There were some peaches in 3 boxes, Q, R and S. 25% of the number of peaches in Box Q was equal to 10% of the number of peaches in Box R. The number of peaches in Box S was 70% of the total number of peaches. After John removed 20% of the peaches in Box S, there were 242 more peaches in Box S than in Box R. In the end, how many peaches should be transferred from Box R to Box S so that the number of peaches in Box Q would be the same as Box R?

Box Q	Box R	Box S
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Box Q	Box R	Box S
Before	6 u	15 u	49 u
Change			- 9.8 u
After	6 u	15 u	39.2 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Box Q = ¹₁₀ Box R

Box Q : Box R
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of peaches in Box Q and Box R is the combined repeated identity.  Make the total number of peaches in Box Q and Box R the same.  LCM of 7 and 3 = 21

Number of peaches removed from Box S
= ²⁰₁₀₀ x 49 u
= 9.8 u

Number of peaches left in Box S
= 49 u - 9.8 u
= 39.2 u

Number of more peaches in Box S than Box R
= 39.2 u - 15 u
= 24.2 u

24.2 u = 242

1 u = 242 ÷ 24.2 = 10

	Box Q	Box R	Box S
Before	6 u	15 u	39.2 u
Change		- 9 u	+ 9 u
After	6 u	6 u	48.2 u

Number of peaches to be transferred from Box R to Box S so that the number of peaches in Box Q would be the same as Box R.
= 15 u - 6 u
= 9 u
= 9 x 10
= 90

Answer(s): 90