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 |
 A's 1 day's work = |
3 | . |
| 50 |
 B's 1 day's work |
|
1 | - | 3 | |
= | 2 | = | 1 | . |
| 10 | 50 | 50 | 25 |
So, B alone could do the work in 25 days.
Discussion:
146 comments Page 11 of 15.
Akil Prakash said:
7 years ago
Thanks for sharing the shortcut method @Awoke.
Gaurav said:
7 years ago
Good, thanks @Shivangi Sharma.
Pavan said:
7 years ago
A = B+C.
A+B = 1/10.
C = 1/50.
A+B+C = 6/50.
we know that A = B+C.
B+C+B+C = 6/50,
B = 1/25,
B takes 25 days to complete the work.
A+B = 1/10.
C = 1/50.
A+B+C = 6/50.
we know that A = B+C.
B+C+B+C = 6/50,
B = 1/25,
B takes 25 days to complete the work.
Bala said:
7 years ago
A=B+C;
IF A+B=1/10,
then.
(B+C)+B=1/10,
c=1/50,
2B+C=1/10,
2B+1/50=1/10,
2B=1/10-1/50,
2B=2/25,
B=1/25.
IF A+B=1/10,
then.
(B+C)+B=1/10,
c=1/50,
2B+C=1/10,
2B+1/50=1/10,
2B=1/10-1/50,
2B=2/25,
B=1/25.
Viknesh said:
7 years ago
Here; C's work A+C = 1/4+1/x = 1/2, 1/4-1/2 =1/4,
B+C 1/3-1/4,
4-3/12 = 12days.
B+C 1/3-1/4,
4-3/12 = 12days.
Swarnali said:
7 years ago
Thanks for the answer @Bala.
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.
Swapnil Mhaske said:
7 years ago
Perfect @Shweta. Thanks.
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.
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.
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