Nicole, Jaslyn and Mary had 1350 coins. Jaslyn won some of the coins from Nicole and as a result, Jaslyn's coins increased by 25%. Mary then won some coins from Jaslyn and Mary's coins increased by 40%. Finally, Mary lost some of her coins to Nicole and Nicole's coins increased by 20%. In the end, they realised that they each had an equal number of coins. How many percent less did Nicole have in the end than what she had at first? Correct your answer to 1 decimal place.

Nicole	Jaslyn	Mary
1350
	4 u
- 1 u	+ 1 u
	5 u
		5 p
	- 2 p	+ 2 p
		7 p
5 boxes
+ 1 box		- 1 box
6 boxes
1 group	1 group	1 group

25% = ²⁵₁₀₀ = ¹₄
40% = ⁴⁰₁₀₀ = ²₅
20% = ²⁰₁₀₀ = ¹₅

Working backwards.

3 groups = 1350

1 group = 1350 ÷ 3 = 450

1 group = 6 boxes
6 boxes = 450

1 box = 450 ÷ 6 = 75

1 group + 1 box = 7 p
450 + 75 = 7 p
7 p = 525

1 p = 525 ÷ 7 = 75

2 p = 2 x 75 = 150

1 group + 2 p = 5 u
450 + 150 = 5 u
5 u = 600

1 u = 600 ÷ 5 = 120

Number of coins that Nicole had at first
= 5 boxes + 1 u
= (5 x 75) + 120
= 375 + 120
= 495

Percent that Nicole had less in the end than at first
= ^{495 - 450}₄₉₅ x 100%
≈ 9.1%

Answer(s): 9.1%