ssc
Exam
Numerical Ability
Time and Work
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
Read Solution (Total 4)
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- 1/a+1/b=1/12
1/b+1/c=1/15
given 2a=c
put the value of c in eqn 2 we get
1/b+1/2a=1/15
then solve the eqn 1 & eqn 3 we get
a=30 b=20 c 60 ans is 20 days alone finish the work
- 10 years agoHelpfull: Yes(5) No(1)
- let say b will complete work in x days..
den b's one day work will be considered as 1/x..
A's one day work =1/12 -1/x.
similarly C's one day work =1/15 -1/x
from question it is given that A's work =2 times (C's work)..
(1/12 -1/x)=2(1/15 -1/x)..
solve u will get x=20..(x is no of B's days to complete work) - 9 years agoHelpfull: Yes(1) No(0)
- A and B together work in 12days i.e;A+B=12
so A and B work in 1 day can be written as A+B=1/12-->(1)
B and C work in 15 days i.e;B+C=15
B and C work together in one day is B+C=1/15-->(2)
But acording to que A,s Efficiency is Twice that of C so A=2C-->(3)
when we submit A=2C in the place of A,Then we modify the eqa(1) as
2C+B=1/12-->(4)
by solving Eqations (2) and (4)
we get C=1/60,B=1/20,A=1/30(these values are for one day)
so B required 20 days to finish the work
- 10 years agoHelpfull: Yes(0) No(0)
- 1/a + 1/b = 1/12
1/b + 1/c = 1/15
1/a = 2*1/c => 2a = c
after solving we get
a = 30, b=20, c = 60
so, B alone finish in 20 days - 4 years agoHelpfull: Yes(0) No(0)
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