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.
Deepika said:
10 years ago
By adding a+b+c i.e. 1/10+1/50.
Esayas said:
10 years ago
If (a+b) = 1/10 and a = b+c, so 2b++c = 1/10.
2b+1/50 = 1/10.
b = 1/25.
So b can finish the work in 25 days.
2b+1/50 = 1/10.
b = 1/25.
So b can finish the work in 25 days.
ESHWAR said:
10 years ago
A = B+C.
Given A+B 1 day work = 1/10;
C 1 day work = 1/50;
Add B on both sides.
A+B = 2B+C;
1/10 = 2B+1/50;
2B = 1/10-1/50;
2B = 4/50;
B = 2/50;
So B's 1 day work is 1/25 and B can finish work in 25 days.
Given A+B 1 day work = 1/10;
C 1 day work = 1/50;
Add B on both sides.
A+B = 2B+C;
1/10 = 2B+1/50;
2B = 1/10-1/50;
2B = 4/50;
B = 2/50;
So B's 1 day work is 1/25 and B can finish work in 25 days.
Atyanand said:
10 years ago
A = B+C......(1).
A+B = 1/10......(2).
C = 1/50......(3).
So work done by A, B, C is.
A+B+C = 1/10 +1/50 = 6/50.
From equation 2.
A+A = 6/50.
A = 3/50.
So B = 1/10-3/50 = 2/50 = 1/25.
So B complete its work in 25 days.
A+B = 1/10......(2).
C = 1/50......(3).
So work done by A, B, C is.
A+B+C = 1/10 +1/50 = 6/50.
From equation 2.
A+A = 6/50.
A = 3/50.
So B = 1/10-3/50 = 2/50 = 1/25.
So B complete its work in 25 days.
Harsha said:
10 years ago
A -> B+C (given).
A+B -> 10.
C -> 50.
Taking L.C.M we get 50.
So A+B -> 10 (5) because 10*5 = 50.
C -> 50 (1) because 50*1 = 50.
A+B+C = 6.
As B+C can be replaced with A.
A+A = 6.
2A = 6.
A = 3.
Then B = 5-3 = 2.
So, 50/2 = 25 days answer.
A+B -> 10.
C -> 50.
Taking L.C.M we get 50.
So A+B -> 10 (5) because 10*5 = 50.
C -> 50 (1) because 50*1 = 50.
A+B+C = 6.
As B+C can be replaced with A.
A+A = 6.
2A = 6.
A = 3.
Then B = 5-3 = 2.
So, 50/2 = 25 days answer.
Sarumathi.S said:
10 years ago
What is the logic in the statement 2* (A's one day work).
Akshay Jaywant said:
9 years ago
10(A + B) = 50 C.
(B + C + B) = 5C.
2B = 4C.
B = 2C.
Therefore, B = 50/2 = 25 days.
(B + C + B) = 5C.
2B = 4C.
B = 2C.
Therefore, B = 50/2 = 25 days.
Anonymous said:
9 years ago
A = B + C.
A + B = 1/10.
A = 1/10 - B.
C = 1/50.
A = B + C.
1/10 - B = B + 1/50.
(1- 10B) /10= (50B + 1) /50.
50 - 500B = 500B + 10.
100B = 40.
B = 40/100.
B = 1/25.
B = 25.
A + B = 1/10.
A = 1/10 - B.
C = 1/50.
A = B + C.
1/10 - B = B + 1/50.
(1- 10B) /10= (50B + 1) /50.
50 - 500B = 500B + 10.
100B = 40.
B = 40/100.
B = 1/25.
B = 25.
Sarang said:
9 years ago
A = B + C.
A = B + 1/50.....(1)
A + B = 1/10.......(2)
PUT EQ (1) IN EQ (2). Then, Eq (2) becomes.
B + 1/50 + B = 1/10.
2B = 1/10 - 1/50.
B = 25.
A = B + 1/50.....(1)
A + B = 1/10.......(2)
PUT EQ (1) IN EQ (2). Then, Eq (2) becomes.
B + 1/50 + B = 1/10.
2B = 1/10 - 1/50.
B = 25.
Sourabh said:
9 years ago
@Awoke.
Your trick is easy to find the solution for these type of question.
Your trick is easy to find the solution for these type of question.
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