There were some guavas in 3 baskets, T, U and V. 25% of the number of guavas in Basket T was equal to 10% of the number of guavas in Basket U. The number of guavas in Basket V was 60% of the total number of guavas. After Lee removed 10% of the guavas in Basket V, there were 178 more guavas in Basket V than in Basket U. In the end, how many guavas should be transferred from Basket U to Basket V so that the number of guavas in Basket T would be the same as Basket U?

Basket T	Basket U	Basket V
2x2	5x2
2x7		3x7
4 u	10 u	21 u

	Basket T	Basket U	Basket V
Before	4 u	10 u	21 u
Change			- 2.1 u
After	4 u	10 u	18.9 u

25% = ²⁵₁₀₀ = ¹₄

10% = ¹⁰₁₀₀ = ¹₁₀

¹₄ Basket T = ¹₁₀ Basket U

Basket T : Basket U
4 : 10
2 : 5

60% = ⁶⁰₁₀₀ = ³₅

The total number of guavas in Basket T and Basket U is the combined repeated identity.  Make the total number of guavas in Box T and Basket U the same.  LCM of 7 and 2 = 14

Number of guavas removed from Basket V
= ¹⁰₁₀₀ x 21 u
= 2.1 u

Number of guavas left in Basket V
= 21 u - 2.1 u
= 18.9 u

Number of more guavas in Basket V than Basket U
= 18.9 u - 10 u
= 8.9 u

8.9 u = 178

1 u = 178 ÷ 8.9 = 20

	Basket T	Basket U	Basket V
Before	4 u	10 u	18.9 u
Change		- 6 u	+ 6 u
After	4 u	4 u	24.9 u

Number of guavas to be transferred from Basket U to Basket V so that the number of guavas in Basket T would be the same as Basket U.
= 10 u - 4 u
= 6 u
= 6 x 20
= 120

Answer(s): 120