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If a number is to be randomly selected from set A above and a number [#permalink]
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13 Dec 2017, 01:49
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49% (01:24) correct 51% (01:08) wrong based on 55 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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13 Dec 2017, 03:49
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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13 Dec 2017, 05:08
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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13 Dec 2017, 05:25
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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13 Dec 2017, 05:30
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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13 Dec 2017, 06:28
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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13 Dec 2017, 07:04
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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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13 Dec 2017, 17:57
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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13 Dec 2017, 19:59
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 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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