IGCSE Mathematics 0580 Paper 41, Oct/Nov 2023

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0580/41 · Oct/Nov 2023

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1
Figure 1 of 1 for question 1
a(i)

Describe fully the single transformation that maps shape A onto shape B

[2]
a(ii)

Describe fully the single transformation that maps shape A onto shape C.

[3]
b(i)

On the grid, draw the image of shape A after a reflection in the line y=2

[2]
b(ii)

On the grid, draw the image of shape A after an enlargement, scale factor −2, centre (0,0).

[2]
2
a

s=12at2 Find the value of s when a=9.8 and t=20.

[2]
b

Solve. 5(4y−3)=15

[3]
c

Expand and simplify. 3(5x−8)−2(3x−7)

[2]
d

Rearrange A=2b2−3c3 to make c the subject.

[3]
e

Factorise completely. 6pq−4q−3p+2

[2]
3

The table shows information about some of the planets in the solar system.

PlanetDiameter (km)Average distance from the Sun (km)
Earth12 8001.496×108
Mars6 8002.279×108
Jupiter143 0007.786×108
Saturn120 5001.434×109
Neptune49 5004.495×109
a(i)

The average distance of Mars from the Sun is 2.279×108 km. Write this distance as an ordinary number.

[1]
a(ii)

The planet Uranus has a diameter that is 35.8% of the diameter of Jupiter. Calculate the diameter of Uranus.

[2]
a(iii)

The ratio diameter of Neptune:diameter of Saturn can be written in the form 1:n. Find the value of n.

[1]
a(iv)

Find the average distance of Neptune from the Sun as a percentage of the average distance of the Earth from the Sun.

[2]
a(v)

Distances within the solar system are also measured in astronomical units (AU). The average distance of Jupiter from the Sun is 5.20 AU. Calculate the average distance of Mars from the Sun in astronomical units.

[2]
a(vi)

The diameter of Mars is 39.2% greater than the diameter of Mercury. Calculate the diameter of Mercury.

[2]
b(i)

One light year is the distance that light travels in a year of 365.25 days. The speed of light is 2.9979×105 kilometres per second. Show that one light year is 9.461×1012 km, correct to 4 significant figures.

[2]
b(ii)

The distance from the Andromeda Galaxy to Earth is 2.40×1019 km. Calculate the time taken for light to travel from this galaxy to Earth. Give your answer in millions of years.

[2]
4

Lucia has two fair spinners. Spinner A is five-sided and is numbered 1, 2, 3, 4, 5. Spinner B is nine-sided and is numbered 3, 3, 3, 4, 4, 4, 4, 5, 5. Lucia spins the two spinners and records whether they land on a prime number.

a(i)

Complete the tree diagram.

[2]
a(ii)(a)

Find the probability that the two numbers are both prime

[2]
a(ii)(b)

Find the probability that the two numbers are not both prime.

[1]
b

Lucia spins Spinner A 120 times. Find the expected number of times the spinner lands on a prime number.

[1]
c

Lucia spins Spinner B twice. Find the probability that the two numbers it lands on add up to 9 or more.

[3]
d

Lucia keeps spinning Spinner B until it lands on a 4. Find an expression, in terms of n, for the probability that this happens on the nth spin.

[2]
5
Figure 1 of 1 for question 5
a

ACD is a right-angled triangle. B is on AC and BC=54.5 m. AD=83.2 m and angle ABD=38∘. Calculate angle ACD.

[5]
b

EFG is a right-angled triangle. A circle can be drawn that passes through the three vertices of the triangle. On the diagram, mark the position of the centre of the circle with a cross. Explain how you decide.

[2]
c

In triangle LMN, the ratio angle L : angle M : angle N=4:5:6. In triangle PQR, PQ=6 cm, PR=4 cm and QR=5 cm. Calculate the difference between the largest angle in triangle PQR and the largest angle in triangle LMN.

[7]
6
a

Complete the table for the three sequences.

Sequence1st term2nd term3rd term4th term5th termnth term
A−7−315
B7132337
C22738142435729
[10]
b

In a sequence, the sum of the first 49 terms is 7644. The sum of the first 50 terms is 7975. Find the 50th term of this sequence.

[1]
7

The frequency table shows the time of each of 42 athletes in a race.

Time (t seconds)Number of athletes
216<t⩽2199
219<t⩽22414
224<t⩽23414
234<t⩽2442
244<t⩽2643
a

Calculate an estimate of the mean time.

[4]
b

Complete the histogram to show the information in the frequency table. Two of the blocks have been drawn for you.

[3]
8
Figure 1 of 2 for question 8
Figure 2 of 2 for question 8
a(i)

The diagram shows a solid formed by joining two hemispheres and a cylinder. The radius of the large hemisphere is 5.4 cm. The radius of the small hemisphere and the radius of the cylinder are both 3.6 cm. The height of the cylinder is 6.5 cm. Show that the volume of the solid is 692 cm3, correct to the nearest cubic centimetre. [The volume, V, of a sphere with radius r is V=43πr3.]

[4]
a(ii)

A mathematically similar solid is made of silver. In this solid, the cylinder has radius 0.6 cm. 1 cm3 of silver has a mass of 10.49 grams. Calculate the total mass of this silver solid.

[4]
b(i)

AOB is a sector of a circle, centre O. AO=10 cm and the sector angle is 216∘. Calculate the length of the arc of this sector. Give your answer as a multiple of π.

[2]
b(ii)

A cone is made from this sector by joining OA to OB. Calculate the volume of the cone. [The volume, V, of a cone with radius r and height h is V=13πr2h.]

[4]
9

f(x)=(3x+1)(x+5)(x−4)g(x)=2x−3h(x)=42x−1

a(i)

Find f(0)

[1]
a(ii)

Find g−1(x)

[2]
a(iii)

Find gh(2).

[2]
b

g(2x)=7 Find the value of x.

[2]
c

Simplify g(x2)+gg(x)+1.

[3]
d

Find h−1(16).

[2]
e

f(x)=(3x+1)(x+5)(x−4) This can be written in the form f(x)=ax3+bx2+cx+d. Find the value of each of a, b, c and d.

[3]
10
Figure 1 of 1 for question 10
a(i)

ABC is a triangle. B is the point (1,−10), A is the point (4,14) and CA⃗=(−118). Find the coordinates of C.

[2]
a(ii)

Find BA⃗.

[1]
a(iii)

Find ∣CA⃗∣.

[2]
b(i)

OMN is a triangle. OM⃗=a and ON⃗=b. R is a point on MN such that MR:RN=3:2. ORT is a straight line. Show that OR⃗=25a+35b.

[3]
b(ii)(a)

NT⃗=4a+kbandOT⃗=cOR⃗. Find the value of k and the value of c.

[4]
b(ii)(b)

Find MT⃗.

[1]
11
a

Differentiate x3−4x2−3x.

[2]
b

A curve has equation y=x3−4x2−3x. Work out the coordinates of the two stationary points. Show all your working.

[5]
c

Determine whether each stationary point is a maximum or a minimum. Show all your working.

[3]

Mark scheme

0580/41 · Oct/Nov 2023

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1a(i) · 2 marks

  • 2 marks

    Translation (−7−1) oe

1a(ii) · 3 marks

  • 3 marks

    Rotation 90º clockwise oe (5, 1)

1b(i) · 2 marks

  • 2 marks

    Image at (2, 6) (3, 6) (3, 8)

1b(ii) · 2 marks

  • 2 marks

    Image at (–4, 4) (–6, 4) (–6, 8)

2a · 2 marks

  • 2 marks

    1960

2b · 3 marks

  • 3 marks

    1.5 or 112 or 32

2c · 2 marks

  • 2 marks

    9x – 10 final answer

2d · 3 marks

  • 3 marks

    2b2−A33 oe final answer

2e · 2 marks

  • 2 marks

    (2q – 1)(3p – 2) or (1 – 2q)(2 – 3p) final answer

3a(i) · 1 mark

  • 1 mark

    227 900 000

3a(ii) · 2 marks

  • 2 marks

    51 200 or 51 190 or 51 194

3a(iii) · 1 mark

  • 1 mark

    2.43 or 2.434…

3a(iv) · 2 marks

  • 2 marks

    3000 or 3004 to 3005

3a(v) · 2 marks

  • 2 marks

    1.52 or 1.522…

3a(vi) · 2 marks

  • 2 marks

    4890 or 4885…

3b(i) · 2 marks

  • M1

    2.9979×105×602×24×365.25

  • A1

    =9.4606…×1012

3b(ii) · 2 marks

  • 2 marks

    2.54 or 2.536 to 2.537

4a(i) · 2 marks

  • 2 marks

    2559,4959,49

4a(ii)(a) · 2 marks

  • 2 marks

    13 oe

4a(ii)(b) · 1 mark

  • 1 mark

    23 oe

4b · 1 mark

  • 1 mark

    72

4c · 3 marks

  • 3 marks

    2081 oe

4d · 2 marks

  • 2 marks

    (59)n−1[×]49 oe final answer

5a · 5 marks

  • 5 marks

    27.3 or 27.32 to 27.33

5b · 2 marks

  • B2

    Centre marked at midpoint of FG. and Angle in a semi-circle is 90

5c · 7 marks

  • 7 marks

    10.8 or 10.81 to 10.82

6a · 10 marks

  • B1

    A 9

  • B2

    4n – 11 oe final answer

  • B1

    B 55

  • B2

    2n2+5 oe final answer

  • B1

    C 62187 oe

  • B3

    n+13n+2 oe final answer

6b · 1 mark

  • 1 mark

    331 cao

7a · 4 marks

  • 4 marks

    226 nfww or 226.2 to 226.3[0] nfww

7b · 3 marks

  • 3 marks

    Blocks with heights 2.8, 1.4, 0.2 and with correct widths

8a(i) · 4 marks

  • M3

    23π(3.6)3+23π(5.4)3+π(3.6)2×6.5

  • A1

    692.1 to 692.2…

8a(ii) · 4 marks

  • 4 marks

    33.6 or 33.60 to 33.62

8b(i) · 2 marks

  • 2 marks

    12π final answer

8b(ii) · 4 marks

  • 4 marks

    302 or 301.5 to 301.6…

9a(i) · 1 mark

  • 1 mark

    –20

9a(ii) · 2 marks

  • 2 marks

    x+32 oe final answer

9a(iii) · 2 marks

  • 2 marks

    125

9b · 2 marks

  • 2 marks

    2.5 oe

9c · 3 marks

  • 3 marks

    2x2+4x−11 final answer

9d · 2 marks

  • 2 marks

    1.5 oe

9e · 3 marks

  • 3 marks

    a = 3 b = 4 c = –59 d = –20

10a(i) · 2 marks

  • 2 marks

    (15, 6)

10a(ii) · 1 mark

  • 1 mark

    (324)

10a(iii) · 2 marks

  • 2 marks

    13.6 or 13.60…

10b(i) · 3 marks

  • M3

    a+35(b−a) or b+25(a−b) leading to 25a+35b with no errors

10b(ii)(a) · 4 marks

  • 4 marks

    k = 5, c = 10

10b(ii)(b) · 1 mark

  • 1 mark

    3a + 6b final answer

11a · 2 marks

  • 2 marks

    3x2−8x−3

11b · 5 marks

  • M1

    3x2−8x−3=0

  • M2

    Correct method to solve their 3-term quadratic (3x+1)(x–3)[=0] −(−8)±(−8)2−4(3)(−3)2(3)

  • B2

    (3,–18) (−13,1427)

11c · 3 marks

  • 3 marks

    (3, –18) minimum with reason (−13,1427) maximum with reason

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  3. 3
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  4. 4
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  5. 5
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  6. 6
    Question 611 marks
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  7. 7
    Question 77 marks
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  8. 8
    Question 814 marks
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  9. 9
    Question 915 marks
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  10. 10
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  11. 11
    Question 1110 marks
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