Maths olympiad question · P5–P6★★★★ Expert
Sum and multiple: olympiad question 7
An expert question on sum and multiple for P5–P6, with a step-by-step solution, diagrams and the answer. Sum-and-multiple and difference-and-multiple questions appear in the Hua Xia Cup's P3 to P5 first rounds, qualifying rounds and finals.
The question
Classes A, B and C each have some books in their reading corners.
Each of the three loans below starts from the books the classes have now: if class A lends class B 17 books, B will have 3 times as many as A; if B lends C 15 books, C will have twice as many as B; if C lends A 15 books, A will have twice as many as C.
How many books does class A have now?
Show the hints (3)
- Hint 1: Call what class A has left after lending 17 one part. How many parts does B have after receiving them — and how many before?
- Hint 2: Keep going: write B's and C's books as "so many parts ± so many books", then look at A and C after C lends 15.
- Hint 3: After lending 15, C has 6 parts less 94 books; 2 times that equals A's 1 part plus 32. Compare: 11 parts = 220 books.
Worked solution
Have a go first, then read the solution step by step.
- 1.Each loan starts afresh from the books now. Say A has 1 part left after lending 17: A now = 1 part + 17.
- 2.With 17 more, B is 3 × A, 3 parts; so B now = 3 parts − 17.
- 3.B lending 15 leaves 3 parts − 32; C with 15 more is 2 times that, 6 parts − 64. So C now = 6 parts − 79.
- 4.C lending 15 leaves 6 parts − 94; A with 15 more has 1 part + 32, which is 2 times that: 1 part + 32 = 12 parts − 188.
- 5.Compare: 12 − 1 = 11 parts = 32 + 188 = 220 books, so 1 part = 220 ÷ 11 = 20.
- 6.Class A has 20 + 17 = 37 books.
- 7.Check: B has 3 × 20 − 17 = 43, C has 41. A lends 17: 20 and 60 = 3 × 20 ✓; B lends 15: 28 and 56 ✓; C lends 15: 52 = 2 × 26 ✓
- ⚠️Common mistake: treating the loans as happening one after another. Each starts from the books now. Another: mixing up "after receiving" with "now" and adding where you should take away.
Answer: 37 books