A bicycle repair shop compares its own labour charge with a mobile repair service's charge for a job needing \(x\) replacement parts, both in pounds, shown ...
Question 1 Report
A bicycle repair shop compares its own labour charge with a mobile repair service's charge for a job needing \(x\) replacement parts, both in pounds, shown on the sign below.
- Find the value of \(x\) for which \(f(x) = g(x)\). (2)
- A customer's bike needs \(x=6\) replacement parts. By evaluating \(f(6)\) and \(g(6)\), state which service is cheaper for this job, giving a reason. (2)
Answer Details
The crossover number of parts is found by setting the two labour-charge functions equal; for a specific job, evaluating both functions directly and comparing the totals shows which service costs less.
- \[ 20+6x=32+2x \] [1 mark] \[ 4x=12 \] so \[ x=3 \] [1 mark]
- \[ f(6)=20+6(6)=20+36=56 \], \[ g(6)=32+2(6)=32+12=44 \] [1 mark]; since \(44\lt56\), the mobile repair service is cheaper for a job needing 6 replacement parts. [1 mark]
The mobile repair service has the higher fixed call-out charge (\(\pounds32\) against \(\pounds20\)) but the lower rate per part (\(\pounds2\) against \(\pounds6\)), so it only becomes the cheaper option once a job needs more than \(3\) parts, which a \(6\)-part job comfortably does.
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