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 12 of 15.
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.
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
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.
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
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)
Dinesh said:
1 decade ago
@AWOKE.
I liked your method. Thank you.
I liked your method. Thank you.
GAURAV said:
1 decade ago
TIME FOR CERTAIN WORK....
A=B+C .....................................i
FOR
A+B=10 DAY.......SO(A+B)_1DAY WORK=1/10
.>A=1/10-B.......................ii
C=50 DAY.........1 DAY C WORK=1/50......................iii
SO ...FOR B=A-C.........................FROM i
B=1/10-B-1/50
2B=1/10-1/50
B=1/25
.........SO B REQUIRE 25 DAYS
A=B+C .....................................i
FOR
A+B=10 DAY.......SO(A+B)_1DAY WORK=1/10
.>A=1/10-B.......................ii
C=50 DAY.........1 DAY C WORK=1/50......................iii
SO ...FOR B=A-C.........................FROM i
B=1/10-B-1/50
2B=1/10-1/50
B=1/25
.........SO B REQUIRE 25 DAYS
Rakesh sonowal said:
1 decade ago
A=B+C (1)
A+B 1 Day work=1/10 (2)
Cs 1 Day work=1/50 (3)
putin eq (1) n (3) in eq (2)
2B +1/50=1/10
2B=1/10-1/50
2B=(5-1)/50
2B=4/50
B=4/50 =1/25
So Bs 1 day work =1/25.
So B alone could do d work in 25 days.
A+B 1 Day work=1/10 (2)
Cs 1 Day work=1/50 (3)
putin eq (1) n (3) in eq (2)
2B +1/50=1/10
2B=1/10-1/50
2B=(5-1)/50
2B=4/50
B=4/50 =1/25
So Bs 1 day work =1/25.
So B alone could do d work in 25 days.
Sowmya said:
1 decade ago
@awoke.
Your method is awesome.
Your method is awesome.
Pavan@9966606261 said:
1 decade ago
A=B+C;
A+B=1/10;
A-B=1/50;
2A=1/10+1/50=6/50;
A=3/50;
A-B=1/50;
B=3/50-1/50;
B=1/25;
B ALONE CAN DO IN "25" DAYS
A+B=1/10;
A-B=1/50;
2A=1/10+1/50=6/50;
A=3/50;
A-B=1/50;
B=3/50-1/50;
B=1/25;
B ALONE CAN DO IN "25" DAYS
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