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yatin kaushik

· started a discussion

· 1 Months ago

taking 2 holes (Question: 273472.0 )

16/93 is the right answer.

Question:

A rectangular field is 22 m long and 10 m wide. Two hemispherical pit holes of radius 2 m are dug at two places and the mud is spread over the remaining part of the field. The rise in the level of the field is –

Options:
A)

893    m

B)

\(\frac{16}{93}\)  m

C)

\(\frac{352}{177}\)   m

D)

2393    m

Solution:

 

manchester united

· commented

· 1 Months ago

We havent paid u for ur 'MINOR TYPOS' moreover ur so called typos can ruin an aspirants life who has sacrificed almost everything to excel in this exam.

Knowledge Expert

· commented

· 1 Months ago

Dear Student we apologize for the wrong explanation
h=704/(3*1188)
In solution
1188 is printed as 118 in the final step
Which is a minor typo

Knowledge Expert

· commented

· 1 Months ago

Dear Student
Please share with us your method and solution
The mud extracted out of the two hemispherical holes is then spread over upon the remaining area resulting in the overall increment of height.
Mud just didn't came from nowhere, but was extracted out from holes. THus the overall volume of holes is contributing in increment of height.
Thanks and Regards
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