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x/y=5.35 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.

I surprised myself at the pace which I solved this one. Looks like a easy one.

Given x/y = 5.35 = 5 + 0.35

5 + 0.35 can be represented as 5 + 35/100 => 5 + 7/20 => 107/20

Hence x/y is 107/20.

Let us look at the statements:

1) Given y/x = some integer + 20 (remainder). We know that x/y = 107/20. Hence y/x = 20/107 -- gives us 20 as the remainder and 0 as the quotient. Sufficient.

2) Given x/y = 7. Yes this is also possible 107/20 gives us a remainder of 7. Sufficient.

Answer D. _________________

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x/y=5.35 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.

x/y = 5.35 = 535/100 = 107/20

x and y will be of the form 107z and 20z

(1) y is less than x so when divided by x the remainder will be y itself. So y=20. If y=20, x=107. Sufficient

(2) x divided by y leaves remainder 7. note that 107z = 5*(20z) + 7z and that 7z < 20z so then the remainder is always 7z. but we know remainder is 7. so z=1 hence x=107 & y=20 Sufficient

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]

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11 Aug 2013, 23:28

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

Therefore,

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]

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26 Jun 2014, 21:30

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20 Aug 2015, 11:28

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That is not correct reto. Although you did end up getting the correct answer, make sure to apply the concepts correctly. This is a very good example of how to interpret the divisbility formula.

When you write the divisbility formula, you get,

x=yp+r , where p is a positive integer.

Thus you get, x/y=p+r/y and this r/y=0.35 (x/y \(\neq\).35. This is where you are making a mistake)

Per statement 2, r=7 and thus y=20, giving you x=107.

Per statement 1, y/x=q+20 , you also know that x/y=5.35 ---> x>y thus when you divide y/x (think 30/67 for example, you will geta quotient of 0 and remainder as 30, the dividend itself!!), you get 0 as the quotient and 20 as the remainder.

Thus, y must be = 20 , again giving you x = 107.

Thus both statements are sufficient individually, giving you D as the correct answer.

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