Types of number — Paper 4 · IGCSE Mathematics 0580

10 For Examiner's Total Use Row 1 1 = 1 Row 2 3 + 5 = 8 Row 3 7 + 9 + 11 = 27 Row 4 13 + 15 + 17 + 19 = 64 Row 5 Row 6 The rows above show sets of consecutive odd numbers and their totals. (a) Complete Row 5 and Row 6. [2] (b) What is the special name given to the numbers 1, 8, 27, 64…? Answer(b) [1] (c) Write down in terms of n, (i) how many consecutive odd numbers there are in Row n, Answer(c)(i) [1] (ii) the total of these numbers. Answer(c)(ii) [1] (d) The first number in Row n is given by n2 − n + 1. Show that this formula is true for Row 4. Answer(d) [1] (e) The total of Row 3 is 27. This can be calculated by (3 × 7) + 2 + 4. For Examiner's The total of Row 4 is 64. This can be calculated by (4 × 13) + 2 + 4 + 6. Use The total of Row 7 is 343. Show how this can be calculated in the same way. Answer(e) [1] (f) The total of the first n even numbers is n(n + 1). Write down a formula for the total of the first (n – 1) even numbers. Answer(f) [1] (g) Use the results of parts (d), (e) and (f) to show clearly that the total of the numbers in Row n gives your answer to part (c)(ii). Answer(g) [2]

10 marks

Mark scheme: 10 (a) 21 + 23 + 25 + 27 + 29 = 125 B1 31 + 33 + 35 + 37 + 39 + 41 = 216 B1 (b) Cubes B1 (c) (i) n oe B1 (ii) n3 oe B1 (d) 42 – 4 + 1 = 13 www E1 Allow 16 for 42, otherwise all must be seen (e) 7 × 43 + 2 + 4 + 6 + 8 + 10 + 12 B1 All must be seen (f) n(n – 1) final answer oe B1 (g) n(n2 – n + 1) + their (f) M1 n3 – n2 + n + n2 – n = n3 E1 All must be seen, no errors or omissions [10]

This question in 0580/41 Oct/Nov 2009

9 (a) 72 = 2 × 2 × 2 × 3 × 3 written as a product of prime factors. (i) Write the number 126 as a product of prime factors. Answer(a)(i) 126 = [2] (ii) Find the value of the highest common factor of 72 and 126. Answer(a)(ii) [1] (iii) Find the value of the lowest common multiple of 72 and 126. Answer(a)(iii) [2] The rest of question 9 is printed on the next page. (b) John wants to estimate the value of π. For He measures the circumference of a circular pizza as 105 cm and its diameter as 34 cm, both Examiner's correct to the nearest centimetre. Use Calculate the lower bound of his estimate of the value of π. Give your answer correct to 3 decimal places. Answer(b) [4] (c) The volume of a cylindrical can is 550 cm3, correct to the nearest 10 cm3. The height of the can is 12 cm correct to the nearest centimetre. Calculate the upper bound of the radius of the can. Give your answer correct to 3 decimal places. Answer(c) cm [5]

14 marks

Mark scheme: 9 (a) (i) 2 × 3 × 3 × 7 oe 2 M1 for prime factors of 2,3,3,7 shown condone 1(‘s) shown as well for method only (ii) 18 1 (iii) 504 2 M1 for other multiples of 504 or 2 × 2 × 2 × 3 × 3 × 7 oe shown If (ii) and (iii) both correct but reversed allow SC1 (b) 3.028 or 3.029 cao 4 B3 for 3.0289(85…) or M1 for their 105/their 34 (their 105 in range 104 to 106 and their 34 in range 33 to 35) and B1 for 104.5 or 34.5 or 34.499.. selected (c) πr2 their h = their V M1 Where V is in range 540 to 560 and h is in range 11 to 13 their V M1 Implies previous method (15.36 implies M2) (r2 =) If using 545 and 12.5 then 13.88 (leading to 3.73) π × their h If using 550 and 12 then 14.59 (leading to 3.82) Sq root M1 Dep on M2, can be implied from answers Selects 555 or 554.99.. and 11.5 B1 Indep 3.919 cao A1 If trials then 5 or 0

This question in 0580/41 Oct/Nov 2011

10 (a) 1 = 1 For Examiner′s Use 1 + 2 = 3 1 + 2 + 3 = 6 1 + 2 + 3 + 4 = 10 (i) Write down the next line of this pattern. Answer(a)(i) … [1] n (ii) The sum of the fi rst n integers is (n + 1). k Show that k = 2. Answer(a)(ii) [2] (iii) Find the sum of the fi rst 60 integers. Answer(a)(iii) … [1] (iv) Find n when the sum of the fi rst n integers is 465. Answer(a)(iv) n = … [2] (n - 8)(n - 7) (v) 1 + 2 + 3 + 4 + … + x = 2 Write x in terms of n. Answer(a)(v) x = … [1] (b) 13 = 1 For Examiner′s Use 13 + 23 = 9 13 + 23 + 33 = 36 13 + 23 + 33 + 43 = 100 (i) Complete the statement. 13 + 23 + 33 + 43 + 53 = … = ( … )2 [2] (ii) The sum of the fi rst n integers is n (n + 1). 2 Find an expression, in terms of n, for the sum of the fi rst n cubes. Answer(b)(ii) … [1] (iii) Find the sum of the fi rst 19 cubes. Answer(b)(iii) … [2] _____________________________________________________________________________________

12 marks

Mark scheme: 10 (a) (i) 1 + 2 + 3 + 4 + 5 = 15 1 n (n + 1) (ii) Correct substitution equating to 2 M1 for using a value of n in k sum 2(2 + 1) 2 (2 + 1) e.g. = 3 and k = 2 stated e.g. = 3 k k with no errors seen or for a verification using k = 2 2 (2 + 1) e.g. = 3 2 (iii) 1830 1 n (n + 1) (iv) 30 2 M1 for = 465 or better 2 (v) n – 8 1 (b) (i) 225, 15 2 B1 either 2 n 2 (n + 1) (ii) oe 1 4 19 2 (19 + 1)2 (iii) 36100 2 M1 for oe or 1902 4

This question in 0580/41 Oct/Nov 2013

9 Layer 1 Layer 2 Layer 3 The diagrams show layers of white and grey cubes. Khadega places these layers on top of each other to make a tower. (a) Complete the table for towers with 5 and 6 layers. Number of layers 1 2 3 4 5 6 Total number of white cubes 0 1 6 15 Total number of grey cubes 1 5 9 13 Total number of cubes 1 6 15 28 [4] (b) (i) Find, in terms of n, the total number of grey cubes in a tower with n layers. Answer(b)(i) … [2] (ii) Find the total number of grey cubes in a tower with 60 layers. Answer(b)(ii) … [1] (iii) Khadega has plenty of white cubes but only 200 grey cubes. How many layers are there in the highest tower that she can build? Answer(b)(iii) … [2] (c) The expression for the total number of white cubes in a tower with n layers is pn2 + qn + 3. Find the value of p and the value of q. Show all your working. Answer(c) p = … q = … [5] (d) Find an expression, in terms of n, for the total number of cubes in a tower with n layers. Give your answer in its simplest form. Answer(d) … [2] __________________________________________________________________________________________

16 marks

Mark scheme: 9 (a) 28 45 1, 1 17 21 1 45 66 1 (b) (i) 4n – 3 oe 2 M1 for 4n + k (ii) 237 1 (iii) 50 2FT FT their (b)(i) = 200 solved and then answer truncated dep on linear expression of form an + k M1 for their 4n – 3 = 200 or their 4n – 3 Y 200 (c) p = 2 and q = –5 with some 5 M2 for any 2 of p + q + 3 = 0 oe, correct supporting working 22 p + 2q + 3 = 1 oe, 32 p + 3q + 3 = 6 oe, leading to the solutions 42 p + 4q + 3 = 15 oe , 52 p + 5q + 3 = their 28 oe, etc. or M1 for any one of these M1 indep for correctly eliminating p or q from pair of linear equations A1 for one correct value If 0 scored SC1 for 2 values that satisfy one of their original equations After M0, 2 correct answers SC1 (d) 2n2 – n or n(2n – 1) 2 B1 for answer 2n2 + k[n] or M1 for their quadratic from (c) + their linear from (b)(i) 1 1 1

This question in 0580/42 Oct/Nov 2014

4 30 students were asked if they had a bicycle (B), a mobile phone (M) and a computer (C). The results are shown in the Venn diagram. B M 2 4 x 7 1 6 3 2 C (a) Work out the value of x. Answer(a) x = … [1] (b) Use set notation to describe the shaded region in the Venn diagram. Answer(b) … [1] (c) Find n(C (M B)). Answer(c) … [1] (d) A student is chosen at random. (i) Write down the probability that the student is a member of the set M  . Answer(d)(i) … [1] (ii) Write down the probability that the student has a bicycle. Answer(d)(ii) … [1] (e) Two students are chosen at random from the students who have computers. Find the probability that each of these students has a mobile phone but no bicycle. Answer(e) … [3]

8 marks

Mark scheme: 4 (a) 5 1 (b) C ∩ M oe 1 Allow e.g. (B ∩ C ∩ M) ∪ (C ∩ M) (c) 3 1 8 (d) (i) oe 1 0.267 or better 30 14 (ii) oe 1 0.467 or better 30 30 6 5 (e) oe 3 M2 for × 272 17 16 6 or M1 for seen 17 0.110[2…] or better 55

This question in 0580/41 May/June 2015

9 (a) Expand and simplify. 3x(x – 2) – 2x(3x – 5) Answer(a) … [3] (b) Factorise the following completely. (i) 6w + 3wy – 4x – 2xy Answer(b)(i) … [2] (ii) 4x2 – 25y2 Answer(b)(ii) … [2] (c) Simplify. 16 - 32 4 c 9 m x Answer(c) … [2] (d) n is an integer. (i) Explain why 2n – 1 is an odd number. Answer(d)(i) … … [1] (ii) Write down, in terms of n, the next odd number after 2n – 1. Answer(d)(ii) … [1] (iii) Show that the difference between the squares of two consecutive odd numbers is a multiple of 8. Answer(d)(iii) [3] __________________________________________________________________________________________

14 marks

Mark scheme: 9 (a) 4x – 3x2 or x(4 – 3x) nfww 3 B2 for 3x2 – 6x – 6x2 + 10x final answer or M1 for 3x2 – 6x or – 6x2 + 10x (b) (i) (2 + y)(3w – 2x) oe final answer 2 M1 for 3w(2 + y) – 2x(2 + y) or 2(3w – 2x) + y(3w – 2x) (ii) (2x + 5y)(2x – 5y) final answer 2 M1 for (2x ± 5y)(2x ± 5y) or (2x + ky)(2x – ky) or (kx + 5y)(kx – 5y), k ≠ 0 or (2x + 5)(2x – 5) or (2 + 5y)(2 – 5y) 27 6x (c) final answer 2 B1 for 2 [out of 3] elements correct in the right 64 form in final answer or final answer contains 27 and 64 and x[–]6 3 2x 729 x 12 or seen or seen 4 4096 (d) (i) 2n is even and subtracting 1 gives 1 Must interpret the 2n as even or not odd and an odd number then the –1 oe (ii) 2n + 1 oe final answer 1 (iii) their(2n + 1)2 – (2n – 1)2 M1 Could use alternate correct expressions for consecutive odd numbers. Allow method and accuracy marks if correct. Could reverse the algebraic terms their(2n – 1)2 – (2n + 1)2 leading to –8n. Allow method and accuracy marks if correct. 4n2 + 4n + 1 – 4n2 + 4n – 1 M1 Dep on M1 for expanding brackets in their expressions. If seen alone and completely correct then implies previous M1 Allow 4n2 + 4n + 1 – (4n2 – 4n + 1) 8n A1 With no errors seen. After 0 scored, allow SC1 for two correctly evaluated numeric examples of subtracting consecutive odd squares isw 2 2

This question in 0580/42 May/June 2015

9 The first three diagrams in a sequence are shown below. The diagrams are made by drawing lines of length 1 cm. Diagram 1 Diagram 2 Diagram 3 (a) The areas of each of the first three diagrams are shown in this table. Diagram 1 2 3 Area (cm2) 1 4 9 (i) Find the area of Diagram 4. Answer(a)(i) … cm2 [1] (ii) Find, in terms of n, the area of Diagram n. Answer(a)(ii) … cm2 [1] (b) The numbers of 1 cm lines needed to draw each of the first three diagrams are shown in this table. Diagram 1 2 3 Number of 1 cm lines 4 13 26 (i) Find the number of 1 cm lines needed to draw Diagram 4. Answer(b)(i) … [1] (ii) In which diagram are 118 lines of length 1 cm needed? Answer(b)(ii) … [1] (c) The total number of 1 cm lines needed to draw both Diagram 1 and Diagram 2 is 17. The total number of 1 cm lines needed to draw all of the first n diagrams is 2 n3 + an2 + bn. 3 Find the value of a and the value of b. Show all your working. Answer(c) a = … b = … [6] __________________________________________________________________________________________ Question 10 is printed on the next page.

10 marks

Mark scheme: 9 (a) (i) 16 1 (ii) n2 1 (b) (i) 43 1 (ii) 7 1 5 5 (c) a = oe, b = oe with 6 M1 for any correct substitution 2 6 2 supporting working eg (2)3 + 22a + 2b 3 A1 for one of 2 eg + a + b = 4 or better 3 16 eg + 4a + 2b = 17 or better 3 54 eg + 9a + 3b = 43 or better 3 A1 for another of 2 eg + a + b = 4 or better 3 16 eg + 4a + 2b = 17 or better 3 54 eg + 9a + 3b = 43 or better 3 M1 for correctly eliminating one variable from two of their equations in a and b 5 A1 for a = oe 2 5 A1 for b = oe 6 After zero scored, SC2 for 2 correct answers without supporting working or SC1 for 2 of 17, 43, 86, 150, 239 seen

This question in 0580/41 Oct/Nov 2015

8 Apples cost x cents each and oranges cost (x + 2) cents each. Dylan spends $3.23 on apples and $3.23 on oranges. The total of the number of apples and the number of oranges Dylan buys is 36. (a) Write an equation in x and show that it simplifies to 18x 2 - 287x - 323 = 0 . [4] (b) (i) Find the two prime factors of 323. … , … [1] (ii) Complete the statement. 18x 2 - 287 x - 323 = (18x … )(x … ) [2] (iii) Solve the equation 18x 2 - 287x - 323 = 0 . x = … or x = … [1] (c) Find the largest number of apples Dylan can buy for $2. … [1]

9 marks

Mark scheme: 323 323 323 323 8 (a) + = 36 oe three term B2 B1 for seen oe or seen oe x x + 2 x x + 2 equation 323(x + 2) + 323x = 36x(x + 2) oe M1 i.e. for clearing the fractions (or all still over common denominator) or reducing the two 323 x + 646 + 323 x or = 36 oe algebraic fractions to one fraction and x ( x + 2) expanding the brackets in the numerator 36 x 2 − 574 x − 646 = 0 A1 answer reached without any omissions or errors 18 x 2 − 287 x − 323 = 0 with at least one intermediate line with brackets expanded after M1 (b) (i) 17, 19 1 (ii) ( ……. + 19)(………. – 17) 2 SC1 for ( ……. + a)(………. + b) where a, b are integers and ab = –323 or a + 18b = –287 19 (iii) 17, − oe 1FT FT their (b)(ii) 18 (c) 11 cao 1

This question in 0580/41 Oct/Nov 2016

10 (a) (i) Write 180 as a product of its prime factors. … [2] (ii) Find the lowest common multiple (LCM) of 180 and 54. … [2] (b) An integer, X, written as a product of its prime factors is a 2 # 7b + 2 . An integer, Y, written as a product of its prime factors is a 3 # 72 . The highest common factor (HCF) of X and Y is 1225. The lowest common multiple (LCM) of X and Y is 42 875. Find the value of X and the value of Y. X = … Y = … [4]

8 marks

Mark scheme: 10(a)(i) 22 × 32 × 5 oe 2 M1 for 3 correct prime factors in a tree or table seen before the first error or for 2, 3, 5 identified 10(a)(ii) 540 2 M1 for 22 × 33 × 5 or 2 × 33 shown or answer 540k 10(b) X = 8575 4 B3 for X = 8575 or Y = 6125 or Y = 6125 B2 for a = 5 or b = 1 soi or B1 for 1225 = 52 × 72 or 42 875 = 53 × 73 or M1 for a² × 7² [= 1225] or a3 × 7b + 2 [= 42 875]

This question in 0580/43 Oct/Nov 2017

9 Car A and car B take part in a race around a circular track. One lap of the track measures 7.6 km. Car A takes 2 minutes and 40 seconds to complete each lap of the track. Car B takes 2 minutes and 25 seconds to complete each lap of the track. Both cars travel at a constant speed. (a) Calculate the speed of car A. Give your answer in kilometres per hour. … km/h [3] (b) Both cars start the race from the same position, S, at the same time. (i) Find the time taken when both car A and car B are next at position S at the same time. Give your answer in minutes and seconds. … min … s [4] (ii) Find the distance that car A has travelled at this time. … km [2]

9 marks

Mark scheme: 9(a) 171 or 171.0… 3 6.7 M2 for × 60 × 60 oe 160 7.6 7.6 7.6 or M1 for or or 160 2 2min 40sec 2 3 If 0 scored, SC1 for answer 189 or 188.6 to 188.7 9(b)(i) 77 [min] 20 [s] 4 32 M3 for × 29 oe 12 58 or B2 for 4640 or 1.29 or 1.288 to 1.289, 45 oe or 32 laps or 29 laps or M2 for 25× 5 × 29 oe or M1 for 2 m 40 sec ÷ (2 m 40 sec – 2 m 25 sec) soi for 2 m 25 sec ÷ (2 m 40 sec – 2 m 25 sec) soi or for an attempt to find LCM or 23 200 seen or correctly find prime factors of 145 or 160 7.6 7.6 7.6 or for or or oe, 145 5 2min 25sec 212 provided SC1 not earned in part (a) 9(b)(ii) 220.4 2 M1 for their (b)(i) ÷ 2min 40 sec [× 7.6] oe or their (a) × their (b)(i) ÷ 60 oe

This question in 0580/42 Oct/Nov 2019

3 (a) Simplify, giving your answer as a single power of 7. (i) 7 5 # 7 6 … [1] (ii) 7 15 ' 7 5 … [1] (iii) 42 + 7 … [1] (b) Simplify. ( 5x 2 # 2xy 4 ) 3 … [3] (c) P = 2 5 # 3 3 # 7 Q = 540 (i) Find the highest common factor (HCF) of P and Q. … [2] (ii) Find the lowest common multiple (LCM) of P and Q. … [2] (iii) P # R is a cube number, where R is an integer. Find the smallest possible value of R. … [2] (d) Factorise the following completely. (i) x 2 - 3x - 28 … [2] (ii) 7 ( a + 2b) 2 + 4a ( a + 2b) … [2] 2 x - 1 1 2 y - x # 3(e) 3 = x 9 Find an expression for y in terms of x. y = … [4]

20 marks

Mark scheme: 3(a)(i) 711 cao 1 3(a)(ii) 710 cao 1 3(a)(iii) 72 cao 1 If answers 11, 10 and 2 in (a) then allow SC1 in this part 3(b) 1000x9y12 final answer 3 B2 for correct answer seen or answer of the form 1000x9yk or 1000xky12 or kx9y12 or B1 for answer with one correct element in product or (10x3y4)[3] seen 3(c)(i) 108 2 M1 for [540 =] 22 [×] 33 [×] 5 or B1 for 108 oe not in prime factor form e.g. 22 × 3 × 9 3(c)(ii) 30 240 2 M1 for (540 × 25 × 33 × 7) ÷ their (c)(i) oe or B1 for answer 30 240 oe not in prime factor form e.g. 25 × 33 × 35 3(c)(iii) 98 2 B1 for 592 704 seen or 26 × 33 × 73 seen or 2 × 72 oe seen 3(d)(i) (x – 7) (x + 4) final answer 2 M1 for x(x – 7) + 4(x – 7) or x(x + 4) – 7 (x + 4) or better or for (x + a)(x + b) where ab = – 28 or a + b = – 3 3(d)(ii) (a + 2b)(11a + 14b) final answer 2 M1 for (a + 2b) (7(a + 2b) + 4a) or (a + pb)(11a + qb) where pq = 28 or 11p + q = 36 If 0 scored, SC1 for a + 2b (11a + 14b) 3(e) 5 x − 1 4 B2 for 2x – 1 = –2x + 2y – x oe [ y = ] oe final answer or B1 for 9x = 32x or better 2 M1dep for correct rearrangement of their 5 term ‘linear’ equation in y and x to make y the subject

This question in 0580/42 May/June 2021

1 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]

12 marks

Mark scheme: Question Answer Marks Partial Marks 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 × 10 4 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

This question in 0580/42 Feb/March 2022

2 (a) Write (i) 2994.99 correct to the nearest 10, … [1] (ii) 0.983 correct to 1 decimal place, … [1] (iii) 2090 correct to 2 significant figures. … [1] (b) Write down a prime number between 90 and 100. … [1] (c) Write 2 -6 as a fraction. … [1] (d) Write 0.007 01 in standard form. … [1] (e) Simplify 1.5 # 10 x + 1 .5 # 10 x - 1 giving your answer in standard form. … [2] (f) Write .037o as a fraction. You must show all your working. … [2]

10 marks

Mark scheme: 2(a)(i) 2990 cao 1 2(a)(ii) 1.0 cao 1 2(a)(iii) 2100 cao 1 2(b) 97 1 2(c) 1 1 final answer 64 2(d) 7.01[0]  10–3 1 2(e) 1.65  10x 2 M1 for final answer figs 165 or for 15  10 x −1 seen or for 0.15  10 x seen 2(f) 37.7... – 3.7... [= 34] oe M1 34 B1 oe fraction 90

This question in 0580/41 Oct/Nov 2022

1 (a) (i) Write 70 as a product of its prime factors. … [2] (ii) Find the highest common factor (HCF) of 70 and 112. … [2] (iii) Find the lowest common multiple (LCM) of 70x 4 y 2 and 112x 3 y 5. … [2] (b) Simplify. (i) a 12 ' a 4 … [1] 5 bc (ii) # 2b 20 … [2] (c) Solve. 4 + 2x = 15 x = … [2] (d) Solve. 34 + 2x = 4 - x 5 x = … [3] 3(e) P = d + m2 (i) Find P when d = 7 and m = -8. P = … [2] (ii) Rearrange the formula to make m the subject. m = … [3]

19 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 2 × 5 × 7 [=70] 2 B1 for 2, 5, 7 1(a)(ii) 14 2 M1 for [112 = ] 24 × 7 oe or for answer 2 × 7 1(a)(iii) 560x4y5 2 B1 for answer kx4y5 or for answer 560xayb or for correct answer seen then spoiled 1(b)(i) a8 1 1(b)(ii) c 2 5 bc final answer M1 for or better 8 40b 1(c) 11 2 15 5.5 or or 5½ M1 for 2x = 15 – 4 oe or 2 + x = oe 2 2 1(d) –2 3 M1 for 34 + 2x = 5(4 – x) oe or better M1 dep for reaching ax = b FT their first step 1(e)(i) 11 2 3 2 M1 for 7 + ( −8) oe 1(e)(ii)  ( P − d )3 oe final answer 3 B1 for P – d = 3 m2 oe M1 for cube both sides M1 for square root leading to final answer

This question in 0580/41 Oct/Nov 2024

3 2 15 23 144 - 2 0.8 5 From this list, write down (a) a natural number … [1] (b) an irrational number. … [1]

2 marks

Mark scheme: 3(a) 23 or 144 1 Accept 12 for 144 3(b) 1 15

This question in 0580/43 May/June 2025