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What is the remainder when the positive integer x is divided

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What is the remainder when the positive integer x is divided [#permalink] New post 14 Apr 2010, 06:04
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What is the remainder when the positive integer x is divided by 3

(1) when x is divided by 6, the remainder is 2
(2) when x is divided by 15, the remainder is 2.


If m,r,x and y are positive, the ratio of m to r equal to the ratio x to y?

(1) the ratio of m to y is equal to the ratio of x to r
(2) the ratio of m + x to r + y is equal to the ratio of x to y
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Re: need explanation for DS question thank you thank you! [#permalink] New post 14 Apr 2010, 07:02
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What is the remainder when the positive integer x is divided by 3

(1) when x is divided by 6, the remainder is 2
(2) when x is divided by 15, the remainder is 2.


Dont know whether this is a very long method to solve.

1) when x is divided by 6, the remainder is 2

so the number will be form of 6*Y+2(k=0,1,2,3,....)

(2) when x is divided by 15, the remainder is 2
so the number will be form of 15*Z+2(k=0,1,2,3,....)

Both 1,2
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Last edited by RaviChandra on 14 Apr 2010, 07:54, edited 1 time in total.
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Re: need explanation for DS question thank you thank you! [#permalink] New post 14 Apr 2010, 07:29
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What is the remainder when the positive integer x is divided by 3

(1) when x is divided by 6, the remainder is 2 --> x=6p+2 --> dividing this by 3 --> 6p is divisible by 3 and 2 divided by 3 leaves remainder of 2. Sufficient.
(2) when x is divided by 15, the remainder is 2 --> x=15q+2 --> dividing this by 3 --> 15q is divisible by 3 and 2 divided by 3 leaves remainder of 2. Sufficient.

Answer: D.

If m,r,x and y are positive, the ratio of m to r equal to the ratio x to y?

Is \frac{m}{r}=\frac{x}{y}? Is my=rx?

(1) the ratio of m to y is equal to the ratio of x to r --> \frac{m}{y}=\frac{x}{r} --> mr=xy. Not sufficient.

(2) the ratio of m + x to r + y is equal to the ratio of x to y --> \frac{m+x}{r+y}=\frac{x}{y} --> cross multiply --> my+xy=rx+xy --> my=rx. Sufficient.

Answer: B.
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Re: need explanation for DS question thank you thank you!   [#permalink] 14 Apr 2010, 07:29
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