Hilda, Min and Abi had 684 coins. Min won some of the coins from Hilda and as a result, Min's coins increased by 50%. Abi then won some coins from Min and Abi's coins increased by 75%. Finally, Abi lost some of her coins to Hilda and Hilda's coins increased by 20%. In the end, they realised that they each had an equal number of coins. How many percent less did Hilda have in the end than what she had at first? Correct your answer to 1 decimal place.

Hilda	Min	Abi
684
	2 u
- 1 u	+ 1 u
	3 u
		4 p
	- 3 p	+ 3 p
		7 p
5 boxes
+ 1 box		- 1 box
6 boxes
1 group	1 group	1 group

50% = ⁵⁰₁₀₀ = ¹₂
75% = ⁷⁵₁₀₀ = ³₄
20% = ²⁰₁₀₀ = ¹₅

Working backwards.

3 groups = 684

1 group = 684 ÷ 3 = 228

1 group = 6 boxes
6 boxes = 228

1 box = 228 ÷ 6 = 38

1 group + 1 box = 7 p
228 + 38 = 7 p
7 p = 266

1 p = 266 ÷ 7 = 38

3 p = 3 x 38 = 114

1 group + 3 p = 3 u
228 + 114 = 3 u
3 u = 342

1 u = 342 ÷ 3 = 114

Number of coins that Hilda had at first
= 5 boxes + 1 u
= (5 x 38) + 114
= 190 + 114
= 304

Percent that Hilda had less in the end than at first
= ^{304 - 228}₃₀₄ x 100%
≈ 25.0%

Answer(s): 25.0%