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01 Dec 2015, 04:30
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Difficulty:
15%
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Question Stats:
77% (01:08) correct
23%
(01:08)
wrong
based on 216
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A number, x is chosen at random from the set of positive integers less than 10. What is the probability that (9/x) > x?
(A) 1/5
(B) 2/9
(C) 1/3
(D) 2/3
(E) 7/9
(A) 1/5
(B) 2/9
(C) 1/3
(D) 2/3
(E) 7/9
01 Dec 2015, 06:12
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Number x has to be chosen from numbers 1-9
(9/x) > x
=> 9 > x^2
=>x^2 - 9 < 0
x can have 2 values only 1 , 2
Therefore , probability = 2 / 9
Answer B
(9/x) > x
=> 9 > x^2
=>x^2 - 9 < 0
x can have 2 values only 1 , 2
Therefore , probability = 2 / 9
Answer B
01 Dec 2015, 14:08
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probability (9/x) > x?
rewrite - 9/x > x i.e. 9 > x^2
means x should be such a number which when squared is less than 9.
x is one of the integers 1 to 9.
only 1 & 2 would be less than 9 when squared. so 2 out of 9 are favorable.
hence "probability that (9/x) > x" = 2/9
Option B is correct.
rewrite - 9/x > x i.e. 9 > x^2
means x should be such a number which when squared is less than 9.
x is one of the integers 1 to 9.
only 1 & 2 would be less than 9 when squared. so 2 out of 9 are favorable.
hence "probability that (9/x) > x" = 2/9
Option B is correct.
