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 7 of 15.
Sachinpachore said:
8 years ago
A=B+C;
C=1/50: A =B+1/50
A+B=1/10,
B = 1/25.
C=1/50: A =B+1/50
A+B=1/10,
B = 1/25.
PRASAD said:
8 years ago
A'S 1 DAY WORK=(B+C) 1 DAY WORK.
(A+B)'S 1 DAY WORK=1/10 =>A'S 1 DAY WORK (1/10-B).
C'S 1 DAY WORK=1/50.
SO,
A'S 1 DAY WORK=(B+C) 1 DAY WORK.
1/10-B=B+1/50.
1/10-1/50=2B.
50-10/500=2B.
1/25=B.
B=25 DAYS (ANS....).
(A+B)'S 1 DAY WORK=1/10 =>A'S 1 DAY WORK (1/10-B).
C'S 1 DAY WORK=1/50.
SO,
A'S 1 DAY WORK=(B+C) 1 DAY WORK.
1/10-B=B+1/50.
1/10-1/50=2B.
50-10/500=2B.
1/25=B.
B=25 DAYS (ANS....).
Tenacious Guy said:
8 years ago
It can also go like this,
Since A=B+C.
So Add B on both sides you get.
A+B = 2B+C.
Solve it. 1/10 = 2B+1/50.
1/10 - 1/50.
4/50 = 2B.
B = 1/25.
25 Days.
Since A=B+C.
So Add B on both sides you get.
A+B = 2B+C.
Solve it. 1/10 = 2B+1/50.
1/10 - 1/50.
4/50 = 2B.
B = 1/25.
25 Days.
Ranganayaki said:
8 years ago
Thank you for explaining this.
SALEEM BALOCH said:
9 years ago
A = B + C
A + B = 1/10
C = 1/50
SO,
A + B + C = 1/10 + 1/50
BECAUSE A = B + C
REPLACE THE VALUE OF A.
SO IT IS,
B + C + B + C = 1/10 + 1/50
2(B+C) = 6/50 OR 2(B + C) = 3/25
B + C = 3/25 * 1/2
B + C = 3/50.
(C=1/50) MINUS C ON BOTH SIDES
B + C - C = 3/50 - 1/50,
B = 2/50 OR 1/25.
25 DAYS FOR B.
A + B = 1/10
C = 1/50
SO,
A + B + C = 1/10 + 1/50
BECAUSE A = B + C
REPLACE THE VALUE OF A.
SO IT IS,
B + C + B + C = 1/10 + 1/50
2(B+C) = 6/50 OR 2(B + C) = 3/25
B + C = 3/25 * 1/2
B + C = 3/50.
(C=1/50) MINUS C ON BOTH SIDES
B + C - C = 3/50 - 1/50,
B = 2/50 OR 1/25.
25 DAYS FOR B.
Sathwik said:
9 years ago
A + B + C = 3/25.
A = B + C.
A = A + A.
A = 3/50.
A + B = 10.
=> 3/50 - 1/10 = 1/25.
Finally B = 25days.
A = B + C.
A = A + A.
A = 3/50.
A + B = 10.
=> 3/50 - 1/10 = 1/25.
Finally B = 25days.
Ayesha said:
9 years ago
Your shortcut is great @Awoke.
MR X said:
9 years ago
Hey, I can explain in a simple way.
1st it says 'A' can do work which is done by 'B' and 'C'
there fore,A=B+C------->eq 1
then,
A + B = 10
c = 50.
Find the LCM for 10 and 50 LCM = 50units.
A + B=5units[50/10].
C = 1units[50/50].
Find B alone?
A + B = 5units
B + C + B = 5 [according to equation 1 A = B + C]
2B + C = 5
2B + 1 = 5
2B = 4
B = 2.
Therefore, 50units/2units = 25days.
I hope u understood this problem solving method.
1st it says 'A' can do work which is done by 'B' and 'C'
there fore,A=B+C------->eq 1
then,
A + B = 10
c = 50.
Find the LCM for 10 and 50 LCM = 50units.
A + B=5units[50/10].
C = 1units[50/50].
Find B alone?
A + B = 5units
B + C + B = 5 [according to equation 1 A = B + C]
2B + C = 5
2B + 1 = 5
2B = 4
B = 2.
Therefore, 50units/2units = 25days.
I hope u understood this problem solving method.
Deepak said:
9 years ago
@Nagu, nice explanation. Thank you.
SubhamChand said:
9 years ago
You can solve it in a very easy way.
A = A + B.
C's 1day work = 1/50.
(A+B) 's 1day work = 1/10.
But according to the question.
A = B + C.
=> A + B = B + B + C (add B on both sides).
=> 1/10 = 2B + 1/50.
Then B = 1/25.
So, B can do it in 25 days.
A = A + B.
C's 1day work = 1/50.
(A+B) 's 1day work = 1/10.
But according to the question.
A = B + C.
=> A + B = B + B + C (add B on both sides).
=> 1/10 = 2B + 1/50.
Then B = 1/25.
So, B can do it in 25 days.
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