Pump A can empty a pool in A minutes, and pump B can empty

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Hope this helps.

I did this question using smart numbers

A = 6
B = 3

Therefore in 1 minute A did 1/6 of the job.
5/6 of the job remains

Together A and B work at a rate of 1/2.

Thus 1/2 (x) = 5/6

x = 5/3 minutes

Now we need to add the minute worked by A so total of 5/3 + 1 = 8/3 minutes

Replacing we can see that in (B)

(6)(4) / 9 = 8/3

Therefore it is the correct answer choice

Gimme some freaking Kudos!
Cheers
J

Method 1:
The time taken by both pumps working together to do the work = AB/(A+B) minutes
Now, since Pump A has already worked for 1 minute, 1*(1/A) of the work has already been done.
Time taken to do the remaining work = [AB/(A+B)]*(1-1/A) = B(A-1)/(A+B) minutes
Total time taken from the start = [B(A-1)/(A+B)] + 1 = A*(B+1)/(A+B)
Option (B).

Method 2:
Let t be the time taken from the start. Then pump A works for t minutes to finish the work and pump B for (t-1) minutes
=> t(1/A) + (t-1)(1/B) = 1
=> t = A*(B+1)/(A+B)
Option (B).

Method 3:
We know that the time taken to do the work by both pumps working together right from the beginning is AB/(A+B) minutes.
Lets flip the question around to say that the two pumps worked together to do the work, except pump B was off in the first minute. So we must add the time it would take the two pumps to do the work that pump B would have done in this one minute to the total time. In this one minute, pump B would have done 1*(1/B) of the work = 1/B of the work.
Total time taken is therefore = AB/(A+B) + AB*(1/B)/(A+B)
= A*(B+1)/(A+B)
Option (B).

Since A empties a pool in A minutes. In 1 minute A empties 1/A fraction of the pool.
B empties a pool in B minutes. In 1 minute B empties 1/B fraction of the pool.

So, 1/A + (1/A + 1/B)n = 1 where is n is the time in minutes for which A and B worked together.
(A+b)/AB *n = 1- 1/A
n = AB/(A+B) - B /(A+B)= (AB- B)/(A+B)

SO, total time taken to empty a pool = 1+ (AB- B)/(A+B) = (A+B+ AB-B)/(A+B) = A(B+1)/(A+B)


Answer B

The fastest time in which tank can be emptied is by working together which is AB /(A+B)

But in this case it would be greater than AB/(A+B) since they have not worked together all time.

out of option B , C , D & E only option "B" is greater than AB /(A+B)

Pump A rate = 1/A
Pumb B rate = 1/B

rate*time = work

The question says A begun 1 minute prior to ( A+B) working together.

So, 1/A + (A+B/AB) ( t-1) = 1

Which gives us , t = AB+A/A+B

If we mark time used as T.
Then A works for T minutes and B for (T-1) minutes => T*(1/A) + (T-1)*(1/B) = 1 => BT + A(T-1) = AB; =>
BT + AT - A = AB => T(B+A) - A = AB; => T = A*(B+1)/(A+B)

­I did this question slightly different, so let me know if it makes sense.

The rate for pump A is \(\frac{(1\ pool)}{(A\ minutes)}\)

The rate for pump B is \(\frac{(1\ pool)}{(B\ minutes)}\)

we can say \(1\ pool\)  = \(\frac{(1\ pool)}{(A\ minutes)}(t \ minutes)\)+ \(\frac{(1\ pool)}{(B\ minutes)} (T - 1\ minutes)  \)

We can then simplify this as 1 = \(\frac{(T)}{(A)} \)+ \(\frac{(T - 1)}{(B)}\)

\(1\) = \(\frac{(BT)}{(AB)} \)+ \(\frac{(AT - A)}{(AB)}\)

\(1\) = \(\frac{(BT + AT - A)}{(AB)} \)

\(AB\) = \(BT + AT - A\)

\(AB + A\) = \(BT + AT\)

\(AB + A\) = \((B + A)T\)

\(\frac{(AB + A)}{(A + B)}\) =\(T\)

\(\frac{A(B + 1)}{(A + B)}\) =\(T\)
 ­