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 10 of 15.
Anandhan said:
1 decade ago
a = a+b;
a+b = 1/10; together;-----------(1).
c = 1/50;
Take 1st equation,
a+b=1/10;
we know a = b+c; so substitute,
b+c+b=1/10;
2b+1/50=1/10;
2b=1/10-1/50;
2b=4/50;
b=4/100;
b=2/50;
b=25;
Thats it.
a+b = 1/10; together;-----------(1).
c = 1/50;
Take 1st equation,
a+b=1/10;
we know a = b+c; so substitute,
b+c+b=1/10;
2b+1/50=1/10;
2b=1/10-1/50;
2b=4/50;
b=4/100;
b=2/50;
b=25;
Thats it.
VRAJ said:
1 decade ago
A = B+C.
A+B = 10.
C = 50.
A+B+C = 1/10+1/50.
A+A = 1/10+1/50.
2A = 6/50.
A = 3/50.
A = B+C.
3/50 = B+1/50.
B = 3/50-1/50.
B = 2/50.
B = 1/25.
B CAN DONE THE WORK IN 25 DAYS.
A+B = 10.
C = 50.
A+B+C = 1/10+1/50.
A+A = 1/10+1/50.
2A = 6/50.
A = 3/50.
A = B+C.
3/50 = B+1/50.
B = 3/50-1/50.
B = 2/50.
B = 1/25.
B CAN DONE THE WORK IN 25 DAYS.
Ram said:
1 decade ago
Number of days taken by B and C to complete a work. with same number of days A alone can do it.
if C can do a work in 50 days. If B joins with C then it becomes 50 days/2 = 25 days for B and C.
A no.of days for work = B&C no.of days for work.
25 days = 25 days.
if C can do a work in 50 days. If B joins with C then it becomes 50 days/2 = 25 days for B and C.
A no.of days for work = B&C no.of days for work.
25 days = 25 days.
Sumasree said:
1 decade ago
@Rohit nice your answer, but we always follow shortcuts for speed up.
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.
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.
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.
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.
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.
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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers