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

Question

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

Bunuel · Accepted Answer

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.

Answer: C.

ak1802 · Answer

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)

Subtract 2/10 from 5/10 and your answer is 3/10 =  30%

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.

registerincog · Answer

Realized from your explanation that this is a disguised Venn Diagram problem. Thanks Bunuel

AccipiterQ · Answer

That was actually much easier than I was trying to make it..great simple explanation

gmatprav · Answer

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

WoundedTiger · Answer

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%

gmatprav · Answer

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?

WoundedTiger · Answer

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

ak1802 · Answer

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.

ash99 · Answer

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%

Zhenek · Answer

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.

Abhishek009 · Answer

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 - Set.png 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% Hence answer is definitely (C)

saurabhsavant · Answer

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

nightblade354 · Answer

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.

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

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