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A number, x is chosen at random from the set of positive integers less

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Bunuel
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A number, x is chosen at random from the set of positive integers less [#permalink] New post   01 Dec 2015, 04:30
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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
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Re: A number, x is chosen at random from the set of positive integers less [#permalink] New post   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
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Re: A number, x is chosen at random from the set of positive integers less [#permalink] New post   01 Dec 2015, 14:08
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.
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Re: A number, x is chosen at random from the set of positive integers less [#permalink] New post   01 Dec 2015, 18:40
0<x<10

let x= 1
(9/x) = 9 >1...satisfy condition
let x= 2
(9/x) = 4.5>2...satisfy condition
let x= 3
(9/x) = 3 is not less tahn 3 ...not satisfying condition
Condition will not be satisfied for x>= 3, thus we can only select values 1 & 2

Thus probability = 2/9

Ans :B
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Re: A number, x is chosen at random from the set of positive integers less [#permalink] New post   25 Mar 2016, 20:55
A number, x is chosen at random from the set of positive integers less than 10. What is the probability that (9/x) > x?
Since x is a positive integer,
9>x^2 => x^2-9<0
-3<x<3
out of this only 1 and 2 satisfy the given constraints i.e. positive integer.

Total universe = 9 i.e. 1,2,3,4,5,6,7,8,9

Probability= favorable cases / total cases =2/9.
hence (B) 2/9.
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Re: A number, x is chosen at random from the set of positive integers less [#permalink] New post   24 May 2019, 08:11
Bunuel wrote:
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


Set- 1, 2, 3...9
Subset(x)- 1, 2.
Probability: 2/9 (Option B)

Interesting to note, 9 > x^2, this statement is valid only because we know x>0.
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Re: A number, x is chosen at random from the set of positive integers less [#permalink]
24 May 2019, 08:11
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