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# M12-36

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:48
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95% (hard)

Question Stats:

39% (01:23) correct 61% (01:10) wrong based on 152 sessions

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How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?

A. 28
B. 29
C. 30
D. 31
E. 32

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Math Expert
Joined: 02 Sep 2009
Posts: 50039

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16 Sep 2014, 00:48
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Official Solution:

How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?

A. 28
B. 29
C. 30
D. 31
E. 32

# of multiples of 13 in the given range $$\frac{(\text{last-first})}{\text{multiple}}+1=\frac{195-13}{13}+1=15$$;

# of multiples of 12 in the given range $$\frac{(\text{last-first})}{\text{multiple}}+1=\frac{192-12}{12}+1=16$$;

# of multiples of both 13 and 12 is 1: $$13*12=156$$. Notice that 156 is counted both in 15 and 16;

So, # of multiples of 13 or 12 but not both in the given range is $$(15-1)+(16-1)=29$$.

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24 Nov 2014, 10:22
1
A humble suggestions.

30 would be a good trap answer here. I got 30 and then realised I'll have to reduce 1 twice. Had 30 been there, I'd got this one wrong.
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JA
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24 Nov 2014, 10:25
Bunuel wrote:
# of multiples of 13 in the given range $$\frac{(\text{last-first})}{\text{multiple}}+1=\frac{195-13}{13}+1=15$$;

# of multiples of 12 in the given range $$\frac{(\text{last-first})}{\text{multiple}}+1=\frac{192-12}{12}+1=16$$;

Bunuel, wouldn't it be easier to simply divide 200 separately by both 13 and 12 and subtract 1 each from 15 and 16 respectively? Is this approach correct?
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JA
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Current Student
Joined: 04 Jul 2014
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GMAT 3: 710 Q49 V37
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WE: Analyst (Accounting)

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25 Nov 2014, 21:04
Bunuel look forward to your views!

joseph0alexander wrote:
Bunuel wrote:
# of multiples of 13 in the given range $$\frac{(\text{last-first})}{\text{multiple}}+1=\frac{195-13}{13}+1=15$$;

# of multiples of 12 in the given range $$\frac{(\text{last-first})}{\text{multiple}}+1=\frac{192-12}{12}+1=16$$;

Bunuel, wouldn't it be easier to simply divide 200 separately by both 13 and 12 and subtract 1 each from 15 and 16 respectively? Is this approach correct?

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13 Oct 2015, 05:01
I think this is a high-quality question and I agree with explanation. Great Q.

Binit.
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25 Feb 2016, 09:36
Bunuel wrote:
How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?

A. 28
B. 29
C. 31
D. 32
E. 33

Hi
Is there a way we can apply Set theory here using the formula n(AUB)=n(A)+n(B)-n(AUB)+none?
IMHO, not using the Set theory will easily make you err in such questions.
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09 Mar 2016, 00:32
1
A humble suggestions.

30 would be a good trap answer here. I got 30 and then realised I'll have to reduce 1 twice. Had 30 been there, I'd got this one wrong.

Exactly ...
I too was getting 30 again and again and then realised the same...
Also if the question would have been either 13 or 12 then 30 is perfect.
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09 Mar 2016, 11:28
Chiragjordan wrote:
A humble suggestions.

30 would be a good trap answer here. I got 30 and then realised I'll have to reduce 1 twice. Had 30 been there, I'd got this one wrong.

Exactly ...
I too was getting 30 again and again and then realised the same...
Also if the question would have been either 13 or 12 then 30 is perfect.

Edited as suggested. Thank you.
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30 Apr 2016, 10:04
I think this is a poor-quality question. If 156 is counted twice , why are we deducting it twice instead of once?
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02 May 2016, 04:13
1
birsencal wrote:
I think this is a poor-quality question. If 156 is counted twice , why are we deducting it twice instead of once?

Because we don;t want the multiples of both 13 and 12 (so multiples of 156) at all.
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12 Jul 2016, 13:08
I think this is a high-quality question and I agree with explanation.
Manager
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05 Aug 2016, 12:12
sinhap07 wrote:
Bunuel wrote:
How many positive integers less than 200 are there such that they are multiples of 13 or multiples of 12 but not both?

A. 28
B. 29
C. 31
D. 32
E. 33

Hi
Is there a way we can apply Set theory here using the formula n(AUB)=n(A)+n(B)-n(AUB)+none?
IMHO, not using the Set theory will easily make you err in such questions.

i tried it this way
>> !!!

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Intern
Joined: 30 Jan 2015
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05 Aug 2016, 12:49
I think there is \frac{195 - 0}{13 + 1 = 16} multiples of 13, since 0 i also a multiple of this number and should be included accordingly.

Same occurs to multiples of 12.
Intern
Joined: 30 Jan 2015
Posts: 3

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05 Aug 2016, 12:50
I think there is 195 - 0/13 + 1 = 16 multiples of 13, since 0 i also a multiple of this number and should be included accordingly.

Same occurs to multiples of 12.
Manager
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05 Aug 2016, 13:52
renanmenezes wrote:
I think there is 195 - 0/13 + 1 = 16 multiples of 13, since 0 i also a multiple of this number and should be included accordingly.

Same occurs to multiples of 12.

The question stem says x>0.

Man, to beat Bunuel we have to watch all the Rocky movies more than twice.
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22 Aug 2016, 09:00
I am sorry but I am really not clear as to why 156 is deducted twice. Can you please elaborate?
Manager
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22 Aug 2016, 13:25
Purelly a "set property".

Since they overlap one another, you must take that out to do not double count them.

Consider to review sets properties from Manhattan or from Gmat Club, it will help you to clarify this concept.

By the way, i shared my reasoning right above.

Posted from my mobile device
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25 Sep 2016, 18:46
WHAT DO YOU MEAN BY 156 IS COUNTED BOTH IN 15 AND 16?
Math Expert
Joined: 02 Sep 2009
Posts: 50039

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25 Sep 2016, 23:39
manchitkapoor wrote:
WHAT DO YOU MEAN BY 156 IS COUNTED BOTH IN 15 AND 16?

Both 15, which is # of multiples of 13 in the given range, as well as 16, which is # of multiples of 12 in the given range, contain 1 number (156), which is a multiples of both 13 and 12.
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# M12-36

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