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Knewton Brutal Math Challenge
A set of positive integers contains twice as many even integers as odd integers. How many integers are in the set?
(1) If two integers are chosen at random from the set, the probability that their sum will be even is 26/75 less than the probability that their product will be even.
(2) The number of integers in the set has fewer than six factors, and one of these factors is 17.
01 Sep 2015, 00:54
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IMO: D
Question Stem:
No. of odd Integers = x
No. of Even Integers = 2x
Total no. of Integers in the set = 3x
St 1:If two integers are chosen at random from the set, the probability that their sum will be even is 26/75 less than the probability that their product will be even
probability that sum will be even = probability that their product will be even - 26/75 --(ii)
probability that sum will be even = prob of picking two even integers + prob of picking two odd integers
= \(\frac{(2xC2 + xC2 )}{3xC2}\) = \(\frac{(2x *(2x-1) + x *(x-1) )}{3x * (3x-1) )}\)
= \(\frac{( 5x^2 - 3x )}{9x^2 - 3x )}\)
probability that their product will be even = 1 - two numbers are odd
= 1 - \(\frac{(xC2 )}{3xC2}\)
= \(\frac{( 8x^2 - 2x )}{9x^2 - 3x )}\)
Sub in eq --(ii) we get
\(9x^2 - 153x = 0\)
9x(x-17)= 0
x = 0 or x = 17 .
Since x cannot be equal to 0. x = 17
Hence suff.
St 2: The number of integers in the set has fewer than six factors, and one of these factors is 17
The number of integers in the set = has less than 6 factors.
i.e The number of integers in the set = \(Prime ^ a\) where is a < 5
OR
The number of integers in the set = \((P_1)^a * (P_2)^b\) (where a =1 & b =1 )
17 is one of the factors
if 3x is the total number of factors then "3" is also one of the factors.
Hence 3,17 are among the factors of total number of integers in the set
Thus condition 2 must be the case, making total no. of integers = 3*17 = 51
Hence suff
Bunuel
It has taken me nearly 4-5 min to solve this question.
What do you advise if we come across such question in GMAT exam?
Solve it even if it takes 4 min and hope that I come across few easy questions to make up the time or not risk and take a guess?
Question Stem:
No. of odd Integers = x
No. of Even Integers = 2x
Total no. of Integers in the set = 3x
St 1:If two integers are chosen at random from the set, the probability that their sum will be even is 26/75 less than the probability that their product will be even
probability that sum will be even = probability that their product will be even - 26/75 --(ii)
probability that sum will be even = prob of picking two even integers + prob of picking two odd integers
= \(\frac{(2xC2 + xC2 )}{3xC2}\) = \(\frac{(2x *(2x-1) + x *(x-1) )}{3x * (3x-1) )}\)
= \(\frac{( 5x^2 - 3x )}{9x^2 - 3x )}\)
probability that their product will be even = 1 - two numbers are odd
= 1 - \(\frac{(xC2 )}{3xC2}\)
= \(\frac{( 8x^2 - 2x )}{9x^2 - 3x )}\)
Sub in eq --(ii) we get
\(9x^2 - 153x = 0\)
9x(x-17)= 0
x = 0 or x = 17 .
Since x cannot be equal to 0. x = 17
Hence suff.
St 2: The number of integers in the set has fewer than six factors, and one of these factors is 17
The number of integers in the set = has less than 6 factors.
i.e The number of integers in the set = \(Prime ^ a\) where is a < 5
OR
The number of integers in the set = \((P_1)^a * (P_2)^b\) (where a =1 & b =1 )
17 is one of the factors
if 3x is the total number of factors then "3" is also one of the factors.
Hence 3,17 are among the factors of total number of integers in the set
Thus condition 2 must be the case, making total no. of integers = 3*17 = 51
Hence suff
Bunuel
It has taken me nearly 4-5 min to solve this question.
What do you advise if we come across such question in GMAT exam?
Solve it even if it takes 4 min and hope that I come across few easy questions to make up the time or not risk and take a guess?
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