Sets M and N contain exactly m and n elements, respectively. : Data Sufficiency (DS)

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

16 Apr 2016, 13:10

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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?

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

16 Apr 2016, 20:52

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jecharte wrote:

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..

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

07 May 2016, 09:21

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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?

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

07 May 2016, 09:29

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broilerc wrote:

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?

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..

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

07 May 2016, 09:33

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 ?

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

12 Jun 2016, 06:33

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?

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

12 Jun 2016, 06:40

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prakhar4 wrote:

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?

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

that is why suff..

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

20 Jun 2016, 17:23

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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.

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

29 Aug 2016, 12:27

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?

Thanks

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

30 Aug 2016, 03:56

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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

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

17 Dec 2017, 11:38

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Hello,

Another way to see the second statement:
(1) m : n = 8 : 7
(2) Think about groups, so: m + (n - 0.4m) = T (total of elements) => 0.6m + n = T

(1-2) Substitute (1) in (2): 0.6 x 8n/7 = T
n and T are variables, but we have just one equation to solve it. Hence, insufficient.

Best,

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

14 Mar 2020, 05:14

Together, when m has to be a multiple of 5 and n has to produce an integer when 7m is divided by 8. Again many answers are possible for this condition, like m = 40, n= 35. or m =80, n=70.

E is the answer.

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

22 Aug 2020, 08:24

minustark wrote:

Together, when m has to be a multiple of 5 and n has to produce an integer when 7m is divided by 8. Again many answers are possible for this condition, like m = 40, n= 35. or m =80, n=70.

E is the answer.

Could please explain this point further?

Why if m is a multiple of 5 ALSO n must be a multiple of 5? VeritasKarishma

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

01 Sep 2020, 01:32

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Genoa2000 wrote:

minustark wrote:

Together, when m has to be a multiple of 5 and n has to produce an integer when 7m is divided by 8. Again many answers are possible for this condition, like m = 40, n= 35. or m =80, n=70.

E is the answer.

Could please explain this point further?

Why if m is a multiple of 5 ALSO n must be a multiple of 5? VeritasKarishma

Given m and n are integers,

If 7m = 8n,
7 * m = 2^3 * n

m must have 2^3 in it and n must have 7. Now, if m is a multiple of any other prime number, n must be a multiple of that prime number too.

If m = 2^3 * 5,
n = 7 * 5

If m = 2^3 * 11
n = 7 * 11

Only then can both sides of the equations be same.

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

08 Mar 2021, 23:33

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.

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

Could any experts please explain the highlighted part? I am confused. I understand that if we were told m= 8*(some integer n), then we can deduce that m is a multiple of '8'. However, in this case, how can we deduce that 'm' is a multiple of 8? if we re-arrange the formula, we get m = 8/7 * n (how is this a multiple of 8?). At best, the equation looks like m = multiple of 8/7.

ScottTargetTestPrep JeffTargetTestPrep BrentGMATPrepNow

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

10 Mar 2021, 11:37

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Gmatboat wrote:

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.

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

Could any experts please explain the highlighted part? I am confused. I understand that if we were told m= 8*(some integer n), then we can deduce that m is a multiple of '8'. However, in this case, how can we deduce that 'm' is a multiple of 8? if we re-arrange the formula, we get m = 8/7 * n (how is this a multiple of 8?). At best, the equation looks like m = multiple of 8/7.

ScottTargetTestPrep JeffTargetTestPrep BrentGMATPrepNow

The key piece of information here is that m and n are integers.
If 7m = 8n, then the prime factorization of both sides should be equal.
That is: (7)(m) = (2)(2)(2)(n)
We can see that, in order for the equation to hold true, m must have at least three 2's in its prime factorization, and n must have at least one 7 in its prime factorization.
If m has at least three 2's in its prime factorization, then m is a multiple of 8
If n has at least one 7 in its prime factorization, then n is a multiple of 7

Does that help?

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

17 Jan 2024, 15:22

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.

can we solve it using below approach

M U N= m+ 7m/8-0.4m +neither

insufficient

gmatclubot

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

17 Jan 2024, 15:22

GMAT Data Sufficiency (DS) Questions

Sets M and N contain exactly m and n elements, respectively.