There were some peaches in 3 bags, K, L and M. 20% of the number of peaches in Bag K was equal to 10% of the number of peaches in Bag L. The number of peaches in Bag M was 60% of the total number of peaches. After Bryan removed 20% of the peaches in Bag M, there were 32 more peaches in Bag M than in Bag L. In the end, how many peaches should be transferred from Bag L to Bag M so that the number of peaches in Bag K would be the same as Bag L?

Bag K	Bag L	Bag M
1x2	2x2
2x3		3x3
2 u	4 u	9 u

	Bag K	Bag L	Bag M
Before	2 u	4 u	9 u
Change			- 1.8 u
After	2 u	4 u	7.2 u

20% = ²⁰₁₀₀ = ¹₅

10% = ¹⁰₁₀₀ = ¹₁₀

¹₅ Bag K = ¹₁₀ Bag L

Bag K : Bag L
5 : 10
1 : 2

60% = ⁶⁰₁₀₀ = ³₅

The total number of peaches in Bag K and Bag L is the combined repeated identity.  Make the total number of peaches in Box K and Bag L the same.  LCM of 3 and 2 = 6

Number of peaches removed from Bag M
= ²⁰₁₀₀ x 9 u
= 1.8 u

Number of peaches left in Bag M
= 9 u - 1.8 u
= 7.2 u

Number of more peaches in Bag M than Bag L
= 7.2 u - 4 u
= 3.2 u

3.2 u = 32

1 u = 32 ÷ 3.2 = 10

	Bag K	Bag L	Bag M
Before	2 u	4 u	7.2 u
Change		- 2 u	+ 2 u
After	2 u	2 u	9.2 u

Number of peaches to be transferred from Bag L to Bag M so that the number of peaches in Bag K would be the same as Bag L.
= 4 u - 2 u
= 2 u
= 2 x 10
= 20

Answer(s): 20