A bicycle repair shop buys chains in bulk. Each chain has a breaking strength of \(8.4\times10^{3}\) newtons. The shop tests a batch of \(2.5\times10^{2}\) ...
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A bicycle repair shop buys chains in bulk. Each chain has a breaking strength of \(8.4\times10^{3}\) newtons. The shop tests a batch of \(2.5\times10^{2}\) chains together, adding their individual breaking strengths, to check against the supplier\'s claimed batch strength of \(2\times10^{6}\) newtons.
- Work out the actual total breaking strength of the batch, giving your answer in standard form. (3)
- State, with a reason, whether the batch meets the supplier\'s claim. (2)
Answer Details
(a) To multiply the two standard-form quantities, multiply the decimal parts and the powers of \(10\) separately. The decimal parts give \(8.4\times2.5=21\). [1 mark] The powers of \(10\) combine by adding their indices: \(10^{3}\times10^{2}=10^{5}\). [1 mark] Combining, the total is \(21\times10^{5}\); since \(21\) is not between \(1\) and \(10\), this must be adjusted to standard form: \(21\times10^{5}=2.1\times10^{6}\) newtons. [1 mark]
(b) Comparing the actual total, \(2.1\times10^{6}\) newtons, with the supplier's claimed \(2\times10^{6}\) newtons: since \(2.1\times10^{6} > 2\times10^{6}\), the batch meets, and slightly exceeds, the supplier's claim. [2 marks]
Writing \(21\times10^{5}\) as \(2.1\times10^{6}\) is not just presentation; a coefficient outside the range \(1\) to \(10\) is not valid standard form, and the shift from \(10^{5}\) to \(10^{6}\) compensates exactly for moving the decimal point in \(21\).
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