Mathematics 0580/42 February/March 2020: question paper and mark scheme
Cambridge IGCSE Mathematics (0580), Paper 4 (Extended), February/March 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Dhanu has a model railway.
(a) He has a train that consists of a locomotive and 4 coaches.
The mass of the locomotive is 87 g and the mass of each coach is 52 g.
(i) Work out the total mass of the train.
…… g [2]
(ii) Work out the mass of the locomotive as a percentage of the total mass of the train.
…… % [1]
(b) The train is 61 cm long and travels at a speed of 18 cm/s.
It takes 4 seconds for the whole of the train to cross a bridge.
Calculate the length of the bridge.
…… cm [2]
(c) A new locomotive costs $64.
Calculate the cost of the locomotive in rupees when the exchange rate is 1 rupee = $0.0154 .
Give your answer correct to the nearest 10 rupees.
…… rupees [2]
(d) The cost of a railway magazine increases by 12.5% to $2.70 .
Calculate the cost of the magazine before this increase.
$ …… [2]
(e) Dhanu plays with his model railway from 06 50 to 11 15.
He then rides his bicycle for 3 hours.
Find the ratio time playing with model railway : time riding bicycle.
Give your answer in its simplest form.
…… : …… [3]
(f) The value of Dhanu’s model railway is $550.
This value increases exponentially at a rate of r % per year.
At the end of 5 years the value will be $736.
Calculate the value of r.
r = …… [3]
Mark scheme
1(a)(i) 295 2 M1 for [87 +] 4 × 52 oe
1(a)(ii) 29.5 or 29.49… 1 87
FT × 100
their(a)(i)
1(b) 11 2 M1 for 18 × 4 [± 61] oe
1(c) 4160 cao nfww 2 M1 for 64 ÷ 0.0154
or B1 for rounding their answer to nearest
10
1(d) 2.4[0] nfww 2 12.5
M1 for 1 + x = 2.7[0] oe
100
1(e) 53 : 36 3 M2 for 265 : 180 oe or for answer 36 : 53
or 53 min: 36 min
or M1 for 4h 25 [mins] or 265 [mins] seen
1(f) 6[.00] or 5.999… 3 736
M2 for 5
550
or M1 for 736 = 550 × (x)5
(a) The table shows some values for y = 2x 3 - 4x 2 + 3.
x -1 - 0.5 0 0.5 1 1.5 2 y -3 1.75 0.75 3
(i) Complete the table. [3]
(ii) On the grid, draw the graph of y = 2x 3 - 4x 2 + 3 for - 1 G x G 2 .
y
3
2
1
–1 0 1 2 x
–1
–2
–3
[4]
(iii) Use your graph to solve the equation 2x 3 - 4x 2 + 3 = 1.5 .
x = …… or x = …… or x = …… [3]
(iv) The equation 2x 3 - 4x 2 + 3 = k has only one solution for - 1 G x G 2 .
Write down a possible integer value of k.
…… [1]
(b) y
6
5
4
3
2
1
–1 0 1 x
–1
–2
–3
–4
(i) On the grid, draw the tangent to the curve at x = 1. [1]
(ii) Use your tangent to estimate the gradient of the curve at x = 1.
…… [2]
(iii) Write down the equation of your tangent in the form y = mx + c .
y = …… [2]
Mark scheme
2(a)(i) 3 2.25 1 3 B1 for each
2(a)(ii) Fully correct smooth curve 4 B3FT for 7 or 6 correct plots
B2FT for 5 or 4 correct plots
B1FT for 3 correct plots
2(a)(iii) −0.6 to −0.51, 0.75 to 0.85, 3 B1 for each
1.7 to 1.85 If 0 scored, SC1 for y = 1.5 drawn
2(a)(iv) −3 or −2 or −1 or 0 1
2(b)(i) Tangent ruled at x = 1 1
2(b)(ii) 4.4 to 5.6 2 Dep on tangent at x = 1 or close attempt
M1 for rise/run for their line
2(b)(iii) y = (4.4 to 5.6)x – (1.8 to 2.2) 2 FT for any line but not horizontal or vertical
or line for 2 marks or B1
[y =] their (b)(ii)x + their(y-intercept)
B1FT for [m =] their 5
or for their y-intercept
(a) Manjeet uses 220 litres of water each day.
She reduces the amount of water she uses by 15%.
Calculate the number of litres of water she now uses each day.
…… litres [2]
(b) Manjeet has two mathematically similar bottles in her bathroom.
The large bottle holds 1.35 litres and is 29.7 cm high.
The small bottle holds 0.4 litres.
Calculate the height of the small bottle.
…… cm [3]
(c) Water from Manjeet’s shower flows at a rate of 12 litres per minute.
The water from the shower flows into a tank that is a cuboid of length 90 cm and width 75 cm.
Calculate the increase in the level of water in the tank when the shower is used for 7 minutes.
…… cm [3]
Mark scheme
3(a) 187 2 15
M1 for 220 × 1 − oe
100
or B1 for 33 seen
3(b) 19.8 3 0.4
M2 for 29.7 × 3 oe
1.35
0.4 1.35
or M1 for 3 or 3 oe seen
1.35 0.4
29.73 1.35
or for = oe
x3 0.4
3(c) 12.4 or 12.44… 3 M1 for 90 × 75 × h = 7 × figs 12
B1 for 1000 cm3 = 1 litre soi
A solid metal cone has radius 1.65 cm and slant height 4.70 cm.
(a) Calculate the total surface area of the cone.
[The curved surface area, A, of a cone with radius r and slant height l is A = rrl .]
…… cm2 [2]
(b) Find the angle the slant height makes with the base of the cone.
…… [2]
(c) (i) Calculate the volume of the cone.
1
[The volume, V, of a cone with radius r and height h is V = rr 2 h .]
3
…… cm3 [4]
(ii) A metal sphere with radius 5 cm is melted down to make cones identical to this one.
Calculate the number of complete identical cones that are made.
4
[The volume, V, of a sphere with radius r is V = rr 3 .]
3
…… [4]
Mark scheme
4(a) 32.9 or 32.91 to 32.92… 2 M1 for π × 1.65 × 4.7 + π × 1.652
4(b) 69.4 or 69.44 to 69.45 2 M1 for cos = 1.65 ÷ 4.7 oe
4(c)(i) 12.5 or 12.54 to 12.55 4 1
M3 for × π × 1.652 × 4.7 2 − 1.652 oe
3
or M2 for 4.7 2 − 1.652 oe
or for 4.7 × sin(their (b)) oe
or M1 for 1.652 + h 2 = 4.7 2 oe
h
or for = sin(their (b)) oe
4.7
4(c)(ii) 41 nfww 4 B3 for 41.7… to 41.9
4
or M2 for × π × 53 ÷ their 12.5
3
4
or M1 for × π × 53
3
After M2 scored, M1 for truncating their
decimal number of cones seen to an integer
answer
(a) Write as a single fraction in its simplest form. x+3 x-2
- x-3 x+2
…… [4]
k
(b) 2 12 ' 2 2 = 32
Find the value of k.
k = …… [2]
(c) Expand and simplify.
( y + 3) ( y - 4) (2y - 1)
…… [3]
(d) Make x the subject of the formula.
3+x x= y
x = …… [3]
Mark scheme
5(a) 10 x 10 x 4 M1 for common denominator ( x − 3)( x + 2 )
or 2
( x − 3)( x + 2 ) x − x − 6 isw
final answer M1 for ( x + 3)( x + 2 ) − ( x − 2 )( x − 3) isw
B1 for correct numerator in terms of x only
5(b) 14 2 k
k
212
M1 for 12 − = 5 or 2 2 = 5 oe
2 2
14
4096
or or 12 – 5 or 212 ÷ 2 2 [= 32] seen
32
5(c) 2 y 3 − 3 y 2 − 23 y + 12 final answer 3 B2 for correct unsimplified expanded
expression
or for simplified four-term expression of
correct form with 3 terms correct
or B1 for correct expansion of 2 of the
brackets with at least 3 terms correct
5(d) 3 3 M1 for xy = 3 + x
[ x =] final answer
x 3
y −1 M1 for xy – x = 3 or x – =
y y
M1 for factorising and dividing
Suleika has six cards numbered 1 to 6.
1 2 3 4 5 6
(a) She takes one card at random, records the number and replaces the card.
(i) Write down the probability that the number is 5 or 6.
…… [1]
(ii) Suleika does this 300 times.
Find how many times she expects the number 5 or 6.
…… [1]
(b) Suleika takes two cards at random, without replacement.
(i) Find the probability that the sum of the numbers on the two cards is 5.
…… [3]
(ii) Find the probability that at least one of the numbers on the cards is a square number.
…… [3]
Mark scheme
6(a)(i) 1 1
oe
3
6(a)(ii) 100 1 FT their (a)(i) × 300 to at least 3 sf or
rounded to the nearest integer
6(b)(i) 2 3 1 1
oe M2 for 4 × × oe
15 6 5
1 1
or M1 for k × oe
6 5
or list or indication of 4 correct pairs
6(b)(ii) 3 3 4 3
oe M2 for 1 – ×
5 6 5
2 4 2 1
or 2 × + × oe
6 5 6 5
2 4 2
or + × oe
6 6 5
4 3 2 4
or M1 for × oe seen or × [×2] oe
6 5 6 5
seen
2 1
or × oe seen
6 5
or correct identification of 18 pairs
or space diagram oe
(a) Naga has n marbles.
Panav has three times as many marbles as Naga.
Naga loses 5 marbles and Panav buys 10 marbles.
Together they now have more than 105 marbles.
Write down and solve an inequality in n.
…… [3]
(b) y is inversely proportional to x2.
When x = 4 , y = 7.5 .
Find y when x = 5.
y = …… [3]
(c) Find the nth term of each sequence.
(i) 4 2 0 -2 -4 …
…… [2]
(ii) 1 7 17 31 49 …
…… [2]
Mark scheme
7(a) n − 5 + 3n + 10 > 105 or better B1
n > 25 final answer B2 M1 for 4n > 100
7(b) 4.8 3 k
M1 for y = or better
x2
their k
M1 for [ y =] 2
5
OR
M2 for y × 52 = 7.5 × 42
7(c)(i) 6 − 2n oe final answer 2 B1 for answer 6 − kn (k ≠ 0) oe
or answer j− 2n oe
or for correct expression shown in working
and then spoilt
7(c)(ii) 2n 2 − 1 oe final answer 2 B1 for 2nd diff = 4 or a quadratic
expression
or for correct expression shown in working
and then spoilt
(a)
S
25°
R
72°
NOT TO
SCALE
6 cm
34°
Q
P 7.4 cm
The diagram shows a quadrilateral PQRS formed from two triangles, PQS and QRS.
Calculate
(i) QR,
QR = …… cm [3]
(ii) PS,
PS = …… cm [3]
(iii) the area of quadrilateral PQRS.
…… cm2 [4]
(b)
X
H
G
E F NOT TO
SCALE
16 cm D C
18 cm
A 20 cm B
The diagram shows an open box ABCDEFGH in the shape of a cuboid.
AB = 20 cm, BC = 18 cm and AE = 16 cm .
A thin rod AGX rests partly in the box as shown.
The rod is 40 cm long.
(i) Calculate GX, the length of the rod which is outside the box.
GX = …… cm [4]
(ii) Calculate the angle the rod makes with the base of the box.
…… [3]
Mark scheme
8(a)(i) 2.67 or 2.666… 3 6 × sin 25
M2 for
sin 72
or M1 for implicit version
8(a)(ii) 4.14 or 4.140… 3 M1 for 62 + 7.42 − 2 × 6 × 7.4 × cos34
A1 for 17.1 to 17.2
8(a)(iii) 20.4 or 20.35 to 20.36… 4 B1 for angle SQR = 83
M1 for
1
× 6 ×their (a)(i) × sin their (180–72–25)
2
oe
1
M1 for × 6 × 7.4 × sin 34 oe
2
8(b)(i) 8.7[0] or 8.695… 4 B3 for 980 oe or 31.3 or 31.30…
or M3 for 40 – 202 + 182 + 162 oe
or M2 for 202 + 182 + 162 oe
or M1 for any correct attempt at
2-dimensional Pythagoras’ e.g. 182 + 162
8(b)(ii) 30.7 or 30.73 to 30.74… 3 16
M2 for [sin =] oe
20 + 182 + 162
2
or B1 for identifying angle GAC
This year, 40 students have each travelled by one or more of plane (P), train (T) or boat (B).
7 have travelled only by plane.
11 have travelled only by train.
9 have travelled only by boat.
n (P + T ) = 8 n (B + T ) = 3 n (B + P ) = 6
P T
B
(a) Complete the Venn diagram. [3]
(b) Find n `(P , B)lj .
…… [1]
(c) Use set notation to complete the statement.
(P , T , B)l = …… [1]
(d) Two students are chosen at random.
Calculate the probability that they both have travelled only by plane.
…… [2]
(e) Two students are chosen at random from those who have travelled by train.
Calculate the probability that they both have also travelled by plane.
…… [2]
Mark scheme
9(a) 3 B2 for 5 correct entries including ‘2’
correctly placed at the intersection of the
P
T 3 sets
6
7 11
or M1 for
2 k + 8 − k + 3 − k + 6 − k = 40 − (7 + 9 + 11) oe
4 1
or for k, 8 – k, 3 – k, 6 – k, seen correctly
placed on diagram with 7, 11 and 9
9 correctly placed
B
9(b) 11 1
9(c) Ø or { } 1
9(d) 7 2 7 6
oe M1 for × oe
260 40 39
9(e) 14 2 FT their Venn diagram
oe
95 8 7
M1 for ×
20 19
f (x) = 4x - 1 g (x) = x 2 h (x) = 3 -x
(a) Find in its simplest form
(i) f (x - 3) ,
…… [1]
(ii) g (5x) .
…… [1]
(b) Find f -1 (x) .
f -1 (x) = …… [2]
(c) Find the value of hh (l) , correct to 4 significant figures.
…… [3]
(d) (i) Show that g (3x - 2) - h (- 3) can be written as 9x 2 - 12x - 23.
[2]
(ii) Use the quadratic formula to solve 9x 2 - 12x - 23 = 0 .
Give your answers correct to 2 decimal places.
x = …… or x = …… [4]
(e) Find x when f (61) = h (x) .
x = …… [2]
Mark scheme
10(a)(i) 4 x − 13 final answer 1
10(a)(ii) 25x 2 final answer 1
10(b) x +1 x 1 2 M1 for correct first step x = 4 y − 1 or
or +
4 4 4 y 1
y + 1 = 4 x or = x −
4 4
10(c) 0.6934 final answer 3 −
1
B2 for 0.69336… or 3 3 oe or 0.693
−x
or M1 for 3−3 oe
10(d)(i) ( 3x − 2 )2 − 3− ( −3) M1
9 x 2 − 6 x − 6 x + 4 − 27 or A1 with no errors seen
9 x 2 − 12 x + 4 − 27
leading to 9 x 2 − 12 x − 23
10(d)(ii) −(−12) ± (−12) 2 − 4(9)(−23) B2 B1 for (−12) 2 − 4(9)( −23) oe
2×9 −(−12) + q −(−12) − q
or better or oe or oe or
2×9 2×9
both
– 1.07, 2.40 final answers B2 B1 for each
If B0, SC1 for answers – 1.1 or –1.06 or
–1.065… to – 1.065 and 2.4 or 2.39 or
2.398 to 2.398…
or – 1.07 and 2.40 seen in working
or for –2.40 and 1.07 as final answer
10(e) −5 final answer 2 M1 for 243 = 3− x
A curve has equation y = x 3 - 3x + 4 .
(a) Work out the coordinates of the two stationary points.
( …… , ……)
( …… , ……) [5]
(b) Determine whether each stationary point is a maximum or a minimum.
Give reasons for your answers.
[3]
Mark scheme
11(a) (1, 2) 5 B2 for [derivative oe = ] 3 x 2 − 3
( −1 , 6)
or B1 for [derivative oe = ] 3x 2 or f( x) − 3
M1 for their derivative = 0
dy
or recognition of = 0 oe
dx
B1 for [x =] −1 , 1 or for one coordinate pair
11(b) (1, 2) minimum with reason 3 Reasons could be e.g.
a reasonable sketch
( −1 , 6) maximum with reason correct use of 2nd derivative = 6x = 6 , 6 > 0,
so (1, 2) minimum oe
2nd derivative = 6x = –6 , –6 < 0 so (–1, 6)
maximum oe,
or finds gradient on each side of both
correct stationary points with correct
conclusion
B2 for 1 correct with reason
or M1 for showing [2nd derivative =] 6x
or gradients for one value on either side of
one correct stationary point
or for reasonable sketch of cubic
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