Latest Elitmus Aptitude Question SOLUTION: What is the approx. value of W, if W=(1.5)*11,Given log2=0.301, log 3=.477. A) 68 B) 86 C) 105 D) 125

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What is the approx. value of W, if W=(1.5)*11,Given log2=0.301, log 3=.477.
A) 68 B) 86 C) 105 D) 125

Asked In Elitmus yatinkarel (10 years ago)

Solved RAKESH Read Solution (8)

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Read Solution (8) :

RAKESH  (10 years ago)

if W =(1.5)*11 then simply,W = 16.5


if, W=(1.5)^11
take log both sides
=> logW = 11*log(1.5) =11*log(3/2)
=> logW = 11(log3-log2)= 11*0.176 = 1.936
=> W = 10^(1.936)
=> W = 86 (approax.)

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ak  (10 years ago)

can anyone tell me how 10^(1.93) is calculated????????????

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JANARTHANAN RAM  (8 years ago)

Ans: 86

Log W=Log(1.5^11)=11*Log(3/2)=11*( Log(3)-Log(2) )
Log W=1.936
W=10^1.936
We know 10^2=100.
So 10^1.936 will be close to and less than 100.
So by options you can get the ans as 86)

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RAHUL KUMAR  (10 years ago)

w=1.5^11
logw=(log1-log2+log3)11
logw=(0-.301+.477)11
=(.176)11
=1.986
w=10^1.986
=86(apprx)

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Yogesh Khichi  (10 years ago)

86
take log both sides, 1.5 is 3/2
than, W=10^(log3-log2)

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Yogesh Khichi  (10 years ago)

hey, please explain the skyscraper problem in 8th june paper

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raushan kumar  (10 years ago)

log(base 10)w = log(3/2)*11= 1.93
w= 10^(1.93)= 86 (approx) [lof base(b)a =x
then a=b^x ]

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Suvankar Saha  (4 years ago)

Given that W=(1.5)^11 ,log2=.301,log=.477
Taking common log on both sides,
logW=log(1.5)^11
logW=11*log(1.5) since log(x) ^a= a*log(x)
logW=11*log(3/2) => 11*(log(3)-log(2))
since log(x/y)=logx-logy
logW=11*(.477-.301)=1.936
Taking antilog on both sides,
W=86.3 (approx).

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