O-LevelMathematicsAlgebra and graphs4024/21 · Oct/Nov 2024 · Q1010 Marks

Bag A contains red balls and green balls. The total number of balls in the bag is x. The number of green balls in the bag is 6 more than the number of red balls.

(a)Show that the fraction of the balls in bag A that are red is (x-6)/(2x).[2]
(b)Bag B also contains red balls and green balls. The number of red balls in bag B is x. The number of green balls in bag B is 4 times the number of green balls in bag A. Show that the fraction of the balls in bag B that are red is x/(3x+12).[2]
(c)x-6/2x = x/3x+12 Show that x²-6x-72 = 0 .[3]
(d)Solve by factorisation x²-6x-72 = 0 .[2]
(e)x is the total number of balls in bag A. Use your answer to part (d) to find the number of green balls in bag A.[1]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

[2] M1 for r+r+6=x oe or r = (x-6)/2 leading to fraction red = (x-6)/2x
[2] M1 Allow green in B = 4((x-6)/2+3) oe or green in A = (x-6)/2+3 oe Total in B = x+4((x-6)/2+3) oe leading to fraction red = x/(3x+12)
[3] M1 Correct removal of fractions
[3] M1 Accept 3 out of 4 terms correct
[3] A1 A0 if any errors or omissions Correct rearrangement to x²-6x-72=0
[2] B1 (x+6)(x-12)[=0]
[2] B1 12,-6
[1] FT their positive solution ÷2+3

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