Mathematics 0580/43 May/June 2022: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), May/June 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Here is part of a bus timetable.
Abbots 06 50 08 25 09 20
Callet 07 12 08 47 09 42
North Moor 07 30 09 05 10 00
South Moor 07 37 09 12 10 07
Centre Point 08 00 09 35 10 30
(a) Rashid catches the 09 20 bus at Abbots.
Find the time the bus arrives at South Moor.
…… [1]
(b) Annisa leaves home at 8.27 am and takes 25 minutes to walk to the bus stop at Callet.
She catches the next bus to Centre Point.
Find the total time, in minutes, for her journey from leaving home to arriving at Centre Point.
…… min [2]
(c) The distance from Abbots to Centre Point is 29.4 km.
Each bus takes the same time for the journey.
Calculate the average speed of a bus for this journey.
Give your answer in kilometres per hour.
…… km / h [2]
(d) On one journey, all 56 seats on the bus are filled.
The ratio of adults to children on this journey is adults : children = 5 : 3.
The cost for an adult ticket is $2.80 .
3
The cost for a child ticket is of the adult cost.
4
Work out the total cost of the tickets for this journey.
$ …… [4]
Mark scheme
1(a) 10 07 1
1(b) 123 2 M1 for 10 30 – 8 27 soi
or 10 30 – 8 52 + 25 soi or 25 + 50 + 48
1(c) 1 2 M1 for figs 29.4 ÷ 70 [× 60] oe
25.2, 25
5
1(d) $142.1[0] cao 4 M2 for [adults =] 56 ÷ 8 × 5
and [child =] 56 ÷ 8 × 3 or better
or M1 for 56 ÷ (5 + 3) × k where k = 1, 3
or 5
M1 for their 35 × 2.80 + their 21 × 2.80 ×
3
oe
4
(a) y
7
6
5
4 A
3
T
2
1
–6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 x
–1
–2
–3
–4
–5
(i) Draw the image of triangle T after a reflection in the line y = x . [2]
-1
(ii) Draw the image of triangle T after a translation by the vector e o. [2]
3
(iii) Describe fully the single transformation that maps triangle T onto triangle A.
…… [3]
(b) A quadrilateral P is enlarged by a scale factor of 1.2 to give quadrilateral Q.
The area of quadrilateral P is 20 cm2.
Calculate the area of quadrilateral Q.
…… cm2 [2]
Mark scheme
2(a)(i) Triangle drawn at (2, – 1), 2 B1 for two correct points
(2, – 4), (3, – 4) If 0 scored, SC1 for reflection of triangle
T in y = – x
2(a)(ii) Triangle drawn at (– 5, 6), 2 −1 k
(– 2, 5), (– 5, 5) B1 for translation by or by
k 3
If 0 scored SC1 for triangle drawn at
(–4.5, 3.5), (–4.5, 4.5) and (–1.5, 3.5)
2(a)(iii) Enlargement 3 B1 for each
[SF] – 1.5 oe
[centre] (0, 3)
2(b) 8 4 2 M1 for 1.2² oe
28.8, 28 , 28
10 5
3(a)(i)(a) 42 1
187 or 186.7 to 186.8 or 186
53
3(a)(i)(b) 2 : 7 : 42 cao 2 B1 for 106 : 371 : 2226 or any equivalent
ratio
If 0 scored, SC1 for 2 : 7 : 42 in the wrong
order
(a) The table shows the numbers of tigers reported to be living in the wild in the year 2014 in some countries.
Country Number
India 2226
Indonesia 371
Nepal 198
Bangladesh 106
(i) Using the table,
(a) find the number of tigers in Nepal as a percentage of the number of tigers in Bangladesh,
…… % [1]
(b) find the ratio tigers in Bangladesh : tigers in Indonesia : tigers in India, giving your answer in its simplest form.
…… : …… : …… [2]
(ii) Five years later, the number of tigers reported in India was 2967.
Find the percentage increase in the population of tigers in India.
…… % [2]
(iii) The number of tigers in India in the year 2014 is approximately 30.48% greater than in the year 2010.
Find the number of tigers in India in the year 2010.
Give your answer correct to the nearest integer.
…… [3]
(b) At the start of June, a hive has a population of 2000 bees.
Three months after the start of June the hive has a population of 2662 bees.
The population of this hive can be calculated using the formula
P = ab x ,
where P is the population of the hive x months after the start of June.
By finding the value of a and the value of b, calculate the population of the hive 7 months after the start of June.
Give your answer correct to the nearest integer.
…… [5]
Mark scheme
3(a)(ii) 33.3 or 33.28 to 33.29 2 2967 − 2226
M1 for [× 100] oe
2226
2967
or × 100 [– 100] oe
2226
3(a)(iii) 1706 cao nfww 3 B2 for 1705 to 1706.0… or 1710
30.48
or M1 for 1 + x = 2226 oe or
100
better
If 0 or M1 scored, SC1 for rounding their
decimal answer seen to nearest integer
3(b) 3897 5 B1 for a = 2000
2662
M2 for [b =] 3
2000
or M1 for 2662 = 2000b3
7
2662
M1 for their 2000 × 3
their 2000
or for their a × (their b)7 provided their a
and their b are clearly identified in the
working
If 0 or M1 scored, SC1 for rounding their
decimal answer seen to nearest integer.
A regular 12-sided polygon has side length 6 cm.
(a) Show that one interior angle of the polygon is 150°.
[1]
(b) The polygon is enclosed by a circle, centre O, so that each vertex touches the circumference of the circle.
A B
6 cm
NOT TO
SCALE
O
(i) Show that the radius, AO, of the circle is 11.6 cm, correct to 1 decimal place.
[3]
(ii) Calculate
(a) the circumference of the circle,
…… cm [2]
(b) the perimeter of the shaded minor segment formed by the chord AB.
…… cm [2]
(c) The regular 12-sided polygon is the cross-section of a prism of length 2 cm.
Calculate the volume of the prism.
…… cm3 [3]
Mark scheme
4(a) (12 − 2) × 180 1 (2 × 12 − 4) × 90
[= 150] oe Accept [= 150]
12 12
360
or 180 – [= 150]
12
4(b)(i) 3 M2 3
oe M1 for = cos75 oe
cos 75 AO
or or
6sin 75 r 6
=
sin 30 sin 75 sin 30
11.59… A1
4(b)(ii)(a) 72.8 or 72.9 or 72.82 to 72.89… 2 M1 for 2 × π × 11.6
4(b)(ii)(b) 12.1 or 12.06 to 12.08 2 M1 for [6 +] their (b)(ii)(a) ÷ 12 oe
4(c) 806 or 807 or 805.9 to 807.4 3 B2 for 402.9… to 403.7
OR
1
M2 for × 6 × 11.6 × sin 75 × 12 × 2 oe
2
1
or M1 for × 6 × 11.6 × sin 75 [×k ] oe
2
The time, t minutes, taken by each of 80 people to travel to work is recorded.
The table shows information about these times.
Time
01tG5 5 1 t G 10 10 1 t G 20 20 1 t G 35 35 1 t G 60
(t minutes)
Frequency 3 7 18 28 24
(a) (i) Write down the class interval containing the median time.
…… 1 t G …… [1]
(ii) Calculate an estimate of the mean time.
…… min [4]
(b) (i) One of these 80 people is chosen at random.
Find the probability that this person took longer than 10 minutes to travel to work.
Give your answer as a fraction in its simplest form.
…… [2]
(ii) Two people are chosen at random from those taking 20 minutes or less to travel to work.
Calculate the probability that one of these people took 5 minutes or less and the other took more than 5 minutes.
…… [3]
(c) (i) Use the frequency table on page 8 to complete the cumulative frequency table.
Time tG5 t G 10 t G 20 t G 35 t G 60
(t minutes)
Cumulative
3 10 80 frequency
[1]
(ii) On the grid, draw a cumulative frequency diagram to show this information.
80
70
60
50
Cumulative frequency 40
30
20
10
0
0 10 20 30 40 50 60 t
Time (minutes)
[3]
(iii) Find an estimate for the 80th percentile.
…… min [2]
(iv) Find an estimate for the percentage of people who took longer than 45 minutes to travel to work.
Show all your working.
…… % [3]
Mark scheme
5(a)(i) 20 < t ⩽ 35 1
5(a)(ii) 28 nfww 4 M1 for midpoints soi
M1 for use of ∑fm with m in correct
interval including both boundaries
M1 (dep on 2nd M1) for ∑fm ÷ 80
5(b)(i) 7 2 18 + 28 + 24
cao M1 for oe
8 80
5(b)(ii) 25 3 3 25
oe M2 for [2 ×] × or [2 ×]
126 28 27
25 3
× oe
28 27
3 25
or M1 for either or or
28 27
25 3
or
28 27
75
If 0 scored, SC1 for answer oe
392
5(c)(i) 28 and 56 1
5(c)(ii) Correct diagram 3 B1FT their (c)(i) for plots at 5 correct
heights
B1 for 5 plots at upper ends of intervals
on correct vertical line
B1FT (dep on at least B1) for increasing
curve or polygon through 5 points
After 0 scored, SC1FT for 4 correct
points plotted
5(c)(iii) Strict FT their reading at 80th percentile 2 B1 for 64 written or a mark at cf = 64 on
for an increasing curve/polygon graph or a mark on curve at (t, 64)
5(c)(iv) Correct integer reading at t = 45 M1 FT their cf graph for all three marks
80 − (their reading at t = 45) M1
[× 100]
80
(their reading at t = 45)
or × 100
80
Percentage consistent with their reading A1 If no working shown then SC1 for a
correct percentage that follows from a
correct reading from their graph.
(a) Simplify. a - 2b - 3a + 7b
…… [2]
(b) Expand and simplify.
4 (x - 5) - (3 - 2x)
…… [2]
(c) Write as a single fraction in its simplest form.
3 7
- x - 5 2x
…… [3]
(d) Solve.
13 - 4x
= 6-x
3
x = …… [3]
(e) Make x the subject of the formula.
5 (p - 2x) y= x
x = …… [4]
Mark scheme
6(a) 5b – 2a final answer 2 B1 for 5b or – 2a in final answer or for
5b – 2a seen
6(b) 6x – 23 final answer nfww 2 M1 for 4x – 20 or –3 + 2x
6(c) 35 − x 35 − x 3 B1 for 3(2x) – 7(x – 5) or better isw
or oe final answer B1 for 2x(x – 5) as common denominator
2 x( x − 5) 2 x 2 − 10 x
isw, allow expanded
nfww
6(d) –5 3 M1 for 13 – 4x = 18 – 3x oe
−4 x 13
or + x =6− oe
3 3
M1FT for ̶ 4x +3x = 18 – 13 oe
−x 5
or for =
3 3
6(e) 5p 4 M1 for correctly clearing the x from the
[x =] oe final answer denominator
y + 10
M1 for correctly expanding the brackets
or (dealing with the 5 correctly
throughout)
M1 for correctly isolating terms in x
M1 for correctly factorising and dividing
by the bracket
Max 3 marks if answer is incorrect
B
82 m
NOT TO
A 76° SCALE
55 m
C
The diagram shows a field ABC.
(a) Calculate BC.
BC = …… m [3]
(b) Calculate angle ACB.
Angle ACB = …… [3]
(c) A gate, G, lies on AB at the shortest distance from C.
Calculate AG.
AG = …… m [3]
(d) A different triangular field PQR has the same area as ABC.
PQ = 90 m and QR = 60 m.
Work out the two possible values of angle PQR.
Angle PQR = …… or …… [5]
Mark scheme
7(a) 87.[0] or 86.98 to 86.99 3 M2 for 822 + 552 − 2 × 82 × 55 × cos 76
oe
OR
M1 for 82 2 + 552 − 2 × 82 × 55 × cos 76 oe
A1 for 7570 or 7566 to 7567
7(b) 66.1 or 66.2 or 66.13 to 66.17 3 82 × sin 76
M2 for oe
their (a)
82 their (a)
or M1 for = oe
sin C sin 76
7(c) 13.3 or 13.30 to 13.31 3 M2 for AG = 55 cos 76 oe
or M1 for recognition that CG is
perpendicular to AB
7(d) 54.1 or 54.13… 5 B4 for 54.1 or 54.13…
and or 125.9 or 125.86 to 125.87
125.9 or 125.86 to 125.87 0.5 × 82 × 55 × sin 76
M3 for [sin Q =] oe
0.5 × 90 × 60
or M2 for
0.5 × 82 × 55 × sin 76 = 0.5 × 60 × 90 ×
sin Q oe
or M1 for 0.5 × 82 × 55 × sin 76 oe
or for 0.5 × 60 × 90 sin Q = their area of
ABC
If B4 not scored then SC1 for two angles
seen that sum to 180 (from use of sine
ratio) but not 0 and 180.
(a) A has coordinates (- 2, 7) , B has coordinates (1, - 5) and C has coordinates (5, 4) .
(i) Find the coordinates of the midpoint of the line AB.
(…… , …… ) [2]
(ii) Find AC .
AC = f p [2]
(iii) Find AC .
…… [2]
(iv) Find the equation of the line AB.
Give your answer in the form y = mx + c .
y = …… [3]
(v) Find the equation of the line perpendicular to AB that passes through C.
Give your answer in the form y = mx + c .
y = …… [3]
(b) The graphs of y + 5x = 8 and y = 2x 2 + 6x - 13 intersect at the points P and Q.
Find the coordinates of P and the coordinates of Q.
Show all your working.
P ( …… , …… )
Q ( …… , …… ) [6]
Mark scheme
8(a)(i) (–0.5, 1) 2 B1 for each
8(a)(ii) 7 2 B1 for each
−3
8(a)(iii) 7.62 or 7.615 to 7.616 2 FT their (a)(ii)
M1 for (their 7)2 + (their –3)2 oe
8(a)(iv) [y =] –4x –1 final answer 3 B2 for answer –4x + c [oe] or for correct
equation in different form or for –4x +–1
or for –4m – 1
OR
−5 − 7
M1 for oe
1 − −2
M1 for correct substitution shown of
(–2, 7) or (1, –5) or their (–0.5, 1) into
y = (their m)x + c oe
OR
M1 for 7 = –2m + c and –5 = m + c
A1 for m = –4 and c = –1
8(a)(v) 1 11 3 1
[y =] x+ final answer M1 for grad = oe nfww soi,
4 4 4
FT negative reciprocal of their gradient
from (iv)
M1 for correct substitution shown of
(5, 4) into y = (their m)x + c oe or, if no
substitution shown, (5, 4) satisfies their
final linear equation.
8(b) 2x2 + 11x – 21 [= 0] M2 or M1 for 8 – 5x = 2x2 + 6x – 13 oe or
better
(2x – 3)(x + 7) [= 0] oe M2 Allow correct method to solve their
or quadratic equation e.g. formula, complete
−11 ± 112 − 4 × 2 × ( −21) the square but not for 2x2 + 6x – 13
2× 2 M1 FT their equation
or for 2x(x+ 7) – 3(x + 7) [= 0]
2
11 21 11
– ± + oe or x(2x – 3) + 7(2x – 3) [= 0]
4 2 4
or (2x + a)(x + b) [= 0]
where ab = – 21 or 2b + a = 11
OR
M1 for 112 − 4 × 2 × −21
−11 + k −11 − k
or for or
2× 2 2× 2
OR
2
11
M1 for x +
4
3 1 B2 B1 for one correct pair or for 2 correct
, and (–7, 43) x-values or 2 correct y-values
2 2
(a) Sketch the graph of y = (x + 1) (3 - x) (3 + x) , indicating the coordinates of the points where the graph crosses the x-axis and the y-axis.
y
O x
[4]
(b) (i) Show that y = (x + 1) (3 - x) (3 + x) can be written as y = 9 + 9x - x 2 - x 3 .
[2]
(ii) Calculate the x–values of the turning points of y = 9 + 9x - x 2 - x 3 .
Show all your working and give your answers correct to 2 decimal places.
x = ……, x = …… [7]
(iii) The equation 9 + 9x - x 2 - x 3 = k has one solution only when k 1 a and when k 2 b , where a and b are integers.
Find the maximum value of a and the minimum value of b.
a = ……
b = …… [3]
Mark scheme
9(a) Correct sketch of negative cubic 4 B1 for any negative cubic shape with two
crossing the x-axis at –3, –1 and 3 turning points
and crossing the y-axis at 9
B2 for three intercepts only with x-axis
labelled at – 3, –1 and 3
or B1 for one or two correctly labelled
x-intercepts
B1 for intercept with y-axis labelled at 9
If no graph drawn, SC1 for all four
intercepts labelled on axes.
9(b)(i) 3 – x + 3x – x2 or better M1 At least 3 of the four terms correct
or
3 + x + 3x + x2 or better or for the correct expansion of all three
or brackets with all 8 terms correct
9 [– 3x + 3x] – x2
Correct completion to A1 with no errors or omissions seen
[y =] 9 + 9x – x2 – x3
9(b)(ii) 9 – 2x – 3x2 = 0 oe B3 B2 for 9 – 2x – 3x2
or B1 for two correct terms
M1 for their derivative = 0 or stating
dy
=0
dx
− − 2 ± (−2) 2 − 4 × −3 × 9 B2 FT their derivative
oe B1FT for (−2) 2 − 4(−3)(9) or better
2 × −3
−(−2) + q −(−2) − q
OR or for or
2 × −3 2 × −3
OR
2
1 9 1 1
2
– ± + oe B1 for x +
3 3 3 3
–2.10 and 1.43 final answer B2 B1 for each
or for answers –2.1 or –2.097 … and 1.4
or 1.430 to 1.431
or SC1 for –2.097... and 1.43[0] to 1.431
seen in working
or for –1.43 and 2.10 as final answer
9(b)(iii) [a =] – 6 3 B2 for either a correct or b correct
[b =] 17 or for [a =] –5.04 or –5.049 to –5.05 and
[b =] 16.9… seen
or M1 for substitution of one of their
solutions into 9 + 9x – x2 – x3 oe
or SC1 for reversed answers, a = 17,
b = –6
H G
NOT TO
E F SCALE
6 cm
24 cm
D C
A B
The diagram shows a cuboid ABCDEFGH.
CG = 6 cm, AG = 24 cm and AB = 2BC.
(a) Calculate AB.
AB = …… cm [4]
(b) Calculate the angle between AG and the base ABCD.
…… [3]
Mark scheme
10(a) 20.8 or 20.76 to 20.79 4 B3 for [BC =] 10.4 or 10.38 to 10.39… or
6 3 oe
or M2 for (2x)2 + x2 + 62 = 242 oe
or M1 for 242 – 62 oe or x2 + 62 oe
or (2x)2 + 62 oe, or x2 + (2x)2 oe
or SC2 for final answer of 12 5 or 26.8
or 26.83…
OR
2
x
M3 for x2 + + 62 = 242 oe
2
2
2 x
or M2 for x +
2
2
x
or M1 for x2 + 62 oe or + 62 oe or
2
2 2
24 – 6 oe
10(b) 14.5 or 14.47 to 14.48 3 6
M2 for sin […] = oe
24
or M1 for recognising the correct angle
GAC
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Mathematics 0580/43 for May/June 2022 is out of 130 marks, with 2 hours 30 minutes to answer it.
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