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Sets M and N contain exactly m and n elements, respectively. What is the value of n?

(1) 7m=8n (2) The intersection of M and N contains exactly 0.4m elements.

Hi,

the Q asks us a numeric value for n..

However I gives us the ratio of m and n and II gives us the common elements in term of m again.. so NO numeric value of any kind to work on.. E

But lets see what each statements tells us..

May be helpful in some other Q.. (1) 7m=8n m is a multiple of 8 and n is a multiple of 7

(2) The intersection of M and N contains exactly 0.4m elements. this means common numbers are 0.4m.. so, m has to be a multiple of 5, otherwise .4m will be decimal..

combined we can say so both m and n are multiples of 5.. m is a multiple of 8*5=40 and n is a multiple of = 7*5=35

if we say we were told m <75.. we would have got our answer _________________

Re: Sets M and N contain exactly m and n elements, respectively. [#permalink]

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16 Apr 2016, 12:10

I'm a little confused - can you please explain what "the intersection of M and N contains exactly 0.4m elements" translates to algebraically? Would that mean N=0.4M? Since these are sets of numbers, it's also inherently assumed that the answer will be an integer, correct?

I'm a little confused - can you please explain what "the intersection of M and N contains exactly 0.4m elements" translates to algebraically? Would that mean N=0.4M? Since these are sets of numbers, it's also inherently assumed that the answer will be an integer, correct?

Hi,

Intersection of M and N consists of 0.4m elements in SET would be written as -- M\(\cap\)N = 0.4M.. so there are 0.4M elements common to M and N.. Since M and N are integers and common elements too should be integer, 0.4M should be integer.. so M has to be a multiple of 5 for 0.4M to be an integer.. _________________

Re: Sets M and N contain exactly m and n elements, respectively. [#permalink]

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17 Apr 2016, 04:33

after combining both statements we still don't get the answer.if we have extra info regarding m<75,then we would have got the answer as c.but correct option is E

Sets M and N contain exactly m and n elements, respectively. [#permalink]

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07 May 2016, 08:21

chetan2u wrote:

Gurshaans wrote:

Sets M and N contain exactly m and n elements, respectively. What is the value of n?

(1) 7m=8n (2) The intersection of M and N contains exactly 0.4m elements.

Hi,

the Q asks us a numeric value for n..

However I gives us the ratio of m and n and II gives us the common elements in term of m again.. so NO numeric value of any kind to work on.. E

But lets see what each statements tells us..

May be helpful in some other Q.. (1) 7m=8n m is a multiple of 8 and n is a multiple of 7

(2) The intersection of M and N contains exactly 0.4m elements. this means common numbers are 0.4m.. so, m has to be a multiple of 5, otherwise .4m will be decimal.. so both m and n are multiples of 5..

combined we can say m is a multiple of 8*5=40 and n is a multiple of = 7*5=35

if we say we were told m <75.. we would have got our answer

Why do both m and n need to be multiples of 5 ? Why also n ? Could you explain the "multiple of 5" part again please?

Sets M and N contain exactly m and n elements, respectively. What is the value of n?

(1) 7m=8n (2) The intersection of M and N contains exactly 0.4m elements.

Hi,

the Q asks us a numeric value for n..

However I gives us the ratio of m and n and II gives us the common elements in term of m again.. so NO numeric value of any kind to work on.. E

But lets see what each statements tells us..

May be helpful in some other Q.. (1) 7m=8n m is a multiple of 8 and n is a multiple of 7

(2) The intersection of M and N contains exactly 0.4m elements. this means common numbers are 0.4m.. so, m has to be a multiple of 5, otherwise .4m will be decimal.. so both m and n are multiples of 5..

combined we can say m is a multiple of 8*5=40 and n is a multiple of = 7*5=35

if we say we were told m <75.. we would have got our answer

Why do both m and n need to be multiples of 5 ? Why also n ? Could you explain the "multiple of 5" part again please?

hi it may have been written in statement 2, But it is meant when both statements are combined.. As 8n = 7m.. so if m is multiple of 5 n has to be a multiple of 5..
_________________

Re: Sets M and N contain exactly m and n elements, respectively. [#permalink]

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07 May 2016, 08:33

chetan2u wrote:

hi it may have been written in statement 2, But it is meant when both statements are combined.. As 8n = 7m.. so if m is multiple of 5 n has to be a multiple of 5..

thanks for the quick help! can you explain again why m is a multiple of 5 because of 0.4m ?

hi it may have been written in statement 2, But it is meant when both statements are combined.. As 8n = 7m.. so if m is multiple of 5 n has to be a multiple of 5..

thanks for the quick help! can you explain again why m is a multiple of 5 because of 0.4m ?

sure.. there are m and n elements.. Also there are 0.4m common elements... Now these elements have to be integer.. MEANING we can't have 4.5 or 5.6 elements common, it has to be an integer say 3,4, or 5 etc.. When will 0.4*m be an integer, ONLY when m is a multiple of 5, say 5x.. so 0.4 *m = 0.4*5*x = 2x, hence an integer.. m could be anything 5,10,35 or 100 but it will surely be a multiple of 5.. Hope it helps
_________________

Re: Sets M and N contain exactly m and n elements, respectively. [#permalink]

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12 Jun 2016, 05:33

Sir,

I have a question regarding the below solution by you. Specifically in the last line where if m,75 was mentioned we would have gotten our answer. How so?

I have a question regarding the below solution by you. Specifically in the last line where if m,75 was mentioned we would have gotten our answer. How so?

Hi Prakhar,

we have got m as a multiple of 40 and n as a multiple of 35... since intersection is 0.4m, we can say that m and n are \(\neq{0}\)..

so if we were given m<75, ONLY 40 would have fit in as next multiple of 40 is 80, which would have >75.. and 7m = 8n.... so 7*40 = 8n......n =35

Re: Sets M and N contain exactly m and n elements, respectively. [#permalink]

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20 Jun 2016, 16:23

The question is asking for a specific value of n. Statement 1: This is a ratio, so clearly insufficient. Statement 2: Another ratio, you don't know m and this statement says nothing about n.

1+2: From statement 1: \(m = \frac{8}{7}n\) and using statement 2: Intersection contains: \(0.4(\frac{8}{7}n)\) elements. Doesn't really tell you anything about n, so answer is E. If you look at it from a higher level both statements give you some sort of ratio info, while the question is asking for a value. You can't get a value from just ratio info.

Sets M and N contain exactly m and n elements, respectively. [#permalink]

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29 Aug 2016, 11:27

chetan2u wrote:

if we say we were told m <75.. we would have got our answer

Great explanation, thank you!

However, I have one question about the above quoted bit...

After combining both statements, we know that M---> Multiple of 7 AND a multiple of 5 N---> Multiple of 8 M/N are in the ratio of 8/7... So for ex if M = 7 x 5, then N= 8 x 5 (giving us values of M=35 and N=40)

Using this logic, we can have M= 7 x 10 , then N= 8 x 10 (giving us values of M=70 and N=80)

Therefore even if we were given m<75, we will still have 2 values..??

I think I am misunderstanding something maybe, can you please help to point out my misunderstanding?

if we say we were told m <75.. we would have got our answer

Great explanation, thank you!

However, I have one question about the above quoted bit...

After combining both statements, we know that M---> Multiple of 7 AND a multiple of 5 N---> Multiple of 8 M/N are in the ratio of 8/7... So for ex if M = 7 x 5, then N= 8 x 5 (giving us values of M=35 and N=40)

Using this logic, we can have M= 7 x 10 , then N= 8 x 10 (giving us values of M=70 and N=80)

Therefore even if we were given m<75, we will still have 2 values..??

I think I am misunderstanding something maybe, can you please help to point out my misunderstanding?

Thanks

Hi, you have understood the method but going wrong on inference from 7m=8n... This tells us that n and NOT m, is multiple of7..... M is s multiple of 8, as m= n*8/7....so n is multiple of 7
_________________

Re: Sets M and N contain exactly m and n elements, respectively. [#permalink]

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30 Aug 2016, 03:57

chetan2u wrote:

18967mba wrote:

chetan2u wrote:

if we say we were told m <75.. we would have got our answer

Great explanation, thank you!

However, I have one question about the above quoted bit...

After combining both statements, we know that M---> Multiple of 7 AND a multiple of 5 N---> Multiple of 8 M/N are in the ratio of 8/7... So for ex if M = 7 x 5, then N= 8 x 5 (giving us values of M=35 and N=40)

Using this logic, we can have M= 7 x 10 , then N= 8 x 10 (giving us values of M=70 and N=80)

Therefore even if we were given m<75, we will still have 2 values..??

I think I am misunderstanding something maybe, can you please help to point out my misunderstanding?

Thanks

Hi, you have understood the method but going wrong on inference from 7m=8n... This tells us that n and NOT m, is multiple of7..... M is s multiple of 8, as m= n*8/7....so n is multiple of 7

Yes! Thank you!! I understand now why if m<75, we would have a unique ans.. Because if M was 8x(2)x5 = 80, that is > 75! Therefore, we are only left with M=8x(1)x5 = 40! (And subsequently N=7x(1)x5 = 35!

Re: Sets M and N contain exactly m and n elements, respectively. [#permalink]

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07 Jan 2017, 23:06

Lettherebelight wrote:

Sets M and N contain exactly m and n elements, respectively. What is the value of n?

(1) 7m=8n (2) The intersection of M and N contains exactly 0.4m elements.

From 1, the LCM of 7 and 8 is 56, so you can plugin either 8 or 7 in the first equation to get the value. However since LCM is 56, the value can be any multiple of 56. For example 7(*8*4) = 8(*7*4), n can be any value 7, 14, 28 etc hence not sufficient.

From 2, intersection of both elements is 0.4m and we know nothing about the value of m hence this not sufficient as well.

Taking 1 and 2, does not help us in answering the question. Hence insufficient.