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If one number is selected at random from set A, and one number is

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If one number is selected at random from set A, and one number is  [#permalink]

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New post 10 Nov 2017, 19:21
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2
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

63% (01:22) correct 37% (01:18) wrong based on 46 sessions

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A: {71,73,79,83,87}
B:{57,59,61,67}
If one number is selected at random from set A, and one number is selected at random from set B, what is the probability that both numbers are prime?

A. \(\frac{9}{20}\)
B. \(\frac{3}{5}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
E. \(1\)

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Re: If one number is selected at random from set A, and one number is  [#permalink]

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New post 10 Nov 2017, 19:46
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Gnpth wrote:
A: {71,73,79,83,87}
B:{57,59,61,67}
If one number is selected at random from set A, and one number is selected at random from set B, what is the probability that both numbers are prime?

A. \(\frac{9}{20}\)
B. \(\frac{3}{5}\)
C. \(\frac{3}{4}\)
D. \(\frac{4}{5}\)
E. \(1\)



Hi...

The Q otherwise is simple but you are not really required to know the prime numbers.
So if you don't know the list what do you do.

All prime numbers are less than 100 and \(√100=10\) so you have to check for divisibility by primes <10, so only 2,3,5, and 7.

1) none is even
2) div by 3 means checking sum of digits by 3..
87=8+7=15 div by 3 AND 57 too is div by 3
3) none is div by 5 as none has 5 or 0 as units digit.
4) div by 7 can be checked as all are close to 70 and 63,70,77,84 are div by 7
None are there in set...

So two are not prime..

Total ways 5(# of numbers in set A)*4(# of numbers in set B)=20
Ways that both are prime =(5-1)*(4-1)=4*3=12
So probability=12/20=3/5
B
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: If one number is selected at random from set A, and one number is &nbs [#permalink] 10 Nov 2017, 19:46
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