Problem Solving (PS) | Re: Pump A can empty a pool in A minutes, and pump B can empty

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

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

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

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

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

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

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

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

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

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

\(\frac{ A(B + 1)}{(A +B)}\) =\(T\)
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Statistics : Posted by jkkids12 • on 24 Jan 2014, 07:13 • Replies 10 • Views 24441

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