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 14 of 15.
Sun said:
1 decade ago
A+b=10..2
c=1/50
sub (1) in (2)
2b+c=1/10
2b+1/50=1/10
2b=1/10-1/50
2b=5-1/50
2b=4/50
2b=2/25
b=25...
Hence proved
c=1/50
sub (1) in (2)
2b+c=1/10
2b+1/50=1/10
2b=1/10-1/50
2b=5-1/50
2b=4/50
2b=2/25
b=25...
Hence proved
Ankit said:
1 decade ago
@Awoke
You are the best.
You are the best.
Jay said:
1 decade ago
@sneha.
Your explanation is so simple, thanks a lot.
Your explanation is so simple, thanks a lot.
Hilal said:
1 decade ago
B's work=x
C's work=y
hence A's work=x+y
A and B together work= 2x+y
given y=1/50
substituting x=1/25
C's work=y
hence A's work=x+y
A and B together work= 2x+y
given y=1/50
substituting x=1/25
Aunjani Kumar said:
1 decade ago
From (i) and (ii), we get: 2 x (A's 1 day's work) = 3/25
Dear i am not able to understand why 2 multiplied with A's day's work??????
Dear i am not able to understand why 2 multiplied with A's day's work??????
Nishant said:
1 decade ago
LET B+C CAN FINISH WORK IN T DAYS
THEN B+C'S 1 day work=1/T.
then,
A's 1 day work=1/T.
so,
B+C's 1 day work=1/T.
now,(A + B)'s 1 day's work =1/10.
then B's 1 day work=1/10-1/T.--------(1)
and C's 1 day's work =1/50.
hence,B's 1 day work=i/T-1/50.-------(2)
now,
from (1) nd (2)
T = 25 days.
then B=50 days.
how it is possible????
THEN B+C'S 1 day work=1/T.
then,
A's 1 day work=1/T.
so,
B+C's 1 day work=1/T.
now,(A + B)'s 1 day's work =1/10.
then B's 1 day work=1/10-1/T.--------(1)
and C's 1 day's work =1/50.
hence,B's 1 day work=i/T-1/50.-------(2)
now,
from (1) nd (2)
T = 25 days.
then B=50 days.
how it is possible????
Kinnari said:
1 decade ago
I like your method Mr. Awoke. :)
Nitesh Nandwana said:
1 decade ago
@Awoke
Your method is best out of all explanation. Good job, keep it up.
Your method is best out of all explanation. Good job, keep it up.
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.
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)
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