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M30-17

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Intern
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Re: M30-17  [#permalink]

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New post 26 Dec 2018, 03:07
Can we take this approach? Bunuel
The equations are
x=3y+z [ Constraints: y>z and z>=0] As Remainder is greater than Quotient.
z=ay+2 [ Constraints: y>2]
Thus,
x=3y+ay+2
x=y(a+3) + 2
x=3y+2 or x =4y+2 or x=5y+2 [a can take values 0,1,2,3...]

Looking at the options:
I. 5 Can only fit if y=1, which breaks constraint of y>2, so NO
II. 8 Can only fit if y=2, which breaks constraint of y>2, so NO
III. 32 Fits the equation x=3*10+2 or x=5*6+2 So YES
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M30-17  [#permalink]

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New post 18 Jan 2019, 09:48
very nice question..
thank you for posting this. decent explanation.

Got a li'l doubt. same line of thinking as i came to y>2 bt substituted z in the first equation and tried to solve.
I got the equation x=3y+z
substituting z=y+2
comes to x=4y+2. (and y>2)
I'm not getting 32 either
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Re: M30-17  [#permalink]

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New post 19 Jan 2019, 00:19
1
venkivety wrote:
very nice question..
thank you for posting this. decent explanation.

Got a li'l doubt. same line of thinking as i came to y>2 bt substituted z in the first equation and tried to solve.
I got the equation x=3y+z
substituting z=y+2
comes to x=4y+2. (and y>2)
I'm not getting 32 either


When z is divided by y, the remainder is 2 does not mean that z=y+2. It means that z =yq + 2.

Check the solution:
When \(z\) is divided by \(y\), the remainder is 2: when divisor (\(y\) in our case) is more than dividend (\(z\) in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, \(z=2\) and \(2 < y\).

So, we have that \(x=3y+2\) and \(2 < y\). This implies that the least value of \(x\) is \(x=3*3+2=11\): \(x\) cannot be 5 or 8.

Could \(x\) be 32? Yes. If \(y=10\), then \(x=3*10+2=32\).
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Joined: 24 Dec 2011
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M30-17  [#permalink]

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New post 19 Jan 2019, 10:20
Bunuel wrote:
venkivety wrote:
very nice question..
thank you for posting this. decent explanation.

Got a li'l doubt. same line of thinking as i came to y>2 bt substituted z in the first equation and tried to solve.
I got the equation x=3y+z
substituting z=y+2
comes to x=4y+2. (and y>2)
I'm not getting 32 either


When z is divided by y, the remainder is 2 does not mean that z=y+2. It means that z =yq + 2.

Check the solution:
When \(z\) is divided by \(y\), the remainder is 2: when divisor (\(y\) in our case) is more than dividend (\(z\) in our case), then the reminder equals to the dividend (for example, 2 divided by 5 gives the remainder of 2). Therefore, \(z=2\) and \(2 < y\).

So, we have that \(x=3y+2\) and \(2 < y\). This implies that the least value of \(x\) is \(x=3*3+2=11\): \(x\) cannot be 5 or 8.

Could \(x\) be 32? Yes. If \(y=10\), then \(x=3*10+2=32\).


thanq very much 4 the reply..
cleared..
GMAT Club Bot
M30-17   [#permalink] 19 Jan 2019, 10:20

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