There were some kiwis in 3 bags, F, G and H. 25% of the number of kiwis in Bag F was equal to 10% of the number of kiwis in Bag G. The number of kiwis in Bag H was 70% of the total number of kiwis. After Wesley removed 10% of the kiwis in Bag H, there were 291 more kiwis in Bag H than in Bag G. In the end, how many kiwis should be transferred from Bag G to Bag H so that the number of kiwis in Bag F would be the same as Bag G?

Bag F	Bag G	Bag H
2x3	5x3
3x7		7x7
6 u	15 u	49 u

	Bag F	Bag G	Bag H
Before	6 u	15 u	49 u
Change			- 4.9 u
After	6 u	15 u	44.1 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Bag F = ¹₁₀ Bag G

Bag F : Bag G
4 : 10
2 : 5

70% = ⁷⁰₁₀₀ = ⁷₁₀

The total number of kiwis in Bag F and Bag G is the combined repeated identity.  Make the total number of kiwis in Box F and Bag G the same.  LCM of 7 and 3 = 21

Number of kiwis removed from Bag H
= ¹⁰₁₀₀ x 49 u
= 4.9 u

Number of kiwis left in Bag H
= 49 u - 4.9 u
= 44.1 u

Number of more kiwis in Bag H than Bag G
= 44.1 u - 15 u
= 29.1 u

29.1 u = 291

1 u = 291 ÷ 29.1 = 10

	Bag F	Bag G	Bag H
Before	6 u	15 u	44.1 u
Change		- 9 u	+ 9 u
After	6 u	6 u	53.1 u

Number of kiwis to be transferred from Bag G to Bag H so that the number of kiwis in Bag F would be the same as Bag G.
= 15 u - 6 u
= 9 u
= 9 x 10
= 90

Answer(s): 90