Time

Topic 1 is not eighteen unrelated skills. It is one habit applied eighteen times: choose the representation that makes the question easy, work exactly for as long as you can, and round only once, at the end. A number can appear as a product of primes, a fraction, a decimal, a percentage, a power, a standard-form pair, a ratio, a rate, a bound or a surd — and every calculation in Chapter 1 becomes short the moment you pick the right one. Every later chapter draws on this: algebra needs indices and exact fractions, mensuration needs ratio, rates and bounds, and statistics needs proportional reasoning.

Every rate question reduces to “how much of this, per one of that?” — so the safe method is always to divide the total of the first quantity by the total of the second. For speed this gives the one rule that matters: average speed is total distance divided by total time, and it is not the mean of the separate speeds unless the journey happened to spend equal times at each.

Learn the multiplier and nine separate percentage “types” collapse into one idea. An increase of \(r\%\) is \(\times\left(1+\frac r{100}\right)\); a decrease is \(\times\left(1-\frac r{100}\right)\); repeated changes multiply their multipliers; \(n\) equal changes give \(\times(1\pm r)^n\); and a reverse percentage divides by the multiplier. Compound interest, depreciation, repeated change, discount and reverse problems are all the same sentence with different numbers.

Time is base 60, so never subtract times as though they were decimals. \(14\,10-13\,50\) is not “\(0.60\)” and \(3.45\) hours is not 3 h 45 min. The safe method is to count forward in stages: from the start to the next whole hour, then whole hours, then the remaining minutes — and if the interval crosses midnight, use midnight itself as one of the boundaries.

This is the same multiplier as 1.13, applied \(n\) times instead of once. The only new decisions are: what is one period, and how many complete periods have passed. Once those are fixed, the calculation is a single line, and the difference between exponential and linear change is that each step is a percentage of the current value, not of the starting value.

\(540=2^2\times3^3\times5\) and \(756=2^2\times3^3\times7\). The common primes are 2 and 3 at powers 2 and 3, so \(\text{HCF}=2^2\times3^3=4\times27=108\). (Check: \(540\div108=5\) and \(756\div108=7\).) \(n(A\cup B)=31+28-15=44\), so the number in neither is \(50-44=6\). Let \(x=0.363636\ldots\); then \(100x=36.363636\ldots\), so \(99x=36\) and \(x=\dfrac{36}{99}=\dfrac{4}{11}\). Same base throughout, so add and subtract the indices: \(-3+7-2=2\), giving \(x^{2}\) for \(x\ne0\). \(8\times3=24\) and \(10^{-4}\times10^{7}=10^{3}\), so the product is \(24\times10^{3}=2.4\times10^{4}\) after renormalising. Half of \(0.1\) is \(0.05\), so \(4.65\le m<4.75\) kg. The lower bound is included and the upper bound excluded. Total parts \(=5+7+14=26\); one part \(=936\div26=36\); shares \(=180\), \(252\) and \(504\). (Check: \(180+252+504=936\).) The increase multiplied the original by \(1.20\), so the original is \(156\div1.20=130\). (Check: \(20\%\) of \(130\) is \(26\), and \(130+26=156\).) \(288=144\times2\) and \(144\) is the largest square factor, so \(\sqrt{288}=12\sqrt2\). \(\dfrac{4}{\sqrt7+1}\times\dfrac{\sqrt7-1}{\sqrt7-1}=\dfrac{4\left(\sqrt7-1\right)}{7-1}=\dfrac{4\left(\sqrt7-1\right)}{6}=\dfrac{2\left(\sqrt7-1\right)}{3}\).