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

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x/y=5.35 If x and y are positive integers, then what is the [#permalink]  05 Oct 2010, 18:39
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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.
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Re: DS [#permalink]  05 Oct 2010, 18:49
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TomB wrote:
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.

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Re: DS [#permalink]  05 Oct 2010, 23:12
TomB wrote:
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

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X/Y = 5.35 If x and y are positive integers, then what is [#permalink]  21 Jun 2011, 00:16
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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.

Master Gmat 22
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Re: confusing arithmetic equation [#permalink]  21 Jun 2011, 00:34
x = 5y + 0.35y

So remainder is 0.35y - (A)

Again,

y/x = 1/5.35 = 100/535 = 20/107 - (B)

(1)
Clearly, from (B), x = 107 as remainder is always less than divisor
Sufficient

(2)

Clearly, from (A), 0.35y = 7

So 0.05Y = 1 => y = 20

Hence X can be found as x = 5.35 * 20 = 107

Sufficient

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Re: confusing arithmetic equation [#permalink]  05 Sep 2011, 09:32
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Quote:
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.

Master Gmat 22

X/Y=5 + 7/20
--> Y is a multiple of 20
--> remainder is a multiple of 7

As such:
Y=20 (or multiple of 20)
remainder=7 (or multiple of 7)
--> lowest possible value of X: 5(20)+7 = 107

From Statement 1
Y/X= Integer, remainder 20
20/107 = 0 + 20
Sufficient

From Statement 2
X = multiple of Y + 7
107=5(20)+7
Sufficient

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x/y =5.35 If x and y are positive integers, then what is the [#permalink]  27 Oct 2011, 02:21
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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.
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Re: Confusing [#permalink]  28 Oct 2011, 05:20
It was confusing but I got D, and waded my way through, although not sure if correctly. This is what I did:

Simplify the given statement:

x/y = 5.35 = 5 + 35/100 = 5 + 7/20 = 107/20

So we want to verify if x = 107 and y = 20.

1) y/x has a remainder of 20. Well, flip the above statement over.
Thus, if x/y = 107/20 => y/x = 20/107

Since it is already in fractional form, the remainder is 20, we can confirm that y=20, and x=107

Sufficient.

2) x/y has a remainder of 7. We have already distilled our given statement to this (5 + 7/20). This confirms it, and thus x = 107 and y = 20.

Sufficient.

The confusing bit is in 1. It is easy to get tripped up over 20/107 and not realize that 20 is the remainder since the Quotient is 0.

You can visualize it as 20/107 = 0 + 20/107 where Quotient = 0, 20 = Remainder, and 107 = Divisor.
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Re: Confusing [#permalink]  28 Oct 2011, 11:01
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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!
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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

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

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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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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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