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 5 of 15.
Shanu Verma said:
1 decade ago
A = B+C-----(1).
A+B = 1/10------(2).
C = 1/50-----(3).
B+B+C = 1/10 (put value of A from equation 1 in equation 2).
2B+C = 1/10 (as C = 1/50).
2B = 1/10-1/50 = 4/50 = 2/25.
B = 2/50 = 1/25.
B alone could do the work in 25 Days.
A+B = 1/10------(2).
C = 1/50-----(3).
B+B+C = 1/10 (put value of A from equation 1 in equation 2).
2B+C = 1/10 (as C = 1/50).
2B = 1/10-1/50 = 4/50 = 2/25.
B = 2/50 = 1/25.
B alone could do the work in 25 Days.
Atyanand said:
10 years ago
A = B+C......(1).
A+B = 1/10......(2).
C = 1/50......(3).
So work done by A, B, C is.
A+B+C = 1/10 +1/50 = 6/50.
From equation 2.
A+A = 6/50.
A = 3/50.
So B = 1/10-3/50 = 2/50 = 1/25.
So B complete its work in 25 days.
A+B = 1/10......(2).
C = 1/50......(3).
So work done by A, B, C is.
A+B+C = 1/10 +1/50 = 6/50.
From equation 2.
A+A = 6/50.
A = 3/50.
So B = 1/10-3/50 = 2/50 = 1/25.
So B complete its work in 25 days.
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.
Tushar said:
4 years ago
1/A+1/B=1/10 ---> EQN1.
Given that,
1/C = 1/50,
1/A=1/B+1/C ---> EQN2.
From EQN1 and EQN2 we get,
1/B + 1/B+1/C = 1/10,
2/B + 1/C = 1/10,
2/B + 1/50 = 1/10,
2/B = 2/25,
1/B = 1/25.
Hence B alone can do it in 25 Days.
Given that,
1/C = 1/50,
1/A=1/B+1/C ---> EQN2.
From EQN1 and EQN2 we get,
1/B + 1/B+1/C = 1/10,
2/B + 1/C = 1/10,
2/B + 1/50 = 1/10,
2/B = 2/25,
1/B = 1/25.
Hence B alone can do it in 25 Days.
(4)
PRASAD said:
8 years ago
A'S 1 DAY WORK=(B+C) 1 DAY WORK.
(A+B)'S 1 DAY WORK=1/10 =>A'S 1 DAY WORK (1/10-B).
C'S 1 DAY WORK=1/50.
SO,
A'S 1 DAY WORK=(B+C) 1 DAY WORK.
1/10-B=B+1/50.
1/10-1/50=2B.
50-10/500=2B.
1/25=B.
B=25 DAYS (ANS....).
(A+B)'S 1 DAY WORK=1/10 =>A'S 1 DAY WORK (1/10-B).
C'S 1 DAY WORK=1/50.
SO,
A'S 1 DAY WORK=(B+C) 1 DAY WORK.
1/10-B=B+1/50.
1/10-1/50=2B.
50-10/500=2B.
1/25=B.
B=25 DAYS (ANS....).
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.
Rakesh sonowal said:
1 decade ago
A=B+C (1)
A+B 1 Day work=1/10 (2)
Cs 1 Day work=1/50 (3)
putin eq (1) n (3) in eq (2)
2B +1/50=1/10
2B=1/10-1/50
2B=(5-1)/50
2B=4/50
B=4/50 =1/25
So Bs 1 day work =1/25.
So B alone could do d work in 25 days.
A+B 1 Day work=1/10 (2)
Cs 1 Day work=1/50 (3)
putin eq (1) n (3) in eq (2)
2B +1/50=1/10
2B=1/10-1/50
2B=(5-1)/50
2B=4/50
B=4/50 =1/25
So Bs 1 day work =1/25.
So B alone could do d work in 25 days.
Sa_1 said:
1 decade ago
@Serina.
Its just LCM nothing else. Simple method.
A + B + C = 3/25.
Now A and B + C together time are same.
A = B + C.
A + B + C = 3/25.
B + C = 3/25 - A.
So we can write
A = 3/25 - A.
2A = 3/25.
A = 3/50.
Its just LCM nothing else. Simple method.
A + B + C = 3/25.
Now A and B + C together time are same.
A = B + C.
A + B + C = 3/25.
B + C = 3/25 - A.
So we can write
A = 3/25 - A.
2A = 3/25.
A = 3/50.
ESHWAR said:
10 years ago
A = B+C.
Given A+B 1 day work = 1/10;
C 1 day work = 1/50;
Add B on both sides.
A+B = 2B+C;
1/10 = 2B+1/50;
2B = 1/10-1/50;
2B = 4/50;
B = 2/50;
So B's 1 day work is 1/25 and B can finish work in 25 days.
Given A+B 1 day work = 1/10;
C 1 day work = 1/50;
Add B on both sides.
A+B = 2B+C;
1/10 = 2B+1/50;
2B = 1/10-1/50;
2B = 4/50;
B = 2/50;
So B's 1 day work is 1/25 and B can finish work 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.
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