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There are X watermelons of 10 Kg each, and Y Watermelons of

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There are X watermelons of 10 Kg each, and Y Watermelons of

by BTGmoderatorAT » Tue Feb 20, 2018 7:18 pm
There are X watermelons of 10 Kg each, and Y Watermelons of R Kg each. The average weight of a watermelon is 12 Kg. What is the value of R?

(1) There are five heavier watermelons more than lighter watermelons.
(2) The weight of the heavier watermelons in Kg is equal to their number

What's the best way to determine whether statement 1 is sufficient? Can any experts help?
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Source: — Data Sufficiency |

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by ErikaPrepScholar » Wed Feb 28, 2018 6:29 am
We'll start by setting up an equation for the average weight of the watermelons:
$$\frac{10x\ +\ Ry}{x+y}=avg.\ weight$$
So we should be able to solve for R if we know the value of x and y.
We should also note that if the average weight of the watermelons is 12, R must be greater than 10.

Statement 1
This means that y = x+5. Plugging this into our equation gives $$\frac{10x\ +\ R\left(x+5\right)}{x+x+5}=avg.\ weight$$ $$\frac{10x\ +\ Rx+5}{2x+5}=avg.\ weight$$ $$\frac{\left(10+R\right)x+5}{2x+5}=avg.\ weight$$ However, since we still don't know the value of x, we can't solve. Insufficient.

Statement 2
This means that R = y. Plugging this into our equation gives: $$\frac{10x\ +\ R^2}{x+R}=avg.\ weight$$ However, like in Statement 1, we still need x to solve. Insufficient.

Both
y=x+5, and R=y, so R=x+5. Rearranging this gives x=R-5. If we plug this into either the equation from Statement 1 or Statement 2, we should be left with an equation containing only R. We should then be able to solve for R. Sufficient.
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