There were some kiwis in 3 baskets, P, Q and R. 25% of the number of kiwis in Basket P was equal to 10% of the number of kiwis in Basket Q. The number of kiwis in Basket R was 60% of the total number of kiwis. After Pierre removed 25% of the kiwis in Basket R, there were 230 more kiwis in Basket R than in Basket Q. In the end, how many kiwis should be transferred from Basket Q to Basket R so that the number of kiwis in Basket P would be the same as Basket Q?

Basket P	Basket Q	Basket R
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Basket P	Basket Q	Basket R
Before	4 u	10 u	21 u
Change			- 5.25 u
After	4 u	10 u	15.75 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Basket P = ¹₁₀ Basket Q

Basket P : Basket Q
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of kiwis in Basket P and Basket Q is the combined repeated identity.  Make the total number of kiwis in Box P and Basket Q the same.  LCM of 7 and 2 = 14

Number of kiwis removed from Basket R
= ²⁵₁₀₀ x 21 u
= 5.25 u

Number of kiwis left in Basket R
= 21 u - 5.25 u
= 15.75 u

Number of more kiwis in Basket R than Basket Q
= 15.75 u - 10 u
= 5.75 u

5.75 u = 230

1 u = 230 ÷ 5.75 = 40

	Basket P	Basket Q	Basket R
Before	4 u	10 u	15.75 u
Change		- 6 u	+ 6 u
After	4 u	4 u	21.75 u

Number of kiwis to be transferred from Basket Q to Basket R so that the number of kiwis in Basket P would be the same as Basket Q.
= 10 u - 4 u
= 6 u
= 6 x 40
= 240

Answer(s): 240