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# In a set of integers, one-half of the numbers are multiples

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In a set of integers, one-half of the numbers are multiples  [#permalink]

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16 Nov 2013, 11:10
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35% (medium)

Question Stats:

74% (01:58) correct 26% (02:21) wrong based on 251 sessions

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In a set of integers, one-half of the numbers are multiples of 7 and one-third of the numbers are multiples of 3. If one-fifth of the numbers are multiples of both 3 and 7, then what percentage of the number are multiples of 7 but not multiples of 21?

A) 9
B) 15
C) 30
D) 33.3
E) 60
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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16 Nov 2013, 11:23
2
5
registerincog wrote:
In a set of integers, one-half of the numbers are multiples of 7 and one-third of the numbers are multiples of 3. If one-fifth of the numbers are multiples of both 3 and 7, then what percentage of the number are multiples of 7 but not multiples of 21?

A) 9
B) 15
C) 30
D) 33.3
E) 60

Say there are 210 number in the set (LCM of 2, 3, 5, and 7).

Multiples of 7 = 1/2*210 = 105;
Multiples of both 3 and 7, so multiples of 21 = 1/5*210 = 42.

Multiples of 7 but not 21 = 105 - 42 = 63.

The percentage = 63/210 = 3/10.

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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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16 Nov 2013, 11:38
Bunuel wrote:
registerincog wrote:
In a set of integers, one-half of the numbers are multiples of 7 and one-third of the numbers are multiples of 3. If one-fifth of the numbers are multiples of both 3 and 7, then what percentage of the number are multiples of 7 but not multiples of 21?

A) 9
B) 15
C) 30
D) 33.3
E) 60

Say there are 210 number in the set (LCM of 2, 3, 5, and 7).

Multiples of 7 = 1/2*210 = 105;
Multiples of both 3 and 7, so multiples of 21 = 1/5*210 = 42.

Multiples of 7 but not 21 = 105 - 42 = 63.

The percentage = 63/210 = 3/10.

Realized from your explanation that this is a disguised Venn Diagram problem. Thanks Bunuel
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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16 Nov 2013, 12:30
Bunuel wrote:
registerincog wrote:
In a set of integers, one-half of the numbers are multiples of 7 and one-third of the numbers are multiples of 3. If one-fifth of the numbers are multiples of both 3 and 7, then what percentage of the number are multiples of 7 but not multiples of 21?

A) 9
B) 15
C) 30
D) 33.3
E) 60

Say there are 210 number in the set (LCM of 2, 3, 5, and 7).

Multiples of 7 = 1/2*210 = 105;
Multiples of both 3 and 7, so multiples of 21 = 1/5*210 = 42.

Multiples of 7 but not 21 = 105 - 42 = 63.

The percentage = 63/210 = 3/10.

That was actually much easier than I was trying to make it..great simple explanation
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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18 Nov 2013, 02:25
Bunuel wrote:
registerincog wrote:
In a set of integers, one-half of the numbers are multiples of 7 and one-third of the numbers are multiples of 3. If one-fifth of the numbers are multiples of both 3 and 7, then what percentage of the number are multiples of 7 but not multiples of 21?

A) 9
B) 15
C) 30
D) 33.3
E) 60

Say there are 210 number in the set (LCM of 2, 3, 5, and 7).

Multiples of 7 = 1/2*210 = 105;
Multiples of both 3 and 7, so multiples of 21 = 1/5*210 = 42.

Multiples of 7 but not 21 = 105 - 42 = 63.

The percentage = 63/210 = 3/10.

Hello Bunuel, Why did you take LCM of 2,3,5 and 7 and not just LCM of 2,3 & 5?
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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Updated on: 19 Nov 2013, 10:01
3
This is similar to Bunuel's approach but it uses the fractions as they are.

Given:

1/2 are multiples of 7
1/3 are multiples of 3
1/5 are multiples of 3 and 7

You want to find the percentage of numbers that are multiples of 7 but not multiples of 21 (3 and 7).

Essentially the 1/3 of numbers being multiples of 3 is extra information in this problem.

You know that 1/2 of the numbers will be multiples of 7. (5/10)

You also know that 1/5 of the numbers are multiples of 3 AND 7. (2/10)

C.

What this does, is removes the percentage of multiples of 21 from the percent of multiples that are 7.

**7 does cross with 21, 1/3 of the time, BUT keep in mind that the overall percentage is also influenced by the factor 3, thus reducing the overall percentage to 30**

p.s. - please double check my methodology, I'm typing this out loud for that reason.

Originally posted by ak1802 on 18 Nov 2013, 20:29.
Last edited by ak1802 on 19 Nov 2013, 10:01, edited 4 times in total.
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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19 Nov 2013, 03:52
gmatprav wrote:
Bunuel wrote:
registerincog wrote:
In a set of integers, one-half of the numbers are multiples of 7 and one-third of the numbers are multiples of 3. If one-fifth of the numbers are multiples of both 3 and 7, then what percentage of the number are multiples of 7 but not multiples of 21?

A) 9
B) 15
C) 30
D) 33.3
E) 60

Say there are 210 number in the set (LCM of 2, 3, 5, and 7).

Multiples of 7 = 1/2*210 = 105;
Multiples of both 3 and 7, so multiples of 21 = 1/5*210 = 42.

Multiples of 7 but not 21 = 105 - 42 = 63.

The percentage = 63/210 = 3/10.

Hello Bunuel, Why did you take LCM of 2,3,5 and 7 and not just LCM of 2,3 & 5?

I think Ans would still be the same

Taking LCM of 2,3 and 5 =30
Multiples of 7: 15
Multiples of 3: 10
Multiples for both or 21 :6

Only 7 and not 21 = 15-6 = 9

% 9/30 or 30%
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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19 Nov 2013, 09:50
WoundedTiger wrote:
gmatprav wrote:
Hello Bunuel, Why did you take LCM of 2,3,5 and 7 and not just LCM of 2,3 & 5?

I think Ans would still be the same

Taking LCM of 2,3 and 5 =30
Multiples of 7: 15
Multiples of 3: 10
Multiples for both or 21 :6

Only 7 and not 21 = 15-6 = 9

% 9/30 or 30%

I know the answer is same in this particular case, however, I took LCM of 2,3 & 5 because 1/2, 1/3 and 1/5th are the fractions mentioned in the question, I am not sure why Bunuel added 7 into the mix and want to know why he did so. Hope I clarified it now. Bunuel could you please comment if my approach is correct or am I doing something wrong?
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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19 Nov 2013, 10:38
I don't think 7 should be in scheme of things and it is only the fractions which should be considered
Anyhow i let the expert respond to you..

Posted from my mobile device
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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19 Nov 2013, 12:31
gmatprav wrote:
WoundedTiger wrote:
gmatprav wrote:
Hello Bunuel, Why did you take LCM of 2,3,5 and 7 and not just LCM of 2,3 & 5?

I think Ans would still be the same

Taking LCM of 2,3 and 5 =30
Multiples of 7: 15
Multiples of 3: 10
Multiples for both or 21 :6

Only 7 and not 21 = 15-6 = 9

% 9/30 or 30%

I know the answer is same in this particular case, however, I took LCM of 2,3 & 5 because 1/2, 1/3 and 1/5th are the fractions mentioned in the question, I am not sure why Bunuel added 7 into the mix and want to know why he did so. Hope I clarified it now. Bunuel could you please comment if my approach is correct or am I doing something wrong?

You don't need anything other than the percentage of the following to solve the problem:

multiples of 7 : 1/2

multiples of 3 AND 7: 1/5

everything else is extra information.

Find a common denominator between these two fractions: 10

multiples of 7 = 5/10 multiples of 3 AND 7 = 2/10

Subtract the two fractions = 3/10 = 30%

This is more about interpreting what the question is asking, as opposed to multiplying every fraction given.

You know that 50% of the numbers in the set are multiples of 7. You also know that 20% of the numbers are multiples of 3 AND 7.
Keep in mind these percentages are also influenced by the 33.33% multiples of 3. Since these are percentages of the set, as a whole, the fractions already reflect this, making the 1/3 multiples of 3 extra information.
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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06 Apr 2015, 10:27
I did this in a different way:

Numbers that are multiples of 7's = 1/2
Numbers that are multiples of 3's = 1/3
Numbers that are multiples of 3's and 7's = 1/5

Required percentage = Numbers that are multiples of 7/ Numbers that are Not Multiples of 21
i.e. 1/2 / (1/2 * 1/3)
1/2 / 1/6
6/2 = 30%
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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06 Apr 2015, 15:50
1
I just drew diagram and concluded that info about 1/3 of numbers being multiple of 3 was redundant. Its easily seen that the answer to the question is easily achived by subtracting 1/5 from 1/2.
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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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15 Apr 2016, 10:57
2
registerincog wrote:
In a set of integers, one-half of the numbers are multiples of 7 and one-third of the numbers are multiples of 3. If one-fifth of the numbers are multiples of both 3 and 7, then what percentage of the number are multiples of 7 but not multiples of 21?

A) 9
B) 15
C) 30
D) 33.3
E) 60

Let the total number of integers in the set be 30 ( LCM of the denominators of the fractions 1/2,1/3 and 1/5)

Multiples of 7 = 15

Multiples of 3 = 10

Multiples of 3&7 = 6

We can safely say that the numbers which are Multiples of 3&7 are actually multiples of 21

Now draw a venn diagram to understand the issue better as below -

Attachment:

Set.png [ 8.1 KiB | Viewed 2070 times ]

We are interested only with the orange part (The numbers that are multiples of 7 but not multiples of 21)

Thus the percentage of the number are multiples of 7 but not multiples of 21 is 9/30*100 = 30%

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Re: In a set of integers, one-half of the numbers are multiples  [#permalink]

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18 Sep 2017, 22:03
registerincog wrote:
In a set of integers, one-half of the numbers are multiples of 7 and one-third of the numbers are multiples of 3. If one-fifth of the numbers are multiples of both 3 and 7, then what percentage of the number are multiples of 7 but not multiples of 21?

A) 9
B) 15
C) 30
D) 33.3
E) 60

A super good question..during reading one gets a feeling taht it is a typical remainder / number properties problem just to realise that it is actually a sets / venn diagram problem.
let total numbers be 30 (so that one half, one third and one fifth can be eaisly calculated)
a= multpple of 3 = 10
b= multiple of 7 = 15
c=multiples of both = 6

only multiple of 3 = 10-6=4
only multiple of 7 = 15-6 = 9
therefore answer = (9/30)*100 = 30 %
Hope it helps
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In a set of integers, one-half of the numbers are multiples  [#permalink]

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19 Sep 2017, 12:34
I created a chart, as I am a visual learner. The colors are as follows: Orange is given information; yellow is information you need to find. In total: 63/210 = 30% C
I used 210 because it is a multiple of 3, 5, and 7, while leaving all numbers integers in my calculations.
Attachments

chart.PNG [ 3.63 KiB | Viewed 1060 times ]

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