SSC CHSL Model Paper 9

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Question : 35 of 100

Marks: +1, -0

If

2^x= 3^y =6^{-z}

then

(1/x+1/y+1/z)

, is equal

0
1
$3/2$
$-1/2$

Solution:

2^x = 3^y = 6^{-z}=k

⇒

2=k^{1/x};3=k^{1/y};6=k^{-1/z}

∵

2×3 = 6

⇒

k^{1/x}×k^{1/y}=k^{-1/z}⇒k^{1/x+1/y}=k^{-1/z}

⇒

1/x+1/y=-1/z⇒1/x+1/y+1/z=0

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