Latest Others Aptitude Question SOLUTION: If 2x = 3y = 6-z, then (1/x + 1/y + 1/z) is equal to:

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(#M40037713) OTHERS question Keep an EYE

If 2x = 3y = 6-z, then (1/x + 1/y + 1/z)
is equal to:

Asked In Others Srishti Sinha (9 years ago)

Unsolved Read Solution (3)

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Read Solution (3) :

Anish Mathur  (9 years ago)

Solution:
2x = 6 -z Þ 2 = 6-z/x -------------(i)
3y = 6 -z Þ 3 = 6-z/y -------------(ii)
Multiplying (i)and(ii)
2x3 = 6-z/x x 6-z/y
6 = 6(1-z/x) *6(1-z/y)
1 = (-z/x) +(-z/y)
1= -z(1/x +1/y)
1/-z = 1/x +1/y
1/x + 1/y + 1/z = 0

Is this solution Helpfull? Yes (7) | No (6)  

AMRUTHA   (9 years ago)

0
2x = 6 -z then 2 = 6-z/x -------------(i)
3y = 6 -z then 3 = 6-z/y -------------(ii)
Multiplying (i)and(ii)
2x3 = 6-z/x x 6-z/y
61 = 6-z/x +-z/y
1 = (-z/x) +(-z/y)
1= -z(1/x +1/y)
1/-z = 1/x +1/y
1/x + 1/y + 1/z = 0

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

SAHIL  (6 years ago)

let 2^x = 3^y = 6^z=k

so , 2 = k^(1/x)
3 = k^(1/y)
6 = k^(-1/z)

since , 2 * 3 = 6

so, 2 * 3 = k ^ (1/x) * k ^ (1/y) = k ^ [ (1/x) + (1/y) ]

also , 6=k ^ ( -1/z )

therefore , k ^ [ (1/x) + (1/y) ]=k ^ ( -1/z )

so [ (1/x) + (1/y) ] = ( -1/z )

==> (1/x) + (1/y) -( -1/z ) = 0

so (1/x) + (1/y) + (1/z ) = 0 ans.

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