Mathematics 0580/41 May/June 2023: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) An orchard has 1250 trees.
The trees are in the ratio apple : pear : cherry = 12 : 9 : 4.
(i) Calculate the number of apple trees.
…… [2]
(ii) Last year in the orchard, the mean mass of fruit produced was 64 kg per tree.
Calculate the total mass of fruit produced last year.
Give your answer in tonnes.
[1 tonne = 1000 kg]
…… tonnes [2]
(iii) Last year, the mean mass of pears produced was 54 kg per tree.
This was a decrease of 10% on the mean mass of pears produced per tree from the year before.
Calculate the mean mass of pears produced by each pear tree the year before.
…… kg [2]
1
(iv) The orchard loses of its total number of trees in a storm.
5
Calculate the number of trees that remain.
…… [2]
(b) Paulo buys some pears from a market.
Pears cost $0.54 each or 0.51 euros each.
(i) Paulo pays in dollars for 12 pears.
Calculate the change he receives from $10.
$ …… [2]
(ii) The exchange rate is $1 = 0.826 euros.
Calculate how much more Paulo pays for each pear when he pays in euros.
Give your answer in dollars, correct to the nearest cent.
$ …… [3]
Mark scheme
1(a)(i) 600 2 1250
M1 for × k where k = 1, 4, 9, 12 oe
12 + 9 + 4
1(a)(ii) 80 2 M1 for 1250 × 64 [÷ 1000]
1(a)(iii) 60 2 10
M1 for x × 1 − = 54 oe
100
1(a)(iv) 1000 2 M1 for 1250 – (1250 ÷ 5) oe
or B1 for 250
1(b)(i) 3.52 2 M1 for [10 –] 12 × 0.54
or B1 for 6.48
1(b)(ii) 0.08 3 B2 for 0.077[4...]
or M1 for 0.51 ÷ 0.826
If 0 or 1 scored award instead SC2 for 0.93 final
answer
OR
If 0 scored SC1 for 0.06 as answer
F
NOT TO
SCALE
D
E C
A B
The diagram shows a solid triangular prism ABCDEF of length 15 cm.
AB = 6.4 cm, EB = 5.7 cm and the volume of the prism is 145 cm 3 .
(a) Show that angle EBA = 32° , correct to the nearest degree.
[3]
(b) Find the length of EA.
…… cm [3]
(c) Calculate the shortest distance from E to AB.
…… cm [3]
(d) Calculate the angle BF makes with the base, ABCD, of the prism.
…… [4]
(e) The prism is made of plastic with density 938 kg/m 3 .
Calculate the mass of the prism in grams.
[Density = mass ' volume ]
…… g [3]
Mark scheme
2(a) 145 M2 M1 for 145 = 12 × 6.4 × 5.7 × sin x × 15 oe
[sin =] 1
2 × 6.4 × 5.7 × 15
1 h
or for × 6.4 × h × 15 = 145 and sinx =
2 5.7
32.0[0] A1 If M0, SC1 for 145 =
0.5 × 6.4 × 5.7 × sin32 ×15 oe
2(b) 3.4[0] or 3.402 to 3.403 nfww 3 M2 for 6.42 + 5.7 2 − 2 × 6.4 × 5.7 × cos ( 32 )
OR
M1 for 6.42 + 5.7 2 − 2 × 6.4 × 5.7 × cos ( 32 )
A1 for 11.6 or 11.57 to 11.58
2(c) 3.02 or 3.020 to 3.021 3 x
M2 for sin ( 32 ) =
5.7
802 + 502 − 2 × 80 × 50 × cos 75
or M1 for recognition that the line from E is
perpendicular to AB e.g. right angle seen or
1
× 6.4 × h
2
2(d) 10.8 or 10.9 or 10.84 to 10.85... 4 their (c)
M3 for [sin =]
152 + 5.7 2
their (c)
or [ tan = ]
( 5.7 × cos32) + 15 )
2 2
or M2 for 152 + 5.7 2 or ( 5.7 × cos32 ) + 152 oe
2
or M1 for recognition of correct angle
2(e) 136 or 136.0... 3 1000
M2 for 938 × 145 × oe
1000000
or M1 for figs 136 or 13601
(a) The table shows information about the mass of each of 1000 eggs.
Mass (m grams) 40 1 m G 50 50 1 m G 56 56 1 m G 64 64 1 m G 70
Frequency 126 520 154 200
(i) Calculate an estimate of the mean.
…… g [4]
(ii) An egg is picked at random from the 1000 eggs.
Find the probability that this egg has a mass greater than 56 g.
Give your answer as a fraction in its simplest form.
…… [2]
(b) One year, a farmer makes a profit of $24 730 selling eggs.
Write this profit
(i) correct to 2 significant figures
$ …… [1]
(ii) in standard form.
$ …… [1]
(c) On a farm, there are 500 hens, correct to the nearest 10.
(i) In one year, the mean number of eggs laid per hen was 320 eggs, correct to the nearest 20.
Calculate the upper bound for the total number of eggs all the hens lay in that year.
…… [3]
(ii) Another farm has 800 hens, correct to the nearest 20.
Calculate the lower bound for the difference between the number of hens on the two farms.
…… [2]
Mark scheme
3(a)(i) 55.87 4
M1 for midpoints soi
M1 for use of Σ fm where m is in the correct
interval including boundaries
M1 (dep on 2nd M1) for Σ fm ÷1000
3(a)(ii) 177 2 154 + 200
cao M1 for oe
500 1000
3(b)(i) 25000 1
3(b)(ii) 2.473 ×104 1
3(c)(i) 166 650 or 165816 nfww 3 M2 for (500 + 5) × ‘320 to 340’
or ‘500 to 510’ × (320 + 10)
or M1 for 500 − 5 or 500 + 5 or 320 −10 or
320 +10
Alternative method
M2 for 504 × ‘320 to 340’
or ‘500 to 510’ × 329
or M1 for 504 or 329
3(c)(ii) 285 or 286 nfww 2 M1 for 800 −10
(a)
P NOT TO
SCALE
8 cm
Q R
24 cm
(i) Calculate the area of triangle PQR.
…… cm 2 [2]
(ii) Calculate angle PRQ.
Angle PRQ = …… [2]
(b)
NOT TO
SCALE
11 cm
6 cm
The diagram shows a half-cylinder of radius 6 cm and length 11 cm.
Calculate the volume of the half-cylinder.
…… cm 3 [2]
(c)
D T C T C
4 cm
S S
O
15 cm NOT TO
X X SCALE
A 20 cm B A B
(i) ABCD is a rectangle with AB = 20 cm and BC = 15 cm.
S, X and T are points on a circle centre O, such that DSA and DTC are tangents to the circle.
The radius of the circle is 4 cm and TX is a diameter of the circle.
The shape DSXT is removed from the corner of the rectangle, leaving the shaded shape shown in the second diagram.
Calculate the area of the shaded shape.
…… cm 2 [5]
(ii) Calculate the perimeter of the shaded shape.
…… cm [3]
Mark scheme
4(a)(i) 96 2 1
M1 for × 24 × 8
2
4(a)(ii) 18.4 or 18.43... 2 8
M1 for tan [ x ] = oe
24
4(b) 622 or 622.0 to 622.1.... 2 1 1
M1 for [ ×] π × 62 × 11 or × π × 62 [×11]
2 2
4(c)(i) 246 or 246.2 to 246.3... 5 270
M4 for 15 × 20 − 4 × 4 − × π × 42 oe
360
OR
270
M2 for × π × 42 oe
360
or M1 for k × π × 4 2 , where k ≤ 1
M1 for 15 × 20 or 4 × 4 oe
4(c)(ii) 80.8 or 80.9 or 80.84 to 80.85... 3
M1 for 15 + 20 + 11 + 16 oe
3
M1 for × 2 × π× 4 oe
4
5(a)(i)(a) 25 1
5(a)(i)(b) 17 to 18 1
5(a)(i)(c) 12 2 B1 for 148 seen
5(a)(i)(d) 30 2 B1 for 104 seen
5(a)(ii)(a) correct diagram or correct for 3 B1 for whiskers at 1 and at 70
their median and LQ
B1 for with median and LQ at their (a)(i)(a) and
(a)(i)(b)
B1 for UQ at 34
Maximum 2 marks if diagram incorrect
If 0 scored SC1 for their 5 correct ages plotted
5(a)(ii)(b) 50 1
(a) There are 160 people in a village.
The cumulative frequency diagram shows information about their ages.
160
140
120
100
Cumulative frequency 80
60
40
20
0
0 10 20 30 40 50 60 70
Age (years)
(i) Find an estimate for
(a) the median age
…… [1]
(b) the lower quartile
…… [1]
(c) the number of people who are 50 or more years of age
…… [2]
(d) the 65th percentile.
…… [2]
(ii) The youngest person in the village is 1 year old and the oldest is 70 years old.
(a) Draw a box-and-whisker plot to show the distribution of ages in the village.
0 10 20 30 40 50 60 70 80
Age (years)
[3]
(b) Write down an estimate of the percentage of people in the village that are younger than the median age.
…… % [1]
(b) The frequency table shows information about the age of each person in another village.
Age (n years) 0 1 n G 20 20 1 n G 30 30 1 n G 50 50 1 n G 80
Frequency 52 37 24 60
On the grid, complete the histogram to show this information.
The first block has been drawn for you.
4
3
Frequency density 2
1
0 n
0 10 20 30 40 50 60 70 80
Age (years)
[3]
Mark scheme
5(b) correct histogram 3 B1 for each correct block
width 10 height 3.7
width 20 height 1.2
width 30 height 2
If 0 scored SC1 for correct frequency densities
3.7, 1.2, 2 oe
(a) In the square ABCD, A has coordinates ( - 2 , 1) and B has coordinates (1, 5).
C has coordinates (a, b), where a and b are both positive integers.
Find the coordinates of C and the coordinates of D.
You may use the grid to help you.
y
6
5
4
3
2
1
–3 –2 –1 0 1 2 3 4 5 6 x
–1
–2
–3
C ( …… , …… )
D ( …… , …… ) [4]
(b) P has coordinates ( - 1, 3) and Q has coordinates (6, 4).
(i) Find the coordinates of the midpoint of PQ.
( …… , …… ) [2]
(ii) Find the length PQ.
…… [3]
(iii) Find the gradient of PQ.
…… [2]
(iv) Find the equation of the line parallel to PQ that crosses the x-axis at x = 2 .
…… [3]
Mark scheme
6(a) (5, 2) 4 B3 for 3 correct values or answers for C and D
(2, − 2) reversed or correct coordinates given on diagram
wrongly labelled
or B2 for one correct coordinate pair correctly
labelled
or M2 for A,B,C and D correctly plotted
or M1 for A and B correctly plotted
If 0 or 1 scored instead award SC2
for answers (–3, 8) and (–6, 4)
or answers (1.5,1.5) and (–2.5, 4.5)
6(b)(i) (2.5, 3.5) oe 2 B1 for each
6(b)(ii) 7.07 or 7.071... 3 M2 for ( 6 − −1)2 + ( 4 − 3)2 oe
or M1 for ( 6 − −1) or ( 4 − 3) oe
6(b)(iii) 1 2 4−3
M1 for oe
7 6 − −1
6(b)(iv) 1 2 3 M1 for gradient = their (iii)
y= x − or 7 y = x − 2 oe
7 7
final answer M1dep for substituting (2, 0) in a linear
equation with their m
allow if (2,0) satisfies y=(their(b)(iii)
gradient)x+c
(a) Factorise fully.
(i) 27y 2 - 3
…… [3]
(ii) 2m - pk + 2k - pm
…… [2]
x-1 6
(b) Solve - = 1. x+1 x-1
x = …… [5]
(c) Solve 4x 2 - 3x - 2 = 0 .
You must show all your working and give your answers correct to 2 decimal places.
x = …… or x = …… [4]
(d) Make k the subject of the formula. k
= 4 + kp m
k = …… [4]
Mark scheme
7(a)(i) 3 ( 3 y − 1)( 3 y + 1) final answer 3 B2 for ( 9 y − 3)( 3 y + 1) or ( 3 y − 1)( 9 y + 3)
or
or M1 for 3 ( 9 y 2 − 1) or [...] ( 3 y − 1)( 3 y + 1)
if 0 scored SC1 for an otherwise correctly
completely factorised expression but with
fractions within the brackets
7(a)(ii) ( 2 − p )( m + k ) final answer 2 M1 for 2 ( m + k ) − p ( m + k )
or m ( 2 − p ) + k ( 2 − p )
7(b) 1 5 B4 −8 x = +4 oe nfww
− oe nfww
2
x2 − 8x − 5
or B3 for = 1 or better
( x − 1)( x + 1)
OR
B2 x 2 − 8 x − 5
or M1 for ( x − 1)( x − 1) − 6 ( x + 1) or better
B1 ( x − 1)( x + 1) as full denominator or on the
right hand side
7(c) M2
− ( −3 ) ± ( −3) − 4 ( 4 )( −2 ) ( −3) − 4 ( 4 )( −2 )
2 2
M1 for
oe
2× 4
− ( −3) + q − ( −3) − q
or for or
2 ( 4) 2 ( 4)
2
3 3 2
or ± + oe
8 8 4
2
3
or for [4] x −
8
−0.43 and 1.18 final ans cao B1 for each
A2
SC1 for −0.4 ,–0.42 or −0.425....
and 1.2 or 1.17 or 1.175....
or answers 0.43 and −1.18
or −0.43 and 1.18 seen in working
7(d) 4m −4m 4
k= or k =
1 − pm pm − 1
final answer M1 for clearing fractions
M1 for collecting terms in k
M1 for factorising
M1 for dividing by bracket
Maximum 3 marks if answer incorrect
A tailor makes x dresses and y shirts in one week.
In one week
• he makes at least 4 dresses
• he makes no more than 7 shirts
• he makes less than 14 dresses and shirts altogether
2
• the number of shirts he makes is more than of the number of dresses.
3
One of the inequalities that shows this information is x H 4 .
(a) Write down the other three inequalities in x and/or y.
…… …… …… [3]
(b) y
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 x
On the grid, draw 4 straight lines and shade the unwanted regions to show these inequalities.
Label the region R that satisfies the 4 inequalities. [6]
(c) Use your diagram to find the smallest number of dresses and the smallest number of shirts the tailor makes in one week.
…… dresses and …… shirts [1]
(d) The profit the tailor makes on one dress is $10 and the profit on one shirt is $6.
Use your diagram to find the largest profit the tailor can make in one week.
$ …… [2]
Mark scheme
8(a) y 7 oe 3 B1 for each
x + y < 14 oe
2
y > x oe
3
8(b) x = 4 solid M4 B1 for each
y = 7 solid
x + y = 14 dashed
2
y= x dashed
3
correct shading everywhere but A2 M1dep (dependent on M4 or B1B1B1B0 where
region R the only error is wrong use of solid/dashed lines)
for shading the correct side of 3 of the 4 lines.
R
8(c) 4 dresses and 3 shirts 1
8(d) 106 2 M1 for 10 x + 6 y evaluated for (x, y) in their
region R
or B1 for (7, 6)
After 0 scored, SC1 for answer 112 or 116
(a) The Venn diagram shows set X and set Y.
X Y r t h e a c s l
(i) List the elements of X.
…… [1]
(ii) Find n (Y l ) .
…… [1]
(b) In each Venn diagram, shade the required region.
P Q P Q
P ,Q P l+ Q
[2]
(c) = {positive integers 1 13}
A = {x : x 1 9 }
B = {x : x is even}
C = {x : x is a multiple of 3}
A B
C
(i) Complete the Venn diagram. [3]
(ii) Find n (Al , (B + C )) .
…… [1]
Question 10 is printed on the next page.
Mark scheme
9(a)(i) r, l, t, e, a 1
9(a)(ii) 2 1
9(b) 1
1
9(c)(i) Fully correct 3 B2 for 7, 6, or 5 sections correct
or B1 for 4, 3 or 2 sections correct
1 2 10
5 4 8
7
6
12
3
11
9
9(c)(ii) 5 1FT strict FT from their diagram
f (x) = x - 4 g (x) = 2x + 5 h (x) = 3 x
(a) Find
(i) f (- 3)
…… [1]
(ii) g -1 (x)
g -1 (x) = …… [2]
(iii) f (x) # g (x) # f (x) .
…… [4]
(b) Find x when h (x) = g (f (2)) .
x = …… [2]
Mark scheme
10(a)(i) −7 1
10(a)(ii) x−5 2 M1 for correct first step e.g. x = 2 y + 5 or
oe final answer
2 2x = y − 5
y 5
or = x +
2 2
10(a)(iii) 2 x 3 − 11x 2 − 8 x + 80 final answer 4 M1 for ( x − 4 ) ( 2 x + 5 ) ( x − 4 ) oe
B2 for
2 x 3 − 8 x 2 − 8 x 2 + 5 x 2 − 20 x − 20 x + 32 x + 80
or for simplified 4 term expression of the correct
form with 3 terms correct in final answer
or B1 for 3 terms correct out of 4 from
x 2 − 4 x − 4 x + 16 or 2 x 2 − 8 x + 5 x − 20
10(b) 0 2 M1 for g(− 2)
or 2(x – 4) + 5 oe
or 3௫ = 1
( )
or g f ( 2 ) = 1
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Mathematics 0580/41 for May/June 2023 is out of 130 marks, with 2 hours 30 minutes to answer it.
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