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:
147 comments Page 7 of 15.
Alok said:
1 decade ago
a = b+c....(i).
a+b = 1/10 (given).
a-b = c.
a-b = 1/50 (c alone can do work in 50 days putting the value of c here)....(ii).
Solving I and II.
a = 1/30.
b = 1/10-1/30.
c = 1/15.
How this one is wrong?
a+b = 1/10 (given).
a-b = c.
a-b = 1/50 (c alone can do work in 50 days putting the value of c here)....(ii).
Solving I and II.
a = 1/30.
b = 1/10-1/30.
c = 1/15.
How this one is wrong?
SYAM said:
1 decade ago
A* = B+C.
A+B = 10 days.
C = 50 days.
C is 1 day of work = 1/50.
A+B 1 day of work = 1/10.
B = 1/10-A.
B is 1 day of work = 1/10-A*.
= 1/10 - (B+C).
= 1/10 - (B+1/50).
2B = 1/10-1/50.
B = 4/100.
= 1/25.
So, 25 is the answer.
A+B = 10 days.
C = 50 days.
C is 1 day of work = 1/50.
A+B 1 day of work = 1/10.
B = 1/10-A.
B is 1 day of work = 1/10-A*.
= 1/10 - (B+C).
= 1/10 - (B+1/50).
2B = 1/10-1/50.
B = 4/100.
= 1/25.
So, 25 is the answer.
Ganesh said:
1 decade ago
How you got 6/50?
Please someone explain.
Please someone explain.
Deepika said:
1 decade ago
By adding a+b+c i.e. 1/10+1/50.
Esayas said:
1 decade ago
If (a+b) = 1/10 and a = b+c, so 2b++c = 1/10.
2b+1/50 = 1/10.
b = 1/25.
So b can finish the work in 25 days.
2b+1/50 = 1/10.
b = 1/25.
So b can finish the work in 25 days.
ESHWAR said:
1 decade 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.
Atyanand said:
1 decade 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.
Harsha said:
1 decade ago
A -> B+C (given).
A+B -> 10.
C -> 50.
Taking L.C.M we get 50.
So A+B -> 10 (5) because 10*5 = 50.
C -> 50 (1) because 50*1 = 50.
A+B+C = 6.
As B+C can be replaced with A.
A+A = 6.
2A = 6.
A = 3.
Then B = 5-3 = 2.
So, 50/2 = 25 days answer.
A+B -> 10.
C -> 50.
Taking L.C.M we get 50.
So A+B -> 10 (5) because 10*5 = 50.
C -> 50 (1) because 50*1 = 50.
A+B+C = 6.
As B+C can be replaced with A.
A+A = 6.
2A = 6.
A = 3.
Then B = 5-3 = 2.
So, 50/2 = 25 days answer.
Saurabh said:
10 years ago
Avoid fractions! they mess up the calculations.
Solution:
Total amount of work = LCM (10, 50) = 50 units.
Given: A's time = (B+C)'s time.
Hence, A's rate = B's rate + C's rate.
(Rate is the rate of doing work, not the monetary rate).
(A+B)'s time = 10 days.
Hence A's rate + B's rate = 50/10 = 5 units/day.
C's time = 50 days.
Hence C's rate = 50/50 = 1 unit/day.
Hence, A's rate + B's rate + C's rate = 5 + 1 = 6 units/day.
Now, using A's rate = B's rate + C's rate,
We get A's rate = 6/2 = 3 units/day.
Hence B's rate = 5-3 = 2 units/day.
Hence B's time = Total work/B's rate = 50/2 = 25 days.
Solution:
Total amount of work = LCM (10, 50) = 50 units.
Given: A's time = (B+C)'s time.
Hence, A's rate = B's rate + C's rate.
(Rate is the rate of doing work, not the monetary rate).
(A+B)'s time = 10 days.
Hence A's rate + B's rate = 50/10 = 5 units/day.
C's time = 50 days.
Hence C's rate = 50/50 = 1 unit/day.
Hence, A's rate + B's rate + C's rate = 5 + 1 = 6 units/day.
Now, using A's rate = B's rate + C's rate,
We get A's rate = 6/2 = 3 units/day.
Hence B's rate = 5-3 = 2 units/day.
Hence B's time = Total work/B's rate = 50/2 = 25 days.
(1)
Sarumathi.S said:
10 years ago
What is the logic in the statement 2* (A's one day work).
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