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There are 35 distinct numbers in set M, there are 28

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antoxavier
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There are 35 distinct numbers in set M, there are 28 [#permalink] New post   14 Jul 2013, 11:07
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There are 35 distinct numbers in set M, there are 28 distinct numbers in set N, and there are 12 distinct numbers that are in both sets M and N. Set H is the set containing the elements that are in at least one of sets M and N. How many elements are in set H?

A. 39
B. 40
C. 51
D. 58
E. 63
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C
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Bunuel
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Re: There are 35 distinct numbers in set M, there are 28 [#permalink] New post   14 Jul 2013, 11:14
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antoxavier wrote:
There are 35 distinct numbers in set M, there are 28 distinct numbers in set N, and there are 12 distinct numbers that are in both sets M and N. Set H is the set containing the elements that are in at least one of sets M and N. How many elements are in set H?

A. 39
B. 40
C. 51
D. 58
E. 63

Official answer not available


{Total} = {M} + {N} - {Both}
{Total} = 35 + 28 - 12 = 51.

Answer: C.
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leroyconje
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Re: There are 35 distinct numbers in set M, there are 28 [#permalink] New post   14 Aug 2013, 09:37
Bunuel wrote:
antoxavier wrote:
There are 35 distinct numbers in set M, there are 28 distinct numbers in set N, and there are 12 distinct numbers that are in both sets M and N. Set H is the set containing the elements that are in at least one of sets M and N. How many elements are in set H?

A. 39
B. 40
C. 51
D. 58
E. 63

Official answer not available


{Total} = {M} + {N} - {Both}
{Total} = 35 + 28 - 12 = 51.

Answer: C.


Are you sure?
at least in one of the sets could mean either M or N or both. therefore 63 should be ok.
Let me know what you think.
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Marco
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Re: There are 35 distinct numbers in set M, there are 28 [#permalink] New post   14 Aug 2013, 11:43
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leroyconje wrote:
Bunuel wrote:
antoxavier wrote:
There are 35 distinct numbers in set M, there are 28 distinct numbers in set N, and there are 12 distinct numbers that are in both sets M and N. Set H is the set containing the elements that are in at least one of sets M and N. How many elements are in set H?

A. 39
B. 40
C. 51
D. 58
E. 63

Official answer not available


{Total} = {M} + {N} - {Both}
{Total} = 35 + 28 - 12 = 51.

Answer: C.


Are you sure?
at least in one of the sets could mean either M or N or both. therefore 63 should be ok.
Let me know what you think.
Regards,
Marco


Yes, I am. {M} and {N} both contain {Both}, when we subtract once we still have one {Both} counted.
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Asifpirlo
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Re: There are 35 distinct numbers in set M, there are 28 [#permalink] New post   14 Aug 2013, 12:27
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antoxavier wrote:
There are 35 distinct numbers in set M, there are 28 distinct numbers in set N, and there are 12 distinct numbers that are in both sets M and N. Set H is the set containing the elements that are in at least one of sets M and N. How many elements are in set H?

A. 39
B. 40
C. 51
D. 58
E. 63

Official answer not available

n(H) = at least one of sets M and N =n(M u N) = n(M) + n(N) - n(M n N)
= 35 + 28 - 12
= 51 (Answer C)
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Re: There are 35 distinct numbers in set M, there are 28 [#permalink] New post   17 Aug 2013, 04:28
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Solution attached:

At least means

Only M + Only N + Both

Hence 23+12+16=51
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solution.JPG [ 14.87 KiB | Viewed 4142 times ]

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unceldolan
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Re: There are 35 distinct numbers in set M, there are 28 [#permalink] New post   28 Jul 2014, 03:53
Since Set H contains the numbers which are AT LEAST once in set M or N, you only need to subtract the 12 numbers once. Hence 35+28-12 = 51

Take the example:
Set M{1,2,3,4,5,6,7,8,9,10,11,12,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120,121,122}
Set N{1,2,3,4,5,6,7,8,9,10,11,12,200,201,202,203,204,205,206,207,208,209,210,211,212,213,214,215}

You'll now have to count the numbers from 1-12 ONCE, and the other numbers. Hence it's 35 (Set M) + 28 (Set N) - 12 (Both)

Hope it helps.
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Re: There are 35 distinct numbers in set M, there are 28 [#permalink] New post   12 Oct 2019, 08:34
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Re: There are 35 distinct numbers in set M, there are 28 [#permalink]
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