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 15 of 15.
VIKRAM said:
6 months ago
Given : AB=10; C=50; find B=?;
LCM of 50, 10 is "50"(total work),
Efficiency of 'AB' =total work : 50/10 = 5,
Efficiency of 'C' =total work : 50/50 = 1.
From efficiency of 'AB & C' is ---> A+B+C : 5 +1 = 6.
Question itself given that : A=B+C;
We know A+B+C = 6--->(1) , in question they said A=B+C -----> (2) ;
Substitute (2) in (1):
We get (1).
B+C+B+C =6;
2B+2C=6; (w.k.t efficiency of C = 1)
2B+2(1)=6;
2B+2=6;
2B=6-2 ; ---> 2B=4; ---> B=2;
Hence, total B alone worked : (50)/2 = '25' days.
LCM of 50, 10 is "50"(total work),
Efficiency of 'AB' =total work : 50/10 = 5,
Efficiency of 'C' =total work : 50/50 = 1.
From efficiency of 'AB & C' is ---> A+B+C : 5 +1 = 6.
Question itself given that : A=B+C;
We know A+B+C = 6--->(1) , in question they said A=B+C -----> (2) ;
Substitute (2) in (1):
We get (1).
B+C+B+C =6;
2B+2C=6; (w.k.t efficiency of C = 1)
2B+2(1)=6;
2B+2=6;
2B=6-2 ; ---> 2B=4; ---> B=2;
Hence, total B alone worked : (50)/2 = '25' days.
(14)
Kayden break said:
5 months ago
1/A + 1/B = 1/10
As A and B + C IS SAME
1/A = 1/B+1/C. (1/C=1/50 AS GIVEN IN QUES)
Equate both equations
1/A + 1/B = 1/10.
(1/B + 1/C) + 1/B = 1/10,
2/B + 1/50 = 1/10,
B = 25.
As A and B + C IS SAME
1/A = 1/B+1/C. (1/C=1/50 AS GIVEN IN QUES)
Equate both equations
1/A + 1/B = 1/10.
(1/B + 1/C) + 1/B = 1/10,
2/B + 1/50 = 1/10,
B = 25.
(1)
Julius Openy said:
5 months ago
Given:
A = B + C.
A + B = 10,
C = 50.
Solution :
A + B = 10.
(A + B)'s one day way is 1/10.
A + B = 1/10 ---> 1
from A = B + C ---> 2
Substitute eqn 2 in 1;
B + C + B = 1/10,
2B + 1/50 = 1/10,
2B = 1/10 - 1/50,
2B = 4/50,
B = 1/25.
B alone can do the work in 25 days.
A = B + C.
A + B = 10,
C = 50.
Solution :
A + B = 10.
(A + B)'s one day way is 1/10.
A + B = 1/10 ---> 1
from A = B + C ---> 2
Substitute eqn 2 in 1;
B + C + B = 1/10,
2B + 1/50 = 1/10,
2B = 1/10 - 1/50,
2B = 4/50,
B = 1/25.
B alone can do the work in 25 days.
(18)
Prathiba said:
4 months ago
a = b+ c ---> (1)
a+ b = 10 = 1/10 for 1 day work ---> (2)
c = 50 = 1/50 for 1 day work
b = 1/10 - a ---> (3)---from eqn(1)
c = 1/50.
a = b +c ---from (1)
a = 1/10 -a + 1/50 ---from (3)
a = 3/50.
b = 1/10 - a = 1/10 -3/50 = 1/25 .
b = 25 days.
a+ b = 10 = 1/10 for 1 day work ---> (2)
c = 50 = 1/50 for 1 day work
b = 1/10 - a ---> (3)---from eqn(1)
c = 1/50.
a = b +c ---from (1)
a = 1/10 -a + 1/50 ---from (3)
a = 3/50.
b = 1/10 - a = 1/10 -3/50 = 1/25 .
b = 25 days.
(9)
K Mohammad Rizwan said:
2 months ago
1/A = 1/B + 1/ C -->1
1/A + I/B = 1/10 -->2
1/C = 1/50 -->3
Subs 1 in 2.
1/B + 1/C + 1/B = 1/10 -->4
Subs 3 in 4.
2/B = 1/10 - 1/50.
2/B = 4/50,
B = 25.
1/A + I/B = 1/10 -->2
1/C = 1/50 -->3
Subs 1 in 2.
1/B + 1/C + 1/B = 1/10 -->4
Subs 3 in 4.
2/B = 1/10 - 1/50.
2/B = 4/50,
B = 25.
(3)
ESNALA SIDDA said:
1 month ago
A = B+C.
A + B = 1/10,
2B + C = 1/10,
C = 1/50,
B = 1/25,
B = 25hrs.
A + B = 1/10,
2B + C = 1/10,
C = 1/50,
B = 1/25,
B = 25hrs.
(8)
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