Latest Elitmus Aptitude Question SOLUTION: a and b together can do a work in 12 days.b and c together do it in 15 days.if a's efficiency is twice that of c,then the days required for b alone to finish the work is how many

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(#M40164076) ELITMUS Numerical Ability question Keep an EYE

a and b together can do a work in 12 days.b and c together do it in 15 days.if a's efficiency is twice that of c,then the days required for b alone to finish the work is how many days?

Asked In Elitmus Sniper (7 years ago)

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Devendra Marghade  (7 years ago)

20 days

1/a + 1/b = 1/12 ---(i)
1/b + 1/c = 1/15 ---(ii)
As a's efficiency is twice that of c, so days required by a is half that of c => a=c/2 or c=2a
Substituting c=2a in (ii), 1/b + 1/2a = 1/15 ---(iii)
On subtracting (iii) from (ii), 1/a=1/30 , so 1/b=1/20 and 1/c=1/60
Hence b alone requires 20 days to finish the work.

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vikalp pandit  (6 years ago)

A+B= 1/ 12------1
A=2C--------------2
FROM 1&2
2C+B=1/12
B=1/12-2C---------3
NOW,B+C= 1/15
(1/12-2C) + C= 1/15
-C= 1/15 -1/12
- C= -1/60
C=1/60
PUTTING THIS VALUE IN B+C= 1/15
B+1/60=1/15
B=1/20
Therefore B finishes the work in 20 days
Ans ->20 days

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jitendra idpache  (6 years ago)

ans : 60 days

one day work of a & b .......................1/(a+b)= 1/12
one day work of b & c .........................1/(b+c) = 1/15
therefore .............one day work of a & c ....................1/(a+c)= [1/(a+b)] - [1/(b+c)] =1/60
given a=2c
[1/(a+c)] = 1/3c=1/60
so 1/c= 1/20 ( c takes 20 days to finish the work)
and [1/(b+c)] - 1/c = 1/60 ( b takes 60 days to finish the same work)

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