Unit 3
7 marks · HardA factory buys two lengths of cable of 168 m and 252 m. They are to be cut into equal-length pieces with no cable left over.
Express 168 and 252 as products of their prime factors.
Find the greatest possible length of each piece.
The total number of pieces obtained from both cables is used to buy identical boxes. Each box costs £x, and the total cost is £30 more than the value of x squared. Write down an equation in x, and solve it.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) 168 = 2^3 × 3 × 7; 252 = 2^2 × 3^2 × 7
(b) 84 m
(c) x^2 + 30 = 5x; x = 6 or x = -5 (ignore negative result)
Common mistakes: Mixing up HCF and LCM; Incorrect prime factorisation
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Attempt correct prime factorisation for at least one number
A1 (a) · 1 mark · Both numbers written correctly as products of prime factors in index form
M2 (b) · 1 mark · Correct method to find HCF using prime factors (follow-through)
A2 (b) · 1 mark · HCF = 84 stated (follow-through)
M3 (c) · 1 mark · Forms correct equation relating number of pieces (5), cost x and x^2+30 (follow-through)
M4 (c) · 1 mark · Attempts to solve resulting quadratic equation (follow-through)
A3 (c) · 1 mark · Fully correct solution: x = 6 (follow-through)
Worked solution
a) 168 = 2 × 2 × 2 × 3 × 7 = 2^3 × 3 × 7; 252 = 2 × 2 × 3 × 3 × 7 = 2^2 × 3^2 × 7
b) The greatest possible length is the HCF of 168 and 252.
The common prime factors are 2^2 × 3 × 7 = 84.
c) Number of pieces = (168 + 252) / 84 = 2 + 3 = 5
Total cost: 5x. Given 5x = x^2 + 30
Solve: x^2 - 5x + 30 = 0. The positive solution is x = 6 (negative solution ignored for cost).