Mathematics 0580/42 February/March 2022: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), February/March 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A company employed 300 workers when it started and now employs 852 workers.
(a) Calculate the percentage increase in the number of workers.
…… % [2]
(b) Of the 852 workers, the ratio part-time workers : full-time workers = 5 : 7.
Calculate the number of full-time workers.
…… [2]
(c) The company makes 40 600 headphones in one year.
Write this number
(i) in words,
…… [1]
(ii) in standard form.
…… [1]
(d) In one month, the company sells 3 000 headphones.
3
Of these, 48% are exported, are sold to shops and the rest are sold online.
8
Calculate the number of headphones that are sold online.
…… [3]
(e) One year, sales increased by 15%.
The following year sales increased by 18%.
Calculate the overall percentage increase in sales.
…… % [3]
1
Mark scheme
1(a) 184 2 852 − 300
M1 for [×100] oe
300
852
or for × 100 [ −100] oe
300
1(b) 497 2 852
M1 for × k oe where k = 1, 5 or 7
5+7
1(c)(i) Forty thousand six hundred 1
1(c)(ii) 4.06 ×104 1
1(d) 435 3 48 3
M2 for 3000 × 1 − − oe
100 8
or B2 for 2565, or 1440 and 1125 or 1875 and
1440 or 1560 and 1125
48 3 48 3
or M1 for 1 − − or 3000 × + oe
100 8 100 8
or B1 for 1440 or 1125 or 1560 or 1875
If 0 scored SC1 for answer 975
1(e) 35.7 3 100 + 15 100 + 18
M2 for × [–1] oe or better
100 100
100 + 15 100 + 18
or M1 for k × × oe
100 100
The table shows some values for y = x 2 - , x ! 0 .
3x
The y-values are rounded to 1 decimal place.
x -2 -1.5 -1 -0.75 -0.5 -0.25 -0.1 y 4.2 2.5 1.3 1.4 3.3
(a) Complete the table. [2]
1
(b) On the grid, draw the graph of y = x 2 - for - 2 G x G - 0.1 .
3x
1
The graph of y = x 2 - for x 2 0 has been drawn for you.
3x y
4
3
2
1
–2 –1 0 1 2 x
–1
–2
[4]
1
(c) By drawing a suitable line on the grid, solve the equation x 2 - + 1 = 0.
3x
x = …… [2]
Mark scheme
2(a) 1[.0] 0.9 2 B1 for each
2(b) correct curve 4 B3 FT for 6 or 7 points
B2 FT for 4 or 5 points
B1 FT for 2 or 3 points
2(c) ruled line at y = − 1 B1
0.3 to 0.32 B1
f (x) = 1 + 4x g (x) = x 2
(a) Find
(i) gf(3),
…… [2]
(ii) fg(x),
…… [1]
(iii) f - 1 f (x) .
…… [1]
(b) Find the value of x when f (x) = 15.
x = …… [2]
Mark scheme
3(a)(i) 169 2 M1 for g(13) or (1 + 4 x )2 or better
3(a)(ii) 1 + 4 x 2 final answer 1
3(a)(iii) x 1
3(b) 7 2 M1 for 1 + 4 x = 15
3.5 or
2
(a)
C
NOT TO
SCALE
29.5 cm 35.3 cm
51.6°
A 45 cm B
In triangle ABC, AB = 45 cm, AC = 29.5 cm, BC = 35.3 cm and angle CAB = 51.6°.
(i) Calculate angle ABC.
Angle ABC = …… [3]
(ii) Calculate the area of triangle ABC.
…… cm 2 [2]
(b)
R
NOT TO
SCALE
S
32 cm
47 cm
P 56°
60°
32 cm
Q
The diagram shows a quadrilateral PQRS formed from two triangles, PQS and QRS.
Triangle PQS is isosceles, with PQ = PS = 32 cm and angle SPQ = 56°.
QR = 47 cm and angle SQR = 60°.
(i) Calculate SR.
SR = …… cm [4]
(ii) Calculate the shortest distance from P to SQ.
…… cm [3]
Mark scheme
4(a)(i) 40.9 or 40.91… 3 29.5sin 51.6
M2 for [sin ABC= ] oe
35.3
35.32 + 452 − 29.52
or for [cos ABC = ]
2 × 35.3 × 45
29.5 35.3
or M1 for = oe
sin ABC sin 51.6
or for correct implicit cosine rule
4(a)(ii) 520 or 520.0 to 520.2… 2 FT their (a)(i) if used provided working shown
M1 for 0.5 × 29.5 × 45 × sin 51.6 oe
or for 0.5 × 35.3 × 45 × sin (their (a)(i))
or for 0.5 × 35.3 × 29.5sin (180–51.6–their(a)(i))
4(b)(i) 41.2 or 41.21 to 41.23 4 1
M1 for SQ = 2 × 32 × sin × 56 oe
2
or 32 2 + 32 2 − 2 × 32 × 32 × cos56 oe
32sin56
or oe
sin((180 - 56) ÷ 2)
M2 for
( )
SR 2 = 47 2 + their SQ 2 − 2 × 47 × their SQ × cos 60
or M1 for implicit form
4(b)(ii) 28.3 or 28.25 to 28.29… 3 M2 for 32 × sin62 oe
or M1 for recognition that line from P is
perpendicular to SQ
The table shows information about the mass, m grams, of each of 120 letters.
Mass (m grams) 0 1 m G 50 50 1 m G 100 100 1 m G 200 200 1 m G 500
Frequency 43 31 25 21
(a) Calculate an estimate of the mean mass.
…… g [4]
(b) Iraj draws a histogram to show this information.
He makes the height of the first bar 17.2 cm.
Calculate the height of each of the remaining bars.
height of bar for 50 1 m G 100 …… cm
height of bar for 100 1 m G 200 …… cm
height of bar for 200 1 m G 500 …… cm [3]
(c) Complete the cumulative frequency table.
Mass (m grams) m G 50 m G 100 m G 200 m G 500
Cumulative frequency
[2]
(d) Draw a cumulative frequency diagram.
120
100
80
Cumulative frequency 60
40
20
0 m
0 100 200 300 400 500
Mass (g)
[3]
(e) Use the cumulative frequency diagram to find an estimate for
(i) the median,
…… g [1]
(ii) the upper quartile,
…… g [1]
(iii) the 40th percentile,
…… g [2]
(iv) the number of letters with a mass m where 250 1 m G 400 .
…… [2]
Mark scheme
5(a) 5 4 M1 for midpoints soi
121 or 120.8… or 120
6
M1 for use of ∑fx with x in correct interval
including both boundaries but not if x is 50, 50, 100
and 300
M1 (dep on 2nd M1) for ∑fx ÷ 120
5(b) 12.4 5 1.4 3 B1 for each
If 0 scored SC1 for fd’s [0.86,] 0.62, 0.25 and 0.07
oe
5(c) 43 74 99 120 2 B1 for 2 or 3 correct
5(d) Correct diagram 3 B1 for correct horizontal placement for 4 plots
B1FT for correct vertical placement for 4 plots
B1FT dep on at least B1 for reasonable increasing
curve or polygon through their 4 points
If 0 scored SC1 FT for 3 out of 4 points correctly
plotted
5(e)(i) Strict FT their median reading 1
5(e)(ii) Strict FT their UQ reading 1
5(e)(iii) Strict FT their reading at 40th 2 B1 for 48 written or mark at cf = 48 on graph
percentile
5(e)(iv) Strict FT their reading at 400 2 B1 for either correct reading at 250 or 400
– their reading at 250
(a) The interior angle of a regular polygon is 156°.
Calculate the number of sides of this polygon.
…… [2]
(b)
NOT TO
SCALE
O C
52°
A B
A, B and C lie on a circle, centre O.
Angle OBA = 52°.
Calculate angle ACB.
Angle ACB = …… [2]
(c)
S R
W
112° NOT TO
SCALE
Q
T
P
P, Q, R, S and T lie on a circle.
WSR is a straight line and angle WSP = 112°.
Calculate angle PTR.
Angle PTR = …… [2]
(d)
K
NOT TO
SCALE
O
M
F G H
G, K and M lie on a circle, centre O.
FGH is a tangent to the circle at G and MG is parallel to OH.
Show that triangle GKM is mathematically similar to triangle OHG.
Give a geometrical reason for each statement you make.
…… [4]
Mark scheme
6(a) 15 2 360 180 ( n − 2 )
M1 for or for = 156 oe
180 − 156 n
6(b) 38 2 B1 for AOB = 76
6(c) 68 2 B1 for RSP = 68 or RQP = 112
6(d) Two pairs of equal angles identified M3 M2 for one pair of equal angles identified with fully
with fully correct reasons correct reasons
KMG = 90 angle in semicircle and OGH = 90
angle between tangent and radius
OR
KMG = OGH alternate segment
OR
GOH = MGK alternate angles
OR
Angle FGM = angle GHO corresponding and angle
FGM = GKM alternate segment and angle H =
angle K
or M1 for KMG = 90, angle in semicircle
or OGH = 90, angle between tangent and radius
Two or three pairs of angles equal A1 Dep on M3 with no incorrect work seen
[so similar] oe
Two rectangular picture frames are mathematically similar.
(a) The areas of the frames are 350 cm 2 and 1134 cm 2 .
The width of the smaller frame is 17.5 cm.
Calculate the width of the larger frame.
…… cm [3]
(b) A picture in the smaller frame has length 15 cm and width 10.5 cm, both correct to the nearest 5 mm.
Calculate the upper bound for the area of this picture.
…… cm 2 [2]
(c) In a sale, the price of a large frame is reduced by 18%.
Parthi pays $166.05 for 5 large frames in the sale.
Calculate the original price of one large frame.
$ …… [2]
(d) Parthi advertises a large frame for a price of $57 or 48.20 euros.
The exchange rate is $1= 0.88 euros.
Calculate the difference between these prices, in dollars and cents, correct to the nearest cent.
$ …… [3]
Mark scheme
7(a) 31.5 3 1134
M2 for 17.5 × oe
350
1134 350
or M1 for oe isw or oe isw
350 1134
2
1134 x
or for = oe
350 17.5
7(b) 15 2 B1 for 15 + 0.25 or 10.5 + 0.25 or better seen
163.937 5 or 163 final answer
16
7(c) 40.5[0] 2 18 166.05
M1 for x × 1 − = oe
100 [ 5]
7(d) $2.23 final answer 3 B2 for 2.227… or 2.23 seen
OR
48.2
M2 for 57 – oe
0.88
48.2
or M1 for oe
0.88
If 0 scored SC1 for 57 × 0.88 oe seen
Darpan runs a distance of 12 km and then cycles a distance of 26 km.
His running speed is x km / h and his cycling speed is 10 km / h faster than his running speed.
He takes a total time of 2 hours 48 minutes.
12
(a) An expression for the time, in hours, Darpan takes to run the 12 km is . x
Write an equation, in terms of x, for the total time he takes in hours.
…… [3]
(b) Show that this equation simplifies to 7x 2 - 25x - 300 = 0 .
[4]
(c) Use the quadratic formula to solve 7x 2 - 25x - 300 = 0 .
You must show all your working.
x = …… or x = …… [4]
(d) Calculate the number of minutes Darpan takes to run the 12 km.
…… min [2]
Mark scheme
8(a) 12 26 3 12 26
+ = 2.8 oe isw B2 for + oe isw
x x + 10 x x + 10
OR
26
B1 for seen
x + 10
168 48
B1 for time = 2.8 or or 2 oe
60 60
8(b) 12 ( x + 10 ) + 26 x = 2.8 x ( x + 10 ) or M2 FT their time, provided 2 algebraic fractions one in
better x and other in ± x ± 10
M1 for 12 ( x + 10 ) + 26 x seen or better
12 x + 120 + 26 x = 2.8 x 2 + 28x M1 FT their equation dep on M2
2.8 x 2 − 10 x − 120 = 0 oe A1
or 30x + 300 + 65x = 7x2 + 70x or
better
with no errors or omissions
leading to 7 x 2 − 25 x − 300 = 0
8(c) B2
[− −]25 ± ([−]25)2 − 4 × 7 × −300 B1 for ([−]25)2 − 4(7)(−300) or better
2×7 [ − −] 25 + q [− −]25 − q
oe or for or
2× 7 2× 7
− 5 and 8.57 or 8.571… B2 B1 for each
or SC1 for final answers 5 and –8.57
8(d) 84 to 84.01… 2 720
FT to 3 sf or better
their positive answer
12
M1 for [× 60 ] oe
their positive answer
(a)
O
O
NOT TO
SCALE x°
A B
2.4 cm AB
The volume of a paper cone of radius 2.4 cm is 95.4 cm 3 .
The paper is cut along the slant height from O to AB.
The cone is opened to form a sector OAB of a circle with centre O.
Calculate the sector angle x°.
1
[The volume, V, of a cone with radius r and height h is V = rr 2 h .]
3
…… [6]
(b) An empty fuel tank is filled using a cylindrical pipe with diameter 8 cm.
Fuel flows along this pipe at a rate of 2 metres per second.
It takes 24 minutes to fill the tank.
Calculate the capacity of the tank.
Give your answer in litres.
…… litres [4]
Mark scheme
9(a) 54[.0] or 53.99 to 54.03… 6 M2 for [h = ] 95.4 × 3 ÷ ( π × 2.42) oe
1
or M1 for 95.4 = × π × 2.42 × h
3
M2 for [slant ht , l = ] ( their h )2 + 2.42
or M1 for (their h)2 + 2.42
x
M1 for × 2 × π × their l = 2 × π × 2.4 oe
360
x 2
or × π × ( their l ) = π × 2.4 × their l
360
9(b) 14500 or 14470 to 14480 4 M3 for 200 × 60 × 24 × π × 42 [÷1000]
or 2 × 60 ×24 ×π × 0.042 [×1000]
or M2 for 200 × π × 42
or for 2 × π × 0.042
or M1 for π × 42 oe or π × 0.042 seen oe isw
or 1000 cm3 = 1 litre soi or 1 m3 = 1000 litres soi
or for 24 × 60 seen oe
(a) Expand and simplify.
(x + 1) (x - 2) (x + 3)
…… [3]
(b) Make g the subject of the formula.
2fg
M= g-c
g = …… [4]
(c) Simplify.
4x 2 - 16x x 2 - 16
…… [3]
1
Mark scheme
10(a) x3 + 2 x 2 − 5x − 6 final answer 3 B2 for correct expansion of three brackets
unsimplified or for simplified expression of correct
form with 3 out of 4 terms correct
or B1 for correct expansion of 2 of the 3 given
brackets with at least 3 terms out of four correct
10(b) Mc − Mc 4 M1 for clearing g – c from denominator
or final answer e.g. M(g – c) = 2fg
M −2f 2f −M
M1 for correctly isolating terms in g in numerator
on one side
M1 for correctly factorising or simplifying, to
single term in g in an equation
M1 for correctly dividing by bracket to final answer
10(c) 4x 3 B1 for 4x(x – 4)
final answer B1 for (x + 4) (x – 4)
x+4
(a) The probability that Shalini is late for school on any day is .
6
(i) Complete the tree diagram for Monday and Tuesday.
Monday Tuesday
Late
Late
Not late
Late
Not late
Not late
[2]
(ii) Calculate the probability that Shalini is late on Monday but is not late on Tuesday.
…… [2]
(b) The Venn diagram shows the number of students in a group of 50 students who wear glasses (G), who wear trainers (T ) and who have a mobile phone (M ).
G
T
0 2 3
2
19 14
1
M 9
(i) Use set notation to describe the region that contains only one student.
…… [1]
(ii) Find n bT l + `G , M jl.
…… [1]
(iii) One student is picked at random from the 50 students.
Find the probability that this student wears trainers but does not wear glasses.
…… [1]
(iv) Two students are picked at random from those wearing trainers.
Find the probability that both students have mobile phones.
…… [3]
Mark scheme
11(a)(i) 1 2 1 5
oe on all late branches B1 for one correct vertical pair oe and oe
6 6 6
5
oe on all not late branches
6
11(a)(ii) 5 2 FT their tree
oe 1 5
36 M1 for their × their
6 6
11(b)(i) ′ 1
( G ∪ T ∪ M ) oe
11(b)(ii) 28 1
11(b)(iii) 17 1
oe
50
11(b)(iv) 4 3 16 15
oe M2 for ×
7 21 20
n n −1 16 15
or M1 for × or for and seen
21 20 21 20
256
If 0 scored SC1 for answer oe
441
(a) Solve the equation tan x = 11.43 for 0° G x G 360° .
x = …… or x = …… [2]
(b) Sketch the curve y = x 3 - 4x .
y
O x
[3]
(c) A curve has equation y = x 3 + ax + b .
The stationary points of the curve have coordinates (2, k) and (-2, 10 - k).
Work out the value of a, the value of b and the value of k.
a = …… , b = …… , k = …… [6]
Mark scheme
12(a) 85[.0], 265[.0] and no others 2 B1 for each
If 0 scored SC1 for two values in the range with a
difference of 180 but not multiples of 90
12(b) correct shape and passes through 3 B1 for any positive cubic shape
origin
B1 for sketch with one max and one min and with 3
roots including zero
If 0 scored, SC1 for x(x + 2)(x – 2) soi
12(c) a = –12 6 B5 for 2 correct
b=5 OR
k = –11 B2 for 3x2 + a
or B1 for 3x2 isw
dy
M1dep on at least B1 for their =0
dx
M1dep on at least B1M1 for x = 2 or x = – 2
dy
substituted in their = 0 equation
dx
M1 for k = 23 + 2 × their a + b and
3
10 − k = ( −2 ) + ( −2 ) × their a + b
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On this page, free: the mark scheme for IGCSE Mathematics 0580/42, February/March 2022 (0580_m22_ms_42), and the question paper (0580_m22_qp_42), to read or download, with each question's marking points written out underneath.
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Mathematics 0580/42 for February/March 2022 is out of 130 marks, with 2 hours 30 minutes to answer it.
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