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From Statement 2 X = multiple of Y + 7 107=5(20)+7 Sufficient
Answer: D _________________
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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
Re: x/y =5.35 If x and y are positive integers, then what is the [#permalink]
11 Aug 2013, 22:28
Gem of a question :
hard to get the statement 1 inference , although statement 2 is easy to infer.
From above post what I can understand is that:
Since X/Y >1 => Y has to be less than X
When Y is divided by X then the quotient will have to be '0', since if remainder is other than '0' than Y will have a greater value than X which will contradict that X>Y as per question stem so when Y is divided by X the quotient's value should be '0' only . Hence Y=20 so we can get the value of X using question stem
Rgds, TGC ! _________________
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Re: X/Y = 5.35 If x and y are positive integers, then what is [#permalink]
26 Jun 2014, 20:30
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