A household budgeting app models monthly savings, in pounds, as \(S = \dfrac{t^{2}-9}{t-3}\), for \(t \gt 3\), where \(t\) is the number of months since the...
Question 1 Report
A household budgeting app models monthly savings, in pounds, as \(S = \dfrac{t^{2}-9}{t-3}\), for \(t \gt 3\), where \(t\) is the number of months since the account opened.
- Simplify \(S\) fully by factorising the numerator. (2)
- Using the simplified form, work out \(S\) when \(t=9\). (1)
- The app adds a fixed joining bonus of \(\pounds15\) to \(S\). Write an expression for the total amount \(T(t)\), in its simplest form. (1)
- The household wants \(T(t)=50\). Solve to find \(t\). (2)
- State, with a reason, whether this value of \(t\) is realistic within a 2-year (24-month) savings plan. (1)
Answer Details
(a) The numerator, \(t^{2}-9\), is a difference of two squares, since \(9=3^{2}\), so it factorises as \((t-3)(t+3)\). Cancelling the common factor \((t-3)\) with the denominator (valid since \(t>3\) means \(t-3\) is never zero) gives \(S=t+3\). [2 marks]
(b) Substituting \(t=9\) into the simplified form: \(S=9+3=12\). [1 mark]
(c) Adding the fixed \(\pounds15\) joining bonus to \(S\): \(T(t)=(t+3)+15=t+18\). [1 mark]
(d) Setting \(T(t)=50\): \(t+18=50\), so \(t=32\). [2 marks]
(e) Since \(t=32\) months is longer than the \(24\)-month savings plan being considered, this value of \(t\) is not realistic within that plan. [1 mark]
The restriction \(t>3\) given in the question is exactly what makes the cancellation in part (a) valid; without it, \(t=3\) would need special treatment, since the original fraction would be \(\dfrac{0}{0}\) there.
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