Latest Elitmus Aptitude Question SOLUTION: 1/x+1/y=1/z where x=12 and y,z positive integer how many possible value of y and z ? a) 4 b) 7 C) 12 d) infinite

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(#M40155227) ELITMUS Numerical Ability question Algebra Keep an EYE

1/x+1/y=1/z where x=12 and y,z positive integer how many possible value of y and z ?
a) 4
b) 7
C) 12
d) infinite

Asked In Elitmus Amit Sankhyan (8 years ago)

Solved anjali gupta Read Solution (12)

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Read Solution (12) :

sahil aggarwal  (8 years ago)

I think the answer would be 7. Y=(12*Z)/(12 - Z). Hence, Z cannot be more than 12. The permissible values of z are 11,10, 9, 8, 6, 4, 3. Hence, the answer is 7. Correct me if I am wrong.

Is this solution Helpfull? Yes (32) | No (8)  

anjali gupta  (8 years ago) Best Solution

infinite values are possible

Is this solution Helpfull? Yes (12) | No (25)  

Gourav Khandelwal  (8 years ago)

INFINITE solution is possible, we can take the several values of of y in 1/12+1/y such that
1/12+1/y=1/z
12y/12+y=z
in this eq. we put positive integer of y and and on many positive integer value of y we will get the many positive value of z so infinite solutions

Is this solution Helpfull? Yes (8) | No (4)  

paras  (7 years ago)

1/x + 1/y = 1/z

or 1/y = 1/z - 1/12

or y = 12z / (12-z)

Solving for y and z ;

For, z = 3 , y = 4

For, z = 4 , y = 6

For, z = 6 , y = 12

For, z = 8 , y = 24

For, z = 9 , y = 36

For, z = 10 , y = 60

For, z = 11 , y = 120

Z can not excceed 11 as for z = 12 , y = infinite and for z > 12 y will be negative

Is this solution Helpfull? Yes (4) | No (1)  

Susmeeta Sau  (6 years ago)

Ans is 7
1/x + 1/y = 1/z

or 1/y = 1/z - 1/12

or y = 12z / (12-z)

Solving for y and z ;

For, z = 3 , y = 4

For, z = 4 , y = 6

For, z = 6 , y = 12

For, z = 8 , y = 24

For, z = 9 , y = 36

For, z = 10 , y = 60

For, z = 11 , y = 120

Z can not excceed 11 as for z = 12 , y = infinite and for z > 12 y will be negative

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Jinal  (8 years ago)

INFINITE VALUES:

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ROHIT KUMAR  (8 years ago)

Ans. (d)infinite
Explanation:
1/12+1/y=1/x
y+12/12y = 1/z
for every values of Y multiple of 12 i.e y= 12,24,36,48 and so Z will be an Integer

Is this solution Helpfull? Yes (2) | No (4)  

Neha Gupta  (7 years ago)

(c) 7 is the correct answer

y = (12*z)/(12-z) => value of z greater than 12 will result into negative value of y. therefore z cannot be greater than 12.

1/y= 1/z-1/12
=>for z=3 ,4,6,8,9,10,11 -------- 7 values
(for z=12 y=0 and 0 is not a positive integer and in question it asks for positive integer so this will not be considered)

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Himanshu Gupta  (6 years ago)

option b(7) is correct answer, infinite solutions are not possible as y and z are positive and integer

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satyaraghavendra reddy Medapati  (8 years ago)

for,
1/12=1/z-1/x and z

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Rizul  (8 years ago)

@Anjali Gupta
Can u pls explain how?

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Satyam  (5 years ago)

from the equation we get,

y = (12 * z) / (12 - z)

Hence , when z=12 the value of y is infinity and beyond z=12 i.e in z=13,14...and so on y becomes a
-ve integer which cannot be the case.
So correct ans is 12.
as the range of z is from 0 - 11.

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