A repair workshop turns steel bushes on a lathe. A... | Mathematics Specification B 4MB1 | Paper 2 Mock 01 | Written Paper 2

Question 1 Report

A repair workshop turns steel bushes on a lathe. A bush is a cylinder of length 45 mm with a smaller cylinder drilled through the middle. The end view and side view are shown, with the two radii marked.

Show that the volume of steel in one bush is \(25920\pi\) mm\(^3\). (3)
Write this volume in cm\(^3\), correct to 3 significant figures. (2)
The steel has density 7.8 g for each cubic centimetre. Work out the mass of one bush, in grams. (2)
Steel costs \(\pounds 4.20\) for each kilogram. Work out the cost of the steel used in one bush. (2)
The workshop sells a bush for \(\pounds 9.50\). Work out the steel cost as a percentage of the selling price. (1)

Answer

A bush is a cylinder with a smaller cylinder removed, so its volume is the difference of the two. Both cylinders have the same length, which lets the length be factored out and keeps the arithmetic exact.

(a) Show the volume is \(25920\pi\) mm³. [3]
Outer cylinder of radius 30 mm minus inner cylinder of radius 18 mm, both 45 mm long:

\[V = \pi \times 30^2 \times 45 - \pi \times 18^2 \times 45\]

Take out the common factor \(45\pi\):

\[V = 45\pi(900 - 324) = 45\pi \times 576 = 25920\pi\ \text{mm}^3\]

Marks: one for each cylinder volume, one for the difference in the required form. Subtracting the radii first, as \(\pi(30 - 18)^2 \times 45\), is a common error and gives a far smaller answer; the squares must be subtracted, not the radii.

(b) Volume in cm³. [2]

\[25920\pi = 81\,430.09\ldots\ \text{mm}^3\]

Since \(1\ \text{cm} = 10\ \text{mm}\), one cubic centimetre is \(10 \times 10 \times 10 = 1000\) mm³:

\[\frac{81\,430.09}{1000} = 81.43009\ \text{cm}^3 = 81.4\ \text{cm}^3\ \text{(3 s.f.)}\]

The conversion factor for volume is 1000, not 10; that is the step this part is testing.

(c) Mass of one bush. [2]
Mass is density times volume:

\[m = 81.43009 \times 7.8 = 635.155\ldots = 635\ \text{g (nearest gram)}\]

(d) Cost of the steel in one bush. [2]
The price is per kilogram, so convert the mass:

\[635.155\ \text{g} = 0.635155\ \text{kg}\] \[\text{cost} = 0.635155 \times 4.20 = 2.6677\ldots\]

The steel costs £2.67 to the nearest penny.

(e) Steel cost as a percentage of the selling price. [1]

\[\frac{2.6677}{9.50} \times 100 = 28.08\ldots = 28.1\%\ \text{(3 s.f.)}\]

Examination point: keep the exact form \(25920\pi\) until a decimal is actually needed. Rounding to 81 400 mm³ at the end of part (a) would shift the mass by about 0.2 g and could change the rounded cost in part (d).

Read lesson note on Measures, Money And Accuracy (PearsonEdexcel)

Answer Details