Linda, Eva and Jen had 720 coins. Eva won some of the coins from Linda and as a result, Eva's coins increased by 50%. Jen then won some coins from Eva and Jen's coins increased by 60%. Finally, Jen lost some of her coins to Linda and Linda's coins increased by 20%. In the end, they realised that they each had an equal number of coins. How many percent less did Linda have in the end than what she had at first? Correct your answer to 1 decimal place.

Linda	Eva	Jen
720
	2 u
- 1 u	+ 1 u
	3 u
		5 p
	- 3 p	+ 3 p
		8 p
5 boxes
+ 1 box		- 1 box
6 boxes
1 group	1 group	1 group

50% = ⁵⁰₁₀₀ = ¹₂
60% = ⁶⁰₁₀₀ = ³₅
20% = ²⁰₁₀₀ = ¹₅

Working backwards.

3 groups = 720

1 group = 720 ÷ 3 = 240

1 group = 6 boxes
6 boxes = 240

1 box = 240 ÷ 6 = 40

1 group + 1 box = 8 p
240 + 40 = 8 p
8 p = 280

1 p = 280 ÷ 8 = 35

3 p = 3 x 35 = 105

1 group + 3 p = 3 u
240 + 105 = 3 u
3 u = 345

1 u = 345 ÷ 3 = 115

Number of coins that Linda had at first
= 5 boxes + 1 u
= (5 x 40) + 115
= 200 + 115
= 315

Percent that Linda had less in the end than at first
= ^{315 - 240}₃₁₅ x 100%
≈ 23.8%

Answer(s): 23.8%