\((A+B)t=1\) or \((A+B)=\frac{1}{t}\)
\(A(t+8)=1\) or \(A=\frac{1}{t+8}\)
\(B(t+4.5)=1\) or \(B=\frac{1}{t+4.5}\)

\(\frac{1}{t+8}+\frac{1}{t+4.5}=\frac{1}{t}\)

\(\frac{1}{t+4.5}=\frac{1}{t}-\frac{1}{t+8}\)

\(\frac{1}{t+4.5}=\frac{t+8-t}{t(t+8)}\)

\(t(t+8)=8(t+4.5)\)

\(t^2+8t=8t+36\)

\(t^2=36\)

\(t=6\)

Answer is (C)
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Question:
Two workers A and B are engaged to do a work. A working alone takes 8 hours more to complete the job than if both worked together. If B worked alone, he would need 4.5 hours more to complete the job than they both working together. What time would they take to do the work together?


Solution:
Let A and B together take t hours to complete the work.
Thus, time taken by A = (t + 8) hrs and time taken by B = (t + 4.5) hrs

Thus, we can say that:
Work done by (A and B) in t hours = Work done by A in (t + 8) hours = Total work
=> (Work done by A in t hrs) + (Work done by B in t hrs) = (Work done by A in t hrs) + (Work done by A in 8 hrs)

Cancelling the (Work done by A in t hrs) from both sides:
(Work done by B in t hrs) = (Work done by A in 8 hrs)
=> Work done by B in 1 hr = Work done by A in (8/t) hrs ... (i)


Similarly, we have:
Work done by (A and B) in t hours = Work done by B in (t + 4.5) hours = Total work
=> (Work done by A in t hrs) + (Work done by B in t hrs) = (Work done by B in t hrs) + (Work done by B in 4.5 hrs)

Cancelling the (Work done by B in t hrs) from both sides:
(Work done by A in t hrs) = (Work done by B in 4.5 hrs)
=> Work done by B in 1 hr = Work done by A in (t/4.5) hrs ... (ii)

Thus, from (i) and(ii):
8/t = t/4.5

=> \(t^2 = 36\)
=> t = 6 hrs

Answer C


Alternate approach
Let time taken by A and B together be t hours
Thus, time taken by A = (t + 8) hrs and time taken by B = (t + 4.5) hrs

thus, equating work done by A and B in 1 hour:
1/(t + 8) + 1/(t + 4.5) = 1/t

Solving, we get: t = 6 hours
Answer C
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A short little formula exists which can solve it in 10 secs -

\(\sqrt{8*4.5}\)

\(= \sqrt{36}\)

\(= 6\) , Answer must be (C)
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r=w/t

(a+b)=1/t
a=1/t+8
b=1/t+4.5
(1/t+8)+(1/t+4.5)=1/t
2t^2+12.5t=t^2+12.5t+36
t^2=36 (t>0) t=6

Ans (C)

Abhishek009

Any theory or links on how this is achievable and in what conditions it usually works?

Thank you!
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Let’s let n = the number of hours it takes to complete the job when A and B work together; thus, their combined rate is 1/n. Since A alone takes 8 hours more than if they work together, then A’s rate is 1/(n + 8). And since B alone takes 4.5 hours more than if they work together, then B’s rate is 1/(n + 4.5).

Thus, , we can create the rate equation:

1/(n + 8) + 1/(n + 4.5) = 1/n

Multiplying by n(n + 8)(n + 4.5), we have:

n(n + 4.5) + n(n + 8) = (n + 8)(n + 4.5)

n^2 + 4.5n + n^2 + 8n = n^2 + 4.5n + 8n + 36

2n^2 + 12.5n = n^2 + 12.5n + 36

n^2 = 36

n = 6 or -6

Since n can’t be negative, n = 6.

Answer: C
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RESPONDING TO A PM IN THIS REGARD -



Time required to complete the task will be \(\sqrt{ab}\)

It can be applied to all such cases without any restriction !!
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let us assume the time taken to do the work together = x hours


Time taken by A to do the work alone = x+8 hours
Time taken by B to do the work alone =x+4.5 hours

L.C.M=(x+8)(x+4.5)=TOTAL WORK
Rate of A = x+4.5 part of work per hour
Rate of B= x+8 part of work per hour
Rate of A & B working together= (x+4.5+x+8)=(2x+12.5) part of work per hour

RATE*TIME TAKEN=TOTAL WORK
=>(2x+12.5)*x=(x+8)(x+4.5)
=>x^2=36
=>x=6

Answer:6 hours.