Working Backwards & Problem-Solving Strategies
16 Class 6 IMO Mathematics Olympiad practice questions on working backwards & problem-solving strategies, part of the Olympiad Thinking topic area. Every question below includes the correct answer and a full explanation.
Practice this topic interactively →1. I think of a number, add 8, then multiply by 3. The result is 45. What is my number?
- A.5
- B.6
- C.7
- D.9
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Answer: C. 7
Reverse the operations: 45 ÷ 3 = 15, then 15 − 8 = 7.
2. A number is doubled, then 6 is subtracted, giving 20. What was the original number?
- A.10
- B.12
- C.13
- D.14
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Answer: C. 13
Reverse the steps: 20 + 6 = 26, then 26 ÷ 2 = 13.
3. A rope is cut into two pieces such that one piece is 3 m longer than the other. If the total length is 17 m, how long is the shorter piece?
- A.6
- B.7
- C.8
- D.9
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Answer: B. 7
Let the shorter piece be s. s+(s+3)=17 → 2s=14 → s=7 m.
4. Using working backwards: a number is divided by 4, then 7 is added, giving 15. What was the original number?
- A.28
- B.30
- C.32
- D.34
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Answer: C. 32
Reverse the steps: 15−7=8, then 8×4=32.
5. A jar has red and blue marbles in the ratio 3:5. If there are 24 red marbles, how many blue marbles are there?
- A.32
- B.36
- C.40
- D.45
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Answer: C. 40
Each 'part' = 24÷3=8 marbles. Blue = 5 parts = 5×8=40.
6. The sum of two numbers is 45 and their difference is 9. Find the larger number.
- A.24
- B.25
- C.27
- D.30
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Answer: C. 27
Larger = (sum+difference)/2 = (45+9)/2 = 27.
7. A number is trebled and then 5 is subtracted to give 22. Find the number.
- A.7
- B.8
- C.9
- D.10
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Answer: C. 9
Reverse: 22+5=27, then 27÷3=9.
8. After spending half his money on a book and then ₹50 on a pen, Arjun had ₹150 left. How much money did he start with?
- A.350
- B.375
- C.400
- D.425
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Answer: C. 400
Before buying the pen he had 150+50=₹200, which was half his original money, so he started with 200×2=₹400.
9. In how many ways can you make 50 paise using only 10 paise and 5 paise coins (using at least one of each)?
- A.3
- B.4
- C.5
- D.6
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Answer: B. 4
Solving 10a+5b=50 with a,b≥1: (a,b) = (1,8),(2,6),(3,4),(4,2) — 4 ways.
10. A farmer has hens and cows. Together they have 35 heads and 98 legs. How many cows are there?
- A.12
- B.14
- C.16
- D.18
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Answer: B. 14
If all 35 were hens, there'd be 70 legs — 28 short. Each cow instead of a hen adds 2 legs, so cows = 28÷2 = 14.
11. Three boxes contain apples. Box B has 5 more apples than Box A. Box C has twice as many apples as Box B. If the total is 63, how many apples are in Box A?
- A.10
- B.11
- C.12
- D.13
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Answer: C. 12
Let Box A = a. Box B = a+5, Box C = 2(a+5). Total: a+(a+5)+2(a+5)=4a+15=63 → 4a=48 → a=12.
12. A man is now 4 times as old as his son. In 5 years, he will be 3 times as old as his son. Find the son's current age.
- A.8
- B.9
- C.10
- D.12
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Answer: C. 10
Let the son's age be s. 4s+5 = 3(s+5) → 4s+5=3s+15 → s=10.
13. A shopkeeper sold 3/5 of his stock on Monday and 40 items on Tuesday, leaving 20 items unsold. How many items did he originally have?
- A.130
- B.140
- C.150
- D.160
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Answer: C. 150
After Monday, 2T/5 remained. Subtracting Tuesday's 40 leaves 20: 2T/5−40=20 → 2T/5=60 → T=150.
14. Reena spends half her pocket money on books, then ₹20 on a pen, and is left with ₹30. How much pocket money did she start with?
- A.80
- B.90
- C.100
- D.110
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Answer: C. 100
Before the pen: 30+20=₹50, which was half her money, so she started with 50×2=₹100.
15. A group of friends went to a restaurant. If each pays ₹60, they are short by ₹40. If each pays ₹70, they have ₹30 extra. How many friends are there?
- A.5
- B.6
- C.7
- D.8
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Answer: C. 7
Let n be the number of friends. 60n+40 = 70n−30 (both equal the total bill) → 70=10n → n=7.
16. A shopkeeper buys pens at ₹8 each and sells them at ₹11 each. He spent a total of ₹960 buying pens. How much profit will he make if he sells all of them?
- A.320
- B.340
- C.360
- D.400
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Answer: C. 360
Number of pens = 960÷8=120. Profit per pen = 11−8=₹3. Total profit = 120×3=₹360.