If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?
A)1/3
B)1/2
C)5/7
D)10/21
E)15/28
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divisors of 64
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j_shreyans
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GMATGuruNY
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P = (good options)/(all options).j_shreyans wrote:If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?
A)1/3
B)1/2
C)5/7
D)10/21
E)15/28
Factors of 64:
1*64
2*32
4*16
8*8.
Total number of distinct factors = 7.
All:
From the 7 distinct factors above, the number of ways to choose 2 =7C2 = (7*6)/(2*1) = 21.
Good = distinct pairs with a sum less than 32:
1+2, 1+4, 1+8, 1+16
2+4, 2+8, 2+16
4+8, 4+16
8+16
Good options = 10.
Result:
P(pair with a sum less than 32) = (good options)/(all options) = 10/21.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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