A community library has two machines that scan the barcodes of returned books. Working on its own, the first machine clears a trolley of books in \(a\) minu...

This formula combines two working rates. The important idea is that the algebra of rearranging is the same whatever the letters mean, but the context lets you check whether the rearranged form makes sense.

(a) Value of \(T\) when \(a = 30\) and \(b = 20\). [2]
Substitute into \(T = \dfrac{ab}{a + b}\), working out the top and the bottom separately before dividing:

\[T = \frac{30 \times 20}{30 + 20} = \frac{600}{50} = 12\ \text{minutes}\]

The answer is sensible: two machines together clear the trolley in 12 minutes, which is faster than either machine on its own.

(b) Make \(a\) the subject. [3]
The letter \(a\) appears twice, once on the top and once on the bottom, so the fraction must be cleared first and the \(a\) terms then collected. Multiply both sides by \((a + b)\):

\[T(a + b) = ab\] \[Ta + Tb = ab\]

Gather the terms containing \(a\) on one side and everything else on the other:

\[Tb = ab - Ta\]

Factorise the right-hand side, since \(a\) is common to both terms:

\[Tb = a(b - T)\] \[a = \frac{Tb}{b - T}\]

Marks: one for clearing the fraction, one for collecting the \(a\) terms on one side, one for factorising and dividing. Factorising is the step that is usually missed; without it \(a\) cannot be isolated at all.

(c) Value of \(a\) when \(T = 8\) and \(b = 24\). [1]

\[a = \frac{8 \times 24}{24 - 8} = \frac{192}{16} = 12\ \text{minutes}\]

Check in the original formula: \(\dfrac{12 \times 24}{12 + 24} = \dfrac{288}{36} = 8\) minutes, as required.

(d) Why \(T\) must always be smaller than \(b\). [1]
In the rearranged formula \(a = \dfrac{Tb}{b - T}\), the value \(a\) is a time taken by a real machine, so it must be positive. The numerator \(Tb\) is positive because both \(T\) and \(b\) are positive times. For the whole fraction to be positive, the denominator \(b - T\) must also be positive, which means

\[b - T > 0 \quad\Rightarrow\quad T < b\]

This matches common sense: adding a second machine can only make the job faster, so the combined time must be less than the time the second machine would take alone.

Examination point: when the subject appears more than once, the routine is always the same: clear fractions, collect that letter on one side, factorise it out, then divide. Recognise that shape early rather than trying to move terms one at a time.