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 9 of 15.
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.
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.
Ranganayaki said:
8 years ago
Thank you for explaining this.
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.
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....).
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.
Shiva krishna kanaparthi said:
8 years ago
Total work = LCM of 10 and 50=50 units.
(A+B)'S 1 DAY WORK = 50/10= 5 UNITS PER DAY.
C'S 1 DAY WORK = 50/50= 1 UNIT PER DAY.
SINCE WE HAVE A=B+C AND A+B=5.
A-B=1----------(1)
A+B=5----------(2)
SOLVING ABOVE EQUATIONS WE GET B= 2 UNITS PER DAY.
SO NUMBER OF DAYS TO COMPLETE THE WORK FOR B= 50/2 = 25 DAYS.
(A+B)'S 1 DAY WORK = 50/10= 5 UNITS PER DAY.
C'S 1 DAY WORK = 50/50= 1 UNIT PER DAY.
SINCE WE HAVE A=B+C AND A+B=5.
A-B=1----------(1)
A+B=5----------(2)
SOLVING ABOVE EQUATIONS WE GET B= 2 UNITS PER DAY.
SO NUMBER OF DAYS TO COMPLETE THE WORK FOR B= 50/2 = 25 DAYS.
Payal said:
8 years ago
B= x days
C = y days
A = x+y days
given: y= 50
& A +B = x+x+y = 2x+y=10.
=>x= 20 ans.
C = y days
A = x+y days
given: y= 50
& A +B = x+x+y = 2x+y=10.
=>x= 20 ans.
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.
Azhar said:
8 years ago
@Sneha best explanations.
(1)
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