Latest Others Aptitude Question SOLUTION: Let (10+x)/(110+x) = (20+y)/(120+y) = (30+z)/(130+z) = 1/n, where x, y, z and n are positive integers. The number of distinct possible value of n is (1) 2 (2) 4 (3) 3 (4) 1 (5)

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(#M40007734) OTHERS question values of n Keep an EYE

Let (10+x)/(110+x) = (20+y)/(120+y) = (30+z)/(130+z) = 1/n, where x, y, z and n are positive integers.

The number of distinct possible value of n is

(1) 2 (2) 4 (3) 3 (4) 1 (5) none of these

Asked In Others Sukham (12 years ago)

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Pargat Pal Singh  (12 years ago)

(1) 2

two values of n are possible .
when n=2, x=90, y=80, z= 70
when n=3, x=40, y=30, z=20

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Rakesh Kumar Kohli  (12 years ago)

Ans.(1) 2
Values of x, y and z should be such that on simplifying fractions
(10+x)/(110+x), (20+y)/(120+y) and (30+z)/(130+z)we should get 1 in the numerator. For this possible values are
(x=90, y=80, z= 70) giving n = 2
and (x=40, y=30, z=20) giving n = 3
Thus number of distinct possible values of n are 2. Hence option (1) is correct

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Sukham  (12 years ago)

correct.
two values of n are possible .

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