Mathematics 0580/41 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.
(a) The list shows 15 midday temperatures, in degrees Celsius, in Suntown.
17 21 21 18 23 22 25 19
21 17 19 18 21 24 23
(i) Complete the stem-and-leaf diagram to show this information.
1 7
2
Key: 1|7 represents 17 °C
[2]
(ii) Find the median.
…… °C [1]
(iii) Find the upper quartile.
…… °C [1]
(iv) Rahul draws a pie chart to show this information.
Calculate the sector angle for the number of days the temperature is 18 °C.
…… [2]
(b)
0 50 100 150 200
Mass (grams)
The box-and-whisker plot shows information about the masses, in grams, of some apples.
(i) Find the median.
…… g [1]
(ii) Find the range.
…… g [1]
(iii) Find the interquartile range.
…… g [1]
(c) (i) The time, t minutes, spent on homework in one week by each of 200 students is recorded.
The table shows the results.
Time (t minutes) 40 1 t G 60 60 1 t G 80 80 1 t G 90 90 1 t G 100 100 1 t G 150
Frequency 6 10 70 84 30
Calculate an estimate of the mean.
…… min [4]
(ii) A new table with different class intervals is completed.
Time (t minutes) 40 1 t G 90 90 1 t G 150
Frequency 86 114
On a histogram the height of the bar for the 40 1 t G 90 interval is 17.2 cm.
Calculate the height of the bar for the 90 1 t G 150 interval.
…… cm [2]
Mark scheme
1(a)(i) 1 7 7 8 8 9 9 2 B1 for one row correctly ordered or for fully correct unordered
stem-and-leaf diagram or for a correct diagram with one error or
2 1 1 1 1 2 3 3 4 5 omission
1(a)(ii) 21 1
1(a)(iii) 23 1
1(a)(iv) 48 2 2 360
M1 for [×360] or [×2]
15 15
1(b)(i) 120 1
1(b)(ii) 130 1
1(b)(iii) 60 1
1(c)(i) 93.4 4 M1 for mid-values soi
M1 for Σfx
M1 dep on second M for Σfx ÷ 200
1(c)(ii) 19 2 86 114
M1 for or
50 60
(a) Alex, Bobbie and Chris share strawberries in the ratio Alex : Bobbie : Chris = 3 : 2 : 2.
Chris receives 12 strawberries.
Calculate the total number of strawberries shared.
…… [2]
(b) In a sale, a shop reduces all prices by 12%.
(i) Dina buys a book which has an original price of $6.50 .
Calculate how much Dina pays for the book.
$ …… [2]
(ii) Elu pays $11 for a toy.
Calculate the original price of the toy.
$ …… [2]
(c) Feri invests some money.
The rate of interest for the first year is 2.5%.
At the end of the second year the overall percentage increase of Feri’s investment is 6.6%.
Find the rate of interest for the second year.
…… % [2]
(d) A radioactive substance decays at an exponential rate of 2% per day.
The initial mass is 80 g.
(i) Find the mass at the end of 5 days.
…… g [2]
(ii) Find how many more whole days, after day 5, it takes for the mass to reduce to less than 67 g.
…… [3]
Mark scheme
2(a) 42 2 M1 for 12 ÷ 2 or better
2(b)(i) 5.72 2 100 − 12
M1 for × 6.50 oe
100
or B1 for 0.88 oe
2(b)(ii) 12.5[0] 2 100 − 12
M1 for × x = 11 or better oe
100
2(c) 4 2 100 + 2.5 100 + 6.6
M1 for × [...] = oe
100 100
2(d)(i) 72.3 or 72.31... 2 100 − 2
5
M1 for 80 × oe
100
2(d)(ii) 4 nfww 3 B2 for answer 9 nfww
or M2 for correct trials with values giving either side of 67
n
100 − 2
or M1 for 80 × = 67
100
k
100 − 2
or their ( i ) × = 67
100
or an evaluated trial with n ⩾ 6 or k ⩾ 1
(a) Geeta buys x apples, (x + 7) oranges and (2x - 1) bananas.
The total number of pieces of fruit Geeta buys is 30.
(i) Find the number of apples Geeta buys.
…… [3]
(ii) The cost of one apple is 15 cents.
The cost of one orange is 18 cents.
The total cost of all the fruit is $5.55 .
Find the cost, in cents, of one banana.
…… cents [3]
(b) (i) Solve.
3w 1
-1 =
16 2
w = …… [2]
3 (2 - y ) 1
(ii) -1 =
16 2
Find the value of y.
y = …… [2]
(c) (i) Solve the simultaneous equations.
2p + q = 2
1 p - q =-
2
p = ……
q = …… [2]
(ii) Hence, for 0° G u G 360° and 0° G v G 360° , solve the simultaneous equations.
2 sin u + cos v = 2
1 sin u - cos v =-
2
u = …… or u = ……
v = …… or v = …… [4]
1
Mark scheme
3(a)(i) 6 3 B2 for 4x + 6 = 30 or better
or M1 for x + x + 7 + 2x – 1 [ = 30]
3(a)(ii) 21 3 M2 for (555 – their x × 15 – their (x + 7) × 18) ÷ their (2x – 1)
or M1 for their x × 15 or their (x + 7) × 18
3(b)(i) 8 2 M1 for isolating the term in w or correctly removing all fractions
3w 1
e.g. =1+ or better or 3w – 16 = 8
16 2
3(b)(ii) −3 2 1
M1 for 2− y = 8 or 2 y = or 2− y = their w or better
8
3(c)(i) 1 2 B1 for each
[p =] oe [q =] 1 If zero scored, SC1 for 2 values satisfying one of the original
2
equations
3(c)(ii) [u =] 30 and 150 4 B1 for each
[v =] 0 and 360 OR
SC1 for sin u = their p and cos v = their q
SC1 if their two different angles for u sum to 180 or if their
different two angles for v sum to 360
f (x) = 2x - 1 g (x) = 3x - 2 h (x) = , x ! 0 j (x) = 5 x x
(a) Find
(i) f (2) ,
…… [1]
(ii) gf (2) .
…… [1]
(b) Find g -1 (x) .
g -1 (x) = …… [2]
(c) Find x when h (x) = j (- 2) .
x = …… [2]
(d) Write f (x) - h (x) as a single fraction.
…… [2]
(e) Find the value of jj (2) .
…… [1]
(f) Find x when j -1 (x) = 4 .
x = …… [2]
Mark scheme
4(a)(i) 3 1
4(a)(ii) 7 1 FT their (i)
3×their (i) −2
4(b) x+2 2 y 2
oe final answer M1 for y + 2 = 3x or = x − or x = 3y – 2
3 3 3
4(c) 25 2 1
M1 for = 5−2 oe
x
4(d) 2 x2 − x − 1 2 1
final answer M1 for 2x – 1 –
x x
4(e) 2.98 × 1017 or 2.980... × 1017 1
4(f) 625 2 M1 for x = j(4)
5(a)(i)(a) ( 8 − 2 ) ×180 oe M2
M1 for
( 8 − 2 ) × 180 or 360 or ( 2 × 8 − 4 ) × 90
8× 2 8 8 8
5(a)(i)(b) 174 or 173.8.... 4 1
M3 for × 6 × OM oe
2
1
× ( OA ) × sin45 oe
2
or
2
1
or × 6 × OA × sin67.5 oe
2
where OA and OM are as in the M2
or M2 for OM = 3 × tan 67.5 oe
3 6 × sin67.5
or for OA = or oe
cos 67.5 sin45
OM
or M1 for = tan 67.5 oe
3
3
or for = cos 67.5 oe
OA
sin45 sin67.5
or for = oe
6 OA
(a) ABCDEFGH is a regular octagon with sides of length 6 cm.
The diagram shows part of the octagon.
O is the centre of the octagon and M is the midpoint of AB.
A M B
NOT TO
SCALE
O
(i) (a) Show that angle OAM is 67.5°.
[2]
(b) Calculate the area of the octagon.
…… cm 2 [4]
(ii) Find the area of the circle that passes through the vertices of the octagon.
…… cm 2 [3]
(b)
NOT TO
SCALE
4m
0.45 m
The diagram shows a horizontal container for water with a uniform cross-section.
The cross-section is a semicircle.
The radius of the semicircle is 0.45 m and the length of the container is 4 m.
(i) Calculate the volume of the container.
…… m 3 [2]
(ii)
NOT TO
SCALE
0.3 m
The greatest depth of the water in the container is 0.3 m.
The diagram shows the cross-section.
Calculate the number of litres of water in the container.
Give your answer correct to the nearest integer.
…… litres [6]
Mark scheme
5(a)(ii) 193 or 193.0 to 193.1 3 3
2
M2 for π × oe
cos 67.5
3 sin45 sin67.5
or M1 for = cos 67.5 or =
r 6 r
5(b)(i) 1.27 or 1.272 to 1.273 2 1
M1 for × π × 0.452 × 4
2
1
or × π × 0.452 [×4]
2
5(b)(ii) 742 or 743 6 M5 for a method leading to the volume of water
0.15
invcos
e.g. 4 × {2 × 0.45 × π × 0.452
360
1 0.15
− × 0.452 × sin 2 invcos } oe
2 0.45
OR
0.15
invcos
M2 [ 2 ×] 0.45 × π × 0.452 oe
360
0.15
90 − invcos
or [ 2 ×] 0.45 × π × 0.452 oe
360
θ
or M1 for use of × π × 0.452 oe
360
1 0.15
M2 for × 0.452 × sin 2 invcos oe
2 0.45
1 0.15
or × 0.15 × 0.45 × sin invcos [×2] oe
2 0.45
5(b)(ii) 1
× 0.452
or M1 for use of 2 × sinθ oe
1
[ 2 ×] × 0.15 × 0.45 × sin β
or 2 oe
If 0 scored,
0.15 0.15
SC1 for invcos 0.45 or invsin 0.45
or 0.452 − 0.152 soi
(a) y
12
11
10
9
8
7
6
5
4
3
2
1
–1 0 1 2 3 4 5 x
–1
–2
–3
–4
–5
–6
–7
–8
–9
– 10
The diagram shows the graph of y = f (x) for -1.5 G x G 5.
(i) Find f (2) .
…… [1]
(ii) Solve the equation f (x) = 0 for -1.5 G x G 5.
x = …… or x = …… or x = …… [3]
(iii) f (x) = k has three solutions for -1.5 G x G 5 where k is an integer.
Find the smallest possible value of k.
k = …… [1]
(iv) On the grid, draw a line y = mx so that f (x) = mx has exactly one solution for -1.5 G x G 5.
[2]
(b) y = 3x 2 - 12x + 7 dy
(i) Find the value of when x = 5. dx
…… [3]
(ii) Find the coordinates of the point on the graph of y = 3x 2 - 12x + 7 where the gradient is 0.
( …… , …… ) [2]
dy
(c) When y = 2x p + qx 2, = 14x 6 + 6x . dx
Find the value of p and the value of q.
p = ……
q = …… [2]
Mark scheme
6(a)(i) –3 1
6(a)(ii) –1 3 B1 for each
1.55 to 1.6
4.4 to 4.45
6(a)(iii) –8 1
6(a)(iv) Ruled line through origin intersecting curve once 2 B1 for ruled line through origin
6(b)(i) 18 3 B2 for 6x – 12
or B1 for 6x or –12
6(b)(ii) (2, –5) 2 B1 for each.
If 0 scored, M1 for their 6x – 12 = 0
dy
or states =0
dx
6(c) [p = ] 7 2 B1 for each
[q = ] 3
D
NOT TO
SCALE
12 km
9 km
14 km
A C
25° 32°
123°
B
(a) Calculate angle ACD.
Angle ACD = …… [4]
(b) Show that BC = 7.05 km , correct to 2 decimal places.
[3]
(c) Calculate the shortest distance from B to AC.
…… km [3]
(d) Calculate the length of the straight line BD.
BD = …… km [4]
(e) C is due east of A.
Find the bearing of D from C.
…… [2]
Mark scheme
7(a) 39.6 or 39.57.... 4 142 + 122 − 92
M2 for [cos =]
2 × 14 × 12
or M1 for 92 = 142 + 122 – 2 × 14 × 12 × cos ACD
37
A1 for 0.7708... or 0.771 or oe
48
7(b) 14sin 25 M2 sin123 sin 25
M1 for = oe
sin123 14 BC
7.054… A1
7(c) 3.74 or 3.735 to 3.739 3 M2 for 7.05 × sin 32
or M1 for recognition that the line from B is perpendicular to AC
7(d) 11.8 or 11.83 to 11.85 4 M1 for 32 + their(a) soi
M2 for
122 + 7.052 − 2 × 12 × 7.05 × cos(their ( a ) + 32)
122 + 7.052 − BD 2
or M1 for cos ( their ( a ) + 32 ) =
2 × 12 × 7.05
7(e) 309.6 or 309.57... 2 FT 270 + their(a)
M1 for 270 + their(a) oe
(a) (i) Use set notation to describe the shaded region in the Venn diagram.
B
A
…… [1]
(ii) Shade the correct region in each Venn diagram.
K L
Q
P
Q , P' M
(K , L ) + M '
[2]
(b)
V E N N D I A G R A M
The diagram shows 11 cards.
(i) One of these cards is chosen at random.
Write down the probability that the letter on the card is not A.
…… [1]
(ii) A card is chosen at random from these 11 cards and then replaced.
A second card is then chosen at random.
Find the probability that exactly one card has the letter N.
…… [3]
(c)
E M
50 students are asked if they like English (E) and if they like mathematics (M).
3 say they do not like English and do not like mathematics.
33 say they like English.
42 say they like mathematics.
(i) Complete the Venn diagram. [2]
(ii) A student is chosen at random.
Find the probability that this student likes English and likes mathematics.
…… [1]
(iii) Two students are chosen at random.
Find the probability that they both like mathematics.
…… [2]
(iv) Two students who like English are chosen at random.
Find the probability that they both also like mathematics.
…… [2]
Mark scheme
8(a)(i) A B 1
8(a)(ii) 2 B1 for each
8(b)(i) 9 1
11
8(b)(ii) 36 3 2 9
oe M2 for 2 × × oe
121 11 11
2 9
or M1 for × oe
11 11
36
If 0 scored SC1 for
110
8(c)(i) 3, 5, 28, 14 correctly placed 2 B1 for 28 in the intersection
8(c)(ii) 28 1 FT their 28 where their 28 < 50
oe
50
8(c)(iii) 123 2 42 41
oe M1 for ×
175 50 49
8(c)(iv) 63 2 FT their 28
oe
88 their 28 their 28 − 1
M1 for ×
33 32
(a)
(x – 1) cm x cm NOT TO
SCALE
(2x + 1) cm x cm
The area of the rectangle is 29 cm 2 greater than the area of the square.
The difference between the perimeters of the two shapes is k cm.
Find the value of k.
You must show all your working.
k = …… [6]
(b)
NOT TO
SCALE
(y + 1) cm y cm
The volume of the larger cube is 5 cm 3 greater than the volume of the smaller cube.
(i) Show that 3y 2 + 3y - 4 = 0 .
[4]
(ii) Find the volume of the smaller cube.
Show all your working and give your answer correct to 2 decimal places.
…… cm 3 [4]
Mark scheme
9(a) x 2 − x − 30 [ = 0] B3 M1 for (2 x + 1)( x − 1) − x 2 = 29 oe
B1 for (2 x + 1)( x − 1) = 2 x 2 − 2 x + x − 1 oe soi
( x − 6)( x + 5) oe M1 or correct factors for their 3 term quadratic equation
or for correct substitution into quadratic formula or correctly
completing the square for their 3 term quadratic equation
x = 6 cao B1
12 or 2 × their x evaluated B1 FT
or k = 2 x stated
9(b)(i) ( y + 1) − y3 = 5 oe M1
3
( y + 1) = y 3 + 3 y 2 + 3 y + 1 soi B2 B1 for ( y + 1)2 = y 2 + y + y + 1 oe soi
3
Completion to 3y2 + 3y – 4 = 0 A1 With no errors or omissions
9(b)(ii) −3 ± 32 − 4(3)(−4) B2 or B1 for 32 − 4(3)(−4)
2×3 −3 + ... −3 − ...
or for or
2×3 2×3
0.44 B2 B1 for 0.758 or 0.7583...
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Mathematics 0580/41 for May/June 2022 is out of 130 marks, with 2 hours 30 minutes to answer it.
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