Easy one for u x,y are non-zero intergers. what is the

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Easy one for u x,y are non-zero intergers. what is the [#permalink]

07 Jun 2006, 01:12

Easy one for u
x,y are non-zero intergers. what is the remainder when x is divided by y
(1) x/2y remainder is 4
(2) (x+y)/y remainder is 4

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07 Jun 2006, 02:43

I think it's D.
(1) x/2y remainder is 4
In this case the remainder when x is divided by y is always the "remainder when 4 is divided by y".
For ex : -
Suppose x = 34, y = 5, So x/2y or 34/10, remainder is 4.
when we divide 4 by y (5 in this case), remainder is 4.
In actual x/y in our case is 34/5; remainder 4.

Another example, suppose x = 22 and y= 3, So x/2y or 22/6, remainder is 4.
when we divide 4 by y (3 in this case), remainder is 1.
In actual x/y in our case is 22/3; remainder 1.

So equation 1 is sufficient.

(2) (x+y)/y remainder is 4
In this case the remainder is always 4.

It is also sufficient.
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07 Jun 2006, 03:33

Sorry, am not sure I understand the logic at all.

For statement (1), leaving aside if x/2y is the same as 4/y - you have taken two examples where x/2y has remained 4, but x/y has had different remainders. Clearly, for different values of x&y, the reminder for x/y is different. Statement (1) therefore is insufficient.

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Re: The remainder [#permalink]

07 Jun 2006, 05:28

quangviet512 wrote:

Easy one for u
x,y are non-zero intergers. what is the remainder when x is divided by y
(1) x/2y remainder is 4
(2) (x+y)/y remainder is 4

D

as from 1) x = 2ny + 4 so 2ny+ 4 / y remainder is always 4/ y

2) remainder is 4 again

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07 Jun 2006, 05:52

Chaps,

Is it only me that can't understand this.

Responding to raja's reply above, the remainer will always be 4/y yes, but we STILL cant tell definately what the value of the remainder will be. for different values of y, the value of the remainer will be different.
If (1) also mentioned that y= 5 (say), then yes, (1) is sufficient. But at the moment, we still dont have enough information to determin the value of the remainder.

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07 Jun 2006, 06:06

Condition #1 x/2y gives a reminder of 4

Eg.1 x = 12, y = 4
x/2y = 12/8 --> reminder = 4
x/y = 12/4 --> reminder =0

Eg.2 x = 14, y = 5
x/2y = 14/10 --> reminder = 4
x/y = 14/5 --> reminder = 4

Condition #1 is not sufficient.
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09 Jun 2006, 19:04

So what is the answer ? is 2) sufficient ?

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10 Jun 2006, 15:54

[quote="giddi77"]Condition #1 x/2y gives a reminder of 4

Eg.1 x = 12, y = 4
x/2y = 12/8 --> reminder = 4
x/y = 12/4 --> reminder =0

Eg.2 x = 14, y = 5
x/2y = 14/10 --> reminder = 4
x/y = 14/5 --> reminder = 4

Condition #1 is not sufficient.[/quote]
:-D

You found the trick here
With (1) only, we often thinnk that x=2y*k+4 so x/y also have remanider of 4. But in fact, there's problem when y<5.
If y=3 and x=10
While x/2y has remainder of 4, x/y has remainder of 1
So (1) is insufficient
The answer for this quest is B.

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10 Jun 2006, 23:38

quang and giddi are right, get B.

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11 Jun 2006, 10:39

B. Stmt 2 is suff.

Given that

(x+y)/y = yxMultiple + 4
x/y +1 = yxMultiple +4
x/y = yxMultiple + (4-1)

=> x/y = yxMultiple + 3

[#permalink] 11 Jun 2006, 10:39

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Easy one for u x,y are non-zero intergers. what is the

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