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 2 of 15.
Zeeshan said:
1 decade ago
From the question,
A = B+C.
(A+B)'s 1 day work = 1/10...........(1).
C's 1 day work = 1/50...............(2).
In eqn.(1) put A=B+C.
Then,
B+C+B = 1/10.
2B+C = 1/10.
2B = 1/10-C.
2B = 1/10-1/50 (since C's 1 day work = 1/50).
2B = (5-1)/50.
2B = 4/50.
B = 4/50*2.
B = 4/100.
B = 1/25 ( ie B's 1 day work).
B's complete work in 25 days.
I HOPE NOW YOU GOT THE SOLUTION.
A = B+C.
(A+B)'s 1 day work = 1/10...........(1).
C's 1 day work = 1/50...............(2).
In eqn.(1) put A=B+C.
Then,
B+C+B = 1/10.
2B+C = 1/10.
2B = 1/10-C.
2B = 1/10-1/50 (since C's 1 day work = 1/50).
2B = (5-1)/50.
2B = 4/50.
B = 4/50*2.
B = 4/100.
B = 1/25 ( ie B's 1 day work).
B's complete work in 25 days.
I HOPE NOW YOU GOT THE SOLUTION.
Kankana ghosh said:
1 decade ago
Let A's 1 day of work be x.
Let B's ,, ,, ,, ,, ,, y.
Let C's ,, ,, ,, ,, ,, z.
Then A.T.P: x = y+z.......................(1).
Given, z = 1/50.............................(2).
And x+y = 1/10.
Which means (y+z)+y =1/10 (using eq.(1)).
2y+z = 1/10.
2y+1/50 = 1/10 (using eq.(2)).
2y = (1/10-1/50).
y = 1/25.
Ans) So, B can alone do the work in 25 days.
Let B's ,, ,, ,, ,, ,, y.
Let C's ,, ,, ,, ,, ,, z.
Then A.T.P: x = y+z.......................(1).
Given, z = 1/50.............................(2).
And x+y = 1/10.
Which means (y+z)+y =1/10 (using eq.(1)).
2y+z = 1/10.
2y+1/50 = 1/10 (using eq.(2)).
2y = (1/10-1/50).
y = 1/25.
Ans) So, B can alone do the work in 25 days.
Dhivyasree said:
3 years ago
Here, 1 day's work of a and b is(50/10) = 5.
1 day work of c is(50/50) = 1
a+b+c=(5+1)=>6->(1).
Given that a's time=sum of b's and c's time so,a=b+c
we can write a=b+c as b=a-c and then substitute this in (1)
that is a+b+c=6.
a+a-c+c=6=>2a=6->a=3.
Then later in that (1).
a+b+c = 6.
b = 6 - a - c,
b = 6 - 3 - 1,
b = 2.
That is 50/2=>25days.
1 day work of c is(50/50) = 1
a+b+c=(5+1)=>6->(1).
Given that a's time=sum of b's and c's time so,a=b+c
we can write a=b+c as b=a-c and then substitute this in (1)
that is a+b+c=6.
a+a-c+c=6=>2a=6->a=3.
Then later in that (1).
a+b+c = 6.
b = 6 - a - c,
b = 6 - 3 - 1,
b = 2.
That is 50/2=>25days.
(21)
Naani said:
1 decade ago
Given that A = B+C.....(1).
A&B completes the work in 10 days so A&B one day work is 1/10.
A+B = 1/10
A = (1/10)-B.
(1/10)-B = B+C..........from(1).
Given that C alone complete the work in 50 days so C's one day's work is = 1/50.
Therefore.
(1/10)-B = B+(1/50).
2B = 4/50.
B = (1/25) works.
So B alone can complete work in 25 days.
A&B completes the work in 10 days so A&B one day work is 1/10.
A+B = 1/10
A = (1/10)-B.
(1/10)-B = B+C..........from(1).
Given that C alone complete the work in 50 days so C's one day's work is = 1/50.
Therefore.
(1/10)-B = B+(1/50).
2B = 4/50.
B = (1/25) works.
So B alone can complete 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
Harsh said:
8 years ago
I have one doubt please help me.
I am doing the solution like this so answer is coming different please correct me.
let A's one day work is = 1\x,
B's=1\y,
c's=1\z,
1\x=1\y+1\z ..(i)
1\x+1\|y=1\10 ....given (ii)
1\z=1\50 ...(iii)
by solving all 3 equation we get,
1\10-2\y=1\50,
(y-20)\10y=1\50,
5(y-20)=y,
6y=100,
y=100\6.
I am doing the solution like this so answer is coming different please correct me.
let A's one day work is = 1\x,
B's=1\y,
c's=1\z,
1\x=1\y+1\z ..(i)
1\x+1\|y=1\10 ....given (ii)
1\z=1\50 ...(iii)
by solving all 3 equation we get,
1\10-2\y=1\50,
(y-20)\10y=1\50,
5(y-20)=y,
6y=100,
y=100\6.
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????
Rohit said:
1 decade ago
I think its easy.
A = B+C ------ (i).
Then,
1 day work of A+B = 1/10.
1 day work of C = 1/50.
Now the Total Work of A+B+C = (1/10+1/50) = 3/25 .... (ii).
From eq (i) put A in place of B+C in eq (ii),
We get 2A = 3/25 = 3/50.
From eq (i),
We know value of A & C.
Put in eq (i),
3/25 = B + 1/50.
B = 1/25.
B= 25 DAYS.
A = B+C ------ (i).
Then,
1 day work of A+B = 1/10.
1 day work of C = 1/50.
Now the Total Work of A+B+C = (1/10+1/50) = 3/25 .... (ii).
From eq (i) put A in place of B+C in eq (ii),
We get 2A = 3/25 = 3/50.
From eq (i),
We know value of A & C.
Put in eq (i),
3/25 = B + 1/50.
B = 1/25.
B= 25 DAYS.
Venkataramireddy said:
8 years ago
1 day work A+B=1/10.
A can do work same as do B+C.
so we can write it as
A=B+C.....
C 1day work =1/50,
1-day work of A+B+C=1/10+1/50
LCM 50.
5+1/50=6/50=3/25;
A+B+C=3/25,
above B+C=A.
So
A+A=3/25.
2A=3/25,
A=3/50,
substitute A value in A+B=1/10.
3/50+B=1/10,
B=1/10-3/50,
lcm 50,
B=5-3/50,
=2/50,
B=1/25,
B Alone 25 days.
A can do work same as do B+C.
so we can write it as
A=B+C.....
C 1day work =1/50,
1-day work of A+B+C=1/10+1/50
LCM 50.
5+1/50=6/50=3/25;
A+B+C=3/25,
above B+C=A.
So
A+A=3/25.
2A=3/25,
A=3/50,
substitute A value in A+B=1/10.
3/50+B=1/10,
B=1/10-3/50,
lcm 50,
B=5-3/50,
=2/50,
B=1/25,
B Alone 25 days.
Gaurabbc said:
4 years ago
Time taken by A and B =10 days and Time taken by C =50 days
Total work=50 units
Efficiency of A and B = 5
The efficiency of C =1
A=B+C is given
A+B=5 => B+C+B=5 => 2B+C=5
2B+1=5 =>2B=4
B=2 is the efficiency
Time taken by B= work /efficiency of B =>50/2 =25.
Therefore, B can complete the work in 25days.
Total work=50 units
Efficiency of A and B = 5
The efficiency of C =1
A=B+C is given
A+B=5 => B+C+B=5 => 2B+C=5
2B+1=5 =>2B=4
B=2 is the efficiency
Time taken by B= work /efficiency of B =>50/2 =25.
Therefore, B can complete the work in 25days.
(45)
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