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If a number is to be randomly selected from set A above and a number

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If a number is to be randomly selected from set A above and a number  [#permalink]

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New post 13 Dec 2017, 00:49
1
2
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

50% (01:57) correct 50% (01:52) wrong based on 67 sessions

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GMAT CLUB'S FRESH CHALLENGE QUESTION:



A = {-10, -9, -8, -7, -6, -5, -4}
B = {-1/2, -1/3, -1/5, 0, 1/5, 1/3, 1/2}

If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?

A. 8/49
B. 12/49
C. 13/49
D. 18/49
E. 21/49

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Re: If a number is to be randomly selected from set A above and a number  [#permalink]

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New post 13 Dec 2017, 02:49
1
I think answer is A

There are 7x7=49 possible products

And 8 oh them yield negative integers

1) -10x1/5
2) -10x1/2
3) -9x1/3
4) -8x1/2
5) -6x1/3
6) -6x1/2
7) -5x1/5
8) -4x1/2
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Re: If a number is to be randomly selected from set A above and a number  [#permalink]

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New post 13 Dec 2017, 04:08
A = {-10, -9, -8, -7, -6, -5, -4}
B = {-1/2, -1/3, -1/5, 0, 1/5, 1/3, 1/2}

If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?

A. 8/49
B. 12/49
C. 13/49
D. 18/49
E. 24/49
--------------
Total number of ways is 7*7 =49

Number of ways to select 1 number from A --> 7C1 -->7
For the product to be negative select a positive number from B > no.of ways --> 3c1 -- >3

Total possibility -- > 7 *3 -- 21

Probability is 21/49

But i am not able to find the answer in the question stem
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Re: If a number is to be randomly selected from set A above and a number  [#permalink]

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New post 13 Dec 2017, 04:25
Munch wrote:
A = {-10, -9, -8, -7, -6, -5, -4}
B = {-1/2, -1/3, -1/5, 0, 1/5, 1/3, 1/2}

If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?

A. 8/49
B. 12/49
C. 13/49
D. 18/49
E. 24/49
--------------
Total number of ways is 7*7 =49

Number of ways to select 1 number from A --> 7C1 -->7
For the product to be negative select a positive number from B > no.of ways --> 3c1 -- >3

Total possibility -- > 7 *3 -- 21

Probability is 21/49

But i am not able to find the answer in the question stem



]The question specifically asks for integers,
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Re: If a number is to be randomly selected from set A above and a number  [#permalink]

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New post 13 Dec 2017, 04:30
kablayi wrote:
Munch wrote:
A = {-10, -9, -8, -7, -6, -5, -4}
B = {-1/2, -1/3, -1/5, 0, 1/5, 1/3, 1/2}

If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?

A. 8/49
B. 12/49
C. 13/49
D. 18/49
E. 24/49
--------------
Total number of ways is 7*7 =49

Number of ways to select 1 number from A --> 7C1 -->7
For the product to be negative select a positive number from B > no.of ways --> 3c1 -- >3

Total possibility -- > 7 *3 -- 21

Probability is 21/49

But i am not able to find the answer in the question stem



]The question specifically asks for integers,


Oh ya!! thanks!! then it should be 8 .. I doubled checked the possibility to match the answer, but still missed the word integer..
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Re: If a number is to be randomly selected from set A above and a number  [#permalink]

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New post 13 Dec 2017, 05:28
A = {-10, -9, -8, -7, -6, -5, -4}
B = {-1/2, -1/3, -1/5, 0, 1/5, 1/3, 1/2}

I am not sure whether we should consider the product numbers of ‘0’.

Considering ‘0’ as an eligible possibility,
A X B will be -ve when, either A or B is -ve. Since all values for A is -ve, then all values of B must be +ve. In such case, only pairs will be -ve when value of B is 1/5 , 1/3 or 1/2.
For each A, there will be 3 Negative Values, i.e
A X B = -10*1/5 = -10/5 ; -10/3 ; -10/2
= -9/5 ; -9/3 ; -9/2
.
.
.
and so on.

Thus possible negative values will be = 7*3 =21

Total Possible product values will be = 7*7=49

Thus P(E) = 21/49. Option E is right.
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Re: If a number is to be randomly selected from set A above and a number  [#permalink]

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New post 13 Dec 2017, 06:04
1
The question specifically asks for -ve integers("0" is not a -ve integer). Now Set A has only -ve integers so we have to select only +ve fractions from set B i.e.{1/5,1/3,1/2}.
With 1/5 we can only select -10 & -5 from set A. So possible combinations : 1/5*-10 & 1/5*-5. Number of possible combinations = 2

With 1/3 we can only select -9 & -6 from set A. Number of possible combinations = 2
With 1/2 we can only select -10,-8,-6,-4 from set A. Number of possible combinations = 4
Thus we have 2+2+4 = 8 combinations
Total Possible combinations = 7C1 * 7C1 = 49
Ans = 8/49
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Re: If a number is to be randomly selected from set A above and a number  [#permalink]

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New post 13 Dec 2017, 16:57
Bunuel wrote:

GMAT CLUB'S FRESH CHALLENGE QUESTION:



A = {-10, -9, -8, -7, -6, -5, -4}
B = {-1/2, -1/3, -1/5, 0, 1/5, 1/3, 1/2}

If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?

A. 8/49
B. 12/49
C. 13/49
D. 18/49
E. 21/49



Since we are looking for negative integer, we have to consider only POSITIVE numbers in set B as all are NEGATIVE in set A..
1) 1/2
This will give integers when multiplied by MULTIPLE of 2
So -4,-6,-8,-10......4ways
2) 1/3
Similarly when multiplied by MULTIPLE of 3
-6,-9.....2ways
3) 1/5
When multiplied by MULTIPLE of 5
So -5,-10.....2ways

Total - 4+2+2=8ways
Possible products = 7*7=49

Ans 8/49

A
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Re: If a number is to be randomly selected from set A above and a number  [#permalink]

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New post 13 Dec 2017, 18:59
chetan2u wrote:
Bunuel wrote:

GMAT CLUB'S FRESH CHALLENGE QUESTION:



A = {-10, -9, -8, -7, -6, -5, -4}
B = {-1/2, -1/3, -1/5, 0, 1/5, 1/3, 1/2}

If a number is to be randomly selected from set A above and a number is to be randomly selected from set B above, what is the probability that the product of the two numbers will be a negative integer?

A. 8/49
B. 12/49
C. 13/49
D. 18/49
E. 21/49


I missed the INTEGER PART :cry:

Since we are looking for negative integer, we have to consider only POSITIVE numbers in set B as all are NEGATIVE in set A..
1) 1/2
This will give integers when multiplied by MULTIPLE of 2
So -4,-6,-8,-10......4ways
2) 1/3
Similarly when multiplied by MULTIPLE of 3
-6,-9.....2ways
3) 1/5
When multiplied by MULTIPLE of 5
So -5,-10.....2ways

Total - 4+2+2=8ways
Possible products = 7*7=49

Ans 8/49

A
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Re: If a number is to be randomly selected from set A above and a number &nbs [#permalink] 13 Dec 2017, 18:59
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If a number is to be randomly selected from set A above and a number

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