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# M30-17

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Math Expert
Joined: 02 Sep 2009
Posts: 51218

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16 Sep 2014, 00:45
1
6
00:00

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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Math Expert
Joined: 02 Sep 2009
Posts: 51218

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16 Sep 2014, 00:45
2
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 $$x=3*10+2=32$$.

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24 Dec 2014, 00:10
I think this question is good and helpful.
Manager
Joined: 17 Mar 2015
Posts: 116

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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"
Intern
Joined: 12 Jul 2015
Posts: 5

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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.
Current Student
Joined: 12 Nov 2015
Posts: 59
Location: Uruguay
Concentration: General Management
Schools: Goizueta '19 (A)
GMAT 1: 610 Q41 V32
GMAT 2: 620 Q45 V31
GMAT 3: 640 Q46 V32
GPA: 3.97

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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
Current Student
Joined: 29 Apr 2015
Posts: 26
Location: Russian Federation
GMAT 1: 710 Q48 V38
GPA: 4

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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.
Senior Manager
Joined: 31 Mar 2016
Posts: 385
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

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01 Aug 2016, 05:39
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 08 Nov 2014
Posts: 4

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30 Sep 2016, 06:08
I think this is a high-quality question and I agree with explanation.
Director
Joined: 18 Aug 2016
Posts: 623
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38

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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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Luckisnoexcuse

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Joined: 12 Sep 2016
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31 Aug 2017, 03:57
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 31 Jul 2013
Posts: 15
Location: Viet Nam
Concentration: General Management, Entrepreneurship
GMAT 1: 650 Q49 V28
GPA: 3.46
WE: Sales (Computer Software)

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12 Oct 2017, 19:42
I think this is a high-quality question and I agree with explanation. very good question
Intern
Joined: 12 Dec 2017
Posts: 48
Location: India
GMAT 1: 770 Q51 V46
GRE 1: Q170 V170
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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.
Bunuel?
Math Expert
Joined: 02 Sep 2009
Posts: 51218

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30 Dec 2017, 04:09
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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Joined: 20 Nov 2017
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09 Feb 2018, 19:04
Excellent Question!!
Intern
Joined: 31 Jul 2018
Posts: 3

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19 Oct 2018, 15:10
I think this is a high-quality question and I agree with explanation. Beautiful question
Intern
Joined: 11 Feb 2013
Posts: 6
GMAT 1: 730 Q47 V42
GMAT 2: 740 Q49 V41

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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!
Math Expert
Joined: 02 Sep 2009
Posts: 51218

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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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Joined: 11 Feb 2013
Posts: 6
GMAT 1: 730 Q47 V42
GMAT 2: 740 Q49 V41

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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!
Manager
Joined: 26 Feb 2017
Posts: 89
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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.

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Re: M30-17 &nbs [#permalink] 29 Oct 2018, 04:06
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# M30-17

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