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 1 of 15.
Ani said:
2 years ago
A = B+C,
A+B = 10,
C = 50.
A = B+C=> A = B+1/50.
A+B = 1/10 => B+1/50+B = 1/10.
2B + 1/50 = 1/10.
2B = 4/50.
B = 1/25.
25 DAYS.
A+B = 10,
C = 50.
A = B+C=> A = B+1/50.
A+B = 1/10 => B+1/50+B = 1/10.
2B + 1/50 = 1/10.
2B = 4/50.
B = 1/25.
25 DAYS.
(120)
Dewesh Kumar said:
3 years ago
A = B+C ,
A - C = B ---> (i).
Then,
A + B + C = 1/10 + 150 = 3/25.
From (i) putting a value of B
A+B+C= 3/25,
A+A - C + C = 3/25.
2A = 3/25.
A = 3/50.
Now, A+B = 1/10,
Putting A = 3/50,
B = 1/10-3/50.
B = 1/25 (B one day work) then total days is 25.
A - C = B ---> (i).
Then,
A + B + C = 1/10 + 150 = 3/25.
From (i) putting a value of B
A+B+C= 3/25,
A+A - C + C = 3/25.
2A = 3/25.
A = 3/50.
Now, A+B = 1/10,
Putting A = 3/50,
B = 1/10-3/50.
B = 1/25 (B one day work) then total days is 25.
(110)
Patil Pranay said:
1 year ago
A = B + C.
LCM of a+b and c =50,
Efficiency of A+B =5 c=1 than A+B+C=6.
A + B + C = 6 BUT A = B + C
A + A = 6
A = 3.
Then B = 2.
Ans = 50/2 = 25.
LCM of a+b and c =50,
Efficiency of A+B =5 c=1 than A+B+C=6.
A + B + C = 6 BUT A = B + C
A + A = 6
A = 3.
Then B = 2.
Ans = 50/2 = 25.
(55)
Gaurabbc said:
4 years ago
Time taken by A and B =10 days and Time taken by C =50 days
Total work=50 units
Efficiency of A and B = 5
The efficiency of C =1
A=B+C is given
A+B=5 => B+C+B=5 => 2B+C=5
2B+1=5 =>2B=4
B=2 is the efficiency
Time taken by B= work /efficiency of B =>50/2 =25.
Therefore, B can complete the work in 25days.
Total work=50 units
Efficiency of A and B = 5
The efficiency of C =1
A=B+C is given
A+B=5 => B+C+B=5 => 2B+C=5
2B+1=5 =>2B=4
B=2 is the efficiency
Time taken by B= work /efficiency of B =>50/2 =25.
Therefore, B can complete the work in 25days.
(45)
Sudip said:
2 years ago
Short and Simple:
a+b=10, c=50, total work 50.
perday a + b = 5, c = 1.
a+b = 5,
b+c+b = 5,
2b+c = 5,
b = 2/d.
b = 50/2 = 25.
a+b=10, c=50, total work 50.
perday a + b = 5, c = 1.
a+b = 5,
b+c+b = 5,
2b+c = 5,
b = 2/d.
b = 50/2 = 25.
(25)
Shubham Uttam Bandgar said:
3 years ago
If A = B+C.
So A-B =C.
i,e A-B = 1/50 ----> eq 1
And
A+B = 1/10 ----> eq 2.
From eq 2 - eq 1 we will get the answer.
.
So A-B =C.
i,e A-B = 1/50 ----> eq 1
And
A+B = 1/10 ----> eq 2.
From eq 2 - eq 1 we will get the answer.
.
(24)
Subhash said:
11 months ago
A = B + C.
Lcm of 10 and 50 = 50 ( Total Work)
Eff of A + B = 5 --->1.
Eff of C = 1.
A = 5- B from 1.
A = B + C.
5 - B = B + 1.
B = 2.
Time= TW/Eff of B.
= 50/2 = 25.
Lcm of 10 and 50 = 50 ( Total Work)
Eff of A + B = 5 --->1.
Eff of C = 1.
A = 5- B from 1.
A = B + C.
5 - B = B + 1.
B = 2.
Time= TW/Eff of B.
= 50/2 = 25.
(24)
Dhivyasree said:
3 years ago
Here, 1 day's work of a and b is(50/10) = 5.
1 day work of c is(50/50) = 1
a+b+c=(5+1)=>6->(1).
Given that a's time=sum of b's and c's time so,a=b+c
we can write a=b+c as b=a-c and then substitute this in (1)
that is a+b+c=6.
a+a-c+c=6=>2a=6->a=3.
Then later in that (1).
a+b+c = 6.
b = 6 - a - c,
b = 6 - 3 - 1,
b = 2.
That is 50/2=>25days.
1 day work of c is(50/50) = 1
a+b+c=(5+1)=>6->(1).
Given that a's time=sum of b's and c's time so,a=b+c
we can write a=b+c as b=a-c and then substitute this in (1)
that is a+b+c=6.
a+a-c+c=6=>2a=6->a=3.
Then later in that (1).
a+b+c = 6.
b = 6 - a - c,
b = 6 - 3 - 1,
b = 2.
That is 50/2=>25days.
(21)
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)
Bunny said:
4 years ago
They gave A = B+C, A+B = 10Days, C=50 Days;
A+B = 1/10;
since A = B+C;
B+C+B = 1/10;
2B+C = 1/10;
2B+1/50 = 1/10;
therefore B = 1/25;
B can complete in 25 days.
A+B = 1/10;
since A = B+C;
B+C+B = 1/10;
2B+C = 1/10;
2B+1/50 = 1/10;
therefore B = 1/25;
B can complete in 25 days.
(14)
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