Olympiad topic · P2–P6
Surplus and shortage
Surplus-and-shortage questions appear in the Hua Xia Cup's P1 to P5 papers; harder ones become questions about distance and time.
Try one
★★ MediumA class is chipping in for a present for their teacher, everyone paying the same.
At $6 each they are $102 short; at $8 each they are still $56 short.
How much does the present cost?
- Hint 1: Going from $6 each to $8 each, how much smaller does the shortfall get?
- Hint 2: Find how many pupils there are first, then the price.
- 1.At $6 each they are $102 short.
- 2.At $8 each, everyone pays $8 − $6 = $2 more, and they are still $56 short.
- 3.The shortfall drops from $102 to $56: 102 − 56 = $46, which is what everyone paid more in all.
- 4.At $2 more each, there are 46 ÷ 2 = 23 pupils.
- 5.Price: 23 × $6 + $102 = $240.
- ⚠️Common mistake: adding 102 + 56. Both times they are short, so subtract.
Answer: 240 dollars
Four levels, step by step
- ★EasyFree2 question types · unlimited questionsOnce exact, once with some over.
- ★★MediumFree2 question types · unlimited questionsOver and short, or short twice.
- ★★★HardFree4 question types · unlimited questions"the last one gets only 3", to turn into a standard one first, or the same idea with distance and time.
- ★★★★Expert (competition level)VIP2 question types · unlimited questionsEmpty rooms, or the speed to arrive exactly on time.
Surplus and shortage: what it is and how to solve it
Sharing things out two ways — once with some left over, once short — to find how many people and how many things.
For example: 3 stickers each leaves 13 over; 6 each is 26 short. How many pupils? (13 + 26) ÷ (6 − 3) = 13.
The key idea: The gap between the two ways ÷ the gap per person = the number of people. Over once and short once: gap = over + short; over both times (or short both times): gap = larger − smaller. The usual method at school uses an equation: 3x + 13 = 6x − 26.
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