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M30-17

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M30-17  [#permalink]

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New post 16 Sep 2014, 00:45
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Difficulty:

  55% (hard)

Question Stats:

59% (01:37) correct 41% (01:21) wrong based on 86 sessions

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When the positive integer is \(x\) divided by the positive integer \(y\), the quotient is 3 and the remainder is \(z\). When \(z\) is divided by \(y\), the remainder is 2. Which of the following could be the value of \(x\)?

I. 5

II. 8

III. 32


A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

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Re M30-17  [#permalink]

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New post 16 Sep 2014, 00:45
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1
Official Solution:


When the positive integer is \(x\) divided by the positive integer \(y\), the quotient is 3 and the remainder is \(z\). When \(z\) is divided by \(y\), the remainder is 2. Which of the following could be the value of \(x\)?

I. 5

II. 8

III. 32


A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


When \(x\) is divided by \(y\), the quotient is 3 and the remainder is \(z\): \(x=3y+z\), where \(0\leq{z} < y\) (the remainder must be less than the divisor).

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 \(x=3*10+2=32\).


Answer: C
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Re M30-17  [#permalink]

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New post 24 Dec 2014, 00:10
I think this question is good and helpful.
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Re: M30-17  [#permalink]

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New post 24 May 2015, 09:48
My approach towards this question was abit different, I found it easy, I guess I might have overlooked something then.
Is there a number "y" that gives us quotinent 3 if you divide 5/y or 8/y? Answer is no.
5/1 = 5, 5/2 = 2 + 1, 5/3 = 1 + 2, 5/4 = 1 + 1 ...
8/1 = 8, 8/2 = 4, 8/3 = 2 + 2, 8/4 = 2, 8/5 = 1 + 3, 8 /6 = 1 + 2 ...

thus the only option is III coz we got no "no options possible"
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Re: M30-17  [#permalink]

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New post 17 Aug 2015, 01:44
Zhenek wrote:
My approach towards this question was abit different, I found it easy, I guess I might have overlooked something then.
Is there a number "y" that gives us quotinent 3 if you divide 5/y or 8/y? Answer is no.
5/1 = 5, 5/2 = 2 + 1, 5/3 = 1 + 2, 5/4 = 1 + 1 ...
8/1 = 8, 8/2 = 4, 8/3 = 2 + 2, 8/4 = 2, 8/5 = 1 + 3, 8 /6 = 1 + 2 ...

thus the only option is III coz we got no "no options possible"


I went through the same thought process.
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Re: M30-17  [#permalink]

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New post 23 Feb 2016, 14:25
CountClaud wrote:
Zhenek wrote:
My approach towards this question was abit different, I found it easy, I guess I might have overlooked something then.
Is there a number "y" that gives us quotinent 3 if you divide 5/y or 8/y? Answer is no.
5/1 = 5, 5/2 = 2 + 1, 5/3 = 1 + 2, 5/4 = 1 + 1 ...
8/1 = 8, 8/2 = 4, 8/3 = 2 + 2, 8/4 = 2, 8/5 = 1 + 3, 8 /6 = 1 + 2 ...

thus the only option is III coz we got no "no options possible"


I went through the same thought process.


Ehm, can someone explain why 5/3= 1+2, 8/5=1+3

thank you
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M30-17  [#permalink]

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New post 28 Feb 2016, 05:00
2
Bunuel wrote:
When the positive integer is \(x\) divided by the positive integer \(y\), the quotient is 3 and the remainder is \(z\). When \(z\) is divided by \(y\), the remainder is 2. Which of the following could be the value of \(x\)?

I. 5

II. 8

III. 32


A. I only
B. II only
C. III only
D. I and II only
E. I, II and III


Alternative approach.

Solving problems of such a type we have to remember 2 fundamental rules:
1. Divident is always equal or greater than remainder.
2. Divisor is always greater than remainder.

We have 2 equations:
1. x/y = 3 + z. Means: x ≥ z, y > z, taking into account the quotient, x ≥ y * 3 + z.
2. z/y = ? + 2. Meаns: z ≥ 2, y > 2.

Make conclusion from both 1 and 2:
I. x ≥ z ≥ 2.
II. y > z ≥ 2; y can be at least 3, because y is integer and y > z = 2.
III. x ≥ 11. Because 3 * 3 + 2 = 11.

Therefore y can't be 5 nor 8.
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Re M30-17  [#permalink]

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New post 01 Aug 2016, 05:39
I think this is a high-quality question and I agree with explanation.
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Re M30-17  [#permalink]

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New post 30 Sep 2016, 06:08
I think this is a high-quality question and I agree with explanation.
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Re M30-17  [#permalink]

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New post 08 Jun 2017, 05:07
I think this is a high-quality question and I agree with explanation. Good Question...Quant questions in Club Tests always have a surprise element
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Re M30-17  [#permalink]

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New post 31 Aug 2017, 03:57
I think this is a high-quality question and I agree with explanation.
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Re M30-17  [#permalink]

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New post 12 Oct 2017, 19:42
I think this is a high-quality question and I agree with explanation. very good question
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Re: M30-17  [#permalink]

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New post 30 Dec 2017, 03:35
How is this question 700-level?
Make no mistake, I get most of my 700 level questions incorrect, but this is surely not more than 500 level.
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Re: M30-17  [#permalink]

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New post 30 Dec 2017, 04:09
gmatinsead2018 wrote:
How is this question 700-level?
Make no mistake, I get most of my 700 level questions incorrect, but this is surely not more than 500 level.
Bunuel?


The difficulty level of a question is calculated automatically based on the timer stats from the users which attempted the question. If you can solve it easily good for you.
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Re: M30-17  [#permalink]

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New post 09 Feb 2018, 19:04
Excellent Question!!
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Re M30-17  [#permalink]

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New post 19 Oct 2018, 15:10
I think this is a high-quality question and I agree with explanation. Beautiful question
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Re: M30-17  [#permalink]

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New post 27 Oct 2018, 09:09
Bunuel

I had a different approach and I do not know why it is wrong. Can you help me?

x = 3y + z (div. by y)
x/y= 3 + z/y
x/y = 3 + 2
x/y = 5
x = 5y

Consequently, x must be a multiple of 5.

Why is this approach wrong?

Thanks!
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Re: M30-17  [#permalink]

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New post 28 Oct 2018, 00:34
ipzamoner wrote:
Bunuel

I had a different approach and I do not know why it is wrong. Can you help me?

x = 3y + z (div. by y)
x/y= 3 + z/y
x/y = 3 + 2
x/y = 5
x = 5y

Consequently, x must be a multiple of 5.

Why is this approach wrong?

Thanks!


How does z/y = 2? When z is divided by y, the remainder is 2 does not mean that y/z = 2.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M30-17  [#permalink]

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New post 28 Oct 2018, 07:11
Bunuel wrote:
ipzamoner wrote:
Bunuel

I had a different approach and I do not know why it is wrong. Can you help me?

x = 3y + z (div. by y)
x/y= 3 + z/y
x/y = 3 + 2
x/y = 5
x = 5y

Consequently, x must be a multiple of 5.

Why is this approach wrong?

Thanks!




How does z/y = 2? When z is divided by y, the remainder is 2 does not mean that y/z = 2.




I see.. Great, thanks!
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Re: M30-17  [#permalink]

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New post 29 Oct 2018, 04:06
Wasted a lot of time to realise that x = 3y + 2
Once I figured that out, then was able to solve it by substituting values of x in the equation.

Posted from my mobile device
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Re: M30-17 &nbs [#permalink] 29 Oct 2018, 04:06
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