Penelope, Shiyun and Joelle had 630 coins. Shiyun won some of the coins from Penelope and as a result, Shiyun's coins increased by 20%. Joelle then won some coins from Shiyun and Joelle's coins increased by 50%. Finally, Joelle lost some of her coins to Penelope and Penelope's coins increased by 25%. In the end, they realised that they each had an equal number of coins. How many percent less did Penelope have in the end than what she had at first? Correct your answer to 1 decimal place.

Penelope	Shiyun	Joelle
630
	5 u
- 1 u	+ 1 u
	6 u
		2 p
	- 1 p	+ 1 p
		3 p
4 boxes
+ 1 box		- 1 box
5 boxes
1 group	1 group	1 group

20% = ²⁰₁₀₀ = ¹₅
50% = ⁵⁰₁₀₀ = ¹₂
25% = ²⁵₁₀₀ = ¹₄

Working backwards.

3 groups = 630

1 group = 630 ÷ 3 = 210

1 group = 5 boxes
5 boxes = 210

1 box = 210 ÷ 5 = 42

1 group + 1 box = 3 p
210 + 42 = 3 p
3 p = 252

1 p = 252 ÷ 3 = 84

1 p = 1 x 84 = 84

1 group + 1 p = 6 u
210 + 84 = 6 u
6 u = 294

1 u = 294 ÷ 6 = 49

Number of coins that Penelope had at first
= 4 boxes + 1 u
= (4 x 42) + 49
= 168 + 49
= 217

Percent that Penelope had less in the end than at first
= ^{217 - 210}₂₁₇ x 100%
≈ 3.2%

Answer(s): 3.2%