A bicycle repair shop manager models the maximum number of punctures her new apprentice can realistically fix in a single working day, \(x\), using \(ax - 7...
Question 1 Report
A bicycle repair shop manager models the maximum number of punctures her new apprentice can realistically fix in a single working day, \(x\), using \(ax - 7 \leq 9\), where \(a\) is a positive integer. She knows the solution to this inequality is \(x \leq 4\).
- Find the value of \(a\). (2)
- Using this value of \(a\), find the greatest integer value of \(x\) satisfying \(ax - 7 \lt 25\). (2)
Answer Details
Solving the inequality \(ax-7\leqslant9\) in terms of \(a\) and matching the result to the known solution \(x\leqslant4\) finds \(a\); that same value of \(a\) is then used to solve a second, related inequality.
(a) Adding 7: \(ax\leqslant16\). Dividing by the positive constant \(a\): \(x\leqslant\dfrac{16}{a}\). Since this must match the given solution \(x\leqslant4\), \(\dfrac{16}{a}=4\), giving \(a=4\). [2 marks]
(b) Substituting \(a=4\) into \(4x-7\lt25\): adding 7 gives \(4x\lt32\), and dividing by 4 gives \(x\lt8\). The greatest integer value satisfying this strict inequality is \(x=7\). [2 marks]
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