Your reasoning worked to reach D in this case, but we have to be careful with setups like this. From the statements alone, we just know that x/y = 107/20. x & y could be any multiple of those numbers, and we don't necessarily want to try to confirm that they fit those particular values.

To solve this, we need to know that the fractional part is equal to the remainder over the divisor. Since statement 2 uses the same setup as our original equation, let’s start there:

2) .35=7/y

.35y=7

(7/20)y=7

y=7(20/7)

y=20

Once we know that y=20, we can plug into our original equation:

x/y=5.35

x/20=5.35

x=107

Sufficient

1) We can’t use the same approach for statement 1, because we don’t know the fractional part when y is divided by x. However, we might notice that since x/y >1, y/x must be less than 1. In other words, if y goes into x more than 5 times, we know that x is bigger than y, so y can’t go into x at all!

If x goes into y zero times, then y/x = 0 r 20. Therefore, the remainder *is* our value for y.

y=20

From there, we can proceed as before to find x=107.

Sufficient.

I hope this helps!

_________________

Dmitry Farber | Manhattan GMAT Instructor | New York

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