Time

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 the compound-interest work in 1.13, applied \(n\) times instead of once — so if you have followed the Extended route this far, you already own the machinery. 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 \(50-44=6\). \(\frac78=7\div8=0.875\), and \(0.875\times100=87.5\%\). \(\frac{17}{5}=3\frac25\), because \(17=3\times5+2\). 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. \(41.3\to40\), \(9.79\to10\) and \(0.765\to0.8\), giving \(\dfrac{40}{10\times0.8}=\dfrac{40}{8}=5\). Half of \(0.1\) is \(0.05\), so \(4.65\le m<4.75\) kg. The lower bound is \(4.65\) kg and the upper bound is \(4.75\) kg; the lower is included and the upper excluded. Total parts \(=5+7+14=26\); one part \(=936\div26=36\); shares \(=180\), \(252\) and \(504\). (Check: \(180+252+504=936\).) \(3\) h \(20\) min \(=3\frac13\) h \(=\frac{10}{3}\) h, so the speed is \(240\div\frac{10}{3}=240\times\frac{3}{10}=72\) km/h. (Check: \(72\times\frac{10}{3}=240\).) A \(15\%\) reduction multiplies by \(0.85\): \(80\times0.85=\$68\). Compound interest multiplies by \(1.05\) each year: \(600\times1.05^2=600\times1.1025=\$661.50\).

Centre \(=5\); pair-only regions \(=7\), \(4\) and \(6\); single-only regions \(=30-(7+6+5)=12\), \(25-(7+4+5)=9\) and \(22-(6+4+5)=7\). Union \(=12+9+7+7+4+6+5=50\), so \(60-50=10\) play none. Let \(x=0.363636\ldots\); then \(100x=36.363636\ldots\), so \(99x=36\) and \(x=\dfrac{36}{99}=\dfrac{4}{11}\). The denominator is the root and the numerator the power: \(81^{3/4}=\left(\sqrt[4]{81}\right)^3=3^3=27\). \(14.5\le L<15.5\) and \(7.5\le W<8.5\). Upper perimeter \(=2(15.5+8.5)=48\) cm. Lower area \(=14.5\times7.5=108.75\ \mathrm{cm^2}\). The increase multiplied the original by \(1.20\), so the original is \(156\div1.20=\$130\). (Check: \(20\%\) of \(130\) is \(26\), and \(130+26=156\).) \(1.08\times0.85=0.918\), so the single multiplier is \(0.918\) — an overall decrease of \(8.2\%\). Depreciation multiplies by \(0.88\) each year: \(8500\times0.88^3=8500\times0.681472=5792.512\), so \(\$5793\) to the nearest dollar. \(288=144\times2\) and \(144\) is the largest square factor, so \(\sqrt{288}=12\sqrt2\). For the second part, \(\dfrac{4}{\sqrt7+1}\times\dfrac{\sqrt7-1}{\sqrt7-1}=\dfrac{4\left(\sqrt7-1\right)}{7-1}=\dfrac{2\left(\sqrt7-1\right)}{3}\).