Aptitude - Time and Work - Discussion

Easiest method:

A = B+C ---> (1)
A+B = 1/10 ---> (2)
C=1/50 ---> (3).

From (2) A = 1/10-B.

By substituting (2) & (3) in (1), we get;

1/10-B = B + 1/50.
Solving for B.

2B = 1/10 - 1/50 = 1/25.

Hi Guys,

In this problem

A's one day work is equal to (b+c)'s one day work
A = (B+C)

So a+b+c become 2A here they taken instead of b+c= a

so A+B+C = A+ (B+C) = A+A becasue B+C=A
3/25 = 2* (A's one day work)

so 2* A's one day work

I THINK YOU ALL UNDER STOOD THE TRICK..

Short cut method:

A + B = 1/10 but A = B + C, then

A + B = B + B + C = 1/10 and C = 1/50

2B + 1/50 = 1/10 gives us

B = 1/2(1/10-1/50)
= 1/2(2/25)
= 1/25

Therefore, B can do the work alone in 25 days.

Avoid fractions! they mess up the calculations.

Solution:

Total amount of work = LCM (10, 50) = 50 units.

Given: A's time = (B+C)'s time.

Hence, A's rate = B's rate + C's rate.

(Rate is the rate of doing work, not the monetary rate).

(A+B)'s time = 10 days.

Hence A's rate + B's rate = 50/10 = 5 units/day.

C's time = 50 days.

Hence C's rate = 50/50 = 1 unit/day.

Hence, A's rate + B's rate + C's rate = 5 + 1 = 6 units/day.

Now, using A's rate = B's rate + C's rate,

We get A's rate = 6/2 = 3 units/day.

Hence B's rate = 5-3 = 2 units/day.

Hence B's time = Total work/B's rate = 50/2 = 25 days.

@Sneha best explanations.

a+b, +c can finished 50 work.
a+b finished in 5 days,
c finished in 1 day,
can say one day work (a+b+c) in 6 days,
a= b+c put,
a+(a) =6,
a=3.

A = b+c.

A+b = 1/10,c=1/50
B+C+B = 1/10.

2b = (1/10)-(1/50).
2b = 2/25,
b = 1/25,
So,b = 25.

A=B+C --- (1).
A+B=1/10 --- (2).
C=1/50 --- (3).

Substitute C in eq 1.
A=B+ (1/50).
From eq 2 we get A= (1/10) -B.
Substitute this in above eq.

(1/10)-B=B+ (1/50).

Solve it.

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.

I'm not getting how 2*(a`s 1day work)=3/25