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 6 of 15.
Logi said:
7 years ago
Thank you @Nagu.
Geeta vandana said:
7 years ago
@Awoke.
Your short cut method is nice. Thanks.
Your short cut method is nice. Thanks.
Imran said:
7 years ago
A+B=1/10. -----> (1)
A+B+C=(1/10 + 1/50) = 3/25.
A+A=3/25......(A = B+C).
A=3/50,
3/50 + B=1/10,
B=1/25,
B alone does it in 25 days.
A+B+C=(1/10 + 1/50) = 3/25.
A+A=3/25......(A = B+C).
A=3/50,
3/50 + B=1/10,
B=1/25,
B alone does it in 25 days.
Shree said:
8 years ago
Thank you for providing Clear and detailed explanation.
PRAVEEN. P said:
8 years ago
A+B=1/10.
C=1/50.
A+B+C=6/50.
But A=B+C.
B+C+B+C=6/50.
2B+2C=6/50.
But C=1/50.
On substitute c in above eqn we get as,
2B=4/50 and,
B=1/25,
Hence B=25 days.
C=1/50.
A+B+C=6/50.
But A=B+C.
B+C+B+C=6/50.
2B+2C=6/50.
But C=1/50.
On substitute c in above eqn we get as,
2B=4/50 and,
B=1/25,
Hence B=25 days.
Venkataramireddy said:
8 years ago
1 day work A+B=1/10.
A can do work same as do B+C.
so we can write it as
A=B+C.....
C 1day work =1/50,
1-day work of A+B+C=1/10+1/50
LCM 50.
5+1/50=6/50=3/25;
A+B+C=3/25,
above B+C=A.
So
A+A=3/25.
2A=3/25,
A=3/50,
substitute A value in A+B=1/10.
3/50+B=1/10,
B=1/10-3/50,
lcm 50,
B=5-3/50,
=2/50,
B=1/25,
B Alone 25 days.
A can do work same as do B+C.
so we can write it as
A=B+C.....
C 1day work =1/50,
1-day work of A+B+C=1/10+1/50
LCM 50.
5+1/50=6/50=3/25;
A+B+C=3/25,
above B+C=A.
So
A+A=3/25.
2A=3/25,
A=3/50,
substitute A value in A+B=1/10.
3/50+B=1/10,
B=1/10-3/50,
lcm 50,
B=5-3/50,
=2/50,
B=1/25,
B Alone 25 days.
Azhar said:
8 years ago
@Sneha best explanations.
(1)
Harsh said:
8 years ago
I have one doubt please help me.
I am doing the solution like this so answer is coming different please correct me.
let A's one day work is = 1\x,
B's=1\y,
c's=1\z,
1\x=1\y+1\z ..(i)
1\x+1\|y=1\10 ....given (ii)
1\z=1\50 ...(iii)
by solving all 3 equation we get,
1\10-2\y=1\50,
(y-20)\10y=1\50,
5(y-20)=y,
6y=100,
y=100\6.
I am doing the solution like this so answer is coming different please correct me.
let A's one day work is = 1\x,
B's=1\y,
c's=1\z,
1\x=1\y+1\z ..(i)
1\x+1\|y=1\10 ....given (ii)
1\z=1\50 ...(iii)
by solving all 3 equation we get,
1\10-2\y=1\50,
(y-20)\10y=1\50,
5(y-20)=y,
6y=100,
y=100\6.
Payal said:
8 years ago
B= x days
C = y days
A = x+y days
given: y= 50
& A +B = x+x+y = 2x+y=10.
=>x= 20 ans.
C = y days
A = x+y days
given: y= 50
& A +B = x+x+y = 2x+y=10.
=>x= 20 ans.
Shiva krishna kanaparthi said:
8 years ago
Total work = LCM of 10 and 50=50 units.
(A+B)'S 1 DAY WORK = 50/10= 5 UNITS PER DAY.
C'S 1 DAY WORK = 50/50= 1 UNIT PER DAY.
SINCE WE HAVE A=B+C AND A+B=5.
A-B=1----------(1)
A+B=5----------(2)
SOLVING ABOVE EQUATIONS WE GET B= 2 UNITS PER DAY.
SO NUMBER OF DAYS TO COMPLETE THE WORK FOR B= 50/2 = 25 DAYS.
(A+B)'S 1 DAY WORK = 50/10= 5 UNITS PER DAY.
C'S 1 DAY WORK = 50/50= 1 UNIT PER DAY.
SINCE WE HAVE A=B+C AND A+B=5.
A-B=1----------(1)
A+B=5----------(2)
SOLVING ABOVE EQUATIONS WE GET B= 2 UNITS PER DAY.
SO NUMBER OF DAYS TO COMPLETE THE WORK FOR B= 50/2 = 25 DAYS.
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