Aptitude - Time and Work - Discussion
Discussion Forum : Time and Work - General Questions (Q.No. 8)
8.
A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:
Answer: Option
Explanation:
(A + B)'s 1 day's work = | 1 |
10 |
C's 1 day's work = | 1 |
50 |
(A + B + C)'s 1 day's work = | ![]() |
1 | + | 1 | ![]() |
= | 6 | = | 3 | . .... (i) |
10 | 50 | 50 | 25 |
A's 1 day's work = (B + C)'s 1 day's work .... (ii)
From (i) and (ii), we get: 2 x (A's 1 day's work) = | 3 |
25 |
![]() |
3 | . |
50 |
![]() |
![]() |
1 | - | 3 | ![]() |
= | 2 | = | 1 | . |
10 | 50 | 50 | 25 |
So, B alone could do the work in 25 days.
Discussion:
146 comments Page 4 of 15.
Jagan said:
6 years ago
Awesome method thanks @Awoke.
Amit Gaikwad said:
6 years ago
@Saurabh.
Thanks for the solution.
Thanks for the solution.
Ameer said:
6 years ago
Suppose A's one day work=1/x=(B+C) one day work.
A+B one day work=1/10,
C's one day work=1/50,
B's work=?
B's one day work= (A+B)'s one-day work-A's one day work.
=1/10-1/x.
Again, for B's one day work=(B+c)'s one-day work- c's one day work.
= 1/x-1/50.
by comparing these, we get;
1/10-1/x = 1/x-1/50,
by solving this we get x=50/3,
Hence A's one day work=3/50,
B's one day work=1/10-3/50,
B's one day work= 1/25,
So, B alone can do it in 25 days.
A+B one day work=1/10,
C's one day work=1/50,
B's work=?
B's one day work= (A+B)'s one-day work-A's one day work.
=1/10-1/x.
Again, for B's one day work=(B+c)'s one-day work- c's one day work.
= 1/x-1/50.
by comparing these, we get;
1/10-1/x = 1/x-1/50,
by solving this we get x=50/3,
Hence A's one day work=3/50,
B's one day work=1/10-3/50,
B's one day work= 1/25,
So, B alone can do it in 25 days.
Surinder kumar said:
6 years ago
a+b, +c can finished 50 work.
a+b finished in 5 days,
c finished in 1 day,
can say one day work (a+b+c) in 6 days,
a= b+c put,
a+(a) =6,
a=3.
a+b finished in 5 days,
c finished in 1 day,
can say one day work (a+b+c) in 6 days,
a= b+c put,
a+(a) =6,
a=3.
(1)
Kingsley said:
6 years ago
Easiest method:
A = B+C ---> (1)
A+B = 1/10 ---> (2)
C=1/50 ---> (3).
From (2) A = 1/10-B.
By substituting (2) & (3) in (1), we get;
1/10-B = B + 1/50.
Solving for B.
2B = 1/10 - 1/50 = 1/25.
A = B+C ---> (1)
A+B = 1/10 ---> (2)
C=1/50 ---> (3).
From (2) A = 1/10-B.
By substituting (2) & (3) in (1), we get;
1/10-B = B + 1/50.
Solving for B.
2B = 1/10 - 1/50 = 1/25.
(2)
Shahin said:
7 years ago
A = B+C.
A+B->10.
C->50.
We can write C as A-B.
Therefore, A-B->50.
LCM of 10 and 50 is 50.
As we know that efficiency x time = work ------> (1)
Here work = lcm of 10 and 50 = 50
Therefore the efficiency of A+B is 50/10 = 5(use formula 1)
The efficiency of C i.e A-B is 50/50 = 1,
Solving both equations we get A=3 B = 2.
Now time is taken by B is=50/2 =25 (by using formula 1).
So, the time taken by B alone is 25days.
A+B->10.
C->50.
We can write C as A-B.
Therefore, A-B->50.
LCM of 10 and 50 is 50.
As we know that efficiency x time = work ------> (1)
Here work = lcm of 10 and 50 = 50
Therefore the efficiency of A+B is 50/10 = 5(use formula 1)
The efficiency of C i.e A-B is 50/50 = 1,
Solving both equations we get A=3 B = 2.
Now time is taken by B is=50/2 =25 (by using formula 1).
So, the time taken by B alone is 25days.
Devesh said:
7 years ago
1/A+1/B=1/10 ------------> 1
GIVEN
1/A = 1/B + 1/C -------------> 2
1/A-1/B = 1/C from eq2.
1/A-1/B = 1/50.
Now solve 1 and
1/A=6/100 put this value in eqn 1.
We get 1/B = 4/100.
So, B can do in 25 days.
GIVEN
1/A = 1/B + 1/C -------------> 2
1/A-1/B = 1/C from eq2.
1/A-1/B = 1/50.
Now solve 1 and
1/A=6/100 put this value in eqn 1.
We get 1/B = 4/100.
So, B can do in 25 days.
Amjad said:
7 years ago
A + B = 10.
C = 50.
Take lcm of 10, 50 which will give total work.
So A+B=50/10=5 ------> (1)
C=50/50=1 -----> (2)
We also know B+C=A ----> (3)
On solving 1,2 & 3.
B=2 unit.
So work done b alone = 50/2 = 25.
C = 50.
Take lcm of 10, 50 which will give total work.
So A+B=50/10=5 ------> (1)
C=50/50=1 -----> (2)
We also know B+C=A ----> (3)
On solving 1,2 & 3.
B=2 unit.
So work done b alone = 50/2 = 25.
Swapnil Mhaske said:
7 years ago
Perfect @Shweta. Thanks.
Shweta Rathi said:
7 years ago
Since we are given with the condition -
A=B+C ------1 (eqn).
C's 1-day work = 1/50.
(A+B)'s 1-day work = 1/10.
Therefore, using eqn 1.
A+B = 2B+C,
1/10 = 2B+ 1/50,
2B=40/500,
B=1/25.
B alone can finish work in 25 days.
Hope this method will help you understand this question easily.
A=B+C ------1 (eqn).
C's 1-day work = 1/50.
(A+B)'s 1-day work = 1/10.
Therefore, using eqn 1.
A+B = 2B+C,
1/10 = 2B+ 1/50,
2B=40/500,
B=1/25.
B alone can finish work in 25 days.
Hope this method will help you understand this question easily.
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