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:
147 comments Page 8 of 15.
Rajni Bala said:
1 decade ago
A=B+C (Given).
Adding A both sides.
2 A = A+B+C.
Now A+B+C 's one day work = 3/25.
On putting these values we get,
A= 3/50 One day's work.
Use the value of A and get the value of B.
So the Answer is 25.
Adding A both sides.
2 A = A+B+C.
Now A+B+C 's one day work = 3/25.
On putting these values we get,
A= 3/50 One day's work.
Use the value of A and get the value of B.
So the Answer is 25.
Sandhya said:
1 decade ago
Given A = B+C ------ Eq.1.
& A+B = 1/10.
& C = 1/50.
NOW, A+B+C = (1/10+1/50).
= 3/25.
But in the question he is asking only 'B',
To get B we have to know the individual of A.
So,
SINCE B+C = A (given).
A+B+C = 3/25;.
A+A = 3/25;.
2A = 3/25.
A = 3/50.
HERE WE KNOW A = 3/50; C = 1/10 then.
A = B+C; 3/50 = B+1/50;.
B = 3/50 - 1/50;.
B = 2/50;.
B = 1/50;.
Therefore B can only do in 25 days.
& A+B = 1/10.
& C = 1/50.
NOW, A+B+C = (1/10+1/50).
= 3/25.
But in the question he is asking only 'B',
To get B we have to know the individual of A.
So,
SINCE B+C = A (given).
A+B+C = 3/25;.
A+A = 3/25;.
2A = 3/25.
A = 3/50.
HERE WE KNOW A = 3/50; C = 1/10 then.
A = B+C; 3/50 = B+1/50;.
B = 3/50 - 1/50;.
B = 2/50;.
B = 1/50;.
Therefore B can only do in 25 days.
Pavan R B said:
1 decade ago
Given 1/A = (1/B)+(1/C)...EQN1.
AND (1/A)+(1/B) = 1/10...EQN2.
ALSO 1/C = 1/50.
.
. . 1/A = (1/B)+(1/50).
SUBSTITUTE ABOVE EQN IN EQN2.
(1/B)+(1/50)+(1/B)=1/10.
THEREFORE 1/B=1/25, B ALONE CAN DO IN 25 DAYS.
AND (1/A)+(1/B) = 1/10...EQN2.
ALSO 1/C = 1/50.
.
. . 1/A = (1/B)+(1/50).
SUBSTITUTE ABOVE EQN IN EQN2.
(1/B)+(1/50)+(1/B)=1/10.
THEREFORE 1/B=1/25, B ALONE CAN DO IN 25 DAYS.
GAURAV said:
1 decade ago
A = B+C => A-B = C.
A+B = 1/10.
C = 1/50.
=> A-B = 1/50.
Solving eqns we get B = 25.
A+B = 1/10.
C = 1/50.
=> A-B = 1/50.
Solving eqns we get B = 25.
Priyanka said:
1 decade ago
A+B = 1/10.
C = 1/50.
A's time is equal to (B+C)'s time,
1/10-B = B+1/50.
2B = 1/10-1/50.
2B = 4/50.
B = 1/25.
B alone can do the work in 25 days.
C = 1/50.
A's time is equal to (B+C)'s time,
1/10-B = B+1/50.
2B = 1/10-1/50.
2B = 4/50.
B = 1/25.
B alone can do the work in 25 days.
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.
Vikram sihag said:
1 decade ago
Efficency of a+b = 10%.
Efficency of c = 2%.
Efficency of a+b+c = 12%.
Put up a = b+c.
2(b+c) = 12%.
b = 4%
Days = 100/4 = 25.
Efficency of c = 2%.
Efficency of a+b+c = 12%.
Put up a = b+c.
2(b+c) = 12%.
b = 4%
Days = 100/4 = 25.
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.
Neil said:
1 decade ago
A=B+C.
A+B+C = 1/10+1/50=6/50.
(B+C)+B+C = 2B+2C=6/50.
B+C = 6/100.
B = 6/100-1/50 = 1/25.
Therefore, B = 25 days.
A+B+C = 1/10+1/50=6/50.
(B+C)+B+C = 2B+2C=6/50.
B+C = 6/100.
B = 6/100-1/50 = 1/25.
Therefore, B = 25 days.
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.
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