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.
GAURAV said:
1 decade ago
A = B+C => A-B = C.
A+B = 1/10.
C = 1/50.
=> A-B = 1/50.
Solving eqns we get B = 25.
A+B = 1/10.
C = 1/50.
=> A-B = 1/50.
Solving eqns we get B = 25.
Pavan R B said:
1 decade ago
Given 1/A = (1/B)+(1/C)...EQN1.
AND (1/A)+(1/B) = 1/10...EQN2.
ALSO 1/C = 1/50.
.
. . 1/A = (1/B)+(1/50).
SUBSTITUTE ABOVE EQN IN EQN2.
(1/B)+(1/50)+(1/B)=1/10.
THEREFORE 1/B=1/25, B ALONE CAN DO IN 25 DAYS.
AND (1/A)+(1/B) = 1/10...EQN2.
ALSO 1/C = 1/50.
.
. . 1/A = (1/B)+(1/50).
SUBSTITUTE ABOVE EQN IN EQN2.
(1/B)+(1/50)+(1/B)=1/10.
THEREFORE 1/B=1/25, B ALONE CAN DO IN 25 DAYS.
Sandhya said:
1 decade ago
Given A = B+C ------ Eq.1.
& A+B = 1/10.
& C = 1/50.
NOW, A+B+C = (1/10+1/50).
= 3/25.
But in the question he is asking only 'B',
To get B we have to know the individual of A.
So,
SINCE B+C = A (given).
A+B+C = 3/25;.
A+A = 3/25;.
2A = 3/25.
A = 3/50.
HERE WE KNOW A = 3/50; C = 1/10 then.
A = B+C; 3/50 = B+1/50;.
B = 3/50 - 1/50;.
B = 2/50;.
B = 1/50;.
Therefore B can only do in 25 days.
& A+B = 1/10.
& C = 1/50.
NOW, A+B+C = (1/10+1/50).
= 3/25.
But in the question he is asking only 'B',
To get B we have to know the individual of A.
So,
SINCE B+C = A (given).
A+B+C = 3/25;.
A+A = 3/25;.
2A = 3/25.
A = 3/50.
HERE WE KNOW A = 3/50; C = 1/10 then.
A = B+C; 3/50 = B+1/50;.
B = 3/50 - 1/50;.
B = 2/50;.
B = 1/50;.
Therefore B can only do in 25 days.
Rajni Bala said:
1 decade ago
A=B+C (Given).
Adding A both sides.
2 A = A+B+C.
Now A+B+C 's one day work = 3/25.
On putting these values we get,
A= 3/50 One day's work.
Use the value of A and get the value of B.
So the Answer is 25.
Adding A both sides.
2 A = A+B+C.
Now A+B+C 's one day work = 3/25.
On putting these values we get,
A= 3/50 One day's work.
Use the value of A and get the value of B.
So the Answer is 25.
Naani said:
1 decade ago
Given that A = B+C.....(1).
A&B completes the work in 10 days so A&B one day work is 1/10.
A+B = 1/10
A = (1/10)-B.
(1/10)-B = B+C..........from(1).
Given that C alone complete the work in 50 days so C's one day's work is = 1/50.
Therefore.
(1/10)-B = B+(1/50).
2B = 4/50.
B = (1/25) works.
So B alone can complete work in 25 days.
A&B completes the work in 10 days so A&B one day work is 1/10.
A+B = 1/10
A = (1/10)-B.
(1/10)-B = B+C..........from(1).
Given that C alone complete the work in 50 days so C's one day's work is = 1/50.
Therefore.
(1/10)-B = B+(1/50).
2B = 4/50.
B = (1/25) works.
So B alone can complete work in 25 days.
Seema duhan said:
1 decade ago
Given, A = B+C.
A+B = 1/10.
C = 1/50.
So, A+B+C = 1/10 + 1/50 = 6/50 = 3/25.
A = B+C = 1/2 of A+B+C.
= 1/2(3/25) = 3/50.
B+C = 3/50.
Where C=1/50.
So, B+1/50 = 3/50.
B = 2/50 = 1/25.
So B alone can do the work in 25 days.
A+B = 1/10.
C = 1/50.
So, A+B+C = 1/10 + 1/50 = 6/50 = 3/25.
A = B+C = 1/2 of A+B+C.
= 1/2(3/25) = 3/50.
B+C = 3/50.
Where C=1/50.
So, B+1/50 = 3/50.
B = 2/50 = 1/25.
So B alone can do the work in 25 days.
Palkin singla said:
1 decade ago
Let work done by A = X.
Same work is also done by B+C = X.
A = X.
B+C = X.
C=1/50 ---->[1].
A+B = 1/10 ---->[2].
Putvalue of a in 2nd eqn and C = X - B in 1st eqn.
We get two equation.
B+X = 1/10.
X-B = 1/50.
By solving we get B = 1/25.
Which is our requirement.
Have a nice day.
Same work is also done by B+C = X.
A = X.
B+C = X.
C=1/50 ---->[1].
A+B = 1/10 ---->[2].
Putvalue of a in 2nd eqn and C = X - B in 1st eqn.
We get two equation.
B+X = 1/10.
X-B = 1/50.
By solving we get B = 1/25.
Which is our requirement.
Have a nice day.
EKTA said:
1 decade ago
1/A=1/B+1/C......I.
1/A+1/B=1/10...II.
1/C=1/50.....III.
PUT 1/A=1/B+1/C IN EQ II.
1/B+1/C+1/B=1/10
2/B+1/50=1/10
2/B=10-1/50
2/B=4/50
1/B=1/25
B=25.
1/A+1/B=1/10...II.
1/C=1/50.....III.
PUT 1/A=1/B+1/C IN EQ II.
1/B+1/C+1/B=1/10
2/B+1/50=1/10
2/B=10-1/50
2/B=4/50
1/B=1/25
B=25.
The great RAMANUJAN said:
1 decade ago
Here comes the simplest method:
GIVEN eq. 1/A=1/B + 1/C
Adding 1/B on both sides
1/A + 1/B = 1/B + 1/B + 1/C
But 1/A + 1/B = 1/10 and 1/C=1/50
So.. 1/10 = 2/B + 1/50
2/B = 2/25
1/B = 1/25
B = 25.
So B alone could do the work in 25 days.
GIVEN eq. 1/A=1/B + 1/C
Adding 1/B on both sides
1/A + 1/B = 1/B + 1/B + 1/C
But 1/A + 1/B = 1/10 and 1/C=1/50
So.. 1/10 = 2/B + 1/50
2/B = 2/25
1/B = 1/25
B = 25.
So B alone could do the work in 25 days.
MITUL said:
1 decade ago
A+B=1/10, C=1/50
A=B+C
> 1/10-B=B+C
> 2B=4/50
> B=1/25
SO B alone could do the work in 25 days.
A=B+C
> 1/10-B=B+C
> 2B=4/50
> B=1/25
SO B alone could do the work in 25 days.
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