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 9 of 15.
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.
Esayas said:
10 years 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.
Deepika said:
10 years ago
By adding a+b+c i.e. 1/10+1/50.
Ganesh said:
1 decade ago
How you got 6/50?
Please someone explain.
Please someone explain.
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.
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?
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.
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.
Serina said:
1 decade ago
How did you get 3/25, if you add 1/10 + 1/50 please explain in brief?
Suresh said:
1 decade ago
a+b = 1/10.
a = b+c eq.1.
c = 1/50.
Then a-b = 1/50 eq.2.
Solving 1 and 2 we get b = 1/25.
a = b+c eq.1.
c = 1/50.
Then a-b = 1/50 eq.2.
Solving 1 and 2 we get b = 1/25.
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