 2. We draw a rectangle $U$ , inside it a circle $S$ and, inside $S$ , a smaller circle $G$ .

b) 1. $G^{'}$ are the cars that are not green sports cars (everything outside $G$ ). 2. $S \cap G^{'}$ is what is at the same time in $S$ and outside $G$ : we shade the part of $S$ that lies outside the circle $G$ (the "ring" between the two circles).

c) These are the cars that are sports cars but are not green, that is, sports cars of any colour other than green.

Venn diagram with U, a circle S and inside it a circle G (G contained in S); the part of S outside G is shaded

Answer: a) b) diagram above c) Sports cars that are not green

Exercise 6 BasicIB · P1 TZ0 November 2001

A committee $U$ has three subcommittees: research $R$ , finance $F$ and purchasing $P$ . No member belongs to both the finance and purchasing subcommittees. Some members belong to both the research and purchasing committees. All members of the finance subcommittee also belong to the research subcommittee. Draw a Venn diagram showing the relationship between the sets $U$ , $R$ , $F$ and $P$ .

View solution

Each condition is interpreted as a relationship between circles:

"No member belongs to finance and purchasing": $F \cap P = \emptyset$ , so the circles $F$ and $P$ do not touch.
"Some belong to research and purchasing": $R \cap P \neq = \emptyset$ , so $R$ and $P$ partly overlap.
"All members of finance are also in research": $F \subset R$ , so $F$ is drawn completely inside $R$ .
The three circles go inside the rectangle $U$ . As $F$ is inside $R$ and $R$ only touches $P$ in its area outside $F$ , the circle $F$ lies inside $R$ without touching $P$ .

Venn diagram with U, two large circles P and R that partly overlap, and a small circle F inside R, not touching P

Answer: diagram above

Exercise 7 BasicIB · P2 TZ0 November 2001

100 students were asked which TV channel (MTV, CNN or BBC) they had watched the previous evening. The results are shown in the Venn diagram below. Venn diagram of MTV, CNN and BBC inside U: only MTV 35, MTV and CNN without BBC 23, only CNN 19, MTV and BBC without CNN 6, all three 5, CNN and BBC without MTV 2, only BBC 3, outside 7 From the information in the Venn diagram, write down the number of students who watched

a) both MTV and BBC;

b) MTV or BBC;

c) CNN and BBC but not MTV;

d) MTV or CNN but not BBC.

Show answer

Answer: a) Watched both MTV and BBC $= 11$ students b) Watched MTV or BBC $= 74$ students c) Watched CNN and BBC but not MTV $= 2$ students d) Watched MTV or CNN but not BBC $= 77$ students

Show working and answer

The regions are read directly from the diagram.

a) MTV and BBC: $MTV \cap BBC$ without CNN ( $6$ ) plus all three ( $5$ ): $6 + 5 = 11$ .

b) MTV or BBC (the union): all the regions of both circles: $35 + 23 + 6 + 5 + 2 + 3 = 74$ . (This equals $100 - 19 - 7 = 74$ , removing only CNN and the outside.)

c) CNN and BBC but not MTV: the region for just those two: $2$ .

d) MTV or CNN, but not BBC: the regions of those two circles outside BBC: $35 + 23 + 19 = 77$ .

Answer: a) Watched both MTV and BBC $= 11$ students b) Watched MTV or BBC $= 74$ students c) Watched CNN and BBC but not MTV $= 2$ students d) Watched MTV or CNN but not BBC $= 77$ students

Exercise 8 MediumIB · P1 TZ2 May 2018

Consider the following Venn diagrams. Three Venn diagrams with universal set U. Diagram 1: circles A and B overlapping, everything in U shaded except circle A. Diagram 2: circle D inside circle C, the part of C outside D shaded. Diagram 3: circles E, F and G overlapping, the intersection of E and F and the whole of circle G shaded

a) Write down an expression, using set notation, that represents the shaded region of

(i) Diagram 1;

(ii) Diagram 2;

(iii) Diagram 3.

b) On each of the following Venn diagrams, shade the region that represents the set

(i) $(H \cup I)^{'}$ ; Venn diagram with universal set U and two overlapping circles, H on the left and I on the right, nothing shaded

(ii) $J \cap K$ . Venn diagram with universal set U, a large circle J and a small circle K completely inside J, nothing shaded

View solution

a) For each diagram we identify which region is shaded and write it using set operations (complement $^{'}$ , union $\cup$ , intersection $\cap$ ).

Diagram 1: everything in the universal set except circle $A$ is shaded, that is, the elements not in $A$ : the complement $A^{'}$ .
Diagram 2: $D$ lies inside $C$ and the part of $C$ outside $D$ is shaded: elements that are in $C$ but not in $D$ , that is, $C \cap D^{'}$ .
Diagram 3: two regions are shaded at the same time, the intersection $E \cap F$ and the whole circle $G$ . "One or the other" means a union: $(E \cap F) \cup G$ .

b) We work out which region has to be coloured.

$(H \cup I)^{'}$ : first, $H \cup I$ is the region covered by the two circles together. Its complement is everything outside both circles, inside the rectangle $U$ . We shade that outside region and no circle.
$J \cap K$ : this is the region common to $J$ and $K$ . Since $K$ lies completely inside $J$ , the common part is the whole of the small circle $K$ . We shade circle $K$ .

Venn diagram with U, overlapping circles H and I; the whole exterior of the union of H and I is shaded, the two circles left blank Venn diagram with U, large circle J and small circle K inside J; the interior of K is shaded (J ∩ K = K)

Answer: a) (i) $A^{'}$ ; (ii) $C \cap D^{'}$ ; (iii) $G \cup (E \cap F)$ . b) (i) the exterior of both circles; (ii) the circle $K$ .

Exercise 9 MediumIB · P1 TZ0 November 2016

The following Venn diagram shows the sets $A$ , $B$ , $C$ and $U$ . In addition, $x$ is an element of $U$ . Venn diagram with U, overlapping circles A and B with the element x in the intersection, and an empty separate oval C below; nothing shaded

a) State whether each of the statements below is True or False.

Statement	True or False
$x \in C$
$x \subset B$
$A \cup B \neq = \emptyset$
$A \cap B \subset C$
$A \cap C = \emptyset$

b) On the Venn diagram, shade the region $A \cap (B \cup C)^{'}$ .

View solution

a) We need to distinguish between belonging to a set ( $\in$ , for an element), being contained in a set ( $\subset$ , for a set) and the empty set.

$x \in C$ : $x$ is in the intersection of $A$ and $B$ , and $C$ is a separate oval. False.
$x \subset B$ : $x$ is an element, not a set, so the correct symbol would be $x \in B$ . Written with $\subset$ it is False.
$A \cup B \neq = \emptyset$ : $A \cup B$ contains at least $x$ , so it is not empty. True.
$A \cap B \subset C$ : $A \cap B$ contains $x$ , which is not in $C$ . False.
$A \cap C = \emptyset$ : $C$ is separate from $A$ , so they have nothing in common. True.

Statement	True or False
$x \in C$	False
$x \subset B$	False
$A \cup B \neq = \emptyset$	True
$A \cap B \subset C$	False
$A \cap C = \emptyset$	True

b) We build the region in parts.

$B \cup C$ is everything that is in $B$ or in $C$ .
$(B \cup C)^{'}$ is everything that is in neither of them.
We want $A \cap (B \cup C)^{'}$ : the part of $A$ that is in neither $B$ nor $C$ . Since $C$ does not touch $A$ , we only need to remove from $A$ the part it shares with $B$ . We shade the part of $A$ that lies outside $B$ (the "crescent" of $A$ ).

Venn diagram with U, overlapping circles A and B with x in the intersection and oval C below; the part of A outside B is shaded

Answer: a) False, False, True, False, True. b) The part of $A$ that does not belong to $B$ .

Exercise 10 MediumIB · P1 TZ1 May 2018

A group of 60 sports enthusiasts visited the PyeongChang 2018 Winter Olympic Games to watch a variety of sporting events. The most popular sports were snowboarding ( $S$ ), figure skating ( $F$ ) and ice hockey ( $H$ ). For this group of 60 people: 4 did not watch any of the most popular sports, $x$ watched all three of the most popular sports, 9 watched only snowboarding, 11 watched only figure skating, 15 watched only ice hockey, 7 watched snowboarding and figure skating, 13 watched figure skating and ice hockey, 11 watched snowboarding and ice hockey.

a) Complete the Venn diagram using the information given. Venn diagram with three sets S, F and H inside U; the letter x is in the central region, the intersection of S, F and H; the other regions are empty

b) Find the value of $x$ .

c) Write down the value of $n ((F \cup H) \cap S^{'})$ .

Show answer

Answer: a) centre $5$ ; $S \cap F$ only $2$ ; $F \cap H$ only $8$ ; $S \cap H$ only $6$ ; only $S$ $9$ ; only $F$ $11$ ; only $H$ $15$ ; outside $4$ . b) $x = 5$ . c) $34$ .

Show working and answer

a) We fill in from the centre outwards, subtracting what has already been counted.

Centre ( $S \cap F \cap H$ ): $x$ .
Only $S \cap F$ (without $H$ ): $7 - x$ , because the 7 people in $S \cap F$ include those in the centre.
Only $F \cap H$ : $13 - x$ .
Only $S \cap H$ : $11 - x$ .
Only $S$ : 9. Only $F$ : 11. Only $H$ : 15.
Outside all three: 4.

b) The total number of people is 60, so the sum of all the regions must be 60:

$9 + 11 + 15 + (7 - x) + (13 - x) + (11 - x) + x + 4 = 60$

$70 - 2 x = 60 \Rightarrow x = 5.$

With $x = 5$ the regions are: $S \cap F$ only, 2; $F \cap H$ only, 8; $S \cap H$ only, 6; centre, 5.

c) $(F \cup H) \cap S^{'}$ are the people who watched figure skating or ice hockey and did not watch snowboarding: only $F$ (11), only $H$ (15) and $F \cap H$ without $S$ (8):

$n = 11 + 15 + 8 = 34.$

Venn diagram with S, F and H inside U: only S 9; only F 11; only H 15; S and F without H 7−x; F and H without S 13−x; S and H without F 11−x; all three x; outside the three circles 4

Answer: a) centre $5$ ; $S \cap F$ only $2$ ; $F \cap H$ only $8$ ; $S \cap H$ only $6$ ; only $S$ $9$ ; only $F$ $11$ ; only $H$ $15$ ; outside $4$ . b) $x = 5$ . c) $34$ .

Exercise 11 MediumIB · P1 TZ1 May 2017

Dune Canyon High School organises its school year into three terms: autumn / fall ( $F$ ), winter ( $W$ ) and spring ( $S$ ). The school offers a variety of sports activities during and outside the school year. The sports activities offered by the school are summarised in the following Venn diagram. Venn diagram with F, W and S inside U and sports in each region. Only F: volleyball, golf. F and W without S: football. Only W: ice hockey, skiing, ice skating. All three: gymnastics, swimming. F and S without W: cycling. W and S without F: mountaineering, basketball. Only S: softball, badminton, athletics, tennis. Outside: archery, baseball, kayaking, surfing

a) Write down the number of sports activities that the school offers during its school year.

b) Determine whether the school offers mountaineering in the autumn / fall term.

c) Write down

(i) the elements of the set $F \cap W^{'}$ ;

(ii) $n (W \cap S)$ .

d) Write down, in terms of $F$ , $W$ and $S$ , an expression for the set containing only archery, baseball, kayaking and surfing.

Show answer

Answer: a) $15$ . b) No. c) (i) ${$ volleyball, golf, cycling $}$ ; (ii) $4$ . d) $(F \cup W \cup S)^{'}$ .

Show working and answer

a) The activities offered during the school year are those inside $F$ , $W$ or $S$ (the 4 outside are not offered in any term). We add the regions: $2 + 1 + 3 + 2 + 1 + 2 + 4 = 15$ .
b) Mountaineering is in $W \cap S$ , outside $F$ . Therefore it is not offered in autumn: No.
c) (i) $F \cap W^{'}$ are the autumn activities that are not in winter: volleyball, golf and cycling. (ii) $W \cap S$ contains gymnastics, swimming, mountaineering and basketball, so $n (W \cap S) = 4$ .
d) Archery, baseball, kayaking and surfing are not in any term: they are what is left outside the union of the three, $(F \cup W \cup S)^{'}$ .

Answer: a) $15$ . b) No. c) (i) ${$ volleyball, golf, cycling $}$ ; (ii) $4$ . d) $(F \cup W \cup S)^{'}$ .

Exercise 12 MediumIB · P1 TZ2 May 2017

At a summer camp, all the children have to practise at least one sport. They can choose between football ( $F$ ) and basketball ( $B$ ). There are 15 children who practise both sports. The number of children who play only football is double the number of children who play only basketball. Let $x$ be the number of children who play only football.

a) Write an expression, in terms of $x$ , to represent the number of children who play only basketball.

b) Complete the Venn diagram using the information above. Venn diagram with two sets F and B inside U, with no numbers written There are 120 children at this summer camp.

c) Find the number of children who play only football.

d) Write down the value of $n (F)$ .

Show answer

Answer: a) $\frac{x}{2}$ . b) $F$ only $x$ , intersection $15$ , $B$ only $\frac{x}{2}$ , outside $0$ . c) $70$ . d) $85$ .

Show working and answer

a) Only football is double only basketball, and only football is $x$ . So only basketball is $\frac{x}{2}$ .
b) We fill in the diagram: $F$ only: $x$ ; $F \cap B$ : $15$ ; $B$ only: $\frac{x}{2}$ . Outside both there are $0$ children, because everyone practises at least one sport.
c) There are 120 children in total and all of them are in some region:

$x + 15 + \frac{x}{2} = 120 \Rightarrow \frac{3 x}{2} = 105 \Rightarrow x = 70.$

d) $F$ includes those who play only football and those who play both: $n (F) = 70 + 15 = 85$ .

Venn diagram with F and B inside U: F only is x; F and B is 15; B only is x/2; outside both circles 0

Answer: a) $\frac{x}{2}$ . b) $F$ only $x$ , intersection $15$ , $B$ only $\frac{x}{2}$ , outside $0$ . c) $70$ . d) $85$ .

Exercise 13 MediumIB · P1 TZ0 November 2014

Tuti has the following polygons to classify: rectangle (R), rhombus (H), isosceles triangle (I), regular pentagon (P) and scalene triangle (T). In the Venn diagram below, set $A$ consists of the polygons that have at least one pair of parallel sides, and set $B$ consists of the polygons that have at least one pair of equal sides. Venn diagram with A and B inside U; the letter R is in the intersection of A and B; all other regions are empty

a) Complete the Venn diagram by placing the letter corresponding to each polygon in the appropriate region. For example, R has already been placed and represents the rectangle.

b) State which of Tuti's polygons are elements of

(i) $A \cap B$ ;

(ii) $(A \cup B)^{'}$ .

View solution

We examine each polygon with the two properties: at least one pair of parallel sides ( $A$ ) and at least one pair of equal sides ( $B$ ).

Rectangle (R): opposite sides are parallel and equal. It is in $A \cap B$ (already placed).
Rhombus (H): opposite sides are parallel and all four sides are equal. It is in $A \cap B$ .
Isosceles triangle (I): no pair of parallel sides (a triangle has no parallel sides), but it has two equal sides. It is in $B$ outside $A$ .
Regular pentagon (P): no parallel sides, but all five sides are equal. It is in $B$ outside $A$ .
Scalene triangle (T): no parallel sides and no equal sides. It is outside $A$ and outside $B$ .

b) $A \cap B$ is the common area: R and H. $(A \cup B)^{'}$ is what lies outside both circles: T.

Venn diagram with A and B inside U: R and H in the intersection of A and B; I and P in the part of B not in A; T outside both circles

Answer: a) $A \cap B$ : R, H; only $B$ : I, P; only $A$ : none; outside: T. b) (i) R, H; (ii) T.

Exercise 14 MediumIB · P1 TZ0 November 2011

A group of 33 people were asked about the passports they hold. 21 have an Australian passport, 15 have a British passport and 3 have neither.

a) Find the number of people who have both an Australian and a British passport. In the Venn diagram below, set $A$ represents the people in the group with Australian passports and $B$ those with British passports. Venn diagram with U and two overlapping circles A and B: p in the part of A only, q in A ∩ B, r in the part of B only, and 3 outside both circles

b) Write down the value of

(i) $q$ ;

(ii) $p$ and $r$ .

c) Find $n (A \cup B^{'})$ .

Show answer

Answer: a) $6$ . b) (i) $q = 6$ ; (ii) $p = 15$ , $r = 9$ . c) $24$ .

Show working and answer

a) There are $33 - 3 = 30$ people with at least one passport. If we add the numbers for each passport, the people who have both are counted twice: $21 + 15 = 36$ . The excess is the number of people who have both: $36 - 30 = 6$ .
b) (i) $q = n (A \cap B) = 6$ . (ii) Australian only: $p = 21 - 6 = 15$ . British only: $r = 15 - 6 = 9$ . Check: $15 + 6 + 9 + 3 = 33$ .
c) $A \cup B^{'}$ are the people who have an Australian passport or who do not have a British passport: everyone except those who have only the British passport. $n (A \cup B^{'}) = 33 - r = 33 - 9 = 24$ (equivalently $p + q + 3 = 15 + 6 + 3$ ).

Answer: a) $6$ . b) (i) $q = 6$ ; (ii) $p = 15$ , $r = 9$ . c) $24$ .

Exercise 15 MediumIB · P1 TZ1 May 2010

A gym has 60 members. 35 of the members attend the club's aerobics class ( $A$ ) and 28 members attend the club's yoga class ( $Y$ ). 17 members attend both classes. A Venn diagram is used to illustrate this situation. Venn diagram with U and two overlapping circles A and Y: p in the part of A only, q in A ∩ Y and 11 in the part of Y only

a) Write down the value of $q$ .

b) Find the value of $p$ .

c) Calculate the number of gym members who attend neither the aerobics class ( $A$ ) nor the yoga class ( $Y$ ).

d) Shade, on your Venn diagram, $A^{'} \cap Y$ .

Show answer

Answer: a) $q = 17$ members b) $p = 18$ members c) 14 members d) diagram above

Show working and answer

a) 1. $q$ is the number of members in the intersection $A \cap Y$ , that is, those who attend both classes. 2. From the question: $q = 17$ .

b) 1. The 35 aerobics members are split between "A only" ( $p$ ) and "A and Y" ( $q$ ): $p + q = 35$ . 2. $p = 35 - 17 = 18$ .

c) 1. Members inside some circle: $p + q + 11 = 18 + 17 + 11 = 46$ . (Check: yoga $= 17 + 11 = 28$ , correct.) 2. Those who attend neither class are the ones outside the circles: $60 - 46 = 14$ .

d) $A^{'} \cap Y$ are those who are in $Y$ and not in $A$ : we shade the part of $Y$ that lies outside $A$ (the region with the $11$ ).

Venn diagram with U and two overlapping circles A and Y; only the part of Y that does not belong to A is shaded

Answer: a) $q = 17$ members b) $p = 18$ members c) 14 members d) diagram above

Exercise 16 MediumIB · P2 TZ1 May 2009

(ii) A survey was carried out in a class of 12-year-olds. The students were asked which pop groups they like from the Rockers (R), the Salseros (S) and the Bluers (B). The results are shown in the following diagram. Venn diagram with U and three circles R, S and B. Only R 2; R and S without B 2x; only S 0; R and B without S x; R, S and B 3; S and B without R 5; only B 4; outside the three circles 7

a) Write down $n (R \cap S \cap B)$ .

b) Find $n (R^{'})$ .

c) Describe which groups the students in the set $S \cap B$ like.

d) Use set notation to describe the group of students who like the Rockers and the Bluers but do not like the Salseros. There are 33 students in the class.

e) (i) Find $x$ .

(ii) Find the number of students who like the Rockers.

Show answer

Answer: a) $n (R \cap S \cap B) = 3$ students b) $n (R^{'}) = 16$ students c) The Bluers and the Salseros d) $R \cap B \cap S^{'}$ e) (i) $x = 4$ students (ii) $n (R) = 17$ students

Show working and answer

a) $R \cap S \cap B$ is the region common to the three circles, which contains $3$ students. Therefore $n (R \cap S \cap B) = 3$ .

b) 1. $R^{'}$ are the students who are not in the circle $R$ . 2. They are the regions: only $S$ ( $0$ ), $S$ and $B$ without $R$ ( $5$ ), only $B$ ( $4$ ) and outside the circles ( $7$ ). 3. $n (R^{'}) = 0 + 5 + 4 + 7 = 16$ .

c) $S \cap B$ are the students who are in $S$ and $B$ at the same time: they like the Salseros and the Bluers (some of them, in addition, like the Rockers).

d) Rockers and Bluers: $R \cap B$ ; but not Salseros: $S^{'}$ . Therefore $R \cap B \cap S^{'}$ .

e) (i) 1. We add all the regions and equate them to the total of $33$ : $2 + 2 x + 0 + x + 3 + 5 + 4 + 7 = 33 \Rightarrow 21 + 3 x = 33.$ 2. $3 x = 12$ , therefore $x = 4$ .

(ii) 1. The four regions of the circle $R$ belong to the Rockers: $2 + 2 x + x + 3 = 5 + 3 x$ . 2. With $x = 4$ : $5 + 12 = 17$ .

Answer: a) $n (R \cap S \cap B) = 3$ students b) $n (R^{'}) = 16$ students c) The Bluers and the Salseros d) $R \cap B \cap S^{'}$ e) (i) $x = 4$ students (ii) $n (R) = 17$ students

Exercise 17 MediumIB · P1 TZ0 November 2002

A survey was carried out on the activities that 90 students do in their free time. 60 students watch TV ( $T$ ), 60 students read ( $R$ ), 70 students go to the cinema ( $C$ ). 26 students watch TV, read and also go to the cinema. 20 students only watch TV and go to the cinema. 18 students only read and go to the cinema. 10 students only read and watch TV.

a) Draw a Venn diagram to illustrate the information given above.

b) Calculate how many students

(i) only watch TV;

(ii) only go to the cinema.

Show answer

Answer: a) diagram above b) (i) Only watch TV $= 4$ (ii) Only go to the cinema $= 6$

Show working and answer

a) We start from the centre and work outwards.

$T \cap R \cap C = 26$ .
Regions of exactly two: $T \cap C$ without $R = 20$ , $R \cap C$ without $T = 18$ , $R \cap T$ without $C = 10$ .
Only $T$ : $60 - 20 - 10 - 26 = 4$ .
Only $R$ : $60 - 18 - 10 - 26 = 6$ .
Only $C$ : $70 - 20 - 18 - 26 = 6$ .
Check the outside: $4 + 6 + 6 + 20 + 18 + 10 + 26 = 90$ , so there are $90 - 90 = 0$ students outside the circles.

b) (i) Only watch TV: $4$ (step 3).

(ii) Only go to the cinema: $6$ (step 5).

Venn diagram with U and three circles R, T and C. Only R 6; only T 4; only C 6; R and T without C 10; R and C without T 18; T and C without R 20; R, T and C 26; outside the circles 0

Answer: a) diagram above b) (i) Only watch TV $= 4$ (ii) Only go to the cinema $= 6$

Exercise 18 MediumIB · P2 TZ0 November 2009

The following Venn diagram represents the students at a school who study Mathematics ( $A$ ), Further Mathematics ( $B$ ) and Physics ( $C$ ). 50 students study Mathematics 38 study Physics 20 study Mathematics and Physics, but not Further Mathematics 10 study Further Mathematics, but not Physics 12 study Further Mathematics and Physics 6 study Physics, but not Mathematics 3 do not study any of these three subjects. Venn diagram with U: large circle A, small circle B inside A on its left, and circle C on the right cutting across A and B. Written values: 20 in A ∩ C outside B, 6 in C outside A and 3 outside all the circles

a) Copy and complete the Venn diagram on your answer sheet.

b) Write down the number of students who study Mathematics but not Further Mathematics.

c) Write down the total number of students in the school.

d) Write $n (B \cup C)$ .

Show answer

Answer: a) diagram above b) 28 students study Mathematics but not Further Mathematics c) Total number of students $= 59$ d) $n (B \cup C) = 38 + 10 = 48$ students

Show working and answer

a) The diagram shows that $B$ is inside $A$ (anyone who studies Further Mathematics studies Mathematics). We use this to fill in.

Since $B \subset A$ , " $B$ and $C$ " is the region $A \cap B \cap C$ : $12$ . " $B$ without $C$ ": $10$ .
$A \cap C$ without $B$ : $20$ . Only $C$ (outside $A$ ): $6$ . Outside the circles: $3$ .
Only $A$ (without $B$ or $C$ ): from the total for $A$ we take away the other regions of $A$ : $50 - 10 - 12 - 20 = 8$ .
Check with Physics: $20 + 12 + 6 = 38$ . Correct.

b) Mathematics but not Further Mathematics: only $A$ plus $A \cap C$ without $B$ : $8 + 20 = 28$ .

c) We add all the regions: $8 + 10 + 12 + 20 + 6 + 3 = 59$ .

d) $B \cup C$ : the regions of $B$ ( $10 + 12$ ) plus the regions of $C$ that are not in $B$ ( $20 + 6$ ): $10 + 12 + 20 + 6 = 48$ . This equals $n (C) + n (B without C) = 38 + 10 = 48$ .

Venn diagram of A, B and C inside U, with B inside A: only A = 8, B without C = 10, B and C = 12, A and C without B = 20, only C = 6, outside the circles = 3

Answer: a) diagram above b) 28 students study Mathematics but not Further Mathematics c) Total number of students $= 59$ d) $n (B \cup C) = 38 + 10 = 48$ students

Exercise 19 MediumIB · P1 TZ0 November 2007

A school offers three activities: basketball (B), choir (C) and drama (D). Every student must take part in at least one activity.

16 students play basketball only.
18 students play basketball and sing in the choir, but do not do drama.
34 students play basketball and do drama, but do not sing in the choir.
27 students are in the choir and do drama, but do not play basketball.

a) Enter the information above in the following Venn diagram.

Venn diagram of B, C and D inside U; the region common to the three circles is labelled x and the other regions are empty

99 of the students play basketball, 88 sing in the choir and 110 do drama.

b) Calculate the number of students $x$ who take part in all three activities.

c) Calculate the total number of students in the school.

Show answer

Answer: a) diagram above b) $x = 31$ students c) Total number of students $= 156$ students

Show working and answer

a) We enter the data as given: only $B = 16$ , $B \cap C$ without $D = 18$ , $B \cap D$ without $C = 34$ , $C \cap D$ without $B = 27$ , and the central region is $x$ . "Only $C$ " and "only $D$ " are calculated in the following parts.

b) 1. In $B$ there are four regions: $16 + 18 + 34 + x = 99$ . 2. $68 + x = 99$ , therefore $x = 31$ .

c) 1. Only $C$ : from the $88$ in the choir we take away the other regions of $C$ : $88 - 18 - 31 - 27 = 12$ . 2. Only $D$ : from the $110$ in drama: $110 - 34 - 31 - 27 = 18$ . 3. As every student takes part in at least one activity, the total is the sum of the seven regions: $16 + 12 + 18 + 18 + 34 + 27 + 31 = 156$ .

Venn diagram of B, C and D inside U: only B = 16, B and C without D = 18, B, C and D = x, B and D without C = 34, C and D without B = 27; only C and only D empty

Answer: a) diagram above b) $x = 31$ students c) Total number of students $= 156$ students

Exercise 20 MediumIB · P1 TZ0 May 2006

At a school there are 90 students enrolled in the IB Diploma Programme. They must study at least one of these subjects: Physics, Biology and Chemistry.

50 students study Physics,
60 students study Biology,
55 students study Chemistry,
30 students study Physics and Biology,
10 students study Biology and Chemistry but not Physics,
20 students study all three subjects.

If $x$ represents the number of students who study Physics and Chemistry but not Biology, then $25 - x$ is the number of students who study only Chemistry.

The following figure shows part of this information and can be used for the calculations.

Venn diagram of Physics, Chemistry and Biology inside U with n(U) = 90: Physics and Chemistry without Biology = x, all three = 20, only Chemistry = 25 − x, Biology and Chemistry without Physics = 10, only Biology = 20; only Physics and Physics with Biology without Chemistry empty

a) Express the number of students who study Physics only, in terms of $x$ .

b) Find $x$ .

c) Determine the number of students who study at least two of these subjects.

Show answer

Answer: a) Study Physics only $= 20 - x$ students b) $x = 15$ students c) Study at least two of these subjects $= 55$ students

Show working and answer

a) 1. The $50$ who study Physics are split into four regions: Physics only, Physics and Biology without Chemistry, Physics and Chemistry without Biology ( $x$ ) and all three ( $20$ ). 2. Physics and Biology total $30$ , and $20$ of them also study Chemistry. So Physics and Biology without Chemistry: $30 - 20 = 10$ . 3. Physics only $= 50 - 10 - x - 20 = 20 - x$ .

b) 1. Everyone studies at least one subject, so the sum of the seven regions is $90$ . Only Biology $= 60 - 10 - 20 - 10 = 20$ (as shown in the figure). 2. Sum: $(20 - x) + 10 + x + 20 + 20 + 10 + (25 - x) = 90$ . 3. $105 - x = 90$ , therefore $x = 15$ .

c) At least two subjects: the three regions of two subjects plus the region of all three: $10 (P and B) + 15 (P and C) + 10 (B and C) + 20 (all three) = 55$ .

Answer: a) Study Physics only $= 20 - x$ students b) $x = 15$ students c) Study at least two of these subjects $= 55$ students

Exercise 21 MediumIB · P2 TZ0 May 2005

The following results were obtained from a survey on students' reading habits.

60% read magazine P
50% read magazine Q
50% read magazine R
30% read magazines P and Q
20% read magazines Q and R
30% read magazines P and R
10% read all three magazines

a) Represent all this information in a Venn diagram.

b) What percentage of students read exactly two magazines?

c) What percentage of students read at least two magazines?

d) What percentage of students do not read any of the magazines?

Show answer

Answer: a) diagram above b) 50% read exactly 2 magazines c) 60% read at least 2 magazines d) 10% read none

Show working and answer

a) We start from the centre. A percentage for " $P$ and $Q$ " includes the part that also reads $R$ .

All three: $10%$ .
Exactly two: $P$ and $Q$ without $R = 30 - 10 = 20$ ; $Q$ and $R$ without $P = 20 - 10 = 10$ ; $P$ and $R$ without $Q = 30 - 10 = 20$ .
Only $P = 60 - 20 - 20 - 10 = 10$ ; only $Q = 50 - 20 - 10 - 10 = 10$ ; only $R = 50 - 10 - 20 - 10 = 10$ .
Sum of the circles: $10 + 10 + 10 + 20 + 10 + 20 + 10 = 90$ , so $100 - 90 = 10%$ is left outside.

b) Exactly two: $20 + 10 + 20 = 50%$ .

c) At least two: exactly two plus all three: $50 + 10 = 60%$ .

d) None: the outside region, $10%$ .

Venn diagram of P, Q and R inside U: only P 10%, only Q 10%, only R 10%, P and Q without R 20%, P and R without Q 20%, Q and R without P 10%, all three 10%, outside the circles 10%

Answer: a) diagram above b) 50% read exactly 2 magazines c) 60% read at least 2 magazines d) 10% read none

Exercise 22 MediumIB · P2 TZ0 May 2003

The sets $U$ , $P$ , $R$ and $S$ are defined as follows:

$U = {all quadrilaterals}$

$P = {all parallelograms}$

$R = {all rectangles}$

$S = {all squares}$

a) Draw a Venn diagram showing the relationships between the sets above.

b) Draw a separate Venn diagram for each of the examples that follow. Indicate, using shading, each of the following:

(i) $(P \cup S)^{'}$

(ii) $(R \cup S) \cap P$

View solution

Every square is a rectangle (4 right angles) and, in addition, has opposite sides parallel: $S \subset R$ .
Every rectangle is a parallelogram (opposite sides parallel): $R \subset P$ .
Every parallelogram is a quadrilateral: $P \subset U$ .
Hence the sets are nested: $S \subset R \subset P \subset U$ . Draw a rectangle $U$ and, inside it, three concentric circles: the outer one is $P$ , the middle one $R$ and the inner one $S$ .

Rectangle U with three unshaded concentric circles: P the outer, R the middle and S the inner

Answer: diagram with $S \subset R \subset P \subset U$ (concentric circles, $P$ the largest and $S$ the smallest).

(i) $(P \cup S)^{'}$

Since $S \subset P$ , the union $P \cup S$ is simply $P$ .
The complement $(P \cup S)^{'} = P^{'}$ is the set of elements of $U$ that are not in $P$ .
Draw two concentric circles ( $P$ outer and $S$ inner) and shade the whole region of $U$ outside the circle $P$ .

Rectangle U with two concentric circles, P outer and S inner; hatched the whole region outside P, the complement of P union S

Answer: everything outside $P$ is shaded.

(ii) $(R \cup S) \cap P$

Since $S \subset R$ , we have $R \cup S = R$ .
Since $R \subset P$ , the intersection $R \cap P = R$ (the common part of $R$ and $P$ is all of $R$ ).
Draw two concentric circles ( $R$ outer and $S$ inner) and shade the whole interior of $R$ , including $S$ . What lies outside $R$ is left unshaded.

Rectangle U with two concentric circles, R outer and S inner; the whole interior of R, including S, is hatched; the exterior of R unshaded

Answer: the whole interior of $R$ , including $S$ , is shaded.

Exercise 23 MediumIB · P2 TZ0 May 2000

(i) A group of 30 children is surveyed to find out which of three sports they play: cricket ( $C$ ), basketball ( $B$ ) or volleyball ( $V$ ). The results are:

3 children do not play any of these sports
2 children play all three sports
6 play volleyball and basketball
3 play cricket and basketball
6 play cricket and volleyball
16 play basketball
12 play volleyball.

a) Draw a Venn diagram to illustrate the relationship between the three sports played.

b) On your Venn diagram, state the number of children in each region.

c) How many children play only cricket?

Show answer

Answer: a) and b) diagram with the regions: only $C$ 5; $C \cap B$ without $V$ 1; only $B$ 9; all three 2; $C \cap V$ without $B$ 4; $B \cap V$ without $C$ 4; only $V$ 2; outside 3. c) $5$ children play only cricket.

Show working and answer

Draw three overlapping circles, $C$ , $B$ and $V$ , inside a rectangle (the 30 children). Fill them in from the inside outwards.
All three sports: $C \cap B \cap V = 2$ .
Exactly two sports (from each intersection of two sets, subtract the central region of 2):
- $B \cap V$ without $C$ : $6 - 2 = 4$ .
- $C \cap B$ without $V$ : $3 - 2 = 1$ .
- $C \cap V$ without $B$ : $6 - 2 = 4$ .
Only one sport:
- Only $B$ : $16 - (4 + 1 + 2) = 9$ .
- Only $V$ : $12 - (4 + 4 + 2) = 2$ .
Outside the circles: 3 children play none, so between the circles there are $30 - 3 = 27$ children.
Only cricket: $27 - (9 + 2 + 4 + 1 + 4 + 2) = 27 - 22 = 5$ .

Venn diagram of C, B and V. Only C: unlabelled (5). C∩B without V: 1. Only B: 9. C∩B∩V: 2. C∩V without B: 4. B∩V without C: 4. Only V: 2. Outside the circles: 3

Exercise 24 MediumIB · P1 TZ2 May 2019

Consider the following sets: The universal set $U$ consists of all positive integers less than 15; $A$ is the set of all numbers that are multiples of 3; $B$ is the set of all even numbers.

a) Write down the elements that belong to $A \cap B$ .

b) Write down

(i) the elements that belong to $A \cap B^{'}$ ;

(ii) $n (A \cap B^{'})$ .

Show answer

Answer: a) $A \cap B = {6, 12}$ b) i) $A \cap B^{'} = {3, 9}$ ii) $n (A \cap B^{'}) = 2$

Show working and answer

Universal set: $U = {1, 2, 3, \dots, 14}$ .
Multiples of 3: $A = {3, 6, 9, 12}$ . Even numbers: $B = {2, 4, 6, 8, 10, 12, 14}$ .

a) $A \cap B$ are the elements that are in $A$ and in $B$ : the multiples of 3 that are even, ${6, 12}$ .

$B^{'}$ are the elements of $U$ that are not even, that is, the odd ones: $B^{'} = {1, 3, 5, 7, 9, 11, 13}$ .
(i) $A \cap B^{'}$ : the elements of $A$ that are odd, ${3, 9}$ .
(ii) $n (A \cap B^{'})$ is the number of elements: 2.

Answer: a) $A \cap B = {6, 12}$ b) i) $A \cap B^{'} = {3, 9}$ ii) $n (A \cap B^{'}) = 2$

Exercise 25 MediumIB · P1 TZ0 November 2013

$U$ is the set of positive integers less than or equal to 10. $A$ , $B$ and $C$ are subsets of $U$ . $A = {even integers}$ $B = {multiples of 3}$ $C = {6, 7, 8, 9}$ .

a) List the elements of $A$ .

b) List the elements of $B$ .

c) Complete the following Venn diagram, including all the elements of $U$ . Empty Venn diagram with the universal set U and three sets A, B and C that overlap, with a region common to all three

View solution

a) The even integers between 1 and 10: $A = {2, 4, 6, 8, 10}$ .

b) The multiples of 3 up to 10: $B = {3, 6, 9}$ .

c) Fill in the diagram from the inside outwards, looking at each element of $U = {1, \dots, 10}$ to see which sets it belongs to:

All three ( $A \cap B \cap C$ ): $6$ is in $A$ , in $B$ and in $C$ .
$A \cap C$ without $B$ : $8$ (even, in $C$ , not a multiple of 3).
$B \cap C$ without $A$ : $9$ (multiple of 3, in $C$ , odd).
$A \cap B$ without $C$ : none (the only element common to $A$ and $B$ is 6, which is already in $C$ ).
Only $A$ : $2, 4, 10$ . Only $B$ : $3$ . Only $C$ : $7$ .
Outside the three circles (within $U$ ): the remaining ones, $1$ and $5$ .
Check: $3 + 1 + 1 + 1 + 1 + 1 + 2 = 10$ elements, all those of $U$ .

Venn diagram of A, B and C within U. Only A: 2, 4, 10. Only B: 3. Only C: 7. A∩B∩C: 6. A∩C without B: 8. B∩C without A: 9. Outside the three: 1 and 5

Answer: a) $A = {2, 4, 6, 8, 10}$ b) $B = {3, 6, 9}$ c) diagram with: only $A$ ${2, 4, 10}$ , only $B$ ${3}$ , only $C$ ${7}$ , $A \cap B \cap C = {6}$ , $A \cap C$ without $B$ ${8}$ , $B \cap C$ without $A$ ${9}$ , outside ${1, 5}$ .

Exercise 26 MediumIB · P1 TZ0 November 2019

Let $U$ (the universal set) be the set of all integers $x$ such that $1 \leq x < 11$ . $A$ , $B$ and $C$ are subsets of $U$ . $A = {1, 2, 3, 4, 6, 8}$ $B = {2, 3, 5, 7}$ $C = {1, 3, 5, 7, 9}$

a) Write down $n (B)$ .

b) Complete the following Venn diagram using all the elements of $U$ . Empty Venn diagram with the universal set U and three sets A, B and C that overlap, with a region common to all three

c) Write down an element that belongs to $(A \cup B)^{'} \cap C$ .

View solution

a) $B = {2, 3, 5, 7}$ has 4 elements: $n (B) = 4$ .

b) Universal set: $U = {1, 2, \dots, 10}$ . For each element, look at the sets it belongs to:

$A \cap B \cap C$ : only $3$ is in all three.
$A \cap B$ without $C$ : $2$ .
$A \cap C$ without $B$ : $1$ .
$B \cap C$ without $A$ : $5$ and $7$ .
Only $A$ : $4, 6, 8$ . Only $B$ : none. Only $C$ : $9$ .
Outside the three: $10$ .
Check: $A$ has $3 + 1 + 1 + 1 = 6$ elements, $B$ has $1 + 1 + 2 = 4$ and $C$ has $1 + 1 + 2 + 1 = 5$ , as in the question.

Venn diagram of A, B and C within U. Only A: 4, 6, 8. A∩B without C: 2. A∩B∩C: 3. A∩C without B: 1. B∩C without A: 5 and 7. Only C: 9. Outside the three: 10

$A \cup B = {1, 2, 3, 4, 5, 6, 7, 8}$ , so $(A \cup B)^{'} = {9, 10}$ (what is left outside $A$ and $B$ ).
Intersect with $C = {1, 3, 5, 7, 9}$ : of ${9, 10}$ , only $9$ is in $C$ .

Answer: a) $n (B) = 4$ b) diagram with: only $A$ ${4, 6, 8}$ , $A \cap B$ without $C$ ${2}$ , all three ${3}$ , $A \cap C$ without $B$ ${1}$ , $B \cap C$ without $A$ ${5, 7}$ , only $C$ ${9}$ , outside ${10}$ c) $9$

Exercise 27 MediumIB · P1 TZ0 May 2007

$B$ and $C$ are subsets of a universal set $U$ , such that $U = {x ∣ x \in Z, 0 \leq x < 10}$ $B = {prime numbers < 10}$ $C = {x ∣ x \in Z, 1 < x \leq 6}$ .

a) List the members of the following sets:

(i) $B$

(ii) $C \cap B$

(iii) $B \cup C^{'}$

Show answer

Answer: i) $B = {2, 3, 5, 7}$ ii) $C \cap B = {2, 3, 5}$ iii) $B \cup C^{'} = {0, 1, 2, 3, 5, 7, 8, 9}$

Show working and answer

Universal set: $U = {0, 1, 2, \dots, 9}$ .

(i) Prime numbers less than 10 (1 is not prime): $B = {2, 3, 5, 7}$ .

(ii) First $C$ : integers with $1 < x \leq 6$ , that is, $C = {2, 3, 4, 5, 6}$ . $C \cap B$ are the common elements: ${2, 3, 5}$ .

(iii)

$C^{'}$ are the elements of $U$ that are not in $C$ : $C^{'} = {0, 1, 7, 8, 9}$ .
$B \cup C^{'}$ gathers all the elements of both sets, without repetition: ${2, 3, 5, 7} \cup {0, 1, 7, 8, 9} = {0, 1, 2, 3, 5, 7, 8, 9}$ .

Answer: i) $B = {2, 3, 5, 7}$ ii) $C \cap B = {2, 3, 5}$ iii) $B \cup C^{'} = {0, 1, 2, 3, 5, 7, 8, 9}$

Exercise 28 MediumIB · P1 TZ0 November 2007

The universal set $U$ is the set of integers from 1 to 20 inclusive. $A$ and $B$ are subsets of $U$ where: $A$ is the set of even numbers between 7 and 17. $B$ is the set of multiples of 3. List the elements of the following sets:

a) $A$ ,

b) $B$ ,

c) $A \cup B$ ,

d) $A \cap B^{'}$ .

Show answer

Answer: a) $A = {8, 10, 12, 14, 16}$ ; b) $B = {3, 6, 9, 12, 15, 18}$ ; c) $A \cup B = {3, 6, 8, 9, 10, 12, 14, 15, 16, 18}$ ; d) $A \cap B^{'} = {8, 10, 14, 16}$ .

Show working and answer

The universal set is $U = {1, 2, \dots, 20}$ .

a) "Even numbers between 7 and 17": the even numbers greater than 7 and less than 17 are 8, 10, 12, 14 and 16.

b) Multiples of 3 up to 20: $3 \cdot 1, 3 \cdot 2, \dots, 3 \cdot 6 = 18$ (the next one, 21, is outside $U$ ).

c) $A \cup B$ gathers the elements that are in $A$ or in $B$ (or in both), listing 12 only once, since it is in both sets. We list them all from smallest to largest.

d) $B^{'}$ are the elements of $U$ that are not multiples of 3. $A \cap B^{'}$ are the elements of $A$ that are not in $B$ : from ${8, 10, 12, 14, 16}$ we remove 12 (which is a multiple of 3).

Answer: a) $A = {8, 10, 12, 14, 16}$ ; b) $B = {3, 6, 9, 12, 15, 18}$ ; c) $A \cup B = {3, 6, 8, 9, 10, 12, 14, 15, 16, 18}$ ; d) $A \cap B^{'} = {8, 10, 14, 16}$ .

Exercise 29 MediumIB · P1 TZ1 May 2013

$U = {x ∣ x an integer, 2 < x < 10}$ . $A$ and $B$ are subsets of $U$ such that $A = {multiples of 3}$ and $B = {factors of 24}$ .

a) List the elements of

(i) $U$ ;

(ii) $B$ .

b) Write the elements of $U$ in the Venn diagram. Empty Venn diagram: rectangle U with two overlapping ellipses, A on the left and B on the right

c) Write down $n (A \cap B)$ .

View solution

a) (i) $U$ is the set of integers $x$ with $2 < x < 10$ (strict inequalities, so neither 2 nor 10 is included): $U = {3, 4, 5, 6, 7, 8, 9}$ .

(ii) The factors of 24 are $1, 2, 3, 4, 6, 8, 12, 24$ . Since $B$ is a subset of $U$ , only those in $U$ count: $B = {3, 4, 6, 8}$ .

b) First write $A$ (the multiples of 3 in $U$ ): $A = {3, 6, 9}$ . Then place each element of $U$ in its region:

3 and 6 are in $A$ and in $B$ : they go in the common region $A \cap B$ .
9 is in $A$ but not in $B$ : only in $A$ .
4 and 8 are in $B$ but not in $A$ : only in $B$ .
5 and 7 are in neither: inside the rectangle $U$ but outside both ellipses.

Venn diagram with A and B: only in A: 9; in A∩B: 3 and 6; only in B: 4 and 8; outside both, inside U: 5 and 7

c) $A \cap B = {3, 6}$ has 2 elements.

Answer: a) i) $U = {3, 4, 5, 6, 7, 8, 9}$ , ii) $B = {3, 4, 6, 8}$ ; b) diagram above; c) $n (A \cap B) = 2$ .

Exercise 30 MediumIB · P1 TZ1 May 2010

The sets $P$ , $Q$ and $U$ are defined as $U = {Real numbers}$ , $P = {Positive numbers}$ and $Q = {Rational numbers}$ . Empty Venn diagram: rectangle U with two overlapping circles, P on the left and Q on the right Write the following numbers in the correct region of the Venn diagram: $\frac{22}{7}, 5 \times 1 0^{- 2}, sin (6 0^{\circ}), 0, 3 - 8, - π$

View solution

Each number is placed according to two questions: is it positive? (does it belong to $P$ ) and is it rational? (does it belong to $Q$ ). All of them are real, so they all lie inside $U$ .

$\frac{22}{7}$ : positive and a fraction, so it is in $P \cap Q$ .
$5 \times 1 0^{- 2} = 0.05 = \frac{5}{100}$ : positive and rational, in $P \cap Q$ .
$sin 6 0^{\circ} = \frac{3}{2} \approx 0.866$ : positive, but irrational (because of the $3$ ): only in $P$ .
$0$ : not positive (it is not greater than 0), but it is rational ( $\frac{0}{1}$ ): only in $Q$ .
$3 - 8 = - 2$ : negative and an integer, so rational: only in $Q$ .
$- π$ : negative and irrational: in $U$ but outside $P$ and $Q$ .

Venn diagram with P and Q: in P∩Q: 22/7 and 5×10⁻²; only in P: sin(60°); only in Q: 0 and ∛−8; outside both circles, inside U: −π

Answer: $P \cap Q$ : $\frac{22}{7}$ and $5 \times 1 0^{- 2}$ ; only $P$ : $sin 6 0^{\circ}$ ; only $Q$ : $0$ and $3 - 8$ ; outside both: $- π$ .

Exercise 31 MediumIB · P1 TZ2 May 2011

$U$ is the set of all positive integers less than or equal to $12$ . $A$ , $B$ and $C$ are subsets of $U$ . $A = {1, 2, 3, 4, 6, 12}$ $B = {odd integers}$ $C = {5, 6, 8}$

a) Write down the number of elements in $A \cap C$ .

b) List the elements of $B$ .

c) Complete the following Venn diagram with all the elements of $U$ . Empty Venn diagram: rectangle U with three overlapping circles, A top left, B top right and C bottom centre

View solution

$U = {1, 2, \dots, 12}$ .

a) $A \cap C$ are the common elements: $A = {1, 2, 3, 4, 6, 12}$ and $C = {5, 6, 8}$ share only 6. Therefore $n (A \cap C) = 1$ .

b) The odd numbers in $U$ are $B = {1, 3, 5, 7, 9, 11}$ .

c) Place each element of $U$ by looking at the sets it belongs to:

1 and 3: in $A$ and $B$ , not in $C$ .
2, 4 and 12: only in $A$ .
5: in $B$ and $C$ , not in $A$ .
6: in $A$ and $C$ , not in $B$ (it is even).
7, 9 and 11: only in $B$ .
8: only in $C$ .
10: in none of the sets, so it goes outside the three circles.

The region $A \cap B \cap C$ is empty (no element is in all three).

Venn diagram with A, B and C: only in A: 2, 4, 12; in A∩B outside C: 1, 3; only in B: 7, 9, 11; in A∩C outside B: 6; in B∩C outside A: 5; only in C: 8; A∩B∩C empty; outside the three circles: 10

Answer: a) $n (A \cap C) = 1$ ; b) $B = {1, 3, 5, 7, 9, 11}$ ; c) diagram above.

Exercise 32 MediumIB · P1 TZ0 November 2010

Consider the universal set $U = {x \in N ∣ 3 < x < 13}$ , and the subsets $A = {multiples of 3}$ and $B = {4, 6, 12}$ .

a) List the elements belonging to the following sets. (i) $A$ (ii) $A \cap B^{'}$

b) Write down an element of $(A \cup B)^{'}$ .

c) One of the statements in the table below is false. Indicate with an X the statement that is false. Give a reasoned answer.


$n (A \cup B) = 4$
$15 \in A^{'}$
$A \subset A \cup B$

View solution

First write the universal set: $U = {x \in N ∣ 3 < x < 13} = {4, 5, 6, 7, 8, 9, 10, 11, 12}$ .

a) (i) The multiples of 3 in $U$ are $A = {6, 9, 12}$ (3 is not included because $3 < x$ ).

(ii) $B^{'}$ are the elements of $U$ that are not in $B = {4, 6, 12}$ . $A \cap B^{'}$ are the elements of $A$ that are not in $B$ : from ${6, 9, 12}$ we remove 6 and 12, leaving ${9}$ .

b) $A \cup B = {4, 6, 9, 12}$ . Its complement is the set of elements of $U$ that are not there: $(A \cup B)^{'} = {5, 7, 8, 10, 11}$ . A valid element is, for example, $5$ .

c) We check each statement:

$n (A \cup B) = 4$ : $A \cup B = {4, 6, 9, 12}$ has 4 elements. True.
$15 \in A^{'}$ : $A^{'}$ consists of elements of $U$ that are not in $A$ , and $15 \in / U$ (because $15 > 13$ ). Therefore $15 \in / A^{'}$ . False.
$A \subset A \cup B$ : every element of $A$ is in the union. True.


$n (A \cup B) = 4$
$15 \in A^{'}$	X
$A \subset A \cup B$

Answer: a) (i) $A = {6, 9, 12}$ , (ii) $A \cap B^{'} = {9}$ ; b) $(A \cup B)^{'} = {5, 7, 8, 10, 11}$ (for example, 5); c) the false statement is " $15 \in A^{'}$ ", because 15 does not belong to $U$ .

Exercise 33 MediumIB · P1 TZ0 May 2003

In the Venn diagram below, $A$ , $B$ and $C$ are subsets of a universal set $U = {1, 2, 3, 4, 6, 7, 8, 9}$ Venn diagram with B (ellipse at the top), A (circle bottom left) and C (circle bottom right): outside the three: 2; in A∩B outside C: 3; in B∩C outside A: 7; in A∩B∩C: 9; only in A: 8 and 4; in A∩C outside B: 1; only in C: 6; only in B: none List the elements of each of the following sets.

a) $A \cup B$

b) $A \cap B \cap C$

c) $(A^{'} \cap C) \cup B$

Show answer

Answer: a) $A \cup B = {1, 3, 4, 7, 8, 9}$ ; b) $A \cap B \cap C = {9}$ ; c) $(A^{'} \cap C) \cup B = {3, 6, 7, 9}$

Show working and answer

First we read from the diagram the elements of each set (each set includes all the regions inside its curve).

$A$ contains: only in $A$ ( $8$ and $4$ ), in $A \cap B$ outside $C$ ( $3$ ), in $A \cap B \cap C$ ( $9$ ) and in $A \cap C$ outside $B$ ( $1$ ). So $A = {1, 3, 4, 8, 9}$ .
$B$ contains: $A \cap B$ outside $C$ ( $3$ ), $B \cap C$ outside $A$ ( $7$ ) and $A \cap B \cap C$ ( $9$ ). So $B = {3, 7, 9}$ .
$C$ contains: $A \cap C$ outside $B$ ( $1$ ), $B \cap C$ outside $A$ ( $7$ ), $A \cap B \cap C$ ( $9$ ) and only in $C$ ( $6$ ). So $C = {1, 6, 7, 9}$ .

a) $A \cup B$ gathers the elements that are in $A$ , in $B$ or in both, without repetition: ${1, 3, 4, 8, 9} \cup {3, 7, 9} = {1, 3, 4, 7, 8, 9}$ .

b) $A \cap B \cap C$ are the elements common to all three sets, that is, the central region: only $9$ is there.

c) Step by step:

$A^{'}$ are the elements of $U$ that are not in $A$ : $A^{'} = {2, 6, 7}$ .
$A^{'} \cap C$ : the elements of $A^{'}$ that are also in $C = {1, 6, 7, 9}$ , that is, ${6, 7}$ .
Union with $B = {3, 7, 9}$ : ${6, 7} \cup {3, 7, 9} = {3, 6, 7, 9}$ .

Answer: a) $A \cup B = {1, 3, 4, 7, 8, 9}$ ; b) $A \cap B \cap C = {9}$ ; c) $(A^{'} \cap C) \cup B = {3, 6, 7, 9}$

Exercise 34 MediumIB · P1 TZ2 May 2008

a) Shade $(A \cup B) \cap C^{'}$ in the following diagram. Empty Venn diagram: rectangle U with three circles in a row, A on the left, B in the centre and C on the right; A overlaps B and B overlaps C, and A and C do not touch

b) In the following Venn diagram, the number of elements in each region is given. Find $n ((P \cap Q) \cup R)$ . Venn diagram with circles P, Q and R in the rectangle U. Only P: 2; P∩Q outside R: 3; only Q: 1; P∩R outside Q: 4; Q∩R outside P: 5; P∩Q∩R: 6; only R: 9

c) $U$ is the set of positive integers, $Z^{+}$ . $E$ is the set of even numbers. $M$ is the set of multiples of $3$ . (i) List the first six elements of the set $M$ . (ii) List the first six elements of the set $E^{'} \cap M$ .

View solution

a) We reason by regions:

$A \cup B$ is everything inside $A$ or $B$ : the two complete circles.
$C^{'}$ is everything outside $C$ .
The intersection $(A \cup B) \cap C^{'}$ asks for the points that are both in $A \cup B$ and outside $C$ . Therefore we shade $A$ and $B$ completely, except the part of $B$ that overlaps with $C$ (the intersection $B \cap C$ is left unshaded). Circle $C$ is left unshaded.

Venn diagram with three circles in a row A, B and C inside rectangle U; A and B shaded except the part inside C (the intersection of B and C is unshaded), and C unshaded

b) We count the elements of each part:

$P \cap Q$ comprises the only- $P \cap Q$ region ( $3$ ) and the central $P \cap Q \cap R$ ( $6$ ): $n (P \cap Q) = 3 + 6 = 9$ .
$R$ comprises: only $R$ ( $9$ ), $P \cap R$ outside $Q$ ( $4$ ), $Q \cap R$ outside $P$ ( $5$ ) and the central region ( $6$ ): $n (R) = 9 + 4 + 5 + 6 = 24$ .
In $(P \cap Q) \cup R$ the central region is already counted within $R$ ; we only add the region $P \cap Q$ outside $R$ ( $3$ ):

$n ((P \cap Q) \cup R) = 24 + 3 = 27.$

(i) $M$ are the multiples of $3$ among the positive integers: $3, 6, 9, 12, 15, 18$ .

(ii) $E^{'}$ is the complement of the even numbers, that is, the odd numbers. $E^{'} \cap M$ are the multiples of $3$ that are also odd. From the list of multiples of $3$ ( $3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, \dots$ ) we keep the odd ones: $3, 9, 15, 21, 27, 33$ .

Answer: a) see the diagram; b) $n ((P \cap Q) \cup R) = 27$ ; c) (i) $M = {3, 6, 9, 12, 15, 18, \dots}$ ; (ii) $E^{'} \cap M = {3, 9, 15, 21, 27, 33, \dots}$

Exercise 35 MediumIB · P1 TZ1 May 2017

In the Canadian city of Ottawa: 97% of the population speaks English, 38% of the population speaks French, 36% of the population speaks English and French.

a) Calculate the percentage of the population of Ottawa that speaks English but not French. The total population of Ottawa is 985,000.

b) Calculate the number of people in Ottawa who speak English and French.

c) Write your answer to part (b) in the form $a \times 1 0^{k}$ where $1 \leq a < 10$ and $k \in Z$ .

Show answer

Answer: a) $61%$ b) $\approx 355000$ c) $3.55 \times 1 0^{5}$

Show working and answer

97% of the population speak English, and 36% of the population speak both English and French. The people who speak English but not French are those who speak English minus the overlap: $97 % - 36 % = 61 %$

36% of the population speaks English and French: $0.36 \times 985000 = 354600$ .
Rounding to 3 significant figures: $\approx 355000$ people.

Converting to scientific notation: $354600 = 3.546 \times 1 0^{5} \approx 3.55 \times 1 0^{5}$ (3 significant figures, as in the previous part).

Answer: a) $61%$ b) $\approx 355000$ c) $3.55 \times 1 0^{5}$

Exercise 36 HardIB · P2 TZ1 May 2012

Beartown has three local newspapers: The Art Journal, The Beartown News and The Currier. A survey shows that 32% of the city's population reads The Art Journal, 46% read The Beartown News, 54% read The Currier, 3% read only The Art Journal and The Beartown News, 8% read only The Art Journal and The Currier, 12% read only The Beartown News and The Currier, and 5% of the population reads all three newspapers.

a) Draw a Venn diagram to represent this information. Label $A$ the set representing the readers of The Art Journal, $B$ the set representing the readers of The Beartown News, and $C$ the set representing the readers of The Currier.

b) Find the percentage of the population that does not read any of the three newspapers.

c) Find the percentage of the population that reads exactly one newspaper.

d) Find the percentage of the population that reads The Art Journal or The Beartown News but not The Currier. A local radio station claims that 83% of the population reads The Beartown News or The Currier.

e) Use your Venn diagram to decide whether the statement is true. Justify your answer. Beartown has a population of 120,000. The local radio station claimed that 34,000 of the city's citizens read at least two of the local newspapers.

f) Find the percentage error in this claim.

Show answer

Answer: a) diagram above b) 1% c) 71% d) 45% e) Yes, it is true f) 1.19%

Show working and answer

a) We fill in the diagram from the inside out: first the central region (all three newspapers) and then the regions of two sets, which already give us the "only" for those two. Finally, we work out what is missing from each circle by subtracting.

Region $A \cap B \cap C$ : $5$ .
Regions of exactly two newspapers: $A \cap B$ without $C$ $= 3$ , $A \cap C$ without $B$ $= 8$ , $B \cap C$ without $A$ $= 12$ .
Only $A$ : from the total for $A$ we take away everything already inside $A$ : $32 - 3 - 8 - 5 = 16$ .
Only $B$ : $46 - 3 - 12 - 5 = 26$ .
Only $C$ : $54 - 8 - 12 - 5 = 29$ .

b) 1. The total population is $100%$ . We add all the regions inside the circles: $16 + 26 + 29 + 3 + 8 + 12 + 5 = 99$ . 2. What lies outside the three circles is $100 - 99 = 1$ .

c) Exactly one newspaper = the regions "only $A$ ", "only $B$ " and "only $C$ ": $16 + 26 + 29 = 71$ .

d) " $A$ or $B$ but not $C$ " are the regions of $A \cup B$ that lie outside $C$ : only $A$ , only $B$ and $A \cap B$ without $C$ . So: $16 + 26 + 3 = 45$ .

e) 1. We add all the regions inside $B$ or $C$ (that is, everything except "only $A$ " and the outside): $26 + 29 + 3 + 8 + 12 + 5 = 83$ . 2. This gives $83%$ , which matches what the radio station says. Equivalently, $100 - 16 - 1 = 83$ . The statement is true.

f) 1. The regions of two and of three sets are the ones who read at least two newspapers: $3 + 8 + 12 + 5 = 28%$ . 2. Exact value: $28%$ of $120000 = 0.28 \cdot 120000 = 33600$ people. 3. Claimed value: $34000$ . Percentage error $= \frac{∣ approximate value - exact value ∣}{exact value} \cdot 100 = \frac{∣34 000 - 33 600∣}{33 600} \cdot 100 = \frac{400}{33 600} \cdot 100 \approx 1.19%$ .

Venn diagram with U and three overlapping circles A, B and C. Percentages: only A 16, only B 26, only C 29, A and B without C 3, A and C without B 8, B and C without A 12, A, B and C 5

Answer: a) diagram above b) 1% c) 71% d) 45% e) Yes, it is true f) 1.19%

Number sets

Exercise 37 BasicIB · P1 TZ2 May 2014

The following Venn diagram shows the relationship between the sets of numbers $Q$ , $N$ , $Z$ and $R$ .

The number $- 3$ belongs to the sets $Z$ , $Q$ and $R$ but not to $N$ , and it is placed in the appropriate area of the Venn diagram as an example.

Venn diagram of number sets: rectangle ℝ with ℚ, ℤ and ℕ nested (labels outside, joined by a line). The number −3 is in ℤ outside ℕ

Write each of the following numbers in the appropriate place in the Venn diagram.

a) $4$

b) $\frac{1}{3}$

c) $π$

d) $0.38$

e) $5$

f) $- 0.25$

View solution

The sets are nested: $N \subset Z \subset Q \subset R$ . Each number is placed in the smallest set to which it belongs.

$4$ is a positive integer: it goes in $N$ .
$\frac{1}{3}$ is a fraction, not an integer: it goes in $Q$ , outside $Z$ .
$π$ has infinitely many non-recurring decimals: it is irrational, so it goes in $R$ , outside $Q$ .
$0.38 = \frac{38}{100}$ : a terminating decimal, hence rational; it goes in $Q$ , outside $Z$ .
$5$ is not an exact root: it is irrational, so it goes in $R$ , outside $Q$ .
$- 0.25 = - \frac{1}{4}$ : a terminating decimal, hence rational; it goes in $Q$ , outside $Z$ .

Venn diagram of ℝ ⊃ ℚ ⊃ ℤ ⊃ ℕ. 4 in ℕ; −3 in ℤ outside ℕ; −0.25, 0.38 and 1/3 in ℚ outside ℤ; π and √5 in ℝ outside ℚ

Answer: $4 \in N$ ; $\frac{1}{3}, 0.38, - 0.25 \in Q$ (outside $Z$ ); $π, 5 \in R$ (outside $Q$ ).

Exercise 38 BasicIB · P1 TZ2 May 2018

The following table shows four different sets of numbers: $N$ , $Z$ , $Q$ and $R$ .

Set	Example of a number belonging to the set
$N$
$Z$
$Q$
$R$

a) Complete the second column of the table by giving an example of a number that belongs to each set. Josh claims: "All integers are natural numbers."

b) State whether Josh's claim is correct. Justify your answer.

View solution

a) Choose one number from each set (any number that satisfies the definition is valid):

$N$ (natural numbers: $0, 1, 2, \dots$ or $1, 2, 3, \dots$ ): $5$ .
$Z$ (integers, including negative ones): $- 3$ .
$Q$ (rational numbers, quotients of integers): $\frac{3}{5}$ .
$R$ (real numbers; for a clear example take an irrational one): $2$ .

b) To disprove a statement, a single counterexample is enough. $- 3$ is an integer, but it is not natural (natural numbers are not negative). Therefore the claim is false; the correct statement is the reverse: every natural number is an integer ( $N \subset Z$ ).

Answer:

Set	Example of a number belonging to the set
$N$	$5$
$Z$	$- 3$
$Q$	$\frac{3}{5}$
$R$	$2$

b) The claim is not correct; the correct statement is the reverse. For example, $- 3$ is an integer but is not natural.

Exercise 39 BasicIB · P1 TZ2 May 2016

Consider the numbers $- 1$ ; $4$ ; $\frac{2}{3}$ ; $2$ ; $0.35$ and $- 2^{2}$ . Complete the following table. In each box, put a tick (✓) if the number belongs to that set of numbers. We have completed the first row as an example.

	$Z$	$Q$	$R$
$- 1$	✓	✓	✓
$4$
$\frac{2}{3}$
$2$
$0.35$
$- 2^{2}$

Show answer

Answer:

	$N$	$Z$	$Q$	$R$
$- 1$		✓	✓	✓
$4$	✓	✓	✓	✓
$\frac{2}{3}$			✓	✓
$2$				✓
$0.35$			✓	✓
$- 2^{2}$		✓	✓	✓

Show working and answer

Simplify each number and mark all the sets it belongs to (remember $N \subset Z \subset Q \subset R$ ).

$4$ : natural, and therefore integer, rational and real.
$\frac{2}{3}$ : a non-integer fraction, so it is rational and real.
$2$ : the root is not exact, so it is irrational: only real.
$0.35 = \frac{35}{100}$ : a terminating decimal, so rational and real.
$- 2^{2} = - (2^{2}) = - 4$ (the exponent acts before the minus sign): a negative integer, so it is an integer, rational and real, but not natural.

Answer:

	$N$	$Z$	$Q$	$R$
$- 1$		✓	✓	✓
$4$	✓	✓	✓	✓
$\frac{2}{3}$			✓	✓
$2$				✓
$0.35$			✓	✓
$- 2^{2}$		✓	✓	✓

Exercise 40 BasicIB · P1 TZ0 November 2015

Complete the table below, placing the ticks (✓) needed to indicate which of the sets of numbers $N$ , $Z$ , $Q$ and $R$ each of these numbers belongs to. We have completed the first row as an example.

	$Q$	$R$
$0.8$	✓	✓
$14$
$sin 3 0^{\circ}$
$4$
$- 3$
$4.12 \times 1 0^{1}$
$3 \frac{1}{3}$

Show answer

Answer: $14$ : only $R$ ; $sin 3 0^{\circ}$ , $4.12 \times 1 0^{1}$ and $3 \frac{1}{3}$ : $Q$ and $R$ ; $4$ : all four sets; $- 3$ : $Z$ , $Q$ and $R$ (table above).

Show working and answer

For each number, first decide what it really is (simplifying if necessary), then look at the sets: $N = {0, 1, 2, \dots}$ (or ${1, 2, \dots}$ ), $Z$ the integers, $Q$ the fractions of integers and $R$ all real numbers.

$14$ : 14 is not a perfect square, so $14$ is irrational (non-recurring infinite decimals). It is only in $R$ .
$sin 3 0^{\circ}$ : it is a known exact value, $sin 3 0^{\circ} = 0.5 = \frac{1}{2}$ . It is a fraction, but not an integer: it is in $Q$ and $R$ .
$4$ : natural, and therefore integer, rational and real ( $4 = \frac{4}{1}$ ). It is in all four sets.
$- 3$ : negative, so it is not natural. It is an integer, rational ( $- \frac{3}{1}$ ) and real.
$4.12 \times 1 0^{1} = 41.2 = \frac{412}{10}$ : it has a finite number of decimals, so it is rational but not an integer. It is in $Q$ and $R$ .
$3 \frac{1}{3} = \frac{10}{3}$ : a fraction that does not give an integer (recurring decimal $3. 3$ ). It is in $Q$ and $R$ .

	$N$	$Z$	$Q$	$R$
$14$				✓
$sin 3 0^{\circ}$			✓	✓
$4$	✓	✓	✓	✓
$- 3$		✓	✓	✓
$4.12 \times 1 0^{1}$			✓	✓
$3 \frac{1}{3}$			✓	✓

Answer: $14$ : only $R$ ; $sin 3 0^{\circ}$ , $4.12 \times 1 0^{1}$ and $3 \frac{1}{3}$ : $Q$ and $R$ ; $4$ : all four sets; $- 3$ : $Z$ , $Q$ and $R$ (table above).

Exercise 41 BasicIB · P1 TZ1 May 2008

Consider the numbers $2$ ; $3$ ; $- \frac{2}{3}$ and the sets $N$ , $Z$ , $Q$ and $R$ . Complete the following table by putting a tick in the box if the number is an element of the set, and a cross if it is not.

		$N$	$Z$	$Q$	$R$
(i)	$2$
(ii)	$3$
(iii)	$- \frac{2}{3}$

Show answer

Answer: $2$ : all four sets; $3$ : only $R$ ; $- \frac{2}{3}$ : $Q$ and $R$ (table above).

Show working and answer

Classify each number and mark ✓ if it belongs to the set and ✗ if it does not.

$2$ is natural, so also an integer, rational ( $\frac{2}{1}$ ) and real.
$3$ is not a perfect square, so it is irrational: it belongs only to $R$ .
$- \frac{2}{3}$ is negative (so not natural) and not an integer, but it is a fraction of integers: rational and real.

	$N$	$Z$	$Q$	$R$
$2$	✓	✓	✓	✓
$3$	✗	✗	✗	✓
$- \frac{2}{3}$	✗	✗	✓	✓

Answer: $2$ : all four sets; $3$ : only $R$ ; $- \frac{2}{3}$ : $Q$ and $R$ (table above).

Exercise 42 BasicIB · P1 TZ2 May 2013

Consider the numbers $3$ , $- 5$ , $7$ , $2^{- 3}$ and $1.75$ . Complete the table below, placing the ticks (✓) needed to indicate which of the sets of numbers $N$ , $Q$ and $R$ each of these numbers belongs to. The first row has been completed as an example.

	$N$	$Q$	$R$
$3$	✓	✓	✓
$- 5$
$7$
$2^{- 3}$
$1.75$

View solution

There is no $Z$ column here, but the integers still have to be considered: every integer is rational.

$- 5$ : negative, so not natural. It is an integer and therefore rational ( $- \frac{5}{1}$ ) and real.
$7$ : 7 is not a perfect square, so it is irrational: only $R$ .
$2^{- 3} = \frac{1}{2 ^{3}} = \frac{1}{8}$ : a fraction that is not an integer, so it is rational and real, not natural.
$1.75 = \frac{175}{100} = \frac{7}{4}$ : a terminating decimal, rational and real.

	$N$	$Q$	$R$
$3$	✓	✓	✓
$- 5$		✓	✓
$7$			✓
$2^{- 3}$		✓	✓
$1.75$		✓	✓

Answer: $- 5$ , $2^{- 3}$ and $1.75$ : $Q$ and $R$ ; $7$ : only $R$ (table above).

Exercise 43 BasicIB · P1 TZ0 November 2002

Consider the numbers $5$ , $0.5$ , $5$ and $- 5$ . Complete the table below, showing which of the sets of numbers $N$ , $R$ and $Q$ these numbers belong to.

	$N$	$R$	$Q$
$5$			✓
$0.5$	✗
$5$	✗
$- 5$		✓

View solution

$5$ is natural, and therefore real and rational ( $\frac{5}{1}$ ).
$0.5 = \frac{1}{2}$ : not natural, but rational and real.
$5$ : 5 is not a perfect square, so it is irrational: real, but neither rational nor natural.
$- 5$ : negative, so not natural, but rational ( $- \frac{5}{1}$ ) and real.

	$N$	$R$	$Q$
$5$	✓	✓	✓
$0.5$	✗	✓	✓
$5$	✗	✓	✗
$- 5$	✗	✓	✓

Answer: $5$ : $N, R, Q$ ; $0.5$ and $- 5$ : $R$ and $Q$ ; $5$ : only $R$ (table above).

Exercise 44 BasicIB · P1 TZ2 May 2008

Consider the numbers $3$ , $6$ , $2 \frac{1}{2}$ , $π$ , $- 5$ and the sets $N$ , $Z$ and $Q$ . Complete the following table by placing a mark in the box if the number is an element of the set.

	$3$	$6$	$2 \frac{1}{2}$	$π$	$- 5$
$N$
$Z$
$Q$

View solution

Classify each number (mark with an X if it belongs to the set):

$3$ : irrational, not in $N$ , $Z$ or $Q$ .
$6$ : natural, and therefore also integer and rational.
$2 \frac{1}{2} = \frac{5}{2}$ : a non-integer fraction, only in $Q$ .
$π$ : irrational, in none of the three.
$- 5$ : negative integer, in $Z$ and $Q$ , not in $N$ .

	$6$	$2 \frac{1}{2}$	$- 5$
$N$	X
$Z$	X		X
$Q$	X	X	X

Answer: $6 \in N, Z, Q$ ; $- 5 \in Z, Q$ ; $2 \frac{1}{2} \in Q$ ; $3$ and $π$ in none of the three (table above).

Exercise 45 MediumIB · P1 TZ1 May 2019

a) Place the numbers $2 π, - 5, 3^{- 1}$ and $2^{\frac{3}{2}}$ in the correct position in the Venn diagram. Empty Venn diagram of number sets: rectangle ℝ with concentric circles ℚ, ℤ and ℕ

b) In the table, state which of the two given statements are true by placing a tick (✓) in the right-hand column.

Statement	True
$Z \subset Q$
$N \subset Q^{'}$
$N \cap Z = N$
$Q \cup R = Z^{'}$

View solution

a) Simplify each number and place it in the smallest set to which it belongs.

$2 π$ : a non-zero multiple of $π$ , hence irrational. It goes in $R$ , outside $Q$ .
$- 5$ : negative integer. It goes in $Z$ , outside $N$ .
$3^{- 1} = \frac{1}{3}$ : a fraction that is not an integer. It goes in $Q$ , outside $Z$ .
$2^{3/2} = 2^{3} = 8 = 22$ : the root is not exact, so it is irrational. It goes in $R$ , outside $Q$ .

Venn diagram ℝ ⊃ ℚ ⊃ ℤ ⊃ ℕ. 3⁻¹ in ℚ outside ℤ; −5 in ℤ outside ℕ; 2π and 2^(3/2) in ℝ outside ℚ; ℕ empty

b) We analyse each statement:

$Z \subset Q$ : every integer can be written as a fraction ( $n = \frac{n}{1}$ ). True.
$N \subset Q^{'}$ : $Q^{'}$ are the irrational numbers, and the natural numbers are rational. False.
$N \cap Z = N$ : since $N \subset Z$ , the common part is all of $N$ . True.
$Q \cup R = Z^{'}$ : since $Q \subset R$ , the union is $R$ , and $Z^{'}$ (the reals that are not integers) is not $R$ : for example $- 5 \in R$ but $- 5 \in / Z^{'}$ . False.

Answer: a) $3^{- 1} \in Q$ outside $Z$ ; $- 5 \in Z$ outside $N$ ; $2 π$ and $2^{3/2}$ in $R$ outside $Q$ . b) The statements $Z \subset Q$ and $N \cap Z = N$ are true.

Statement	True
$Z \subset Q$	✓
$N \subset Q^{'}$
$N \cap Z = N$	✓
$Q \cup R = Z^{'}$

Exercise 46 MediumIB · P1 TZ0 May 2004

Let $U = {- 4; - \frac{2}{3}; 1; π; 13; 26.7; 69; 1 0^{33}}$ . $A$ is the set of all integers that belong to $U$ . $B$ is the set of all rational numbers that belong to $U$ .

a) Write down all the prime numbers that belong to $U$ .

b) Write down all the elements of $A$ .

c) Write down all the elements of $B$ .

d) Write down all the elements of the set $A \cap B$ .

Show answer

Answer: a) Primes of $U = {13}$ b) $A = {- 4; 1; 13; 69; 1 0^{33}}$ c) $B = {- 4; - \frac{2}{3}; 1; 13; 26.7; 69; 1 0^{33}}$ d) $A \cap B = {- 4; 1; 13; 69; 1 0^{33}}$

Show working and answer

Elements of $U$ : $- 4; - \frac{2}{3}; 1; π; 13; 26.7; 69; 1 0^{33}$ .

a) A prime is a natural number greater than 1 with only two divisors (1 and itself).

$1$ is not prime. $13$ is. $69 = 3 \cdot 23$ is not prime. $1 0^{33}$ is a multiple of 2 and is not prime. Negative numbers and non-integers cannot be prime.
Only $13$ .

b) $A$ : the integers in $U$ are $- 4, 1, 13, 69, 1 0^{33}$ (we discard $- \frac{2}{3}$ , $π$ and $26.7$ , which are not integers).

c) $B$ : rational numbers (they can be written as a fraction of integers). All the elements are rational except $π$ , which is irrational. Integers are rational, and $26.7 = \frac{267}{10}$ .

d) Every integer is rational, so $A \subset B$ and $A \cap B = A$ .

Answer: a) Primes of $U = {13}$ b) $A = {- 4; 1; 13; 69; 1 0^{33}}$ c) $B = {- 4; - \frac{2}{3}; 1; 13; 26.7; 69; 1 0^{33}}$ d) $A \cap B = {- 4; 1; 13; 69; 1 0^{33}}$

Exercise 47 MediumIB · P1 TZ0 November 2009

The following diagram shows the sets of numbers $N$ , $Z$ , $Q$ and $R$ . Place each of the following numbers in the appropriate area of the Venn diagram. $\frac{1}{4}; - 3; π; cos 12 0^{\circ}; 2.7 \times 1 0^{3}; 3.4 \times 1 0^{- 2}$ Diagram of nested number sets, empty: rectangle ℝ containing a large circle ℚ, inside a medium circle ℤ and inside a small circle ℕ, all concentric

View solution

The sets are nested: $N \subset Z \subset Q \subset R$ . Each number is placed in the smallest set to which it belongs.

$\frac{1}{4}$ : a fraction that is not an integer, so it is in $Q$ , outside $Z$ .
$- 3$ : negative integer, so it is in $Z$ , outside $N$ .
$π$ : irrational, so it is in $R$ , outside $Q$ .
$cos 12 0^{\circ} = - \frac{1}{2}$ : a rational non-integer, so it is in $Q$ , outside $Z$ .
$2.7 \times 1 0^{3} = 2700$ : positive integer, so it is in $N$ .
$3.4 \times 1 0^{- 2} = 0.034 = \frac{34}{1000}$ : a rational non-integer, so it is in $Q$ , outside $Z$ .

Nested sets ℕ ⊂ ℤ ⊂ ℚ inside ℝ: in ℕ: 2.7×10³; in ℤ outside ℕ: −3; in ℚ outside ℤ: cos120°, 1/4 and 3.4×10⁻²; in ℝ outside ℚ: π

Answer: $N$ : $2.7 \times 1 0^{3}$ ; $Z$ (outside $N$ ): $- 3$ ; $Q$ (outside $Z$ ): $\frac{1}{4}$ , $cos 12 0^{\circ}$ and $3.4 \times 1 0^{- 2}$ ; $R$ (outside $Q$ ): $π$ .

Exercise 48 MediumIB · P1 TZ0 May 2005

a) Assuming that $x = 2.6 \times 1 0^{4}$ and $y = 5.0 \times 1 0^{- 8}$ , calculate the value of $w = x \cdot y$ . Express the answer in the form $a \times 1 0^{k}$ , with $1 \leq a < 10$ , $k \in Z$ .

b) Two of the following statements about the nature of $x$ , $y$ and $w$ are incorrect. State which they are.

(i) $x \in N$

(ii) $y \in Z$

(iii) $y \in Q$

(iv) $w < y$

(v) $x + y \in R$

(vi) $\frac{1}{w} < x$

Show answer

Answer: a) $w = 1.3 \times 1 0^{- 3}$ ; b) the incorrect ones are (ii) and (iv)

Show working and answer

Multiply: $w = x \cdot y = (2.6 \times 1 0^{4}) (5.0 \times 1 0^{- 8})$ .
Multiply the coefficients and add the exponents: $2.6 \cdot 5.0 = 13$ and $1 0^{4} \cdot 1 0^{- 8} = 1 0^{- 4}$ , so $w = 13 \times 1 0^{- 4}$ .
To have $1 \leq a < 10$ : $13 \times 1 0^{- 4} = 1.3 \times 1 0^{- 3}$ .

b) Values: $x = 26000$ , $y = 0.00000005$ , $w = 0.0013$ .

(i) $26000$ is a natural number: correct.
(ii) $y$ is not an integer (it is a decimal between 0 and 1): incorrect.
(iii) $y$ is a terminating decimal, that is, a fraction: $y \in Q$ , correct.
(iv) $w < y$ would mean $0.0013 < 0.00000005$ , which is not true: incorrect.
(v) The sum of two reals is real: correct.
(vi) $\frac{1}{w} = \frac{1}{0.0013} \approx 769 < 26000 = x$ : correct.

Answer: a) $w = 1.3 \times 1 0^{- 3}$ ; b) the incorrect ones are (ii) and (iv)

Approximation, rounding and scientific notation

Exercise 49 BasicIB · P1 TZ2 May 2011

The planet Earth takes one year to go around the Sun. Assume that a year is 365 days and that the path of the Earth around the Sun is the circumference of a circle of radius 150000000 km. Circular orbit with the Sun, a grey dot, at the centre and the Earth, a black dot, on the circumference at the top, slightly to the left of the vertical

a) Calculate the distance travelled by the Earth in one day.

b) Give your answer to part (a) in the form $a \times 1 0^{k}$ where $1 \leq a < 10$ and $k \in Z$ .

Show answer

Answer: a) Distance travelled in one day $\approx 2580000$ km; b) $2.58 \times 1 0^{6}$

Show working and answer

Distance in one year (one complete revolution): $2 π \times 150000000 = 9.42478 \times 1 0^{8}$ km (approx.).
Divide by the $365$ days of the year: $\frac{9.42478 \times 1 0 ^{8}}{365} \approx 2582131$ km per day.
To three significant figures: $\approx 2580000$ km.

b) Writing with $1 \leq a < 10$ : $2580000 = 2.58 \times 1 0^{6}$ km.

Answer: a) Distance travelled in one day $\approx 2580000$ km; b) $2.58 \times 1 0^{6}$

Exercise 50 BasicIB · P1 TZ0 May 2003

Let $x = 6.4 \times 1 0^{7}$ and $y = 1.6 \times 1 0^{8}$ . Find

a) $\frac{x}{y}$ .

b) $y - 2 x$ , expressing your answers in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) $\frac{x}{y} = 4 \times 1 0^{- 1}$ ; b) $y - 2 x = 3.2 \times 1 0^{7}$

Show working and answer

Divide the coefficients and subtract the exponents: $\frac{x}{y} = \frac{6.4 \times 1 0 ^{7}}{1.6 \times 1 0 ^{8}} = \frac{6.4}{1.6} \times 1 0^{7 - 8}$ .
$\frac{6.4}{1.6} = 4$ , so $\frac{x}{y} = 4 \times 1 0^{- 1}$ (it is $0.4$ , and $1 \leq 4 < 10$ ).

Calculate $2 x = 2 \cdot 6.4 \times 1 0^{7} = 12.8 \times 1 0^{7} = 1.28 \times 1 0^{8}$ .
Subtract with the same exponent: $y - 2 x = 1.6 \times 1 0^{8} - 1.28 \times 1 0^{8} = 0.32 \times 1 0^{8}$ .
Adjust so that $1 \leq a < 10$ : $0.32 \times 1 0^{8} = 3.2 \times 1 0^{7}$ .

Answer: a) $\frac{x}{y} = 4 \times 1 0^{- 1}$ ; b) $y - 2 x = 3.2 \times 1 0^{7}$

Exercise 51 BasicIB · P1 TZ0 May 2002

A rectangle has a length of $2.6 \times 1 0^{4}$ and a width of $1.9 \times 1 0^{4}$ . Find each of the following quantities, giving your answer in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

a) The area of the rectangle;

b) The perimeter of the rectangle.

Show answer

Answer: a) Area $= 4.94 \times 1 0^{8}$ ; b) Perimeter $= 9.0 \times 1 0^{4}$

Show working and answer

Area = length $\times$ width $= (2.6 \times 1 0^{4}) (1.9 \times 1 0^{4})$ .
Coefficients: $2.6 \cdot 1.9 = 4.94$ . Powers: $1 0^{4} \cdot 1 0^{4} = 1 0^{8}$ .
Area $= 4.94 \times 1 0^{8}$ , which already satisfies $1 \leq a < 10$ .

Perimeter $= 2 (length + width) = 2 (2.6 \times 1 0^{4} + 1.9 \times 1 0^{4})$ .
Add with the same exponent: $2.6 + 1.9 = 4.5$ , so the sum is $4.5 \times 1 0^{4}$ .
Multiply by 2: $9.0 \times 1 0^{4}$ (that is $90000$ ).

Answer: a) Area $= 4.94 \times 1 0^{8}$ ; b) Perimeter $= 9.0 \times 1 0^{4}$

Exercise 52 BasicIB · P1 TZ2 May 2009

A rectangle measures 2680 cm long and 1970 cm wide.

a) Find the perimeter of the rectangle, giving your answer in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ and $k \in Z$ .

b) Find the area of the rectangle, giving your answer correct to the nearest thousand square centimetres.

Show answer

Answer: a) Perimeter $= 9.30 \times 1 0^{3}$ cm; b) Area $= 5280000 cm^{2}$

Show working and answer

Perimeter $= 2 (length + width) = 2 (2680 + 1970) = 2 \cdot 4650 = 9300$ cm.
Write in scientific notation: the decimal point moves three places to the left, $9300 = 9.30 \times 1 0^{3}$ .

Area $= 2680 \cdot 1970 = 5279600 cm^{2}$ .
To the nearest thousand: we look at the hundreds digit, which is $6$ (greater than or equal to 5), so we round up the thousands: $5280000 cm^{2}$ .

Answer: a) Perimeter $= 9.30 \times 1 0^{3}$ cm; b) Area $= 5280000 cm^{2}$

Exercise 53 MediumIB · P1 TZ1 May 2013

$z = \frac{17 x ^{2}}{a - b} .$

a) Find the value of $z$ when $x = 12.5$ , $a = 0.572$ and $b = 0.447$ . Write down the full value shown on your calculator display.

b) Write your answer to part (a) (i) rounded to the nearest 1,000; (ii) rounded to three significant figures.

c) Write your answer to part (b)(ii) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$

Show answer

Answer: a) $z = 21250$ ; b) (i) $z \approx 21000$ ; (ii) $z \approx 21300$ ; c) $z \approx 2.13 \times 1 0^{4}$

Show working and answer

We substitute into the formula: $z = \frac{17 \times 12. 5 ^{2}}{0.572 - 0.447}$ .
Numerator: $12. 5^{2} = 156.25$ and $17 \times 156.25 = 2656.25$ .
Denominator: $0.572 - 0.447 = 0.125$ .
$z = \frac{2656.25}{0.125} = 21250$ (this is the full value on the display).

(i) Rounding to the nearest thousand: $21250$ lies between $21000$ and $22000$ , and its hundreds digit is $2$ (less than $5$ ), so it rounds down to $21000$ .

(ii) To three significant figures: the first three digits are $2, 1, 2$ and the next one is $5$ , so we round up: $21300$ .

c) To write it in the form $a \times 1 0^{k}$ with $1 \leq a < 10$ , we move the decimal point $4$ places to the left: $21300 = 2.13 \times 1 0^{4}$ .

Answer: a) $z = 21250$ ; b) (i) $z \approx 21000$ ; (ii) $z \approx 21300$ ; c) $z \approx 2.13 \times 1 0^{4}$

Exercise 54 MediumIB · P1 TZ2 May 2012

Consider $c = 5200$ and $d = 0.0000037$ .

a) Write down the value of $r = c \times d$ .

b) Write down the value of $r$ in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ and $k \in Z$ .

c) Consider the following statements about $c$ , $d$ and $r$ . Only three of these statements are true. Circle the true statements.


$c \in N$
$d \in Z$
$d \in Q$
$r < d$
$c + d \in R$
$\frac{1}{r} > c$

Show answer

Answer: a) $r = 0.01924$ ; b) $r = 1.924 \times 1 0^{- 2}$ ; c) The true statements are: $c \in N$ , $d \in Q$ and $c + d \in R$ .

Show working and answer

a) $r = c \times d = 5200 \times 0.0000037 = 0.01924$ .

b) To have $1 \leq a < 10$ we move the decimal point two places to the right to obtain $1.924$ ; this is equivalent to multiplying by $1 0^{- 2}$ : $r = 1.924 \times 1 0^{- 2}$ .

c) We check each statement:

$c \in N$ : $5200$ is a positive integer. True.
$d \in Z$ : $d = 0.0000037$ is not an integer. False.
$d \in Q$ : it is a terminating decimal, so it can be written as the fraction $\frac{37}{1 0 ^{7}}$ . True.
$r < d$ : $r = 0.01924$ and $d = 0.0000037$ , so $r > d$ . False.
$c + d \in R$ : the sum of two real numbers is real. True.
$\frac{1}{r} > c$ : $\frac{1}{0.01924} \approx 51.97$ , which is less than $5200$ . False.

Answer: a) $r = 0.01924$ ; b) $r = 1.924 \times 1 0^{- 2}$ ; c) The true statements are: $c \in N$ , $d \in Q$ and $c + d \in R$ .

Exercise 55 MediumIB · P1 TZ1 May 2012

Ross is a star that is $82414080000000$ km from the Earth. A spacecraft launched from the Earth travels at $48000$ kmh $^{- 1}$ towards Ross.

a) Calculate the exact time, in hours, for the spacecraft to reach the star Ross.

b) Give your answer to part (a) in years. (Assume that 1 year = 365 days.)

c) Give your answer to part (b) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ and $k \in Z$ .

Show answer

Answer: a) $1716960000$ hours; b) $196000$ years; c) $1.96 \times 1 0^{5}$

Show working and answer

a) We use $time = \frac{distance}{speed}$ :

$t = \frac{82414080000000}{48000} = 1716960000 hours .$

One year of $365$ days has $365 \times 24 = 8760$ hours.
Converting hours to years by dividing: $\frac{1716960000}{8760} = 196000$ years.

c) Moving the decimal point $5$ places to the left: $196000 = 1.96 \times 1 0^{5}$ .

Answer: a) $1716960000$ hours; b) $196000$ years; c) $1.96 \times 1 0^{5}$

Exercise 56 MediumIB · P1 TZ1 May 2011

A satellite travels around the Earth in a circular orbit $500$ kilometres above the surface of the Earth. The radius of the Earth is taken as $6400$ kilometres.

a) Write down the radius of the satellite's orbit.

b) Calculate the distance travelled by the satellite in one orbit of the Earth. Give your answer correct to the nearest km.

c) Write your answer to part (b) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) $r = 6900$ km; b) Distance of orbit $\approx 43354$ km; c) $4.3354 \times 1 0^{4}$

Show working and answer

a) The radius of the orbit is the radius of the Earth plus the height of the satellite: $r = 6400 + 500 = 6900$ km.

b) The distance in one orbit is the circumference of the circle, $2 π r$ :

$2 π \times 6900 = 13800 π \approx 43353.98 \dots$

Rounding to the nearest km: $43354$ km.

c) Moving the decimal point $4$ places to the left: $43354 = 4.3354 \times 1 0^{4}$ km.

Answer: a) $r = 6900$ km; b) Distance of orbit $\approx 43354$ km; c) $4.3354 \times 1 0^{4}$

Exercise 57 MediumIB · P1 TZ2 May 2019

The shape of the Moon can be modelled by a sphere of diameter 3,474,000 metres.

a) Use this model to calculate the circumference of the Moon in kilometres. Write down the full value shown on the calculator display.

b) Give the answer you obtained in part (a) rounded to three significant figures.

c) Write the answer you obtained in part (b) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) $10913.89288$ km; b) $\approx 10900$ km; c) $\approx 1.09 \times 1 0^{4}$ km

Show working and answer

We convert the diameter to kilometres: $3474000$ m $= 3474$ km (divide by $1000$ ).
The circumference of a sphere (great circle) is $C = π d = π \times 3474$ .
On the calculator: $C = 10913.89288 \dots$ km.

b) To three significant figures: the first three digits are $1, 0, 9$ and the next one is $1$ (less than $5$ ), so we keep the $9$ and fill with zeros: $10900$ km.

c) Moving the decimal point $4$ places to the left: $10900 = 1.09 \times 1 0^{4}$ km.

Answer: a) $10913.89288$ km; b) $\approx 10900$ km; c) $\approx 1.09 \times 1 0^{4}$ km

Exercise 58 MediumIB · P1 TZ2 May 2014

The average radius of the orbit described by the Earth around the Sun is equal to 150 million kilometres. Circular orbit of the Earth around the Sun: the Sun is a grey circle at the centre, a double-headed arrow marks the radius up to the circumference, and the Earth is a black dot on the circumference, top right

a) Write this radius, in kilometres, in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ . The Earth travels around the Sun along an orbit, taking one year to complete this orbit.

b) Calculate the distance, in kilometres, travelled by the Earth around the Sun in one orbit, assuming the orbit is a circle. Today Anna celebrates her 17th birthday.

c) Calculate the total distance Anna has travelled around the Sun since she was born.

Show answer

Answer: a) $1.5 \times 1 0^{8}$ ; b) Distance of one orbit $= 9.42 \times 1 0^{8}$ ; c) Distance travelled by Anna $= 1.60 \times 1 0^{10}$

Show working and answer

a) $150$ million $= 150000000 = 1.5 \times 1 0^{8}$ km (the decimal point moves $8$ places to the left).

b) The orbit is a circle of radius $1.5 \times 1 0^{8}$ km:

$2 π r = 2 π \times 1.5 \times 1 0^{8} = 3 π \times 1 0^{8} \approx 9.42 \times 1 0^{8} km .$

c) Anna has completed $17$ full orbits around the Sun (one per year lived). We multiply:

$17 \times 9.424778 \times 1 0^{8} \approx 1.602 \times 1 0^{10} \approx 1.60 \times 1 0^{10} km .$

Answer: a) $1.5 \times 1 0^{8}$ ; b) Distance of one orbit $= 9.42 \times 1 0^{8}$ ; c) Distance travelled by Anna $= 1.60 \times 1 0^{10}$

Exercise 59 MediumIB · P1 TZ1 May 2010

The volume of a sphere is $V = \frac{S ^{3}}{36 π}$ , where $S$ is its surface area. The surface area of the sphere is $500 cm^{2}$ .

a) Calculate the volume of the sphere. Give your answer correct to two decimal places.

b) Write your answer to part (a) correct to the nearest integer.

c) Write your answer to part (b) in the form $a \times 1 0^{n}$ , where $1 \leq a < 10$ and $n \in Z$ .

Show answer

Answer: a) $V = 1051.31 cm^{3}$ ; b) $V \approx 1051 cm^{3}$ ; c) $V \approx 1.051 \times 1 0^{3}$

Show working and answer

Substitute $S = 500$ : $V = \frac{50 0 ^{3}}{36 π}$ .
$50 0^{3} = 125000000$ and $36 π \approx 113.097$ , so the radicand is $\approx 1105 242.7$ .
Square root with the calculator: $V \approx 1051.305 \dots$ , which to two decimal places is $1051.31$ cm $^{3}$ .

b) To the nearest integer: the decimal part $0.31$ is less than $0.5$ , so $V \approx 1051$ cm $^{3}$ .

c) Moving the decimal point $3$ places to the left: $1051 = 1.051 \times 1 0^{3}$ .

Answer: a) $V = 1051.31 cm^{3}$ ; b) $V \approx 1051 cm^{3}$ ; c) $V \approx 1.051 \times 1 0^{3}$

Exercise 60 MediumIB · P1 TZ0 November 2010

Consider these four numbers shown below. $p = 0.00314; q = 0.00314 \times 1 0^{2}; r = \frac{π}{1000}; s = 3.14 \times 1 0^{- 2}$

a) One of these numbers is written in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ and $k \in Z$ . Write down this number.

b) Write down the smallest of the four numbers shown.

c) Write down the value of $q + s$ .

d) Give your answer to part (c) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ and $k \in Z$ .

Show answer

Answer: a) $s = 3.14 \times 1 0^{- 2}$ ; b) The smallest is $p$ ; c) $q + s = 0.3454$ ; d) $q + s = 3.454 \times 1 0^{- 1}$

Show working and answer

a) The form $a \times 1 0^{k}$ requires $1 \leq a < 10$ :

$p = 0.00314$ is not in that form.
$q = 0.00314 \times 1 0^{2}$ has $a = 0.00314 < 1$ : not valid.
$r = \frac{π}{1000}$ does not have that form.
$s = 3.14 \times 1 0^{- 2}$ has $a = 3.14$ and $k = - 2$ : yes.

b) We convert all of them to decimals:

$p = 0.00314$
$q = 0.314$
$r = \frac{π}{1000} = 0.00314159 \dots$
$s = 0.0314$

The smallest is $p = 0.00314$ (it is slightly smaller than $r$ ).

c) $q + s = 0.314 + 0.0314 = 0.3454$ .

d) Moving the decimal point one place to the right to have $a = 3.454$ : $0.3454 = 3.454 \times 1 0^{- 1}$ .

Answer: a) $s = 3.14 \times 1 0^{- 2}$ ; b) The smallest is $p$ ; c) $q + s = 0.3454$ ; d) $q + s = 3.454 \times 1 0^{- 1}$

Exercise 61 MediumIB · P1 TZ0 November 2008

Given that $h = l^{2} - \frac{d ^{2}}{4}$ .

a) Calculate the exact value of $h$ when $l = 0.03625$ and $d = 0.05$ .

b) Write your answer to part (a) correct to three decimal places.

c) Write your answer to part (a) correct to three significant figures.

d) Write your answer to part (a) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) $h = 0.02625$ ; b) $h \approx 0.026$ ; c) $h \approx 0.0263$ ; d) $h = 2.625 \times 1 0^{- 2}$

Show working and answer

Substitute the data into the formula: $h = 0.0362 5^{2} - \frac{0.0 5 ^{2}}{4}$ .
Calculate each term: $0.0362 5^{2} = 0.0013140625$ and $\frac{0.0 5 ^{2}}{4} = \frac{0.0025}{4} = 0.000625$ .
Subtract: $0.0013140625 - 0.000625 = 0.0006890625$ .
Square root: $0.0006890625 = 0.02625$ (check: $0.0262 5^{2} = 0.0006890625$ ). It is an exact value, with few decimal places.

b) Three decimal places: we look at the fourth decimal. In $0.02625$ the first three decimals are $0.026$ and the next digit is $2$ (less than 5), so we do not round up: $0.026$ .

c) Three significant figures: the first significant figure is the $2$ (leading zeros do not count). We take $2, 6, 2$ and look at the next digit, which is $5$ : we round up, $0.0263$ .

d) Moving the decimal point until only one non-zero digit is in front: $0.02625 = 2.625 \times 1 0^{- 2}$ (two places to the right, exponent $- 2$ ).

Answer: a) $h = 0.02625$ ; b) $h \approx 0.026$ ; c) $h \approx 0.0263$ ; d) $h = 2.625 \times 1 0^{- 2}$

Exercise 62 MediumIB · P1 TZ0 November 2006

Let $x = 7.94$ .

a) Calculate the value of $\frac{2 x + 1}{x ^{3}}$ .

b) (i) Give your answer correct to three decimal places.

(ii) Write your answer to (b) (i) as a percentage.

c) Give your answer to part (b) (i) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) $0.0337$ ; b) (i) $0.034$ ; (ii) $3.4%$ ; c) $3.4 \times 1 0^{- 2}$

Show working and answer

Numerator: $2 x + 1 = 2 \cdot 7.94 + 1 = 16.88$ .
Denominator: $x^{3} = 7.9 4^{3} = 500.566184$ .
Quotient (with the calculator): $\frac{16.88}{500.566184} = 0.03372 \dots \approx 0.0337$ .

(i) Three decimal places: $0.03372 \dots \to 0.034$ (the fourth decimal is $7$ , so we round up).
(ii) To convert to a percentage we multiply by $100$ , using the value from part (i): $0.034 \cdot 100 = 3.4%$ .

c) $0.034 = 3.4 \times 1 0^{- 2}$ (the decimal point moves two places to the right).

Answer: a) $0.0337$ ; b) (i) $0.034$ ; (ii) $3.4%$ ; c) $3.4 \times 1 0^{- 2}$

Exercise 63 MediumIB · P1 TZ0 November 2005

Given the equation $p = r^{2} + 2 q r$ .

a) Calculate the exact value of $p$ when $q = 3.6$ and $r = 24$ .

b) Write your answer to part (a) correct to two significant figures.

c) Express your answer to part (b) in the form $a \times 1 0^{k}$ , with $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) $p = 748.8$ ; b) $p \approx 750$ ; c) $p \approx 7.5 \times 1 0^{2}$

Show working and answer

Substitute: $p = 2 4^{2} + 2 \cdot 3.6 \cdot 24$ .
Calculate: $2 4^{2} = 576$ and $2 \cdot 3.6 \cdot 24 = 172.8$ .
Add: $p = 576 + 172.8 = 748.8$ .

b) Two significant figures: $7$ and $4$ ; the next digit is $8$ , so we round up to $75$ . As the number has three integer digits, we fill in with a zero: $750$ .

c) $750 = 7.5 \times 1 0^{2}$ (the decimal point moves two places to the left).

Answer: a) $p = 748.8$ ; b) $p \approx 750$ ; c) $p \approx 7.5 \times 1 0^{2}$

Exercise 64 MediumIB · P1 TZ0 November 2004

The total weight of 256 identical pencils is 4.24 kg. Calculate the weight, in kg, of one pencil.

a) Give the exact answer.

b) Express the answer with three significant figures.

c) Express your answer to part (b) in the form $a \times 1 0^{k}$ , with $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) $0.0165625$ kg; b) $\approx 0.0166$ kg; c) $1.66 \times 1 0^{- 2}$ kg

Show working and answer

Weight of one pencil = total weight divided by the number of pencils (all are identical).
$\frac{4.24}{256} = 0.0165625$ kg, exact value (the division terminates).

b) Three significant figures: they start at the $1$ (leading zeros do not count): $1, 6, 5$ ; the next digit is $6$ , so we round the $5$ up to $6$ : $0.0166$ kg.

c) $0.0166 = 1.66 \times 1 0^{- 2}$ (the decimal point moves two places to the right).

Answer: a) $0.0165625$ kg; b) $\approx 0.0166$ kg; c) $1.66 \times 1 0^{- 2}$ kg

Exercise 65 MediumIB · P1 TZ0 November 2003

Using the formula $V = π r^{2} (H - h)$ , and the value of $π$ given by the calculator, find the value of $V$ when $r = 4.26$ ; $H = 21.58$ and $h = 14.35$ .

a) Write down the full result shown on your calculator.

b) Express the answer correct to two decimal places.

c) Express the answer correct to two significant figures.

d) Write down your answer to part (c) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) $V = 412.1994123$ ; b) $V \approx 412.20$ ; c) $V \approx 410$ ; d) $V \approx 4.10 \times 1 0^{2}$

Show working and answer

Substitute: $V = π \cdot 4.2 6^{2} \cdot (21.58 - 14.35)$ .
Calculate $H - h = 7.23$ and $4.2 6^{2} = 18.1476$ .
Multiply: $18.1476 \cdot 7.23 = 131.207148$ .
Multiply by $π$ (using the calculator's value): $V = 412.1994123 \dots$

b) Two decimal places: the third decimal is $9$ , so we round up: $412.20$ .

c) Two significant figures: $4$ and $1$ ; the next digit is $2$ , so we do not round up, and we fill with zeros up to the units position: $410$ .

d) $410 = 4.10 \times 1 0^{2}$ (the decimal point moves two places to the left).

Answer: a) $V = 412.1994123$ ; b) $V \approx 412.20$ ; c) $V \approx 410$ ; d) $V \approx 4.10 \times 1 0^{2}$

Exercise 66 MediumIB · P1 TZ0 November 2001

The speed of sound in air is $300 m s^{- 1}$ .

a) How many metres does sound travel in air in one hour?

b) Express your answer to part (a)

(i) correct to two significant figures;

(ii) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) $1080000$ m; b) (i) $1100000$ m; (ii) $1.08 \times 1 0^{6}$ m

Show working and answer

One hour is $60 \cdot 60 = 3600$ s.
Distance = speed $\times$ time $= 300 \cdot 3600 = 1080000$ m.

(i) Two significant figures: $1$ and $0$ ; the next digit is $8$ , so we round up: $1100000$ m.
(ii) Starting from the value in part (a), the decimal point moves six places to the left: $1080000 = 1.08 \times 1 0^{6}$ m.

Answer: a) $1080000$ m; b) (i) $1100000$ m; (ii) $1.08 \times 1 0^{6}$ m

Exercise 67 MediumIB · P1 TZ0 November 2018

The volume, $V$ , of a hemisphere is given by the formula $V = \frac{4 S ^{3}}{243 π}$ , where $S$ is the total surface area. The total surface area of a given hemisphere is equal to $350 cm^{2}$ .

a) Calculate the volume in $cm^{3}$ of this hemisphere. Give your answer correct to one decimal place.

b) Write your answer to part (a) correct to the nearest integer.

c) Write your answer to part (b) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ and $k \in Z$ .

Show answer

Answer: a) $V \approx 474.0 cm^{3}$ ; b) $V \approx 474 cm^{3}$ ; c) $V = 4.74 \times 1 0^{2} cm^{3}$

Show working and answer

Substitute $S = 350$ : $V = \frac{4 \cdot 35 0 ^{3}}{243 π}$ .
Calculate the numerator: $35 0^{3} = 42875000$ , and multiplied by 4 this gives $171500000$ .
With the calculator (using $π$ from the calculator), the inner quotient is $\approx 224651$ and its square root gives $V \approx 473.97$ .
One decimal place: the second decimal is $7$ , so we round up, giving $V \approx 474.0 cm^{3}$ .

b) To the nearest integer: $474 cm^{3}$ .

c) $474 = 4.74 \times 1 0^{2} cm^{3}$ (the decimal point moves two places to the left).

Answer: a) $V \approx 474.0 cm^{3}$ ; b) $V \approx 474 cm^{3}$ ; c) $V = 4.74 \times 1 0^{2} cm^{3}$

Exercise 68 MediumIB · P1 TZ0 November 2017

The speed of light is equal to 300,000 kilometres per second. The average distance between the Sun and the Earth is equal to 149.6 million km.

a) Calculate the time, in minutes, that light from the Sun takes to reach the Earth. A light year is the distance travelled by light in one year and is equal to 9,467,280 million km. Polaris is a bright star that is visible from the northern hemisphere. The distance between the Earth and Polaris is equal to 323 light years.

b) Find the distance between the Earth and Polaris in millions of km. Give the answer in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ and $k \in Z$ .

Show answer

Answer: a) $\approx 8.31$ minutes b) $\approx 3.06 \times 1 0^{9}$ million km

Show working and answer

Convert the distance to km: $149.6$ million km $= 149.6 \times 1 0^{6}$ km.
Time $= \frac{distance}{speed}$ , so in seconds: $t = \frac{149.6 \times 1 0 ^{6}}{300 000} \approx 498.67 s$
A minute has 60 seconds, so we divide by 60: $t = \frac{498.67}{60} \approx 8.311 min$

The distance is $323$ light years and each light year is $9467280$ million km, so we multiply: $323 \times 9467280 = 3057931440 million km$
To write it in the form $a \times 1 0^{k}$ with $1 \leq a < 10$ , the decimal point moves 9 places to the left: $3057931440 = 3.05793 \dots \times 1 0^{9}$
Rounding to 3 significant figures: $3.06 \times 1 0^{9}$ .

Answer: a) $\approx 8.31$ minutes b) $\approx 3.06 \times 1 0^{9}$ million km

Exercise 69 MediumIB · P1 TZ2 May 2018

Last year, a South American sweet factory sold $4.8 \times 1 0^{8}$ spherical sweets. Each sweet has a diameter of 2.5 cm. The factory is preparing an advertisement showing all these sweets placed in a row, forming a straight line. A row of equal circles touching each other, continued with dots on both sides; below each circle a double-headed arrow shows its diameter: 2.5 cm

a) Find the length, in cm, of this line. Give your answer in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ and $k \in Z$ . The advertisement claims that the length of this line is equal to $x$ times the length of the Amazon River. The Amazon River is 6,400 km long.

b) (i) Write down the length of the Amazon River in cm.

(ii) Find the value of $x$ .

Show answer

Answer: a) $1.2 \times 1 0^{9}$ cm b) i) $6.4 \times 1 0^{8}$ cm ii) $x = 1.88$

Show working and answer

The length of the row is the number of sweets times the length of each one (they are touching, tangent): $L = 4.8 \times 1 0^{8} \times 2.5 = 12 \times 1 0^{8} cm$
For $a$ to be between 1 and 10, we write $12 = 1.2 \times 10$ : $L = 1.2 \times 1 0^{9} cm$

(i) We convert km to cm: $1 km = 1000 m = 100000 cm = 1 0^{5}$ cm. Then: $6400 \times 1 0^{5} = 6.4 \times 1 0^{3} \times 1 0^{5} = 6.4 \times 1 0^{8} cm$
(ii) The length of the row is $x$ times that of the Amazon, so $x = \frac{1.2 \times 1 0 ^{9}}{6.4 \times 1 0 ^{8}} = 1.875 \approx 1.88$

Answer: a) $1.2 \times 1 0^{9}$ cm b) i) $6.4 \times 1 0^{8}$ cm ii) $x = 1.88$

Exercise 70 MediumIB · P1 TZ0 November 2017

The speed of light is equal to 300,000 kilometres per second. The average distance between the Sun and the Earth is equal to 149.6 million km.

b) Find the distance between the Earth and Polaris in millions of km. Give the answer in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ and $k \in Z$ .

Show answer

Answer: a) $\approx 8.31$ minutes b) $\approx 3.06 \times 1 0^{9}$ million km

Show working and answer

Convert the distance to km: $149.6$ million km $= 149.6 \times 1 0^{6}$ km.
Time $= \frac{distance}{speed}$ , so in seconds: $t = \frac{149.6 \times 1 0 ^{6}}{300 000} \approx 498.67 s$
A minute has 60 seconds, so we divide by 60: $t = \frac{498.67}{60} \approx 8.311 min$

The distance is $323$ light years and each light year is $9467280$ million km, so we multiply: $323 \times 9467280 = 3057931440 million km$
To write it in the form $a \times 1 0^{k}$ with $1 \leq a < 10$ , the decimal point moves 9 places to the left: $3057931440 = 3.05793 \dots \times 1 0^{9}$
Rounding to 3 significant figures: $3.06 \times 1 0^{9}$ .

Answer: a) $\approx 8.31$ minutes b) $\approx 3.06 \times 1 0^{9}$ million km

Exercise 71 MediumIB · P1 TZ0 November 2007

a) Calculate $\frac{77.2 \times 3 ^{3}}{3.60 \times 2 ^{2}}$ .

b) Express your answer to part (a) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

c) Juan estimates that the length of a carpet is 12 metres and the width is 8 metres. He then estimates the area of the carpet.

(i) Write down his estimated area of the carpet. When the carpet is measured accurately, it is found to have an area of 90 square metres.

(ii) Calculate the percentage error made by Juan.

Show answer

Answer: a) $144.75$ b) $1.4475 \times 1 0^{2}$ c) i) $96 m^{2}$ ii) $6.67 %$

Show working and answer

Numerator: $77.2 \times 3^{3} = 77.2 \times 27 = 2084.4$ .
Denominator: $3.60 \times 2^{2} = 3.60 \times 4 = 14.4$ .
Divide: $\frac{2084.4}{14.4} = 144.75$ .

Move the decimal point two places to the left so that $1 \leq a < 10$ : $144.75 = 1.4475 \times 1 0^{2}$

(i) Estimated area $=$ length $\times$ width $= 12 \times 8 = 96 m^{2}$ .
(ii) The exact value is $90 m^{2}$ and the estimate is $96 m^{2}$ : $\frac{96 - 90}{90} \times 100 = \frac{6}{90} \times 100 \approx 6.67 %$

Answer: a) $144.75$ b) $1.4475 \times 1 0^{2}$ c) i) $96 m^{2}$ ii) $6.67 %$

Exercise 72 MediumIB · P1 TZ0 May 2006

A problem has an exact answer of $x = 0.1265$ .

a) Write the exact value of $x$ in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

b) State the value of $x$ correct to two significant figures.

c) Calculate the percentage error if the value of $x$ is approximated to two significant figures.

Show answer

Answer: a) $x = 1.265 \times 1 0^{- 1}$ b) $x \approx 0.13$ c) $2.77 %$

Show working and answer

Move the decimal point one place to the right to get a number between 1 and 10, which forces a negative exponent: $0.1265 = 1.265 \times 1 0^{- 1}$

The first two significant figures are 1 and 2 (the leading zero does not count). The next digit is 6, so we round up: $x \approx 0.13$ .

Exact value $0.1265$ , approximate value $0.13$ : $\frac{0.13 - 0.1265}{0.1265} \times 100 = \frac{0.0035}{0.1265} \times 100 \approx 2.77 %$

Answer: a) $x = 1.265 \times 1 0^{- 1}$ b) $x \approx 0.13$ c) $2.77 %$

Percentage error

Exercise 73 MediumIB · P1 TZ1 May 2018

Each year, the Peterson United football team plays 25 matches at its home stadium. The owner of Peterson United claimed that last year the average attendance per match at its home stadium was 24,500.

a) Based on the owner's information, calculate the total attendance at Peterson United's home stadium last year. The actual total attendance last year was 617,700.

b) Calculate the percentage error in the owner's claim.

c) Write your answer to part (b) in the form $a \times 1 0^{k}$ where $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) $612500$ b) $0.842 %$ c) $8.42 \times 1 0^{- 1}$

Show working and answer

Estimated total attendance = matches $\times$ average attendance: $25 \times 24500 = 612500$

The approximate (claimed) value is $612500$ and the exact value is $617700$ .
Percentage error $= \frac{v _{A} - v _{E}}{v _{E}} \times 100$ : $\frac{612 500 - 617 700}{617 700} \times 100 = \frac{5200}{617 700} \times 100 \approx 0.842 %$

We write $0.842 = 8.42 \times 1 0^{- 1}$ (the decimal point moves one place to the right, so the exponent is $- 1$ ).

Answer: a) $612500$ b) $0.842 %$ c) $8.42 \times 1 0^{- 1}$

Exercise 74 MediumIB · P1 TZ1 May 2008

a) Calculate exactly $\frac{( 3 \times 2.1 ) ^{3}}{7 \times 1.2}$ .

b) Write your answer to part (a) correct to 2 significant figures.

c) Calculate the percentage error when the answer to part (a) is written as an approximation to 2 significant figures.

d) Write your answer to part (c) in the form $a \times 1 0^{k}$ where $1 \leq a < 10$ and $k \in Z$ .

Show answer

Answer: a) $29.7675$ b) $30$ c) $0.781 %$ d) $7.81 \times 1 0^{- 1}$

Show working and answer

Calculate the bracket: $3 \times 2.1 = 6.3$ , and raise it to the cube: $6. 3^{3} = 250.047$ .
Calculate the denominator: $7 \times 1.2 = 8.4$ .
Divide: $\frac{250.047}{8.4} = 29.7675$

$29.7675$ rounded to 2 significant figures (the first two significant figures are 2 and 9, and the next digit, 7, means we round up): $30$ .

The exact value is $29.7675$ and the approximate value is $30$ : $\frac{30 - 29.7675}{29.7675} \times 100 = \frac{0.2325}{29.7675} \times 100 \approx 0.781 %$

$0.781 = 7.81 \times 1 0^{- 1}$ .

Answer: a) $29.7675$ b) $30$ c) $0.781 %$ d) $7.81 \times 1 0^{- 1}$

Exercise 75 MediumIB · P1 TZ0 May 2007

Five pipes labelled "6 metres long" were delivered to a building site. The contractor measured each pipe to check its length (in metres), and recorded the following values: $5.96; 5.95; 6.02; 5.95; 5.99.$

a) (i) Find the mean of the values measured by the contractor.

(ii) Calculate the percentage error between the mean and the stated (approximate) length of 6 metres.

b) Calculate $3.8 7^{5} - 8.7 3^{- 0.5}$ , and express your answer

(i) rounded to the nearest integer,

(ii) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10$ , $k \in Z$ .

Show answer

Answer: a) i) mean $= 5.97$ m ii) $0.435 %$ (with the rounded mean $5.97$ : $0.503 %$ ) b) i) $29$ ii) $2.95 \times 1 0^{1}$

Show working and answer

(i) The mean is the sum of the values divided by how many there are (5): $x ˉ = \frac{5.96 + 5.95 + 6.02 + 5.95 + 5.99}{5} = \frac{29.87}{5} = 5.974 \approx 5.97 m$
(ii) The exact value is the mean, $5.974$ m, and the approximate value is $6$ m: $\frac{6 - 5.974}{5.974} \times 100 = \frac{0.026}{5.974} \times 100 \approx 0.435 %$
Note: if the rounded mean $5.97$ is used, the calculation is $\frac{0.03}{5.97} \times 100 \approx 0.503 %$ .

Calculate using the GDC, respecting the order of operations: $3.8 7^{5} \approx 868.07$ and $8.7 3^{- 0.5} = \frac{1}{8.73} \approx 0.338$ .
The subtraction gives $868.07 - 0.338 \approx 867.73$ and the square root is: $3.8 7^{5} - 8.7 3^{- 0.5} \approx 29.457$
(i) To the nearest integer: $29$ .
(ii) In the form $a \times 1 0^{k}$ : $29.457 = 2.9457 \times 1 0^{1} \approx 2.95 \times 1 0^{1}$ (3 significant figures).

Answer: a) i) mean $= 5.97$ m ii) $0.435 %$ (with the rounded mean $5.97$ : $0.503 %$ ) b) i) $29$ ii) $2.95 \times 1 0^{1}$

Linear equations, inequalities and systems

Exercise 76 BasicIB · P1 TZ0 May 2002

a) Solve the inequality $3 - 2 (x + 5) > x - 6$ .

b) Plot the solution of part (a) on the following number line. Empty number line labelled −3, −2, −1, 0, 1, 2 and 3

Show answer

Answer: a) $x < - \frac{1}{3}$ b) See figure.

Show working and answer

We remove the bracket: $3 - 2 x - 10 > x - 6$ .
We simplify the left-hand side: $- 2 x - 7 > x - 6$ .
We collect the $x$ terms on one side and the numbers on the other: we add $2 x$ and add $6$ to both sides: $- 1 > 3 x$ .
We divide by $3$ (positive, so the direction does not change): $- \frac{1}{3} > x$ , i.e. $x < - \frac{1}{3}$ .

We locate $- \frac{1}{3} \approx - 0.33$ between $- 1$ and $0$ , closer to $0$ .
As the inequality is strict ( $<$ ), the point is marked with an open circle.
The valid values are less than $- \frac{1}{3}$ , so we draw a thick line to the left with an arrow.

Number line with marks −3, −2, −1, 0, 1, 2, 3: open circle at x = −1/3 and thick line to the left with an arrow (x < −1/3)

Answer: a) $x < - \frac{1}{3}$ b) See figure.

Exercise 77 BasicIB · P1 TZ0 November 2002

The following figures show inequalities in $x$ . Four number lines: (a) x > −1, open circle at −1 and a line to the right with an arrow; (b) x ≤ 2, closed dot at 2 and a line to the left with an arrow; (c) −1 < x ≤ 2, open circle at −1 and closed dot at 2 joined by a line; (d) x < −1 or x ≥ 1, two lines with arrows, one from −1 (open) to the left and another from 1 (closed) to the right Write down the inequality represented in each of the figures.

Show answer

Answer: a) $x > - 1$ b) $x \leq 2$ c) $- 1 < x \leq 2$ d) $x < - 1$ or $x \geq 1$

Show working and answer

Read each number line by looking at the type of point and the direction of the line: an open circle means $<$ or $>$ (the endpoint is not included), a closed dot means $\leq$ or $\geq$ (the endpoint is included).

(a) Open circle at $- 1$ and a line to the right: the values greater than $- 1$ , not including it: $x > - 1$ .
(b) Closed dot at $2$ and a line to the left: the values less than or equal to $2$ : $x \leq 2$ .
(c) Open circle at $- 1$ and closed dot at $2$ joined by a line: the values between both, excluding $- 1$ and including $2$ : $- 1 < x \leq 2$ .
(d) Two separate lines: one to the left from $- 1$ (open) and another to the right from $1$ (closed). They are joined by "or": $x < - 1$ or $x \geq 1$ .

Answer: a) $x > - 1$ b) $x \leq 2$ c) $- 1 < x \leq 2$ d) $x < - 1$ or $x \geq 1$

Exercise 78 BasicIB · P1 TZ0 May 2001

a) Solve the inequality $2 x + 3 \leq - 5$ , $x \in R$ .

b) Represent its solution on the following number line. Empty number line with an arrow pointing right and six equally spaced marks, unlabelled

Show answer

Answer: a) $x \leq - 4$ b) See figure.

Show working and answer

We subtract $3$ from both sides: $2 x \leq - 8$ .
We divide by $2$ (positive, so the direction does not change): $x \leq - 4$ .

On the unlabelled line, we choose one of the marks for $- 4$ and label it. As the marks are one unit apart, the values to its right would be $- 3, - 2, \dots$
As the inequality includes equality ( $\leq$ ), the point at $- 4$ is marked as closed.
The valid values are less than $- 4$ : a thick line to the left with an arrow.

Number line with six marks; the fourth is labelled −4, with a closed dot and a thick line to the left with an arrow (x ≤ −4)

Answer: a) $x \leq - 4$ b) See figure.

Exercise 79 MediumIB · P1 TZ0 May 2004

a) Solve the inequality $3 x - 7 \leq 2$ .

b) Solve the inequality $- x < - 2$ .

c) The inequalities in (a) and (b) describe two regions on the number line. Plot the intersection of these two regions on the number line given below. Empty number line with an arrow pointing right and unit marks labelled −2, −1, 1, 2, 3 and 4 (0 unlabelled)

Show answer

Answer: a) $x \leq 3$ b) $x > 2$ c) $2 < x \leq 3$

Show working and answer

a) We isolate $x$ as in an equation:

We add $7$ to both sides: $3 x \leq 9$ .
We divide by $3$ (positive, so the direction of the inequality does not change): $x \leq 3$ .

We multiply both sides by $- 1$ . When multiplying or dividing by a negative number, the direction of the inequality is reversed: $x > 2$ .

c) The intersection is the set of values that satisfy both conditions at the same time: $x \leq 3$ and $x > 2$ .

We mark $2$ with an open circle ( $2$ does not satisfy $x > 2$ ).
We mark $3$ with a closed dot ( $3$ does satisfy $x \leq 3$ ).
We join both points with a thick line: it is the region common to both, $2 < x \leq 3$ .

Number line with a thick line between x = 2 and x = 3: open circle at 2 (not included) and closed dot at 3 (included)

Answer: a) $x \leq 3$ b) $x > 2$ c) $2 < x \leq 3$

Quadratic equation

Exercise 80 BasicIB · P1 TZ0 November 2005

a) Factorise the expression $x^{2} - 25$ .

b) Factorise the expression $x^{2} - 3 x - 4$ .

c) Using the answer to part (b), or otherwise, solve the equation $x^{2} - 3 x - 4 = 0$

Show answer

Answer: a) $x^{2} - 25 = (x - 5) (x + 5)$ b) $x^{2} - 3 x - 4 = (x + 1) (x - 4)$ c) $x = - 1$ , $x = 4$

Show working and answer

It is a difference of squares, $a^{2} - b^{2} = (a - b) (a + b)$ , with $a = x$ and $b = 5$ : $x^{2} - 25 = (x - 5) (x + 5)$

We look for two numbers that multiply to $- 4$ and add to $- 3$ : they are $- 4$ and $1$ .
Therefore $x^{2} - 3 x - 4 = (x + 1) (x - 4)$ .

Use the factorisation from part (b): $(x + 1) (x - 4) = 0$ .
A product is zero when one of its factors is zero: $x + 1 = 0$ or $x - 4 = 0$ .

Answer: a) $x^{2} - 25 = (x - 5) (x + 5)$ b) $x^{2} - 3 x - 4 = (x + 1) (x - 4)$ c) $x = - 1$ , $x = 4$

Exercise 81 BasicIB · P1 TZ0 May 2002

a) Find the solution of the equation $x^{2} - 5 x - 24 = 0$ .

b) The equation $a x^{2} - 9 x - 30 = 0$ has the solutions $x = 5$ and $x = - 2$ . Find the value of $a$ .

Show answer

Answer: a) $x = - 3$ , $x = 8$ b) $a = 3$

Show working and answer

We look for two numbers that multiply to $- 24$ and add to $- 5$ : they are $- 8$ and $3$ .
We factorise: $(x - 8) (x + 3) = 0$ .
Each factor can be zero: $x = 8$ or $x = - 3$ .

(The GDC can also be asked to solve the equation, and it gives the same roots.)

b) If $x = 5$ is a solution, substituting it into the equation must satisfy it.

We substitute $x = 5$ : $a \cdot 5^{2} - 9 \cdot 5 - 30 = 0$ .
We simplify: $25 a - 45 - 30 = 0 \Rightarrow 25 a = 75$ .
We solve for $a$ : $a = 3$ .
Check with $x = - 2$ : $3 \cdot 4 + 18 - 30 = 0$ . It holds.

Answer: a) $x = - 3$ , $x = 8$ b) $a = 3$

Exercise 82 BasicIB · P1 TZ0 November 2001

a) Factorise the expression $2 x^{2} - 3 x - 5$ .

b) Using the above, or otherwise, solve the equation $2 x^{2} - 3 x - 5 = 0$ .

Show answer

Answer: a) $2 x^{2} - 3 x - 5 = (2 x - 5) (x + 1)$ b) $x = \frac{5}{2} = 2.5$ , $x = - 1$

Show working and answer

We look for two numbers that multiply to $a \cdot c = 2 \cdot (- 5) = - 10$ and add to $- 3$ : they are $- 5$ and $2$ .
We split the middle term: $2 x^{2} - 5 x + 2 x - 5$ .
We factor out common factors by grouping: $x (2 x - 5) + 1 (2 x - 5)$ .
We factor out the common factor $(2 x - 5)$ : $(2 x - 5) (x + 1)$ .

We set the factorised expression equal to zero: $(2 x - 5) (x + 1) = 0$ .
A product is zero if one of its factors is zero: $2 x - 5 = 0$ or $x + 1 = 0$ .
We solve for $x$ : $x = \frac{5}{2}$ or $x = - 1$ .

Answer: a) $2 x^{2} - 3 x - 5 = (2 x - 5) (x + 1)$ b) $x = \frac{5}{2} = 2.5$ , $x = - 1$

Exercise 83 MediumIB · P1 TZ0 November 2007

a) Solve the following equation for $x$ : $3 (2 x + 1) - 2 (3 - x) = 13$ .

b) Factorise the expression $x^{2} + 2 x - 3$ .

c) Find the positive solution of the equation $x^{2} + 2 x - 6 = 0$ .

Show answer

Answer: a) $x = 2$ b) $x^{2} + 2 x - 3 = (x - 1) (x + 3)$ c) $x = - 1 + 7 \approx 1.65$

Show working and answer

Expand the brackets (be careful with the minus sign that multiplies the second one): $6 x + 3 - 6 + 2 x = 13$
Collect like terms: $8 x - 3 = 13$ .
Add 3: $8 x = 16$ , and divide by 8: $x = 2$ .

We look for two numbers that multiply to $- 3$ and add to $2$ : they are $3$ and $- 1$ .
Therefore $x^{2} + 2 x - 3 = (x + 3) (x - 1)$ .

Apply the general formula with $a = 1$ , $b = 2$ , $c = - 6$ : $x = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2} = \frac{- 2 \pm 28}{2} = - 1 \pm 7$
The two solutions are $x \approx 1.65$ and $x \approx - 3.65$ (they can also be found with the GDC). The positive one is $x \approx 1.65$ .

Answer: a) $x = 2$ b) $x^{2} + 2 x - 3 = (x - 1) (x + 3)$ c) $x = - 1 + 7 \approx 1.65$

Currency conversion

Exercise 84 BasicIB · P1 TZ1 May 2012

Sasha travelled from the United States to Mexico and converted 650 US dollars (USD) into Mexican pesos (MXN). The bank offered an exchange rate of 1 USD = 12.50 MXN.

a) Find the amount of MXN that Sasha received. Before returning to the USA, Sasha exchanged 2,300 MXN back into USD. The bank charges a 1% commission. The exchange rate was still 1 USD = 12.50 MXN.

b) Write down the commission charged by the bank, in MXN.

c) Calculate the amount of USD that Sasha received after the commission. Give your answer to the nearest US dollar.

Show answer

Answer: a) $8125 MXN$ b) $23 MXN$ c) $182 USD$

Show working and answer

We multiply the USD by the exchange rate: $650 \times 12.50 = 8125 MXN$

The commission is $1 %$ of $2300$ MXN: $2300 \times 0.01 = 23 MXN$

The amount exchanged is $2300 - 23 = 2277$ MXN.
Each USD is $12.50$ MXN, so we divide: $2277 \div 12.50 = 182.16 \to 182 USD$

Answer: a) $8125 MXN$ b) $23 MXN$ c) $182 USD$

Exercise 85 BasicIB · P1 TZ0 November 2009

The following table shows the exchange rate between the pound sterling (GBP) and the US dollar (USD), and the exchange rate between the USD and the euro (EUR).

1 GBP	2.034 USD
1 USD	0.632 EUR

a) Find the exchange rate between GBP and EUR in the form 1 GBP = k EUR, where k is a constant. Give your answer correct to two decimal places. Isabella exchanges 400 USD into euros and is charged a 2% commission for this.

b) Calculate how many euros Isabella receives. Give your answer correct to two decimal places.

Show answer

Answer: a) $1 GBP = 1.29 EUR$ b) $247.74 EUR$

Show working and answer

We convert GBP to EUR via USD: $1 GBP = 2.034 USD$ and each USD is $0.632 EUR$ .
$k = 2.034 \times 0.632 = 1.285488$ .
Rounding to two decimal places: $1 GBP = 1.29 EUR$ .

Exchange without commission: $400 \times 0.632 = 252.80 EUR$ .
The 2% commission means she receives 98% of that amount: $252.80 \times 0.98 = 247.744$ .
Rounded: $247.74 EUR$ .

Answer: a) $1 GBP = 1.29 EUR$ b) $247.74 EUR$

Exercise 86 BasicIB · P2 TZ2 November 2009

The following table shows the exchange rate from US dollars to euros and from US dollars to Japanese yen. Give all answers to this question correct to two decimal places.

1 USD	0.6337 euros
1 USD	99.7469 yen

a) Enrico has 475 USD.

(i) How many euros is this equivalent to? Enrico goes to a bank to exchange his dollars. The bank charges a 3% commission.

(ii) How many euros does Enrico receive?

b) Find the exchange rate from euros to yen.

Show answer

Answer: a) i) $301.01 EUR$ ii) $291.98 EUR$ b) $1 EUR = 157.40 YEN$

Show working and answer

(i) We multiply by the USD to EUR rate: $475 \times 0.6337 = 301.0075$ , i.e. $301.01 EUR$ .
(ii) With a 3% commission he receives 97% of that amount: $301.0075 \times 0.97 = 291.977 \dots$ , i.e. $291.98 EUR$ .

We know that $1 USD = 0.6337 EUR = 99.7469 YEN$ , so $0.6337 EUR = 99.7469 YEN$ .
For 1 EUR we divide by 0.6337: $99.7469 \div 0.6337 = 157.40 \dots$
Therefore, $1 EUR = 157.40 YEN$ .

Answer: a) i) $301.01 EUR$ ii) $291.98 EUR$ b) $1 EUR = 157.40 YEN$

Exercise 87 BasicIB · P2 TZ0 November 2007

Robert, who lives in the United Kingdom, travels to Belgium. The exchange rate is 1.37 euros to one pound sterling (GBP), with a commission of 3 pounds sterling, which is subtracted before the exchange takes place. Robert gives the bank 120 pounds sterling.

a) Calculate, correct to 2 decimal places, the number of euros Robert receives. He buys 1 kilogram of Belgian chocolates at 1.35 euros per 100 g.

b) Calculate the cost of his chocolates in pounds sterling, correct to 2 decimal places.

Show answer

Answer: a) $160.29 EUR$ b) $9.85 GBP$

Show working and answer

The commission is subtracted before the exchange: $120 - 3 = 117 GBP$ .
We exchange them: $117 \times 1.37 = 160.29 EUR$ .

$1 kg = 1000 g = 10$ times $100 g$ , so the chocolates cost $10 \times 1.35 = 13.50 EUR$ .
Since $1 GBP = 1.37 EUR$ , we divide: $13.50 \div 1.37 = 9.854 \dots$
To two decimal places: $9.85 GBP$ .

Answer: a) $160.29 EUR$ b) $9.85 GBP$

Exercise 88 BasicIB · P1 TZ0 May 2002

The exchange rate from US dollars (USD) to French francs (FFR) is given by 1 USD = 7.5 FFR. Give your answers to the following questions correct to 2 decimal places.

a) Convert 115 US dollars to French francs.

b) Roger receives 600 Australian dollars (AUD) for 2,430 FFR. Calculate the value of the US dollar in Australian dollars.

Show answer

Answer: a) $862.50 FFR$ b) $1 USD = 1.85 AUD$

Show working and answer

Using $1 USD = 7.5 FFR$ , we multiply: $115 \times 7.5 = 862.5$ .
To two decimal places: $862.50 FFR$ .

Convert the francs to US dollars by dividing by $7.5$ : $\frac{2430}{7.5} = 324 USD$ .
These $324 USD$ are equivalent to $600 AUD$ , so $1 USD = \frac{600}{324} = 1.8518 \dots AUD$ .
To two decimal places: $1.85 AUD$ .

Answer: a) $862.50 FFR$ b) $1 USD = 1.85 AUD$

Exercise 89 MediumIB · P1 TZ0 November 2018

Harry travelled from the USA to Mexico and exchanged 700 US dollars (USD) into Mexican pesos (MXN). The exchange rate applied was 1 USD = 18.86 MXN.

a) Calculate how many MXN Harry received. On the way back, Harry still had 2,400 MXN and wanted to exchange them back into USD. A commission of 3.5% had to be paid for the currency exchange.

b) Calculate, in MXN, the value of the commission Harry paid. The exchange rate applied to this currency exchange was 1 USD = 17.24 MXN.

c) Calculate how many USD Harry received. Give your answer rounded to the nearest cent.

Show answer

Answer: a) $\approx 13200$ MXN b) $84$ MXN c) $134.34$ USD

Show working and answer

Each dollar is exchanged for $18.86$ pesos, so we multiply: $700 \times 18.86 = 13202$ MXN.
Rounding to 3 significant figures: approximately $13200$ MXN.

The commission is $3.5 %$ of the $2400$ MXN: $2400 \times \frac{3.5}{100} = 84$ MXN.

We deduct the commission from the pesos being exchanged: $2400 - 84 = 2316$ MXN.
To convert MXN to USD we divide by the exchange rate: $\frac{2316}{17.24} = 134.338 \dots$
Rounding to cents (2 decimal places): $134.34$ USD.

Answer: a) $\approx 13200$ MXN b) $84$ MXN c) $134.34$ USD

Exercise 90 MediumIB · P1 TZ1 May 2017

Daniela is going on holiday to South America. She flies from the United States to Argentina, stopping over in Peru on the way. In Peru she exchanges 85 US dollars (USD) for Peruvian nuevo sol (PEN). The exchange rate is 1 USD = 3.25 PEN and a fixed fee of 5 USD is charged as commission.

a) Calculate the amount of PEN she receives. At the end of Daniela's holiday she has 370 Argentine pesos (ARS). She converts this back into USD at a bank that charges a 4% commission on the exchange. The exchange rate is 1 USD = 9.60 ARS.

b) Calculate the amount of USD she receives.

Show answer

Answer: a) $260$ PEN b) $37$ USD

Show working and answer

The commission is fixed: $5$ USD. What is actually exchanged is $85 - 5 = 80$ USD.
We multiply by the exchange rate: $80 \times 3.25 = 260$ PEN.

We convert ARS to USD by dividing by the exchange rate: $\frac{370}{9.60} = 38.5416 \dots$ USD.
The bank charges $4 %$ , so Daniela receives $96 %$ : $38.5416 \dots \times 0.96 = 37$ USD.

Answer: a) $260$ PEN b) $37$ USD

Exercise 91 MediumIB · P1 TZ0 November 2017

In this question, give all answers rounded to 2 decimal places. José travels from Buenos Aires to Sydney. He uses Argentine pesos, ARS, to buy 350 Australian dollars, AUD, at a bank. The exchange rate applied is 1 ARS = 0.1559 AUD.

a) Use this exchange rate to calculate how many ARS is equivalent to 350 AUD. The bank charges José a commission of 2%.

b) Calculate the total amount of ARS that José pays to obtain these 350 AUD. José uses his credit card to pay the bill for his hotel in Sydney. The bill is 585 AUD. The credit card company charges José 4,228.38 ARS for this payment. The exchange rate used by the credit card company is 1 AUD = $x$ ARS. No commission is charged.

c) Find the value of $x$ .

Show answer

Answer: a) $2245.03$ ARS b) $2289.93$ ARS c) $x = 7.23$ ARS

Show working and answer

The exchange rate is $1 ARS = 0.1559 AUD$ . To convert AUD to ARS we divide: $\frac{350}{0.1559} = 2245.028 \dots$
Rounding to 2 decimal places: $2245.03$ ARS.

José pays that amount plus the 2% commission, i.e. $102 %$ : $2245.03 \times 1.02 = 2289.93$ ARS.

The exchange rate is what is paid for each AUD: $x = \frac{4228.38}{585}$ .
We calculate: $x = 7.228$ , which rounded to 2 decimal places is $7.23$ .

Answer: a) $2245.03$ ARS b) $2289.93$ ARS c) $x = 7.23$ ARS

Exercise 92 MediumIB · P1 TZ1 May 2016

In this question, give all answers correct to the nearest whole number. Loic travelled from China to Brazil. At the airport he exchanged 3,100 Chinese yuan, CNY, for Brazilian real, BRL, at an exchange rate of 1 CNY = 0.3871 BRL. No commission was charged.

a) Calculate the amount of BRL he received. When he returned to China, Loic exchanged his remaining BRL at a bank. The exchange rate at the bank was 1 CNY = 0.3756 BRL and a 5% commission was charged. He received 285 CNY.

b) (i) Calculate the amount of CNY Loic would have received if no commission had been charged.

(ii) Calculate the amount of BRL that Loic exchanged when he returned to China.

Show answer

Answer: a) $1200$ BRL b) (i) $300$ CNY (ii) $113$ BRL

Show working and answer

Without commission, we multiply by the exchange rate: $3100 \times 0.3871 = 1200.01$ .
Rounding to the nearest whole number: $1200$ BRL.

b) (i)

The $285$ CNY he received is what remains after removing the $5 %$ commission, i.e. $95 %$ of the total.
Without commission he would have received $x$ , with $0.95 x = 285$ , so $x = \frac{285}{0.95} = 300$ CNY.

(ii) 1. Those $300$ CNY (without commission) correspond to the BRL exchanged. We convert CNY to BRL by multiplying by the bank's exchange rate: $300 \times 0.3756 = 112.68$ . 2. Rounding to the nearest whole number: $113$ BRL.

Answer: a) $1200$ BRL b) (i) $300$ CNY (ii) $113$ BRL

Exercise 93 MediumIB · P1 TZ0 November 2016

In this question, give all answers rounded to two decimal places. Javier is going on a business trip to Venezuela and takes 5,000 US dollars (USD) with him. When he arrives he exchanges USD 3000 into Venezuelan bolívares (VEF). The exchange rate applied is USD 1 = VEF 6.3021.

a) Calculate the amount of VEF that Javier receives. During his stay in Venezuela, Javier spends USD 1,250 and VEF 12,000. On returning home, Javier exchanges the VEF he has left over into USD. The exchange rate applied is USD 1 = VEF 8.7268.

b) Calculate the total amount of USD that Javier has left over from his initial USD 5,000 after his trip to Venezuela.

Show answer

Answer: a) $18906.30$ VEF b) He has $1541.39$ USD left over in total.

Show working and answer

We multiply by the exchange rate: $3000 \times 6.3021 = 18 906.30$ VEF.

USD neither exchanged nor spent: $5000 - 3000 - 1250 = 750$ USD.
VEF left over: $18 906.30 - 12000 = 6906.30$ VEF.
We convert to USD by dividing by the return exchange rate: $\frac{6906.30}{8.7268} = 791.39$ USD.
We add the dollars: $750 + 791.39 = 1541.39$ USD.

Answer: a) $18906.30$ VEF b) He has $1541.39$ USD left over in total.

Exercise 94 MediumIB · P1 TZ1 May 2015

Pietro arrives in Singapore and, at the airport, exchanges 800 euros (EUR) into Singapore dollars (SGD). The bank rates quoted at the airport for exchanging EUR into SGD are shown in the following table. The exchange rates for SGD against pounds sterling (GBP) and US dollars (USD) are also shown. No commission is charged on the exchanges.

The bank buys	The bank sells
1 EUR = 1.55 SGD	1 EUR = 1.75 SGD
1 GBP = 1.92 SGD	1 GBP = 2.05 SGD
1 USD = 1.15 SGD	1 USD = 1.28 SGD

a) Calculate the number of SGD that Pietro receives. Pietro also has 100 GBP that he wishes to exchange into USD for a trip to Cambodia. To carry out this transaction, the GBP must first be converted into SGD and then into USD.

b) Calculate the amount of USD that Pietro receives.

Show answer

Answer: a) $1240$ SGD b) $150$ USD

Show working and answer

Pietro hands over euros and the bank buys them, so we use the "The bank buys" column: $1 EUR = 1.55 SGD$ .
We multiply the amount by the exchange rate: $800 \times 1.55 = 1240 SGD$

First, GBP to SGD. Pietro hands over pounds and the bank buys them: $1 GBP = 1.92 SGD$ . $100 \times 1.92 = 192 SGD$
Then, SGD to USD. Pietro hands over SGD and receives USD, i.e. the bank sells USD: $1 USD = 1.28 SGD$ . To find how many USD correspond to $192$ SGD we divide: $192 \div 1.28 = 150 USD$

Answer: a) $1240$ SGD b) $150$ USD

Exercise 95 MediumIB · P1 TZ0 November 2015

Sebastian travels from Fiji to Brazil. He exchanges 5,000 Fijian dollars (FJD) into Brazilian reals (BRL), at an exchange rate of 1 FJD = 1.1824 BRL.

a) Calculate the exact amount of BRL that Sebastian receives. Sebastian had estimated that he would receive 6,000 BRL.

b) Find the percentage error in his estimate. While in Brazil, Sebastian spends 4,000 BRL. On the return trip to Fiji he exchanges the BRL he has left back into FJD, at the same exchange rate: 1 FJD = 1.1824 BRL.

c) Calculate the amount of FJD that he receives. Give your answer rounded to two decimal places.

Show answer

Answer: a) $5912$ BRL b) Percentage error $= 1.49 %$ c) $1617.05$ FJD

Show working and answer

We multiply the FJD by the exchange rate: $5000 \times 1.1824 = 5912 BRL$

The percentage error is calculated relative to the exact value: $error = \frac{∣ approximate value - exact value ∣}{exact value} \times 100$
We substitute: $\frac{∣6000 - 5912∣}{5912} \times 100 = \frac{88}{5912} \times 100 \approx 1.49 %$

He has $5912 - 4000 = 1912$ BRL left over.
To convert BRL to FJD we divide by the exchange rate (1 FJD costs $1.1824$ BRL): $1912 \div 1.1824 \approx 1617.05 FJD$

Answer: a) $5912$ BRL b) Percentage error $= 1.49 %$ c) $1617.05$ FJD

Exercise 96 MediumIB · P1 TZ1 May 2014

In this question, give all answers correct to two decimal places. Dumisani has received a scholarship of 5,000 US dollars (USD) to study in Singapore. He has to travel from South Africa and must exchange USD to pay his airfare of 6,600 South African rand (ZAR). The exchange rate is 1 USD = 8.2421 ZAR.

a) Calculate the amount of USD that Dumisani must exchange to pay for his airfare. On arrival in Singapore, Dumisani exchanges 3,000 USD into Singapore dollars (SGD) to pay his school fees. A commission of 2.8% is charged on the exchange.

b) Calculate the value, in USD, of the commission Dumisani must pay. The exchange rate is 1 USD = 1.29903 SGD.

c) Calculate the number of SGD that Dumisani receives.

Show answer

Answer: a) $800.77$ USD b) $84$ USD c) $3787.97$ SGD

Show working and answer

Each USD is worth $8.2421$ ZAR. To find how many USD are equivalent to $6600$ ZAR we divide: $6600 \div 8.2421 \approx 800.77 USD$

The commission is $2.8 %$ of $3000$ USD: $3000 \times 0.028 = 84.00 USD$

The amount actually exchanged is what remains after the commission: $3000 - 84 = 2916$ USD.
We multiply by the exchange rate: $2916 \times 1.29903 \approx 3787.97 SGD$

Answer: a) $800.77$ USD b) $84$ USD c) $3787.97$ SGD

Exercise 97 MediumIB · P1 TZ2 May 2014

In this question, give all answers rounded to the nearest whole number.

a) Fumie is going on holiday to the United Kingdom. She exchanges 100,000 Japanese yen (JPY) into pounds sterling (GBP) and no commission is charged. The exchange rate between GBP and JPY is 1 GBP = 129 JPY. Calculate the value in GBP of the 100,000 JPY.

b) After her holiday in the United Kingdom, Fumie has 239 GBP left over. She exchanges this amount into JPY at a bank that charges no commission, and is given 30,200 JPY.

(i) Find the exchange rate applied in this second transaction.

(ii) Determine whether, when exchanging GBP into JPY, the exchange rate found in part (b)(i) is better for Fumie than the exchange rate in part (a). Justify your answer.

Show answer

Answer: a) $775$ GBP b) (i) $1 GBP = 126$ JPY (ii) It is not better, because $126 < 129$

Show working and answer

Each GBP is $129$ JPY, so to convert JPY to GBP we divide: $100000 \div 129 \approx 775.19 \to 775 GBP$

b) (i)

The exchange rate is what is received for each pound: $30200 \div 239 \approx 126.36 \to 1 GBP = 126 JPY$

b) (ii)

When exchanging GBP into JPY, Fumie would prefer to receive more yen for each pound.
In part (a) the bank gave $129$ JPY per GBP and in the second transaction only $126$ JPY per GBP. As $126 < 129$ , the rate found in (b)(i) is not better for Fumie.

Answer: a) $775$ GBP b) (i) $1 GBP = 126$ JPY (ii) It is not better, because $126 < 129$

Exercise 98 MediumIB · P1 TZ0 November 2014

In this question, give all answers correct to two decimal places. Albena travels to Bulgaria on a business trip. She is paid 3,550 Bulgarian leva (BGN) at the end of her trip. She converts all her BGN into euros (EUR) at a bureau de change which sells 1 EUR for 1.95 BGN and charges a 3% commission.

a) Calculate the amount in EUR that Albena receives. In the airport shop, Albena buys chocolates that cost 34.50 BGN. She uses 20 EUR to pay for the chocolates but receives her change in BGN. The exchange rate in the shop is 1 EUR = 1.90 BGN.

b) Find how many BGN she receives as change.

Show answer

Answer: a) $1765.90$ EUR b) $3.50$ BGN

Show working and answer

The bureau de change charges $3 %$ , so Albena only exchanges $97 %$ of her BGN: $3550 \times 0.97 = 3443.5 BGN$
Each EUR costs $1.95$ BGN, so we divide: $3443.5 \div 1.95 \approx 1765.90 EUR$

We convert the $20$ EUR to BGN at the shop's rate: $20 \times 1.90 = 38 BGN$
The change is what is left after paying for the chocolates: $38 - 34.50 = 3.50 BGN$

Answer: a) $1765.90$ EUR b) $3.50$ BGN

Exercise 99 MediumIB · P1 TZ2 May 2013

Yoshi is going to spend a year travelling: first from Japan to Italy, and then to the United States. Before leaving Japan, Yoshi exchanges 100,000 Japanese yen (JPY) into euros (EUR). The exchange rate is 1 JPY = 0.006 EUR.

a) Calculate how much money Yoshi receives, in EUR. Yoshi spends 426.70 EUR in Italy. At a US bank, Yoshi exchanges the money left over into US dollars (USD), at the following exchange rate: 1 USD = 0.673 EUR. The bank charges a 1.5% commission.

b) Calculate how much money Yoshi receives in USD, after the commission has been deducted. Give your answer rounded to the nearest whole number of USD.

Show answer

Answer: a) $600$ EUR b) $254$ USD

Show working and answer

We multiply the JPY by the exchange rate: $100000 \times 0.006 = 600 EUR$

The money left over after spending in Italy is: $600 - 426.70 = 173.30 EUR$
After the $1.5 %$ commission Yoshi exchanges $98.5 %$ : $173.30 \times 0.985 = 170.7005 EUR$
Each USD costs $0.673$ EUR, so we divide: $170.7005 \div 0.673 \approx 253.64 \to 254 USD$

Answer: a) $600$ EUR b) $254$ USD

Exercise 100 MediumIB · P1 TZ2 May 2012

In this question, give all answers rounded to two decimal places. Chiara is an Italian tourist visiting Sweden. The exchange rate from euros (€) to Swedish kronor (SEK) is 1€ = 10.275 SEK. Chiara exchanges 350 euros into Swedish kronor at a bank which charges a 2% commission.

a) Calculate, in SEK, the commission charged.

b) Write down how much money the bank gives Chiara once it has deducted the commission. Chiara returns to Italy with 296 SEK. She exchanges this money back into euros at a bank, and receives 32€. The bank charges no commission.

c) Calculate the value in SEK of 1€.

Show answer

Answer: a) $71.93 SEK$ b) $3524.33 SEK$ c) $1€ = 9.25 SEK$

Show working and answer

We convert the euros to kronor without commission: $350 \times 10.275 = 3596.25 SEK$
The commission is $2 %$ of that amount: $3596.25 \times 0.02 = 71.925 \to 71.93 SEK$

The bank gives her what remains after the commission: $3596.25 - 71.925 = 3524.325 \to 3524.33 SEK$

With $296$ SEK she receives $32$ €, so $1$ € equals: $296 \div 32 = 9.25 SEK$

Answer: a) $71.93 SEK$ b) $3524.33 SEK$ c) $1€ = 9.25 SEK$

Exercise 101 MediumIB · P1 TZ0 November 2012

Give all answers to this question correct to two decimal places. Isabel travels from Geneva to Toronto via Amsterdam. She exchanges 1,240 Swiss francs (CHF) into euros (EUR). The exchange rate is 1 CHF = 0.7681 EUR.

a) Calculate the amount of euros that Isabel receives. Isabel then exchanges 750 EUR into Canadian dollars (CAD) and a 3.12% commission is charged. The exchange rate is 1 CAD = 0.7470 EUR.

b) Calculate the amount of Canadian dollars she receives.

Show answer

Answer: a) $952.44 EUR$ b) $972.69 CAD$

Show working and answer

We multiply the CHF by the exchange rate: $1240 \times 0.7681 = 952.444 \approx 952.44 EUR$

The commission is $3.12 %$ of $750$ EUR: $750 \times 0.0312 = 23.40 EUR$
The amount exchanged is $750 - 23.40 = 726.60$ EUR.
Each CAD costs $0.7470$ EUR, so we divide: $726.60 \div 0.7470 \approx 972.69 CAD$

Answer: a) $952.44 EUR$ b) $972.69 CAD$

Exercise 102 MediumIB · P1 TZ2 May 2011

A manufacturer in England makes 16,000 garden statues. 12% are defective and cannot be sold.

a) Find the number of statues that have no defects. The manufacturer sells each non-defective statue for 5.25 pounds sterling (GBP) to a US company. The exchange rate from GBP to US dollars (USD) is 1 GBP = 1.6407 USD.

b) Calculate the amount in USD that the US company paid for all the non-defective statues. Give your answer correct to two decimal places. The US company sells one of the statues to an Australian customer for 12 USD. The exchange rate from Australian dollars (AUD) to USD is 1 AUD = 0.8739 USD.

c) Calculate the amount that the Australian customer pays, in AUD, for this statue. Give your answer correct to two decimal places.

Show answer

Answer: a) $14080$ statues b) $121280.54 USD$ c) $13.73 AUD$

Show working and answer

$12 %$ are defective, so the proportion without defects is $100% - 12% = 88%$ .
We calculate 88% of the total: $16000 \times 0.88 = 14080$ .

Income in GBP: $14080 \times 5.25 = 73920 GBP$ .
To convert to USD we multiply by the exchange rate (1 GBP = 1.6407 USD): $73920 \times 1.6407 = 121280.544 USD$ .
Rounding to two decimal places: $121280.54 USD$ .

Since $1 AUD = 0.8739 USD$ , to convert USD to AUD we divide by 0.8739.
$12 \div 0.8739 = 13.7317 \dots$ , which rounded is $13.73 AUD$ .

Answer: a) $14080$ statues b) $121280.54 USD$ c) $13.73 AUD$

Exercise 103 MediumIB · P1 TZ0 November 2011

In this question, give all answers to 2 decimal places. George travelled from the USA to Europe and exchanged 1,200 US dollars (USD) for euros (EUR). The exchange rate was 1 USD = 0.8154 EUR.

a) Calculate the amount of EUR that George received. On his return, George had 160 EUR to exchange back into USD. A commission of 4.5% was charged on the exchange. The exchange rate was 1 USD = 0.8202 EUR.

b) Calculate the value, in EUR, of the commission George paid.

c) Calculate the amount of dollars George received.

Show answer

Answer: a) $978.48 EUR$ b) Commission $= 7.2 EUR$ c) $186.30 USD$

Show working and answer

We multiply the dollars by the exchange rate: $1200 \times 0.8154 = 978.48 EUR$ .

The commission is 4.5% of the 160 EUR being exchanged.
$160 \times 0.045 = 7.20 EUR$ .

After the commission, $160 - 7.20 = 152.80 EUR$ remain to be exchanged.
Since $1 USD = 0.8202 EUR$ , to convert EUR to USD we divide: $152.80 \div 0.8202 = 186.296 \dots$
Rounded: $186.30 USD$ .

Answer: a) $978.48 EUR$ b) Commission $= 7.2 EUR$ c) $186.30 USD$

Exercise 104 MediumIB · P1 TZ1 May 2010

Susi travels from Singapore to Thailand and exchanges 1,500 Singapore dollars (SGD) into Thai baht (THB). The exchange rate is 1 SGD $=$ 21.03464 THB.

a) Calculate the number of Thai baht that Susi receives. Give your answer to the nearest baht. Susi leaves Thailand and travels to Indonesia. She has 20,000 THB and uses them to buy Indonesian rupiah (IDR). The exchange rate is 3.28352 THB per 1,000 IDR.

b) Calculate the total number of Indonesian rupiah that Susi receives, rounded to the nearest thousand rupiah. Susi wants to find the approximate exchange rate between Singapore dollars and Indonesian rupiah, and uses the exchange rates for the Thai baht to do this.

c) Calculate Susi's exchange rate between Singapore dollars and Indonesian rupiah. Give your answer in the form 1 SGD buys x IDR, where x is given correct to the nearest rupiah.

Show answer

Answer: a) $\approx 31552 THB$ b) $6091000 IDR$ c) $1 SGD = 6406 IDR$

Show working and answer

We multiply by the exchange rate: $1500 \times 21.03464 = 31551.96 THB$ .
To the nearest baht: $31552 THB$ .

The rate is $3.28352 THB$ for every $1000 IDR$ . Therefore, the number of groups of 1,000 IDR that she buys is $20000 \div 3.28352 = 6091.02 \dots$
In rupiah: $6091.02 \times 1000 \approx 6091024 IDR$ .
To the nearest thousand rupiah: $6091000 IDR$ .

We convert SGD to THB and THB to IDR. First, $1 THB = \frac{1000}{3.28352} = 304.5512 \dots IDR$ .
Then $1 SGD = 21.03464 THB = 21.03464 \times 304.5512 \dots = 6406.12 \dots IDR$ .
To the nearest rupiah: $6406$ .

Answer: a) $\approx 31552 THB$ b) $6091000 IDR$ c) $1 SGD = 6406 IDR$

Exercise 105 MediumIB · P1 TZ1 May 2008

Jane plans to travel from Amsterdam to Chicago. She exchanges 1,500 euros (EUR) into US dollars (USD) at an exchange rate of 1 EUR to 1.33 USD. Give all answers in this question correct to two decimal places.

a) Calculate the amount of USD that Jane receives. Jane spends 1,350 USD and then decides to convert the rest back into EUR at a rate of 1 EUR to 1.38 USD.

b) Calculate the amount of EUR that Jane receives. If Jane had waited until returning to Amsterdam, she could have exchanged her USD at a rate of 1 EUR to 1.36 USD, but the bank would have charged a 0.8% commission.

c) Calculate the amount of EUR that Jane gained or lost by exchanging her money in Chicago.

Show answer

Answer: a) $1995 USD$ b) $467.39 EUR$ c) Jane lost $3.08 EUR$ by exchanging her money in Chicago.

Show working and answer

$1500 \times 1.33 = 1995 USD$ .

The amount left is: $1995 - 1350 = 645 USD$ .
Since $1 EUR = 1.38 USD$ , we divide: $645 \div 1.38 = 467.39 EUR$ .

In Amsterdam: $645 \div 1.36 = 474.26 \dots EUR$ .
With the 0.8% commission she would receive 99.2%: $474.26 \dots \times 0.992 = 470.47 EUR$ .
We compare with what she received in Chicago: $467.39 - 470.47 = - 3.08$ . She would have had 3.08 EUR more by waiting, so she lost that amount.

Answer: a) $1995 USD$ b) $467.39 EUR$ c) Jane lost $3.08 EUR$ by exchanging her money in Chicago.

Exercise 106 MediumIB · P1 TZ2 May 2008

The following table shows some exchange rates for the Japanese yen (JPY).

Currency	1 JPY
Canadian dollar	0.010406
Chinese yuan	0.07127
Euro	0.0072591
Norwegian krone	0.057319

Minbin has 1,250 Japanese yen which they want to change into Chinese yuan.

a) Calculate how many yuan Minbin will receive. Give your answer to the nearest yuan. Rupert has 855 Canadian dollars which he wants to exchange for Japanese yen.

b) Calculate how many yen Rupert will receive. Give your answer to the nearest yen.

c) Find how many Norwegian krone there are per euro. Give your answer correct to 2 decimal places.

Show answer

Answer: a) $89 Yuan$ b) $82164 JPY$ c) $1 EUR = 7.90$ Norwegian krone

Show working and answer

$1250 \times 0.07127 = 89.0875$ .
To the nearest yuan: $89$ yuan.

Since $1 JPY = 0.010406 CAD$ , to convert from CAD to JPY we divide: $855 \div 0.010406 = 82164.1 \dots$
To the nearest yen: $82164 JPY$ .

Using 1 JPY: $0.0072591 EUR = 0.057319 NOK$ .
To find the amount for 1 EUR we divide: $0.057319 \div 0.0072591 = 7.896 \dots$
To two decimal places: $7.90$ .

Answer: a) $89 Yuan$ b) $82164 JPY$ c) $1 EUR = 7.90$ Norwegian krone

Exercise 107 MediumIB · P1 TZ0 November 2008

The exchange rate between Indian rupees (INR) and Singapore dollars (S$) is as follows: 100 INR = S$3.684. Kwai Fan exchanges S$500 into rupees.

a) Calculate how many Indian rupees Kwai Fan will receive using this exchange rate. Give your answer to the nearest whole number of rupees. When Kwai Fan returns to Singapore, they have 2,500 rupees left over from the trip. They want to exchange these rupees into Singapore dollars. This transaction has a 3% commission, and the exchange rate applied is 100 INR = S$3.672.

b) Calculate, in Indian rupees, the commission charged on this exchange.

c) Calculate, in Singapore dollars, the amount of money received, giving your answer rounded to 2 decimal places.

Show answer

Answer: a) $13572 INR$ b) Commission $= 75 INR$ c) $89.05 S$$

Show working and answer

$100 INR = 3.684 S$$ , so for each S$ they receive $100 \div 3.684$ rupees.
$500 \times \frac{100}{3.684} = 13572.2 \dots$
To the nearest whole number: $13572 INR$ .

The commission is 3% of 2,500 INR: $2500 \times 0.03 = 75 INR$ .

The amount exchanged is $2500 - 75 = 2425 INR$ .
Using $100 INR = 3.672 S$$ : $2425 \times \frac{3.672}{100} = 89.046$ .
To two decimal places: $89.05 S$$ .

Answer: a) $13572 INR$ b) Commission $= 75 INR$ c) $89.05 S$$

Exercise 108 MediumIB · P1 TZ0 May 2006

Sven plans to travel to Europe. He withdraws $ 800 (dollars) from his savings account and converts them into euros. The bank buys euros at $1: € 0.785 and sells them at $1: € 0.766.

a) Use the appropriate rate to calculate the amount in euros that Sven will receive.

b) Suppose the trip is cancelled. How much will Sven receive if the amount in euros from part (a) is changed back into dollars?

c) How much money has Sven lost through the two transactions? Express your answer as a percentage of Sven's original amount of $ 800.

Show answer

Answer: a) $\approx 613 EUR$ b) $781 $$ c) $2.38%$

Show working and answer

a) Sven gives dollars and wants to receive euros, that is, the bank sells him euros. The appropriate rate is the selling rate: $1 $ = 0.766 €$ .

Multiply the amount in dollars by the rate: $800 \times 0.766 = 612.8 €$ .
Rounding to the nearest euro: $\approx 613 €$ .

b) Now Sven gives euros and receives dollars, that is, the bank buys the euros from him. The appropriate rate is the buying rate: $1 $ = 0.785 €$ .

Since $1 $$ is equivalent to $0.785 €$ , to convert from euros to dollars we divide by $0.785$ .
Using the rounded amount from part (a): $\frac{613}{0.785} = 780.89 \dots \approx 781 $$ .

c) The loss is the difference between what he had and what he gets back.

Loss: $800 - 781 = 19 $$ .
As a percentage of the original amount: $\frac{19}{800} \times 100 = 2.375% \approx 2.38%$ .

(If the decimals are carried through without rounding, $612.8/0.785 = 780.64 $$ and the percentage is $\approx 2.42%$ . The difference comes only from rounding the intermediate values to $613$ and $781$ .)

Answer: a) $\approx 613 EUR$ b) $781 $$ c) $2.38%$

Exercise 109 MediumIB · P1 TZ0 November 2006

Clara, visiting Great Britain from the United States, exchanges 1,000 US dollars (USD) into pounds (GBP). The exchange rate is 1 USD = 0.543 GBP. The bank charges a commission of 2% on each transaction.

a) Calculate how many pounds sterling she receives. Clara then wants to travel to France. She exchanges 150 GBP into euros (€) at a rate of 1 GBP = 1.35 €. The bank charges a commission and then gives Clara 200 €.

b) Find the total commission in GBP.

Show answer

Answer: a) $532 GBP$ b) Commission $= 1.85 GBP$

Show working and answer

a) First the money is exchanged, and then the 2% commission is deducted.

Exchange without commission: $1000 \times 0.543 = 543 GBP$ .
Commission of 2%: $543 \times 0.02 = 10.86 GBP$ .
What she receives: $543 - 10.86 = 532.14 GBP$ (equivalent to multiplying by $0.98$ ), that is, $\approx 532 GBP$ .

b) The commission is found by comparing what she should receive with what she actually receives.

Without commission, $150 GBP$ would be: $150 \times 1.35 = 202.5 €$ .
Clara receives $200 €$ , so the commission in euros is $202.5 - 200 = 2.5 €$ .
To convert it to pounds we divide by the rate ( $1 GBP = 1.35 €$ ): $\frac{2.5}{1.35} = 1.85 \dots GBP$ .

Answer: a) $532 GBP$ b) Commission $= 1.85 GBP$

Exercise 110 MediumIB · P1 TZ0 May 2003

Zog, from the planet Mars, wants to exchange some Martian dollars (MD) for US dollars (USD). The exchange rate is 1 MD = 0.412 USD. The bank charges a commission of 2%.

a) How many US dollars will Zog receive if they pay 3,500 MD? Zog meets Zania, from Venus, whose currency is the Venusian rupee (VR). They want to exchange money and avoid bank charges. The exchange rate is 1 MD = 1.63 VR.

b) How many Martian dollars, rounded to the nearest dollar, will Zania receive if Zania gives Zog 2,100 VR?

Show answer

Answer: a) $\approx 1413 USD$ b) $1288 MD$

Show working and answer

Exchange without commission: $3500 \times 0.412 = 1442 USD$ .
Commission of 2%: $1442 \times 0.02 = 28.84 USD$ .
Zog receives: $1442 - 28.84 = 1413.16 USD \approx 1413 USD$ .

b) Zog and Zania exchange directly with each other, without a bank, so there is no commission.

Rate: $1 MD = 1.63 VR$ . To convert from rupees to Martian dollars we divide by $1.63$ .
$\frac{2100}{1.63} = 1288.34 \dots$
Rounding to the nearest dollar: $1288 MD$ .

Answer: a) $\approx 1413 USD$ b) $1288 MD$

Exercise 111 HardIB · P1 TZ1 May 2013

Neung returns to Vietnam after working in Singapore. Neung has 5,000 Singapore dollars (SGD) and exchanges them into US dollars (USD) to take home. The exchange rate between Singapore dollars (SGD) and US dollars (USD) is 1 USD = 1.2945 SGD. There is also a 2.4% commission on the exchange.

a) Calculate the amount of the commission on the exchange, in SGD.

b) Calculate the amount of US dollars (USD) that Neung takes home. Give your answer correct to 2 decimal places. At the airport in Vietnam, Neung exchanges 150 USD into Vietnamese dong (VND) to pay for transport home. The exchange rate between US dollars (USD) and Vietnamese dong (VND) is 1 USD = 19,495 VND. There is no commission.

c) Calculate the number of Vietnamese dong that Neung receives. Give your answer to the nearest thousand dong.

Show answer

Answer: a) $120$ SGD b) $3769.80$ USD c) $2924000$ VND

Show working and answer

The commission is $2.4 %$ of $5000$ SGD: $5000 \times 0.024 = 120 SGD$

The amount exchanged is $5000 - 120 = 4880$ SGD.
Each USD is worth $1.2945$ SGD, so we divide: $4880 \div 1.2945 \approx 3769.80 USD$

We multiply the USD by the exchange rate: $150 \times 19495 = 2924250 VND$
To the nearest thousand: $2924250 \approx 2924000 VND$

Answer: a) $120$ SGD b) $3769.80$ USD c) $2924000$ VND

Loans and amortisation

Exercise 112 MediumIB · P1 TZ0 November 2013

The table shows the monthly repayment (instalment) to be paid for each $10,000 borrowed, depending on the nominal annual interest rate and the term of the loan.

Loan term (years)	3%	5%	7%
5	179.69	188.72	198.02
10	96.57	106.07	116.11
15	69.06	79.08	89.89
20	55.46	66.00	77.53
25	47.43	58.46	70.68

(Monthly repayment (instalment) in $, for each $10,000; the percentages are the nominal annual interest rate.)

Zachary takes out a loan of $80,000 at a nominal annual interest rate of 5%. Zachary must repay it over a term of 20 years.

a) Using the table, calculate the total interest that Zachary will have to pay. Finley borrows an amount of money at a nominal annual interest rate of 7%. The loan must be repaid over a term of 10 years. The monthly repayment (instalment) cannot exceed $500.

b) Using the table, calculate the maximum amount that Finley can borrow. Give your answer rounded to the nearest $100.

Show answer

Answer: a) $46720 $$ b) The maximum amount Finley can borrow is $43000 $$

Show working and answer

For $10,000 the monthly repayment at $5 %$ over $20$ years is $66.00. A loan of $80,000 is $8$ times that amount: $8 \times 66.00 = 528 $ per month$
There are $20 \times 12 = 240$ payments: $528 \times 240 = 126720 $ repaid in total$
The interest is the amount repaid minus the amount borrowed: $126720 - 80000 = 46720 $$

For $7 %$ and $10$ years the repayment is $116.11 per $10,000.
If Finley borrows $x$ blocks of $10,000, the repayment is $116.11 x \leq 500$ , so $x \leq \frac{500}{116.11} \approx 4.306$
Maximum amount: $4.306 \times 10000 \approx 43063 $$ .
Rounded to the nearest hundred this gives $43,100, but that amount would give a repayment of about $500.4, which exceeds the limit. To avoid going over $500 we round down instead: $43,000.

Answer: a) $46720 $$ b) The maximum amount Finley can borrow is $43000 $$

Exercise 113 MediumIB · P1 TZ1 May 2008

The following table shows the monthly repayments needed to repay a loan of $ 1,000 at various interest rates and terms.

Loan term (months)	5%	5.5%	6%	6.5%
12	87.50	87.92	88.34	88.75
18	59.74	60.21	60.62	61.12
24	45.94	46.38	46.84	47.25
30	37.66	38.11	38.57	39.04
36	32.16	32.62	33.09	33.56
42	28.25	28.72	29.20	29.70
48	25.33	25.81	26.30	26.80

(Table of monthly repayments per annual interest rate for $1,000.)

Sarah takes out a personal loan of $ 24,000 to buy a car. She negotiates a loan over three years at 6% annual interest.

a) Calculate the exact monthly repayment she will make.

b) Find the exact total of the repayments she will make. Beryl took out a loan of $ 10,000 over 18 months. The total she paid for the loan was $ 10,837.80.

c) Find the interest rate charged on the loan.

Show answer

Answer: a) $794.16 $$ b) $28589.76 $$ c) Interest rate $5.5%$

Show working and answer

The table is for loans of $1,000. In the row for 36 months (3 years) and the 6% column, the monthly repayment is $33.09 per $1,000.
As Sarah borrows $24,000, that is 24 times that amount: $24 \times 33.09 = 794.16$ .

She will make 36 monthly repayments: $36 \times 794.16 = 28589.76$ .

The 10,000 borrowed is 10 times 1,000, so we divide the total paid by 10 to find the amount corresponding to $1,000: $10837.80 \div 10 = 1083.78$ .
We divide by the 18 months: $1083.78 \div 18 = 60.21$ per month.
In the 18-month row, 60.21 appears in the 5.5% column.

Answer: a) $794.16 $$ b) $28589.76 $$ c) Interest rate $5.5%$

Exercise 114 MediumIB · P1 TZ0 May 2007

The table below shows the monthly amortization payments for a loan of $ 10,000 at various nominal annual interest rates.

Loan term (years)	7%	8%	9%
5	198.0112	202.7634	207.5836
10	116.1085	121.3276	126.6758
15	89.8828	95.5652	101.4267
20	77.5299	83.6440	89.9726
25	70.6779	77.1816	83.9196

(Table of monthly amortization payments in $, for a loan of $ 10,000, according to the annual interest rate.)

In order to buy a flat, Leticia takes out a loan of $ 150,000 to be repaid over 20 years at an interest rate of 8%.

a) Calculate the exact monthly amortization payment for Leticia.

b) Find the exact amount of interest she will pay on the loan over the 20 years.

Show answer

Answer: a) Monthly amortization $= 1254.66 $$ b) Interest paid $= 151118.4 $$

Show working and answer

From the table, for 20 years and 8%, the monthly amortization payment per $10,000 is $83.6440.
$150,000 is 15 times $10,000: $15 \times 83.6440 = 1254.66$ .

Total paid over 20 years (20 × 12 = 240 payments): $240 \times 1254.66 = 301118.40$ .
The interest is the amount paid minus the amount borrowed: $301118.40 - 150000 = 151118.40$ .

Answer: a) Monthly amortization $= 1254.66 $$ b) Interest paid $= 151118.4 $$

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