There were some oranges in 3 baskets, Q, R and S. 25% of the number of oranges in Basket Q was equal to 10% of the number of oranges in Basket R. The number of oranges in Basket S was 60% of the total number of oranges in Basket Q and R. After Oscar removed 10% of the oranges in Basket S, there were 89 more oranges in Basket S than in Basket R. In the end, how many oranges should be transferred from Basket R to Basket S so that the number of oranges in Basket Q would be the same as Basket R?
Let's call the number of oranges in Basket Q "Q", the number in Basket R "R", and the number in Basket S "S".
From the first sentence, we know that:
0.25Q = 0.1R
S = 0.6(Q+R)
After Oscar removes 10% of the oranges in S, there are (0.9S) oranges left. We know that this is 89 more than the number of oranges in R, so:
0.9S - R = 89
We want to end up with the same number of oranges in Q and R, so let's call the number we need to transfer from R to S "x". After the transfer, we'll have:
Q = R - x
S = S + x
Now we can substitute these values into the equations we had before:
0.25(Q) = 0.1(R-x) (from the first sentence)
S = 0.6(Q+R) + x (from the second sentence)
0.9(S) - R = 89 (from the third sentence)
Let's simplify the first two equations by distributing:
0.25Q = 0.1R - 0.1x
S = 0.6Q + 0.6R + x
Now we can substitute S and Q into the third equation:
0.9(0.6Q + 0.6R + x) - R = 89
0.54Q + 0.54R + 0.9x - R = 89
0.54Q + 0.46R + 0.9x = 89
Substitute Q = R - x into this equation:
0.54(R-x) + 0.46R + 0.9x = 89
0.54R - 0.54x + 0.46R + 0.9x = 89
1R + 0.36x = 89
Solve for x:
0.36x = 89 - R
x = (89 - R) / 0.36
Now we can substitute this value for x into Q = R - x to find the final answer:
Q = R - (89-R)/0.36
Q = 1.64R - 247
So the number of oranges that should be transferred from R to S is (89-R)/0.36, and the final number of oranges in Q will be 1.64 times the number of oranges in R, minus 247.