O-LevelMathematicsStatistics4024/21 · May/June 2023 · Q1010 Marks

Bags of sweets are packed into boxes.

(a)(i)A box is opened and the number of sweets in each bag is counted. The results are shown in the table. Number of sweets | 11 | 12 | 13 | 14 | 15 Frequency | 15 | 26 | 38 | p | 9 Explain why the total number of bags in the box cannot be 87.[1]
(a)(ii)The mean number of sweets per bag in this box is 12.8. Find the value of p.[3]
(b)(i)Another box is opened and the number of sweets in each bag is counted. The results are shown in the table. Number of sweets | 11 | 12 | 13 | 14 | 15 Frequency | 12 | 28 | 39 | r | 9 A bag of sweets is chosen at random from this box and not replaced. A second bag of sweets is then chosen at random from the same box. The probability that both bags contain 15 sweets is 4/539. Show that r² +175r-2046 = 0.[3]
(b)(ii)Solve r²+175r-2046 = 0 to find the value of r. Show your working.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

[1] (a)(i) p cannot be negative oe
[3] (a)(ii) 17 M2 for 14p + 1106 = 1126.4 + 12.8p oe or M1 for (11×15+12×26+13×38+14p+15×9)/(15+26+38+p+9) oe
[1] (b)(i) 9/(88+r) × 8/(87+r) = 4/539 oe M1
[1] (b)(i) 72 × 539 = 4(88+ r)(87 + r) oe M1 Correct removal of fractions with algebraic denominators of the form k±r
[1] (b)(i) Correct expansion of brackets leading to r² + 175r-2046 = 0 A1 With no errors or omissions including = 0
[2] (b)(ii) (r-11)(r +186) B2 M1 for brackets giving two out of three terms correct or B1 for √(175² – 4×1×-2046) oe or (-175±√k)/2 or (-175±√(175²-4×1×-2046))/2×1 or r + (175/2) = ±√(175/2)² - (-2046) oe
[1] (b)(ii) 11 cao B1

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