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There are 1600 jelly beans divided between two jards, X and

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by Ian Stewart » Wed Jun 05, 2019 9:09 am
The wording is awkward, but if we have x beans in jar X, and y beans in jar Y, the question tells us x = 3y - 100. Since x+y = 1600, substituting for 'x' we get

3y - 100 + y = 1600
4y = 1700
y = 425

and the rest of beans, 1600-425 = 1175 beans, are in jar X.
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by Scott@TargetTestPrep » Thu Jun 06, 2019 5:01 pm
BTGmoderatorLU wrote:Source: Veritas Prep

There are 1600 jelly beans divided between two jards, X and Y. If there are 100 fewer jelly beans in jar X than three times the number of beans in jar Y, how many beans are in jar X?

A. 375
B. 950
C. 1150
D. 1175
E. 1350

The OA is D
We can let y = the number of jelly beans in jar Y; thus, there are 3y - 100 jelly beans in jar X. We can now create the equation:

3y - 100 + y = 1600

4y = 1700

y = 425

Therefore, there are 3(425) - 100 = 1275 - 100 = 1175 jelly beans in jar X.

Answer: D

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