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.
Dinesh said:
1 decade ago
@AWOKE.
I liked your method. Thank you.
I liked your method. Thank you.
Piyush said:
1 decade ago
Hey there is also a easy way..
A'1 day work = (B+C)'1 day work
(A+B) TOGETHER DO IT IN= 1/10.... equ(1)
C= 1/50..... equ(2)
put the value of A in equ(1)
that is (B+C+B)= 1/10
THEN (2B+C)=1/10
put the value of C in this equation
2B+1/50=1/10
then it becomes 2B= 1/10-1/50
2B= 4/50
2B= 2/25
B= 1/25( the value 2 is cancel by divided by 2)
A'1 day work = (B+C)'1 day work
(A+B) TOGETHER DO IT IN= 1/10.... equ(1)
C= 1/50..... equ(2)
put the value of A in equ(1)
that is (B+C+B)= 1/10
THEN (2B+C)=1/10
put the value of C in this equation
2B+1/50=1/10
then it becomes 2B= 1/10-1/50
2B= 4/50
2B= 2/25
B= 1/25( the value 2 is cancel by divided by 2)
Varun said:
1 decade ago
First calculate the one day work of A+B+C = 1/10 + 1/50 = 3/25
Then given is A's work time = B+C's Work time...
So we can put "A = B+C" in (A + B + C = 3/25)
B+C+B+C = 3/25
2B+2C=3/25
Put C's one day work=1/50
2B+2*1/50=3/25
B=1/25
So B Time value=25 Day
Then given is A's work time = B+C's Work time...
So we can put "A = B+C" in (A + B + C = 3/25)
B+C+B+C = 3/25
2B+2C=3/25
Put C's one day work=1/50
2B+2*1/50=3/25
B=1/25
So B Time value=25 Day
Sudhakar said:
1 decade ago
a=b+c............(1).
a+b=1/10.
c=1/50.
From (1),
[(1/10)-b=b+(1/50)].
2b=4/50.
b=1/25.
25 days.
a+b=1/10.
c=1/50.
From (1),
[(1/10)-b=b+(1/50)].
2b=4/50.
b=1/25.
25 days.
Ananya said:
1 decade ago
2 x (A's 1 day's work) = 3/25.
how this get? pls tell me
how this get? pls tell me
Shruthi said:
1 decade ago
In general we get A+B+C=1/50,
But here we have A=B+C.
So, A+A=2A i.e 1/50*2 = 1/25
i.e B's 1 day work.
Finally B alone could do the work in 25 days.
But here we have A=B+C.
So, A+A=2A i.e 1/50*2 = 1/25
i.e B's 1 day work.
Finally B alone could do the work in 25 days.
MITUL said:
1 decade ago
A+B=1/10, C=1/50
A=B+C
> 1/10-B=B+C
> 2B=4/50
> B=1/25
SO B alone could do the work in 25 days.
A=B+C
> 1/10-B=B+C
> 2B=4/50
> B=1/25
SO B alone could do the work in 25 days.
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.
EKTA said:
1 decade ago
1/A=1/B+1/C......I.
1/A+1/B=1/10...II.
1/C=1/50.....III.
PUT 1/A=1/B+1/C IN EQ II.
1/B+1/C+1/B=1/10
2/B+1/50=1/10
2/B=10-1/50
2/B=4/50
1/B=1/25
B=25.
1/A+1/B=1/10...II.
1/C=1/50.....III.
PUT 1/A=1/B+1/C IN EQ II.
1/B+1/C+1/B=1/10
2/B+1/50=1/10
2/B=10-1/50
2/B=4/50
1/B=1/25
B=25.
Palkin singla said:
1 decade ago
Let work done by A = X.
Same work is also done by B+C = X.
A = X.
B+C = X.
C=1/50 ---->[1].
A+B = 1/10 ---->[2].
Putvalue of a in 2nd eqn and C = X - B in 1st eqn.
We get two equation.
B+X = 1/10.
X-B = 1/50.
By solving we get B = 1/25.
Which is our requirement.
Have a nice day.
Same work is also done by B+C = X.
A = X.
B+C = X.
C=1/50 ---->[1].
A+B = 1/10 ---->[2].
Putvalue of a in 2nd eqn and C = X - B in 1st eqn.
We get two equation.
B+X = 1/10.
X-B = 1/50.
By solving we get B = 1/25.
Which is our requirement.
Have a nice day.
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