A household's savings after \(t\) years under scheme P are modelled by \(S = 200\sqrt{t} + 50\) pounds, and under scheme Q by \(S = 40t + 90\) pounds. Find ...
Question 1 Report
A household's savings after \(t\) years under scheme P are modelled by \(S = 200\sqrt{t} + 50\) pounds, and under scheme Q by \(S = 40t + 90\) pounds.
- Find the value of \(S\) under each scheme when \(t=4\). (2)
- Show that scheme P gives more than scheme Q at \(t=4\), and find by how much. (2)
- By setting the two expressions equal, show that the equation simplifies to \(t - 5\sqrt{t} + 1 = 0\). (2)
Answer Details
Comparing two savings schemes at a fixed time means substituting that value of \(t\) into each formula separately, while finding when the schemes are ever equal means setting the two formulas equal to each other.
(a) Substituting \( t=4 \) into scheme P: \( 200\sqrt{4}+50 = 200(2)+50 = 400+50=450 \). Substituting \( t=4 \) into scheme Q: \( 40(4)+90 = 160+90=250 \) [2 marks].
(b) Since \( 450 > 250 \), scheme P gives more than scheme Q at \( t=4 \); the difference is \( 450-250=200 \) pounds [2 marks].
(c) Setting the two formulas equal gives \( 200\sqrt{t}+50 = 40t+90 \) [1 mark]. Rearranging by moving every term to one side (subtracting \( 200\sqrt t \) and \( 50 \) from both sides, then reordering) gives \( 40t - 200\sqrt t + 40 = 0 \), and dividing every term by \( 40 \) gives \( t - 5\sqrt t + 1 = 0 \), as required [1 mark].
The equation in part (c) mixes \(t\) with \(\sqrt t\), which is why it cannot be solved by the usual quadratic formula directly; substituting \( u = \sqrt t \) would turn it into an ordinary quadratic in \(u\), though that further step is not asked for here.
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