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Ssc Aspirant

· started a discussion

· 1 Months ago

solution is "1" (Question: 197117.0 )

X=2Y=3Z
=> 1/3X + 1/6Y + 1/9Z = 1/3
=> 1/9Z + 1/9Z + 1/9Z = 1/3
=> 3/9Z = 1/3
=> Z =1

Question:
If 3x = 9y = 27z and  \(\left ( \cfrac {1}{3x} + \cfrac{1}{6y} + \cfrac{1}{9z} \right ) = \cfrac{13}{2}\) , then find the value of z.
Options:
A) \(\cfrac{39}{2}\)
B) \(\cfrac{1}{39}\)
C) \(\cfrac{2}{39}\)
D) 1
Solution:
Ans: (c) 


Knowledge Expert

· commented

· 1 Months ago

Dear Student,

In this Question 1/3x+1/6y+1/9z = 13/2, where put the value x and y in terms of Z, using the X=2Y=3Z, after solving we get Z=2/39. try again.

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