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 4 of 15.
SubhamChand said:
9 years ago
You can solve it in a very easy way.
A = A + B.
C's 1day work = 1/50.
(A+B) 's 1day work = 1/10.
But according to the question.
A = B + C.
=> A + B = B + B + C (add B on both sides).
=> 1/10 = 2B + 1/50.
Then B = 1/25.
So, B can do it in 25 days.
A = A + B.
C's 1day work = 1/50.
(A+B) 's 1day work = 1/10.
But according to the question.
A = B + C.
=> A + B = B + B + C (add B on both sides).
=> 1/10 = 2B + 1/50.
Then B = 1/25.
So, B can do it in 25 days.
Sneha said:
2 decades ago
We can explain much easier than this:
Here we know A = B + C
So in the equation A + B + C = 3/25, substitute B + C for A.
We get 2(B + C) = 3/25 --->2A
Hence A's 1 day work is = 3/25*2 = 3/50
Then do the remaining part and find work done by B.
Here we know A = B + C
So in the equation A + B + C = 3/25, substitute B + C for A.
We get 2(B + C) = 3/25 --->2A
Hence A's 1 day work is = 3/25*2 = 3/50
Then do the remaining part and find work done by B.
Dewesh Kumar said:
3 years ago
A = B+C ,
A - C = B ---> (i).
Then,
A + B + C = 1/10 + 150 = 3/25.
From (i) putting a value of B
A+B+C= 3/25,
A+A - C + C = 3/25.
2A = 3/25.
A = 3/50.
Now, A+B = 1/10,
Putting A = 3/50,
B = 1/10-3/50.
B = 1/25 (B one day work) then total days is 25.
A - C = B ---> (i).
Then,
A + B + C = 1/10 + 150 = 3/25.
From (i) putting a value of B
A+B+C= 3/25,
A+A - C + C = 3/25.
2A = 3/25.
A = 3/50.
Now, A+B = 1/10,
Putting A = 3/50,
B = 1/10-3/50.
B = 1/25 (B one day work) then total days is 25.
(110)
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.
Prathiba said:
4 months ago
a = b+ c ---> (1)
a+ b = 10 = 1/10 for 1 day work ---> (2)
c = 50 = 1/50 for 1 day work
b = 1/10 - a ---> (3)---from eqn(1)
c = 1/50.
a = b +c ---from (1)
a = 1/10 -a + 1/50 ---from (3)
a = 3/50.
b = 1/10 - a = 1/10 -3/50 = 1/25 .
b = 25 days.
a+ b = 10 = 1/10 for 1 day work ---> (2)
c = 50 = 1/50 for 1 day work
b = 1/10 - a ---> (3)---from eqn(1)
c = 1/50.
a = b +c ---from (1)
a = 1/10 -a + 1/50 ---from (3)
a = 3/50.
b = 1/10 - a = 1/10 -3/50 = 1/25 .
b = 25 days.
(9)
Awoke said:
2 decades ago
Short cut method:
A + B = 1/10 but A = B + C, then
A + B = B + B + C = 1/10 and C = 1/50
2B + 1/50 = 1/10 gives us
B = 1/2(1/10-1/50)
= 1/2(2/25)
= 1/25
Therefore, B can do the work alone in 25 days.
A + B = 1/10 but A = B + C, then
A + B = B + B + C = 1/10 and C = 1/50
2B + 1/50 = 1/10 gives us
B = 1/2(1/10-1/50)
= 1/2(2/25)
= 1/25
Therefore, B can do the work alone in 25 days.
(1)
Stevan said:
4 years ago
A = B + C,
A+B = 10 or A and B's one day work A+B=1/10.
C=50 or C=1/50.
A+B=1/10 (Since A=B+C),
B+C+B=1/10,
2B+1/50=1/10,
2B=1/10-1/50,
2B=2/25,
B=2/25*1/2.
B=1/25 (Which is one day's work).
B would take 25 days to complete certain work.
A+B = 10 or A and B's one day work A+B=1/10.
C=50 or C=1/50.
A+B=1/10 (Since A=B+C),
B+C+B=1/10,
2B+1/50=1/10,
2B=1/10-1/50,
2B=2/25,
B=2/25*1/2.
B=1/25 (Which is one day's work).
B would take 25 days to complete certain work.
(3)
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.
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.
Aswathi Radhakrishnan said:
4 years ago
Total work is = 50
C's efficiency = 50/50=1.
A+B's Efficiency = 50/10=5.
A+B+C's Efficiency = 5 + 1 = 6.
A = B+C,
B+C+B+C=6,
2(B+C)=6,
B+C= 6/2 =3,
A's Efficiency = 3,
So B's Efficiency =5-3 =2.
Hence B alone do it in 50/2 = 25 Days.
C's efficiency = 50/50=1.
A+B's Efficiency = 50/10=5.
A+B+C's Efficiency = 5 + 1 = 6.
A = B+C,
B+C+B+C=6,
2(B+C)=6,
B+C= 6/2 =3,
A's Efficiency = 3,
So B's Efficiency =5-3 =2.
Hence B alone do it in 50/2 = 25 Days.
(11)
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