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## x/y=5.35 If x and y are positive integers, then what is the

tagged by: Mo2men

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Mo2men Legendary Member
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#### x/y=5.35 If x and y are positive integers, then what is the

Mon Feb 26, 2018 1:39 am
x/y=5.35 and If x and y are positive integers, then what is the value of x?

(1) When y is divided by x, the remainder is 20.

(2) When x is divided by y, the remainder is 7.

OA: D
Source: Economist

### GMAT/MBA Expert

Jay@ManhattanReview GMAT Instructor
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Mon Feb 26, 2018 6:30 am
Mo2men wrote:
x/y=5.35 and If x and y are positive integers, then what is the value of x?

(1) When y is divided by x, the remainder is 20.

(2) When x is divided by y, the remainder is 7.

OA: D
Source: Economist
We are given that x/y = 5.35 and x and y are positive integers.

We have to determine the value of x.

x/y = 5.35 => x/y = 535/100 = 107/20; after cancelling 535 and 100 by 5.

=> 107 and 20 would be the multiples of the same common factor.

Say the common factor is n, then we have x/y = (107n) / (20n).

If we get the value of n, we would get the value of x = 107n.

Let's take each statement one by one.

(1) When y is divided by x, the remainder is 20.

=> y = xq + 20

=> 20n = (107n)q + 20; where q is the quotient

=> n(20 - 107q) = 20

Since n is postive quality, (20 - 107q) must be positive. Since q is a non-negative integer, q = 0.

=> n*(20 - 0) = 20 => 20n = 20 => n = 1.

Thus, x = 107n = 107*1 = 107. Sufficient.

(2) When x is divided by y, the remainder is 7.

=> 107n = (20n)q + 7

=> n(107 - 20q) = 7

We see that 7, a prime number, is a product of two positive integers n and (107 - 20q).

Thus, one between n and (107 - 20q) should be 1 and the other 7,

Case 1: Say n = 7 and 107 - 20q = 1 => q = 106/20 = fraction. It's not possible as q is an integer.

Case 2: Say n = 1 and 107 - 20q = 7 => q = 100/20 = 5. It's possible as q is an integer.

=> n = 1.

Thus, x = 107n = 107*1 = 107. Sufficient.

Hope this helps!

-Jay
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