📐 Problems in Real-World Contexts — Revision Notes

Walkthrough 3 — how big an air-conditioner does the room need? Basic

1. 3900 mm = 3.9 m and 3000 mm = 3.0 m, so the floor area is 3.9 × 3.0 = 11.7 m².
   The formula is written in m² and the room is given in mm, so the conversion comes first: divide each length by 1000, not the area. In millimetres the area would be 11 700 000 and the capacity would run into the millions.
2. With direct sunlight, k = 450 × 1.12 = 504.
   "Increases by 12%" means multiply by 1.12, not add 12. The sunlight condition is one clause in the stem, so it is worth checking whether the stem has changed a constant before the constant is used.
3. Required capacity = 504 × 11.7 = 5896.8 BTU
   Substitute both worked values into the formula exactly as it is printed. Keep 5896.8 in full: the rounding instruction applies to the answer, not to the working.
4. To the nearest thousand, Nurul needs an air-conditioner of at least 6000 BTU.
   5896.8 is nearer to 6000 than to 5000, so it rounds up — and the rounding makes physical sense, since a unit rated below what the room needs will not cool it. This number filters the table: any model below 6000 BTU is out.

Walkthrough 4 — the life-cycle cost of one model Intermediate

1. Price paid = 869 × 1.09 = $947.21, and installation is free, so the whole up-front cost is $947.21.
   Model F's price is quoted excluding GST, so it must be brought up to what Nurul actually pays before it can join anything else. Multiplying by 1.09 does that in one step. 869 × 1.09 = 947.21 exactly, so no rounding is involved.
2. Annual energy cost = 731 × $0.32 = $233.92
   32 cents is $0.32; multiplying by 32 would give an answer a hundred times too big. Working in dollars from the start keeps every later line in dollars too.
3. Energy cost over the lifespan = $233.92 × 8 = $1871.36
   This is the definition the question supplied, used exactly as printed: annual energy cost × average lifespan. The full $233.92 is carried forward; rounded to $234 first, the eight-year total would read $1872.00, 64 cents adrift before the comparison starts.
4. Life-cycle cost = $947.21 + $1871.36 = $2818.57
   Purchase price (including installation) plus energy cost over the lifespan, as defined. Notice how little of it is the price: two thirds of what Model F really costs Nurul is electricity she has not bought yet. That is the whole reason the question exists.

Walkthrough 5 — which model should Nurul buy? Advanced

1. Model B is only 5000 BTU, so it is too small and is rejected. Models A, C, D, E and F all meet the 6000 BTU requirement.
   Filter before costing. Model B is the cheapest in the table at $689 and the one model that cannot do the job, so the question is testing whether the requirement was read. Writing down why it is rejected is worth a mark.
2. Put every price on the same footing (all inclusive of GST):
   A $749.00  ·  C $899 + $155 = $1054.00  ·  D 815 × 1.09 + 160 = $1048.35  ·  E (699 + 140) × 1.09 = $914.51  ·  F 869 × 1.09 = $947.21
   Four of these five need work: D's price excludes GST while its installation fee includes it, so only the price is multiplied; E has GST missing from both. One deliberate conversion line keeps an inclusive price from being added to an exclusive one.
3. Annual energy cost = kWh × $0.32:
   A $270.40  ·  C $224.64  ·  D $309.76  ·  E $242.56  ·  F $233.92
   One multiplication per model, all at the same rate. Doing them together, in a row, makes the pattern obvious: the models are close on price but spread over $85 a year on electricity, and eight years multiplies that spread by eight.
4. Life-cycle cost = up-front + 8 × annual:
   A 749.00 + 2163.20 = $2912.20  ·  C 1054.00 + 1797.12 = $2851.12  ·  D 1048.35 + 2478.08 = $3526.43
   E 914.51 + 1940.48 = $2854.99  ·  F 947.21 + 1871.36 = $2818.57
   All five, not just the ones that look promising. The order has completely changed: Model A, much the cheapest to buy at $749, finishes fourth, because it is the thirstiest model that meets the requirement. Nothing in the price column predicted that.
5. $2818.57 < $2851.12 < $2854.99 < $2912.20 < $3526.43
   Rank them, so the comparison is on the page as a comparison. It also shows how close the top three are — $36 apart over eight years — which is exactly why this cannot be settled by eye.
6. Nurul should buy Model F. Over 8 years it costs $2818.57, the lowest of the five models that provide at least 6000 BTU; Model B was rejected as too small.
   The full-mark sentence: the name, the number, the comparison it won, and the option that was ruled out and why. Everything the marker needs is in one sentence, and none of it is left for them to infer.

Walkthrough 6 — how long before the efficient model wins? Advanced

1. For y years, total cost = up-front + y × annual energy cost:
   C: 1054.00 + 224.64y  ·  F: 947.21 + 233.92y  ·  E: 914.51 + 242.56y  ·  A: 749.00 + 270.40y  ·  D: 1048.35 + 309.76y
   Replacing the 8 by a variable y turns each model into a straight line: a fixed part and a part that grows every year. Model C's line starts highest of the three contenders and grows slowest, which is how it can end up lowest.
2. C is cheaper than F when 1054.00 + 224.64y < 947.21 + 233.92y, i.e. 106.79 < 9.28y, so y > 11.5 (3 s.f.).
   Model F is the one to beat, because Walkthrough 5 showed it winning at 8 years. Rearranging gives a linear inequality: C costs $106.79 more up front and saves $9.28 a year, so it needs about 11.5 years of saving to catch up.
3. Against the others, C wins sooner: over E after $\frac{139.49}{17.92}$ = 7.78 years, and over A after $\frac{305.00}{45.76}$ = 6.67 years; it is cheaper than D from the first year.
   "Lowest among the models that meet her requirement" means lowest against all of them, so every rival is checked, not just the current leader. All three thresholds here are below 11.5, so Model F is the binding one.
4. y must be a whole number of years greater than 11.5, so the least is 12 years. Check: at 12 years C costs $3749.68 and F costs $3754.25, so C is cheaper; at 11 years C costs $3525.04 and F costs $3520.33, so F is still ahead.
   The inequality gives 11.5, and a whole number of years is wanted, so it rounds up to 12 — 11 years is not enough. Checking both 11 and 12 puts the boundary beyond doubt, which is what makes this a full answer.

Walkthrough 7 — instalment plan or pay cash? Intermediate

1. Cash price = $899 + $155 = $1054.00
   Both figures already include GST, so they simply add — no conversion is needed here. Establishing the cash price first gives the comparison something to be measured against.
2. Total paid on the plan = $154 + 12 × $85 = 154 + 1020 = $1174.00
   The deposit is part of what she pays, so it goes into the total. A plan's headline is always the monthly figure, because $85 sounds small; the total is the number that can be compared.
3. Extra paid = $1174.00 − $1054.00 = $120.00
   The difference of the two totals — the price of spreading the payments out. It is a plain subtraction, and it is the figure the rest of the question is about.
4. As a percentage of the cash price: $\frac{120}{1054}\times 100$% = 11.4% (3 s.f.)
   "As a percentage of the cash price" names the denominator, so use it. The answer is not exact, so the paper's default applies and it is given to 3 significant figures.
5. Nurul is understating it. She only borrows $1054 − $154 = $900, and $120 on $900 is $\frac{120}{900}\times 100$% = 13.3% (3 s.f.) — and that is for one year, not for the eight years she will own the machine.
   "Comment on her statement" wants a judgement backed by a number. The deposit is money she already had, so it is not borrowed; measuring the charge against what she actually borrowed gives the fairer figure, and naming the time turns a calculation into a comment.

Walkthrough 8 — the obvious arrangement Basic

1. Layers: $\frac{19}{9.2}$ = 2.06…, so 2 complete layers fit.
   The tins stand upright, so it is the height of the tin that divides into the height of the carton. Round down: two thirds of a third layer is no layer, and the leftover 0.6 cm of headroom is wasted.
2. Along the 58 cm side: $\frac{58}{6.4}$ = 9.06…, so 9 tins (using 57.6 cm and leaving 0.4 cm).
   Seen from above a tin occupies a 6.4 cm by 6.4 cm square, so its diameter divides into each side of the carton. Round down again, and write down the 0.4 cm left over: leftover space is what the next arrangement will try to use.
3. Across the 44 cm side: $\frac{44}{6.4}$ = 6.875, so 6 rows (using 38.4 cm and leaving 5.6 cm).
   The same division on the other side. This is the interesting one: 5.6 cm of the carton's breadth is unused, nearly a whole extra row, and a seventh row would need 44.8 cm. That wasted strip is exactly what Walkthrough 10 goes after.
4. Tins per layer = 9 × 6 = 54, so the carton holds 2 × 54 = 108 tins.
   Multiply the three counts, not the three lengths. A check: the carton is 58 × 44 × 19 = 48 488 cm³ and 108 tins occupy 108 × π × 3.2² × 9.2 ≈ 31 964 cm³, 66% of it, because round tins leave gaps.

Walkthrough 9 — how far apart are staggered rows? Intermediate

1. Each side of △ABC joins the centres of two touching tins, so AB = BC = CA = 6.4 cm.
   Two circles that touch have their centres a full diameter apart — radius plus radius. This is the step that converts a picture of tins into a triangle with known sides; everything after it is ordinary geometry.
2. Since AD ⊥ BC in an equilateral triangle, D is the midpoint of BC, so BD = $\frac{6.4}{2}$ = 3.2 cm.
   The perpendicular from a vertex of an equilateral triangle bisects the opposite side. Without that fact you would have an unknown at both ends of BC; with it, △ABD is right-angled with two sides known.
3. By Pythagoras' theorem in △ABD: AD² = 6.4² − 3.2² = 40.96 − 10.24 = 30.72
   AB is the hypotenuse — it is opposite the right angle — so the square of the shorter side is subtracted. Getting this the wrong way round gives AD² = 51.2 and a row spacing bigger than the tins themselves, which is impossible.
4. AD = $\sqrt{30.72}$ = 5.543 cm (4 s.f.)
   Four significant figures because the question said so, overriding the paper's usual three. Keep the unrounded 5.5425… for the counting in Walkthrough 10: over six row gaps the difference from 5.543 is small, but carrying the full value is the habit that pays elsewhere.

Walkthrough 10 — which arrangement packs the most? Advanced

1. In Arrangement 2, n rows need 6.4 + (n − 1) × 5.543 cm across the 44 cm breadth.
   The first row needs a full diameter, 6.4 cm: its centre sits 3.2 cm from the wall and 3.2 cm from the row above. Each extra row adds only the row spacing, not another diameter; counting 6.4 cm per row gives the grid answer back.
2. 6.4 + (n − 1) × 5.543 ≤ 44 gives n − 1 ≤ 6.78, so n = 7 rows (using 39.66 cm; an eighth row would need 45.20 cm).
   Solve it as an inequality and round down, then confirm the next row really does not fit — 45.20 cm is more than 44 cm, so 7 is the maximum. Stating the failed eighth row is what turns "7" into a justified 7.
3. Odd-numbered rows hold 9 tins as before. An offset row starts 3.2 cm further along, so it needs 3.2 + 8 × 6.4 = 54.4 cm for 8 tins; 9 tins would need 3.2 + 9 × 6.4 = 60.8 cm > 58 cm.
   The stagger buys space in one direction and spends it in the other: offset rows are pushed half a tin along, so they lose their last tin. A solution that assumes 9 tins in every row is over-counting.
4. Seven rows alternate 9, 8, 9, 8, 9, 8, 9, so a layer holds 4 × 9 + 3 × 8 = 36 + 24 = 60 tins.
   Four full rows and three offset rows, because seven is odd and both the first and the last row are full ones. Starting and ending on a full row is worth an extra tin, so the count is done row by row.
5. Arrangement 2 holds 2 × 60 = 120 tins; Arrangement 1 holds 108 tins.
   Both arrangements are two layers deep, since the stagger changes nothing about the height. Now both totals are on the page, in the same units, ready to be compared.
6. Arrangement 2 packs more: 120 tins against 108, an extra 12 tins — about 11% more — in exactly the same carton.
   Name the arrangement, give both totals, and quantify the gain. The reason is worth one more clause: staggering makes the rows only 5.543 cm apart instead of 6.4 cm, so a seventh row fits, and that gains more than the offset rows lose.

Walkthrough 11 — how many lots, and can you do better? Advanced

1. A module is 5 + 6 + 5 = 16 m deep and can be as long as the aisle it serves.
   Naming the repeating unit first turns a design problem into two divisions: how many modules fit across the plot, and how many lots fit along each one. Every layout below is the same calculation with the 36 and the 26 swapped.
2. Layout A. The two accessways take 3 + 3 = 6 m out of the 36 m, so the aisle is 30 m long and each row holds $\frac{30}{2.5}$ = 12 lots.
   The accessways sit at the ends of the aisle here, so they eat into its length, not into the depth of the plot. 30 ÷ 2.5 = 12 exactly, with nothing wasted along that side.
3. Modules across the 26 m: $\frac{26}{16}$ = 1.625, so only 1 module fits. Layout A holds 2 × 12 = 24 lots, and 10 m of the plot is unused.
   Round down — 0.625 of a module is not a module — and then say what happens to the remainder. Ten metres is not enough for a second module, so the whole strip is dead ground.
4. Layout B. The accessways take 6 m out of the 26 m, so each aisle is 20 m long and holds $\frac{20}{2.5}$ = 8 lots per row; modules across the 36 m: $\frac{36}{16}$ = 2.25, so 2 modules. Layout B holds 4 × 8 = 32 lots, with 4 m spare.
   Exactly the same two divisions with the sides exchanged. Turning the aisles through 90° costs 4 lots per row but buys a whole second module, and the second module is worth 16 lots — so the swap gains 8 overall.
5. 32 > 24, so Layout B holds more, by 8 lots.
   The comparison, stated as a comparison, with the difference. Both totals are on the page and both were reached the same way, so a marker can see the method as well as the answer.
6. Layout C. Make the aisles two-way (6.6 m) so one accessway serves everything: modules are 5 + 6.6 + 5 = 16.6 m and $\frac{36}{16.6}$ = 2.16… still gives 2 modules, while each aisle is now 26 − 3 = 23 m long and holds $\frac{23}{2.5}$ = 9.2 → 9 lots. Layout C holds 4 × 9 = 36 lots, four more than Layout B.
   Widening the aisles by 0.6 m costs 1.2 m of the plot, which Layout B's 4 m of slack absorbs; dropping one accessway lengthens every aisle by 3 m, an extra lot per row. A proposal needs the count and the check that it fits.

Walkthrough 12 — hitting, and then holding Basic

1. P(hit Disc 4) = $\frac{1}{5}$ × 0.75 = 0.15
   "Proportional to area" means the probability is multiplied by exactly the same factor as the area. Disc 4 has one fifth of the area, so one fifth of the probability. No circle formula is needed — only the ratio, which the question has already given.
2. The three multipliers are: Disc 4 → 0.9, orange → 0.5, gold ring → 0.8.
   Read one value from each of the three tables, checking each comes from the right row. Writing them out before multiplying catches the kind of error that picks 0.45 (Disc 1) instead of 0.9 (Disc 4).
3. P(hit and hold) = 0.15 × 0.9 × 0.5 × 0.8
   Four independent stages, so all four probabilities multiply — the chance of hitting, then the three separate reasons the ring might come off. Leaving the hitting probability out answers a different question: "given that she hits, does it hold?".
4. = 0.135 × 0.5 × 0.8 = 0.0675 × 0.8 = 0.054 (shown)
   A "show that" is answered only if every intermediate value is on the page: the marker is checking the route, since the destination was printed in the question. Finish with the word "shown".

Walkthrough 13 — when should she throw? Advanced

1. Both cycles have length 4, so from a red Disc 3 the states are:
   t = 0 red Disc 3  ·  t = 1 green Disc 4  ·  t = 2 yellow Disc 1  ·  t = 3 orange Disc 2  ·  t = 4 red Disc 3 again.
   Both wheels turn once per second and both have four positions, so they stay locked together and the whole pattern repeats every 4 seconds. That is what makes the question finite: there are only four different situations to test, not infinitely many.
2. At t = 0, red Disc 3: P = 0.375 × 0.75 × 0.35 × 0.7 = 0.0689 (3 s.f.)
   The full product for the first state: P(hit Disc 3) = 0.5 × 0.75 = 0.375, then the size multiplier 0.75, red's colour multiplier 0.35, and the silver ring's 0.7. Setting the pattern out clearly once saves time on the other three states.
3. At t = 1, green Disc 4: P = 0.15 × 0.9 × 0.85 × 0.7 = 0.0803 (3 s.f.)
   Green is the kindest colour and Disc 4 the best at holding a ring — but Disc 4 is also the hardest to hit, at only 0.15. The two effects nearly cancel, which is exactly why this state has to be computed rather than judged.
4. At t = 2, yellow Disc 1: P = 0.75 × 0.45 × 0.65 × 0.7 = 0.154 (3 s.f.)
   The biggest disc is much the easiest to hit at 0.75, and yellow is the second-best colour. Its holding multiplier is the worst at 0.45, but 0.75 is five times Disc 4's hitting probability while 0.45 is only half of 0.9 — so hitting wins.
5. At t = 3, orange Disc 2: P = 0.6 × 0.6 × 0.5 × 0.7 = 0.126
   The fourth and last state. All four are now on the page, worked the same way, which is what "show your calculations clearly" is asking for — a single winning line with no rivals beside it is not a justification.
6. 0.154 > 0.126 > 0.0803 > 0.0689, so t = 2 is the best of the four.
   Rank them so the comparison is explicit. The best chance is more than twice the worst, so the timing genuinely matters — worth saying, because it tells the reader the decision was not a coin toss.
7. Rania should throw at t = 2 seconds, when the disc is a yellow Disc 1, giving a probability of about 0.154. Because the pattern repeats every 4 seconds, t = 6, 10, 14, … are equally good.
   Name the time, describe the state it corresponds to, and give the probability. The question asked when, and the honest answer is a repeating set of times rather than a single instant she might already have missed.

Walkthrough 14 — which shoe should the company make? Advanced

1. Shoe R: ∑fx = 0 + 40 + 1200 + 1480 + 50 = 2770 and ∑fx² = 0 + 80 + 3600 + 5920 + 250 = 9850, with ∑f = 800.
   Both formulae on the paper's MATHEMATICAL FORMULAE page need exactly these three totals, so build them first, in a column. ∑fx² uses f × x² — square the rating, then multiply by the frequency.
2. Shoe R: mean = $\frac{2770}{800}$ = 3.4625 = 3.46 (3 s.f.), and s.d. = $\sqrt{\frac{9850}{800}-{3.4625}^2}$ = $\sqrt{\mathrm{0.32359\dots }}$ = 0.569 (3 s.f.)
   Substitute the unrounded mean, 3.4625, into the standard deviation — using 3.46 changes the third figure of the answer. Round both results only at the end, to the 3 significant figures the question asked for.
3. Shoe S: ∑fx = 210 + 100 + 45 + 1060 + 1300 = 2715 and ∑fx² = 210 + 200 + 135 + 4240 + 6500 = 11 285, so mean = 3.39 (3 s.f.) and s.d. = 1.61 (3 s.f.).
   The identical calculation on the second table. The two standard deviations are the striking part: 1.61 against 0.569, nearly three times as spread out, which is the split-opinion shape showing up as a number.
4. 3.46 > 3.39, so on the means Shoe R scores higher, by 0.07 of a rating point.
   Answer the question that was asked — "by comparing the means" — and state the comparison with its size. The gap is tiny, and saying so is honest reporting: the mean alone is a thin basis for a decision.
5. The company should make Shoe R, because its mean rating of 3.46 is higher than Shoe S's 3.39, and its standard deviation of 0.569 is much smaller, so opinions of it are far more consistent.
   Two sentences of comparison — one about the average, one about the spread, both in context — is the standard shape of a full-mark comparison in em-c13, and it is the shape expected here too.
6. Rated 4 or 5: Shoe R has 370 + 10 = 380, which is $\frac{380}{800}\times 100$% = 47.5%; Shoe S has 265 + 260 = 525, which is 65.6% (3 s.f.).
   A different criterion means a fresh calculation from the same table: count the two top rows rather than averaging all five. The neutral 3s that propped up Shoe R's mean count for nothing, because a neutral person does not buy a shoe.
7. On this criterion the company should make Shoe S: 525 people (65.6%) would buy it against only 380 (47.5%) for Shoe R — 145 more customers. The recommendation is the opposite of the one based on the means.
   The decision flips, and saying so is the answer. Shoe R wins on the mean because 400 people sat on the fence; Shoe S wins on sales because far more were enthusiastic. Each answers its own question, so state which criterion you used.

Walkthrough 15 — from a table to a picture, and then to a sentence Intermediate

1. Year 2: every one of its ten weekly figures is larger than every figure in either other year.
   This one really can be read off the table, and the reason must be given — not "the numbers look bigger" but "every value exceeds every value in the other two rows". When one row dominates another value by value, no arithmetic is needed.
2. Year 1 and Year 3 cannot be separated by eye: Year 1 is higher in weeks 2, 4, 7 and 9, and Year 3 is higher in the other six weeks.
   The honest answer to "can you see this easily?" is no: the two rows swap places repeatedly. Counting the weeks each year wins suggests Year 3 — it wins 6 of the 10 — and that is wrong, so week-counting is not a method.
3. Total the rows: Year 1 = 1542, Year 3 = 1523, so Year 3 had the fewest, by only 19 hires.
   The totals settle it, and they contradict the week-counting: Year 1 wins fewer weeks but wins them by more. A gap of 19 in over 1500 is under 1.3%, so it was never going to be visible in ten separate comparisons.
4. A better picture is a graph of the running total — the cumulative number of hires — for each year, plotted against week.
   This is the representation-choosing mark. A running-total graph turns a comparison of ten pairs of numbers into a comparison of where three lines finish, and running totals can never fall, so the lines cannot cross back and forth the way the weekly ones do.
5. Year 2 rose then levelled off. Hires climbed steadily from 231 in week 1 to 281 in week 6, then stayed between 274 and 292 for the rest of the period — an increase of about 50 over the first six weeks and almost none after that.
   A trend is described in words plus evidence: the shape ("rose then levelled off"), the figures that show it, and the period they cover. "It increased" misses the flattening, which is the interesting half.

Walkthrough 16 — building M, and reading M² Intermediate✕ Not in G2

1. $$M=\left(\begin{array}{cccc}0& 2& 1& 0\\ 1& 0& 0& 1\\ 1& 1& 0& 1\\ 0& 1& 1& 0\end{array}\right)$$ with rows and columns in the order A, B, C, D.
   Read the diagram arrow by arrow, filling the row of the island each arrow leaves. Saying which order the rows and columns are in is part of the answer: the same ferries in a different order are a different matrix.
2. The entry 2 in row A, column B means there are two different sailings from Marina to Kusu that day.
   An incidence matrix counts services, not merely records whether a connection exists, so entries larger than 1 are meaningful. This will matter later: two sailings on one leg double the number of complete journeys through it.
3. $$M^2=MM=\left(\begin{array}{cccc}3& 1& 0& 3\\ 0& 3& 2& 0\\ 1& 3& 2& 1\\ 2& 1& 0& 2\end{array}\right)$$
   Ordinary matrix multiplication — row of the first times column of the second. The AD entry, for instance, is (0)(0) + (2)(1) + (1)(1) + (0)(0) = 3, each term pairing a sailing out of A with a sailing on to D.
4. The entry in row i, column j of M² is the number of ways to travel from island i to island j using exactly two sailings. So there are 3 ways from A to D with two sailings, and none from A to C.
   Say what the entries count, in the language of the question — "ways to travel using exactly two sailings", not "the square of the connections". The zero at row A, column C says there is no two-sailing route from Marina to Lazarus.

Walkthrough 17 — how many routes, and which ones? Advanced✕ Not in G2

1. One sailing: the AD entry of M is 0. Two sailings: the AD entry of M² is 3.
   "At most three" means one, two or three, so each has to be counted separately and then added. The zero confirms what the diagram showed — no direct sailing — so every journey must change ferries at least once.
2. Three sailings: the AD entry of M³ = M²M is
   (3)(0) + (1)(1) + (0)(1) + (3)(0) = 1.
   Only one entry of M³ is needed, so only row A of M² is multiplied by column D of M — computing the whole of M³ would be sixteen times the work for the same mark.
3. Total ways with at most three sailings = 0 + 3 + 1 = 4, so Dinesh is not correct.
   Adding the three entries is the same as reading the AD entry of M + M² + M³. Four is more than the two he claimed, so the rest of the question is to find which routes he has missed.
4. Listing them: A → B → D counts twice, because there are two sailings from A to B; A → C → D counts once; and the third sailing gives A → C → B → D.
   The two things Dinesh missed are of different kinds. One is a whole route he never thought of; the other is that a route he did think of can be travelled two ways, because two ferries make the first leg.
5. There are 4 ways, not 2: A → B → D (on either of the two A–B sailings), A → C → D, and A → C → B → D.
   The count and the list, together. The matrix gives the number and guarantees nothing has been missed; the diagram gives the routes themselves. A route he has now been shown turns out in the next section to be the cheapest of the lot.

Walkthrough 18 — two routes for a tunnel Advanced

1. By Pythagoras' theorem, PQ² = 480² + 315² = 230 400 + 99 225 = 329 625, so PQ = 574 m (3 s.f.).
   △PRQ is right-angled at R, so PQ is the hypotenuse and the two squares are added. Keep the unrounded 574.129… in the calculator: it is about to be divided and then multiplied by a four-figure rate, where rounding early would show.
2. The direct tunnel lies entirely in the granite, so time = $\frac{\mathrm{574.129\dots }}{1.5}$ = 383 hours (3 s.f.).
   The straight line from P to Q starts on the boundary and immediately enters the granite, so all 574 m is drilled at the slow rate. Time = distance ÷ speed, so divide by 1.5.
3. Its cost is 574.129… × $1600 = $919 000 (3 s.f.).
   With cost per metre already worked out, the cost is one multiplication. As 383 hours × $2400 it gives $919 200, which rounds the same to 3 significant figures here but is $592 out — an error that grows each time it is reused.
4. P to R: 480 m of shale at $500/m = $240 000 (taking 480 ÷ 2.5 = 192 hours).
   PR runs along the boundary, so this leg is all in the soft rock — which is both faster and cheaper. Its length is given, so no geometry is needed here at all.
5. R to Q: 315 m of granite at $1600/m = $504 000 (taking 315 ÷ 1.5 = 210 hours).
   The second leg is the full width of the granite, at the slow rate. The total distance is now 480 + 315 = 795 m against the direct route's 574 m, nearly 40% further, so this route can only win on the cheaper rock.
6. Total = $240 000 + $504 000 = $744 000, taking 192 + 210 = 402 hours. This is $175 000 cheaper (3 s.f.) than drilling direct, although it takes about 19 hours longer.
   Both totals, the comparison, and the size of it. The cheaper route is also the slower one, so a company in a hurry might still drill direct: a question that gives you both time and cost usually wants that tension named.