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 |
![]() |
3 | . |
50 |
![]() |
![]() |
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.
Suresh said:
1 decade ago
From given statement
A+B+C = 6/50 ( A+B = 1/10 and C = 1/50)
Substitute A= B+C in above equation to get A= 3/50
3/50+B+1/50 = 6/50
B = 1/25
25 days
A+B+C = 6/50 ( A+B = 1/10 and C = 1/50)
Substitute A= B+C in above equation to get A= 3/50
3/50+B+1/50 = 6/50
B = 1/25
25 days
Mohanapriyadarshini said:
1 decade ago
@awoke
How come 2 b's ?
A + B = 1/10 but A = B + C, then
A + B = B + B + C = 1/10 and C = 1/50
a value is a+b ,what about b+c, why you are adding b+b+c ?
How come 2 b's ?
A + B = 1/10 but A = B + C, then
A + B = B + B + C = 1/10 and C = 1/50
a value is a+b ,what about b+c, why you are adding b+b+c ?
Nikhil said:
1 decade ago
Mr.Awoke your method is awesome and moreover its applicable to all such type of probs...thanks!
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
Sowmya said:
1 decade ago
@awoke.
Your method is awesome.
Your method is awesome.
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.
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
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
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