Cambridge IGCSE Mathematics 0580 — 2021 Oct/Nov Paper 2 · Variant 2
0580/22/O/N/21 · 23 questions · 70 marks · ≈79 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper12 pages
Mark scheme7 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q1 · The temperature at midnight is -8.5 °C
1 The temperature at midnight is -8.5 °C. The temperature at 11 am is -1 °C. Work out the difference between the temperature at midnight and the temperature at 11 am. ............................................. °C [1]
Mark scheme: Question Answer Marks Partial Marks 1 7.5 1
Q2 · The stem-and-leaf diagram shows the age, in years, of each of 15 women
2 The stem-and-leaf diagram shows the age, in years, of each of 15 women. 3 1 5 8 9 4 1 1 2 3 5 6 9 5 0 2 3 8 Key: 3 | 1 represents 31 years Complete these statements. The modal age is ............................... The median age is ............................... The percentage of women that are older than 51 years is ............................... %. [3]
Mark scheme: 2 41 3 B1 for each 43 20
Q3 · Change 2.15 hours into minutes
3 Change 2.15 hours into minutes. .......................................... min [1]
Mark scheme: 3 129 1
Q4 · A 86° B x° NOT TO SCALE C 58° D Triangle ABC and triangle ACD are isosceles
4 A 86° B x° NOT TO SCALE C 58° D Triangle ABC and triangle ACD are isosceles. Angle DAB = 86° and angle ADC = 58°. Find the value of x. x = ................................................ [3]
Mark scheme: 4 79 nfww 3 M2 for x + x + 58 + 58 + 86 = 360 oe or 86 – (180 – 2 × 58) implied by CAB = 22 or B1 for DCA = 58 or BCA = x or DAC = 64
Q5 · Angelique rents a room for a party
5 Angelique rents a room for a party. The cost of renting the room is $15.50 for the first hour and then $7.25 for each additional hour. She pays $95.25 in total. Work out the total number of hours she rents the room for. ........................................ hours [3]
Mark scheme: 5 12 3 M2 for (95.25 – 15.5) ÷ 7.25 oe or (95.25 – (15.5 – 7.25)) ÷ 7.25 oe or M1 for 95.25 – 15.5 or B1 for 79.75
Q6 · 7 16 Without using a calculator, work out ' +
1 7 16 Without using a calculator, work out ' + . 3 6 5 You must show all your working and give your answer as a fraction in its simplest form. ................................................. [4]
Mark scheme: 6 1 6 2 7 M1 × oe or ÷ oe 3 7 6 6 2 A1 oe 7 2 1 M1 10 7 their + with a correct method to find e.g. + oe 7 5 35 35 fractions with a common denominator 17 A1 If order of operations not correct SC2 for cao 35 10 answer with correct working for 41 1 7 1 ÷ + 3 6 5 35 6 or SC1 for + oe 30 30
Q7 · Katy has 5 white flowers, x red flowers and ( 2x + 1) yellow flowers
7 Katy has 5 white flowers, x red flowers and ( 2x + 1) yellow flowers. She picks a flower at random. 1 The probability that it is white is . 12 Find the probability that it is yellow. ........................................................................... [4]
Mark scheme: 7 37 4 B3 for x = 18 or 37 [yellow] oe 60 5 or SC2 for answer 12 1 5 or M2 for = oe 12 5 + x + 2 x + 1 or M1 for 5 + x + 2x + 1 oe or [total number of flowers =] 60
Q8 · Calculate 4 39 161
8 Calculate 4 39 161 . ................................................. [1]
Mark scheme: 8 2.5 oe 1
Q9 · -1 249 .21 # 10 .02o 22% .02 1000 Write these values in order of size, starting with the…
-1 249 .21 # 10 .02o 22% .02 1000 Write these values in order of size, starting with the smallest. ........................ 1 ........................ 1 ........................ 1 ........................ 1 ........................ [2] smallest
Mark scheme: 9 24 2 M1 for four values in the correct order < 2.1 × 10–1 < 22% < 0.2 < 0.2 1000 or for conversion to consistent comparable form e.g. 0.21, 0.22…, 0.22, 0.4.., 0.024
Q10 · The interior angle of a regular polygon is 156°
10 The interior angle of a regular polygon is 156°. Work out the number of sides of this polygon. ................................................. [2]
Mark scheme: 10 15 2 M1 for 360 ÷ (180 – 156) or 180 ( n – 2 ) = 1 56 oe n
Q11 · A car starts its journey by accelerating from rest at a constant rate of 0.7 m/s2 for 20…
11 A car starts its journey by accelerating from rest at a constant rate of 0.7 m/s2 for 20 seconds, before reaching a constant speed of 14 m/s. It then travels at 14 m/s for a distance of 210 m. The car then decelerates at a constant rate of 1.4 m/s2, before coming to a stop. On the grid, complete the speed–time graph for the car’s journey. 16 14 12 10 Speed 8 (m/s) 6 4 2 0 0 5 10 15 20 25 30 35 40 45 50 Time (s) [3]
Mark scheme: 11 Straight line from (20, 14) to (35, 14) 3 M1 for 210 ÷ 14 soi and straight line from (35, 14) to (45, 0) M1 for 14 ÷ 1.4 or any line with gradient –1.4 ending at x axis
Q12 · The table shows the first five terms of sequences A, B and C
12 The table shows the first five terms of sequences A, B and C. 1st term 2nd term 3rd term 4th term 5th term nth term Sequence A 8 3 -2 -7 -12 3 4 5 6 Sequence B 2 2 3 4 5 1 Sequence C 1 2 4 8 2 Complete the table to show the nth term of each sequence. [5]
Mark scheme: 12 13 – 5n oe final answer 5 B2 for 13 – 5n oe final answer n + 1 or B1 for – 5n + c or 13 – kn k ≠ 0 oe final answer or 13 – 5n seen then spoilt n 2n – 2 oe final answer n + 1 B1 for oe final answer n B2 for 2n –2 oe final answer or B1 for 2n – k oe k can be 0
Q13 · Write 243 # 27 2 n as a single power of 3 in terms of n
13 (a) Write 243 # 27 2 n as a single power of 3 in terms of n. ................................................. [2] (b) k = 2 # 32 # p 3 , where p is a prime number greater than 3. Write 6k 2as a product of prime factors in terms of p. ........................................................................... [2]
Mark scheme: 13(a) 36n + 5 final answer 2 B1 for 35 or (33)2n or better or answer 6n + 5 13(b) 23 × 35 × p6 final answer 2 B1 for two parts correct or 2 × 3 × 2 × 32 × p3 × 2 × 32 × p3 or 1944p6 or k2 = 22 × 34 × p6
Q14 · Q 65° S x° NOT TO 55° 60° SCALE P R T P, Q and R are points on a circle
14 Q 65° S x° NOT TO 55° 60° SCALE P R T P, Q and R are points on a circle. ST is a tangent to the circle at R. (a) Write down the value of x. Give a geometrical reason for your answer. x = ........................ because ......................................................................................................... .................................................................................................................................................... [2] (b) Another tangent from the point S touches the circle at V. Give a geometrical reason why triangle SVR is isosceles. ..................................................................................................................................................... ..................................................................................................................................................... [1]
Mark scheme: 14(a) 55 2 B1 for 55 Alternate segment theorem 14(b) Tangents from an external point are equal 1 in length
Q15 · A is the point (3, 16) and B is the point (8, 31)
15 (a) A is the point (3, 16) and B is the point (8, 31). Find the equation of the line that passes through A and B. Give your answer in the form y = mx + c . y = ................................................ [3] (b) The line CD has equation y = 0.5x - 11. Find the gradient of a line that is perpendicular to the line CD. ................................................. [1]
Mark scheme: 15(a) [y =] 3x + 7 final answer 3 31 − 16 M1 for . oe 8 − 3 M1 for correct substitution of (3, 16) or (8, 31) into y = (their m)x + c 15(b) –2 1
Q16 · Sachin picks a number at random from the first three multiples of 3
16 Sachin picks a number at random from the first three multiples of 3. He then picks a number at random from the first three prime numbers. He adds the two numbers to find a score. (a) Complete the table. Multiples of 3 3 9 2 5 11 Prime 3 6 numbers [2] (b) Given that the score is even, find the probability that one of the numbers he picks is 9. ................................................. [2]
Mark scheme: 16(a) Multiples of 3 2 B1 for at least 4 correct entries + 3 6 9 2 5 8 11 Prime numbers 3 6 9 12 5 8 11 14 16(b) 2 2 their 2 oe B2FT for 5 their 5 their 2 or B1FT for k is any integer in the k range 1 ⩽ k ⩽ 7 c or c is 0, 1 or 2 their 5
Question 17
17 Solve. ( 5x - 3)( 2x + 7) = 0 x = ........................ or x = ........................ [1]
Mark scheme: 17 3 7 1 oe and – oe 5 2
Q18 · Solve the simultaneous equations
18 Solve the simultaneous equations. You must show all your working. y = x 2 - 9x + 21 y = 2 x - 3 x = ........................ y = ........................ x = ........................ y = ........................ [5]
Mark scheme: 18 x² – 11x + 24 [= 0] M2 M1 for x² – 9x + 21 = 2x – 3 oe or 2 y + 3 y + 3 y² – 16y + 39 [= 0] or y = − 9 + 21 oe 2 2 (x – 8)(x – 3) [= 0] M1 or for correct factors for their quadratic or equation (y – 13)(y – 3) [= 0] or for correct use of quadratic formula for their equation [x =] 3 [y =] 3 B2 B1 for one correct pair or two correct [x =] 8 [y =] 13 x values or two correct y values. If B0 scored and at least 2 method marks scored SC1 for correct substitution of both of their x values or their y values into y = x2 – 9x + 21 or y = 2x – 3
Q19 · In these Venn diagrams, shade the given regions
19 In these Venn diagrams, shade the given regions. A B C D E Al , B l ( C , D ) + E l [2]
Mark scheme: 19 2 B1 for each C D E
Q20 · X - 3 520 f ( )x = 2 g ( )x = 2 x - 1 h ( )x = x - 4 (a) Find ff(6)
x - 3 520 f ( )x = 2 g ( )x = 2 x - 1 h ( )x = x - 4 (a) Find ff(6). ................................................. [2] (b) Find g -1 g ( x + 21) . ................................................. [1] (c) Find x when f ( x) = h ( 84) . x = ................................................ [2]
Mark scheme: 20(a) 32 2 M1 for f(6) = 8 ( 2 x – 3 ) – 3 or ff( x ) = 2 oe 20(b) x + 21 1 20(c) –1 2 1 M1 for oe or 2–4 oe 16
Question 21
21 Expand and simplify. ( x - 3) 2 ( 2x + 5) ........................................................................... [3]
Mark scheme: 21 2 x 3 − 7 x 2 − 12 x + 45 3 B2 for unsimplified expansion of the three brackets with at most one error final answer or for simplified four-term expression of correct form with three terms correct or B1 for correct expansion of two of the given brackets with at least three terms out of four correct
Q22 · Solve the equation 7 sinx + 2 = 0 for 0° G x G 360°
22 Solve the equation 7 sinx + 2 = 0 for 0° G x G 360° . ........................................................................... [3] Question 23 is printed on the next page.
Mark scheme: 22 196.6 or 196.60… 3 B2 for one correct angle and 343.4 or 343.39… 2 or M1 for sin x = − or better 7 If 0 scored SC1 for two angles that sum to 540°
Question 23
23 Simplify. 3 xy + 36 y - 5 x - 60 2 x 2 - 288 ................................................. [4]
Mark scheme: 23 3 y – 5 3 y – 5 4 3 y – 5 or final answer SC3 for answer 2 ( x –1 2 ) 2 x – 24 x –12 or B3 for (3y – 5)(x + 12) and 2(x – 12)(x + 12) or (2x – 24)(x + 12) or B2 for (3y – 5)(x + 12) or 2(x – 12)(x + 12) or (2x – 24)(x + 12) or (2x + 24)(x – 12) or B1 for 3y(x + 12) – 5(x + 12) or x(3y – 5) + 12(3y – 5) or 2(x2 – 144) or (x – 12)(x + 12)
What was in this paper
The subtopics covered by these 23 questions, and how many questions each got. Open one in a new tab to see every Cambridge question on it.
2Introduction to probability2Algebraic fractions1Algebraic manipulation1Angles1Averages and measures of spread1Circle theorems I1Equations of linear graphs1Fractions, decimals and percentages1Functions1Graphs in practical situations1Indices I1Money1Ordering1Powers and roots1Ratio and proportion1Sequences1Sets1The four operations1Time1Trigonometric functions1What you needed in this session
Cambridge’s own grade thresholds for 2021 Oct/Nov, Paper 2 · Variant 2. A higher threshold means an easier paper — the bar moves with how the cohort did.