Unit 18 · Trapped Between Bounds

Module 13 of 45Lesson 1 of 14 · Unit 18 · Trapped Between Bounds

Number · Unit N18 · Extended

Trapped Between Bounds

A rounded measurement is not a number, it is a trap with two walls. This unit finds both walls exactly, writes them as an inequality, draws them on a number line, and then does the hard part: pushes them through a sum, a difference, a product, a quotient and a formula, so you can say what a calculation built on rounded data is guaranteed to give.

Extended · MYP 4–5~55 minutes60 practice questionsCriteria A · B · C · D

Coming from Unit 17 Accuracy on Trial and Unit 4 Round It Right: you already know that a rounded figure stands for an interval one rounding unit wide — this unit makes that interval formal, gives its two ends names, and calculates with them.

By the end of this unit you can

Write down the lower and upper bounds of a value rounded to a unit, to decimal places or to significant figures, and explain why the interval is half-open
Represent those bounds as an inequality and on a number line
Calculate the bounds of a sum and a difference, using the cross rule
Calculate the bounds of a product and a quotient, using the flip — the divisor takes the bound at the opposite end
Calculate the bounds of a quantity given by a formula with several rounded inputs
Justify a yes-or-no decision from the worst case rather than the nominal case

The one idea behind all of it

Every rounded figure you are handed is really a pair of numbers. “15 m to the nearest metre” is not a length, it is the promise 14.5 ≤ d < 15.5, and the whole of this unit is one habit: carry BOTH walls of that promise through the calculation, decide at every step which wall helps and which hurts, and report the answer as an interval — because that is all the data ever supported.

LB = value − ½ unit · UB = value + ½ unit

and the interval is LB ≤ x < UB — closed at the bottom, open at the top

Four facts do almost all the work in this unit. A rounded value is trapped between two exact bounds half a unit either side. Adding and multiplying pair like with like; subtracting and dividing cross over. Bounds are exact and are never rounded — only the final derived answer is. And a guarantee compares the upper bound of what must fit with the lower bound of what it fits into, never the two published figures.

Give every derived answer to 3 significant figures, unless it is exact — and give every amount of money in pula to 2 decimal places. Round, never truncate. A bound that is exact, such as 14.5 or 3 550 000, is written in full and used in full.

Where this unit sits

This unit stands on N04 Round It Right and N17 Accuracy on Trial. N04 taught you to round to a place value, to decimal places and to significant figures. N17 showed you informally that a rounded figure stands for an interval one rounding unit wide, and that the error travels through a calculation. Both are assumed here and neither is re-taught.

What is new is that the interval becomes the object you calculate with. N17 asked “how far out could this be?” and answered with a percentage. N18 asks “what is the largest the answer could possibly be, and the smallest?” and answers with two exact numbers — the upper bound and the lower bound — which is what an Extended paper asks for and what a tolerance decision actually needs.

It also stands on N08 The Number Line and N14 Between the Lines. LB ≤ x < UB is a double inequality with a closed circle at one end and an open circle at the other, exactly as those units drew it. If the open and closed circles are shaky, go back to N08 first.

The two bounds, and why the interval is half-open

A rounded value is trapped between a lower bound it can reach and an upper bound it cannot

When a figure has been rounded, the value you are given is not the value that was measured: the truth is trapped somewhere in an interval half a rounding unit either side. The bottom of that interval is the lower bound and the top is the upper bound. The lower bound is a value the measurement could really take, so it is included; the upper bound would round to the next figure up, so it is excluded. That is why the interval is written LB ≤ x < UB and is called half-open.

The filled circle is included and the hollow circle is not, which is the whole meaning of “half-open”. 14.5 m rounds to 15 m, so it is allowed; 15.5 m rounds to 16 m, so it is not.

Worked example 1 · What “15 m, to the nearest metre” traps the length between

Name the rounding unit that was used

1 m

Half the unit is the most the figure can be out by

0.5 m

The lower bound: half a unit below

15 − 0.5 = 14.5 m

The upper bound: half a unit above

15 + 0.5 = 15.5 m

Write it as one inequality

14.5 ≤ d < 15.5

Why the second sign is strict, not ≤

15.5 rounds to 16, not to 15

Two bounds and two different signs. The lower bound is a value the length can actually take: 14.5 m rounds to 15 m, so it is allowed, and that is why it gets ≤. The upper bound is a value the length can never take: 14.5 rounds up but so does 15.5, straight past 15 to 16, so 15.5 is excluded and gets <. The interval is half-open, it is written [14.5, 15.5), and it is exactly one rounding unit wide. Write both bounds down before you calculate anything, every single time.

The same 12 m, rounded three ways. Each bar is everything the figure allows the true length to be. The filled circle at the left is the lower bound, which the length can actually reach; the hollow circle at the right is the upper bound, which it cannot. The bar is exactly as wide as the unit it was rounded to.

As published	Rounded to	Rounding unit	The interval	Width
15 m	nearest metre	1 m	14.5 ≤ d < 15.5	1 m
4.7 kg	nearest 0.1 kg	0.1 kg	4.65 ≤ m < 4.75	0.1 kg
12 m	nearest 0.5 m	0.5 m	11.75 ≤ r < 12.25	0.5 m
2 500 people	nearest 100	100	2 450 ≤ n < 2 550	100
7.30 s	2 decimal places	0.01 s	7.295 ≤ t < 7.305	0.01 s
3 600 000 m²	2 significant figures	100 000 m²	3 550 000 ≤ A < 3 650 000	100 000 m²

EXAMPLEA rope is 12 m to the nearest 0.5 m. Half the unit is 0.25 m, so 11.75 ≤ r < 12.25. The same rope to the nearest metre would only be pinned between 11.5 m and 12.5 m — twice as loose. A finer rounding unit is a stronger claim, and the width of the interval is always exactly the rounding unit, never half of it.

Practice · The two bounds, and why the interval is half-open · 10 questions

P1A length is given as 15 m, to the nearest metre. Write down the lower bound.Fluency

Answer14.5 m

P2Write down the upper bound of that same length.Fluency

Answer15.5 m

P3Write down that length as an inequality, using L for the true length.Fluency

Answer14.5 ≤ L < 15.5

P4State why the upper bound itself is not a possible value of L.Fluency

Answerbecause 15.5 would round up to 16, not to 15

P5A mass is recorded as 4.7 kg, to the nearest 0.1 kg. Determine its bounds and the width of the interval.Applied

Answer4.65 kg ≤ m < 4.75 kg; width 0.1 kgHalf the rounding unit is 0.05 kg. Take it off 4.7 for the lower bound and add it on for the upper bound: 4.65 kg and 4.75 kg. The width is 4.75 − 4.65 = 0.1 kg, which is the rounding unit itself — always.

P6A crowd is reported as 2 500 people, to the nearest 100. Determine the bounds.Applied

Answer2 450 ≤ n < 2 550Half of 100 is 50, so the true number is at least 2 450 and less than 2 550. Note that 2 550 is excluded: it would round to 2 600. The interval is 100 people wide, so the report can be 50 people out either way.

P7A rope is measured as 12 m, to the nearest 0.5 m. Determine its bounds.Applied

Answer11.75 m ≤ r < 12.25 mHalf of 0.5 is 0.25. So r is at least 12 − 0.25 = 11.75 m and less than 12 + 0.25 = 12.25 m. A finer rounding unit buys a narrower claim: the same rope to the nearest metre would only be pinned between 11.5 m and 12.5 m.

P8A distance is given as 8.0 km, to 1 decimal place. Determine its bounds, and explain what the zero is doing there.Applied

Answer7.95 km ≤ d < 8.05 kmTo 1 decimal place the rounding unit is 0.1 km, so half of it is 0.05 km and the bounds are 7.95 km and 8.05 km. The zero after the point is not decoration: “8.0 km” claims the distance to within 50 m, while “8 km” to the nearest kilometre only claims it to within 500 m.

P9The lower bound of a length rounded to the nearest 0.2 m is 6.3 m. Determine the length as it was published, and its upper bound.Challenge

Answer6.4 m; upper bound 6.5 mHalf the rounding unit is 0.1 m, and the lower bound sits half a unit below the published figure, so the published length is 6.3 + 0.1 = 6.4 m. The upper bound is half a unit above it: 6.4 + 0.1 = 6.5 m. Check: the interval runs 6.3 to 6.5, which is 0.2 m wide, as it must be.

P10A value x satisfies 24.5 ≤ x < 25.5. Determine what was published and to what unit, and determine how many values written to 1 decimal place lie in the interval.Challenge

Answer25, to the nearest whole number; 10 valuesThe interval is 1 wide, so the rounding unit is 1 and the published value is the midpoint, 25. The 1-decimal-place values inside are 24.5, 24.6, …, 25.4 — ten of them. 25.5 is not one of them, because the upper bound is excluded: it rounds to 26.

Bounds from decimal places and significant figures

Find the rounding unit first; everything else is half of it

Decimal places and significant figures are not two new rules. Both of them are just a way of telling you the rounding unit, and once you have the unit the bounds are half of it, each way. To n decimal places the unit is 10 to the power −n. To n significant figures, find the LAST significant digit and ask what it is worth: in 3 600 000 to 2 significant figures the 6 is worth six hundred thousand, so the unit is 100 000.

Decimal places and significant figures are not two rules. Both of them just tell you the rounding unit, and the bounds are half of it, each way.

Worked example 2 · Bounds when the figure is given to significant figures: 3 600 000 m² to 2 s.f.

Find the last SIGNIFICANT digit

the 6, standing for 6 hundred thousand

That digit fixes the rounding unit

100 000 m²

Half the unit

50 000 m²

Lower bound

3 600 000 − 50 000 = 3 550 000 m²

Upper bound

3 600 000 + 50 000 = 3 650 000 m²

As a percentage, the most it can be out by

50 000 ÷ 3 600 000 × 100 = 1.39%

Significant figures are not a different rule, they are the same rule with one extra step: find the rounding unit first. The last significant digit tells you what the figure was rounded to, and everything after that is “half of it, each way”. Here 2 significant figures on a seven-digit number means the unit is 100 000, so an area printed as a single tidy number is really a claim about a hundred-thousand-square-metre band. The same method reads 0.0250 kg to 3 s.f. as 0.02495 to 0.02505 kg — leading zeros are never significant, so they never change the unit.

Finding the unit, safely

to n d.p.: the unit is 0.1, 0.01, 0.001, …
to n s.f.: find the last significant digit and ask what it is worth
leading zeros are never significant, so 0.0250 to 3 s.f. has a unit of 0.0001
halve the unit, then take it off and add it on

The ways it goes wrong

halving the VALUE instead of the unit
counting leading zeros as significant figures
using a whole unit each way, making the interval twice as wide
rounding the bound you just found

EXAMPLEAn area is published as 3 600 000 m² to 2 significant figures. The unit is 100 000 m², so the area is trapped between exactly 3 550 000 m² and 3 650 000 m². The greatest possible error is 50 000 m², which is 1.39% of the published figure — and both bounds are exact numbers, not estimates.

Practice · Bounds from decimal places and significant figures · 10 questions

P11An area is given as 3 600 000 m², to 2 significant figures. Write down the rounding unit that implies.Fluency

Answer100 000 m²

P12Write down the lower bound of that area.Fluency

Answer3 550 000 m²

P13Write down the upper bound of that area.Fluency

Answer3 650 000 m²

P14Write down the bounds of a mass given as 0.0250 kg, to 3 significant figures.Fluency

Answer0.02495 kg and 0.02505 kg

P15A time is recorded as 7.30 s, to 2 decimal places. Determine its bounds.Applied

Answer7.295 s ≤ t < 7.305 sTo 2 decimal places the rounding unit is 0.01 s, so half of it is 0.005 s. The bounds are 7.30 − 0.005 = 7.295 s and 7.30 + 0.005 = 7.305 s. The trailing zero is significant and is what makes this claim ten times tighter than “7.3 s”.

P16A population is given as 48 000, to 2 significant figures. Determine the bounds and the width of the interval.Applied

Answer47 500 ≤ P < 48 500; width 1 000With 2 significant figures the last significant digit of 48 000 is the 8, which stands for 8 thousands, so the rounding unit is 1 000. Half of it is 500, giving 47 500 and 48 500. The width is the rounding unit, 1 000 people.

P17One speed is quoted as 25 m/s to 2 significant figures and another as 25.0 m/s to 3 significant figures. Determine both intervals and state which is the stronger claim.Applied

Answer24.5 to 25.5, and 24.95 to 25.05; the second is ten times strongerTo 2 significant figures the unit is 1, so the bounds are 24.5 m/s and 25.5 m/s. To 3 significant figures the unit is 0.1, so the bounds are 24.95 m/s and 25.05 m/s. The second interval is a tenth as wide, so the extra written zero is a claim about the instrument, not a decoration.

P18Calculate the greatest possible error in 3 600 000 m² as a percentage of the published area, to 3 significant figures.Applied

Answer1.39%The rounding unit is 100 000 m², so the greatest possible error is half of it: 50 000 m². As a percentage of the published figure, 50 000 ÷ 3 600 000 × 100 = 1.3888…%, which is 1.39% to 3 significant figures.

P19A number rounded to 3 significant figures is 0.0504. Determine its bounds, and explain why the leading zeros make no difference.Challenge

Answer0.05035 ≤ x < 0.05045The three significant figures are 5, 0 and 4; the zeros in front of the 5 only place the number and are not significant. The last significant digit is the 4, which stands for 4 ten-thousandths, so the rounding unit is 0.0001 and half of it is 0.00005. The bounds are 0.05035 and 0.05045.

P20A value v satisfies 0.4835 ≤ v < 0.4845. Determine what was published, and to how many significant figures.Challenge

Answer0.484, to 3 significant figuresThe interval is 0.001 wide, so the rounding unit is 0.001 and the published value is the midpoint, 0.484. For 0.484 the last significant digit is the 4, standing for 4 thousandths, so the rounding unit 0.001 means the figure was given to 3 significant figures.

Sums and differences: the cross rule

Add like for like; subtract crossways, because taking less away leaves more

Once both inputs are intervals, the answer is an interval too, and the only question is which end of each input produces which end of the answer. For a sum it is obvious: smallest with smallest and largest with largest. For a difference the second input crosses over, because taking away less leaves more: the largest answer is UB(a) − LB(b). What is worth noticing is that the difference is exactly as wide as the sum. Uncertainty never cancels.

Two operations keep like with like, and two cross over. Adding and multiplying pair the lower bounds together and the upper bounds together. Subtracting and dividing swap the second input to the opposite end, because taking away less leaves more and dividing by less gives more.

Worked example 3 · a − b when both are rounded: why the bounds cross

The two intervals, written out first

8.55 ≤ a < 8.65 and 3.15 ≤ b < 3.25

For a SUM you add like for like

8.55 + 3.15 = 11.7 and 8.65 + 3.25 = 11.9

For a DIFFERENCE, the largest answer takes away the least

8.65 − 3.15 = 5.5

and the smallest answer takes away the most

8.55 − 3.25 = 5.3

So the interval is

5.3 ≤ a − b < 5.5

Its width, like the sum’s, is the two units added

0.1 + 0.1 = 0.2

The cross rule. For a + b you add like for like; for a − b you swap the second input over: max = UB(a) − LB(b) and min = LB(a) − UB(b). The reason is one sentence long: taking less away leaves more. And notice what does not happen — the uncertainties do not cancel. A difference is exactly as uncertain as the corresponding sum, 0.2 m wide either way. A student who subtracts like for like gets 5.4 for both bounds, an interval of zero width, which would mean the difference of two inexact measurements was known exactly.

Operation	Lower bound	Upper bound	Why	For a = 8.6, b = 3.2
a + b	LB(a) + LB(b)	UB(a) + UB(b)	like with like	11.7 to 11.9
a − b	LB(a) − UB(b)	UB(a) − LB(b)	crossed — taking less away leaves more	5.3 to 5.5
a × b	LB(a) × LB(b)	UB(a) × UB(b)	like with like (for positive quantities)	26.9325 to 28.1125
a ÷ b	LB(a) ÷ UB(b)	UB(a) ÷ LB(b)	flipped — dividing by less gives more	2.63 to 2.75

EXAMPLEA tank holds 250 litres to the nearest 10 litres and 85 litres to the nearest 5 litres is drawn off. The most left is 255 − 82.5 = 172.5 litres and the least left is 245 − 87.5 = 157.5 litres. The interval is 15 litres wide — the two rounding units added, 10 + 5, exactly as it would be for a sum.

Practice · Sums and differences: the cross rule · 10 questions

P21Two lengths are a = 8.6 m and b = 3.2 m, each given to 1 decimal place. Write down the upper bound of a + b.Fluency

Answer11.9 m

P22Write down the lower bound of a + b.Fluency

Answer11.7 m

P23Write down the upper bound of a − b.Fluency

Answer5.5 m

P24Write down the lower bound of a − b.Fluency

Answer5.3 m

P25Determine the interval for a + b and its width.Applied

Answer11.7 m ≤ a + b < 11.9 m; width 0.2 ma lies between 8.55 m and 8.65 m and b between 3.15 m and 3.25 m. For a sum you add like for like: smallest with smallest, 8.55 + 3.15 = 11.7 m, and largest with largest, 8.65 + 3.25 = 11.9 m. The width is 0.2 m, which is exactly the two input widths added: 0.1 + 0.1.

P26Determine the interval for a − b and its width, and explain the crossing.Applied

Answer5.3 m ≤ a − b < 5.5 m; width 0.2 mFor a difference the bounds cross. The largest answer comes from the largest a and the SMALLEST b: 8.65 − 3.15 = 5.5 m — taking less away leaves more. The smallest answer comes from the smallest a and the largest b: 8.55 − 3.25 = 5.3 m. The width is 0.2 m, the same as for the sum: subtracting does not cancel the uncertainty, it adds it.

P27Two planks, 2.4 m and 1.8 m each to the nearest 0.1 m, are laid end to end. Determine the bounds of the total length.Applied

Answer4.1 m ≤ T < 4.3 mEach plank has bounds half a unit either side: 2.35 m to 2.45 m and 1.75 m to 1.85 m. The least total is 2.35 + 1.75 = 4.1 m and the greatest is 2.45 + 1.85 = 4.3 m. The published total, 4.2 m, is carrying two roundings, so its interval is 0.2 m wide rather than 0.1 m.

P28A tank holds 250 litres, to the nearest 10 litres. 85 litres, to the nearest 5 litres, is drawn off. Determine the bounds of what is left.Applied

Answer157.5 litres ≤ V < 172.5 litresThe tank holds between 245 and 255 litres; the amount drawn off is between 82.5 and 87.5 litres. Cross over: the most left is 255 − 82.5 = 172.5 litres and the least left is 245 − 87.5 = 157.5 litres. The interval is 15 litres wide, which is 10 + 5 — the two rounding units added.

P29Three lengths, each 6.0 m to 1 decimal place, are joined. Determine the bounds of the total and state the width.Challenge

Answer17.85 m ≤ T < 18.15 m; width 0.3 mEach length lies between 5.95 m and 6.05 m. Three smallest give 3 × 5.95 = 17.85 m and three largest give 3 × 6.05 = 18.15 m. The width is 0.3 m: three roundings, three tenths. Every measurement you add to a total adds its own rounding unit to the width of the answer.

P30A student finds the bounds of a − b by subtracting like for like: 8.55 − 3.15 and 8.65 − 3.25. Determine what they get, and explain why the answer is nonsense.Challenge

Answerboth give 5.4, so they get an interval of zero width8.55 − 3.15 = 5.4 and 8.65 − 3.25 = 5.4. The two “bounds” are the same number, so the method claims the difference is known exactly — from two measurements that are not known exactly. That is impossible, and it is the giveaway that the bounds must cross: the true interval is 5.3 to 5.5, and it is 0.2 m wide.

Products and quotients: the flip

The divisor takes the bound at the opposite end — the single most-failed step in the topic

A product of two positive quantities behaves like a sum: lower with lower and upper with upper. A quotient does not. Dividing by a smaller number gives a bigger answer, so the greatest value of a ÷ b pairs the greatest a with the least b. This is the flip, and it is the single most-failed step in the whole topic — the instinct “upper bound means upper bounds everywhere” is right for + and × and wrong for − and ÷.

Every one of the four values is a value a ÷ b really could take. Only two of them are bounds, and the two in the middle are the ones a careless answer reaches for. Test all four corners whenever you are not sure.

Worked example 4 · a ÷ b when both are rounded: the flip

Try all four corner combinations

8.65÷3.15, 8.65÷3.25, 8.55÷3.15, 8.55÷3.25

The greatest: biggest top, SMALLEST bottom

8.65 ÷ 3.15 = 2.746… = 2.75

The least: smallest top, LARGEST bottom

8.55 ÷ 3.25 = 2.630… = 2.63

The tempting wrong answer, upper over upper

8.65 ÷ 3.25 = 2.66 — inside, not the bound

So the interval is

2.63 ≤ a ÷ b < 2.75 (3 s.f.)

A product does NOT flip: both bounds stay on the same side

8.55×3.15 = 26.9325 and 8.65×3.25 = 28.1125

The flip. Dividing by a smaller number gives a bigger answer, so the upper bound of a quotient is UB(a) ÷ LB(b) and the lower bound is LB(a) ÷ UB(b). This is the single most-failed step in the topic: the instinct is “upper bound means upper bounds everywhere”, and for a sum and a product that instinct is right, while for a difference and a quotient it is wrong. Two operations keep like with like (+ and ×) and two operations cross over (− and ÷). If you cannot remember which, test the four corners: it takes twenty seconds and it is never wrong.

The published quotient sits inside a band about 0.115 wide. Neither end of that band comes from dividing an upper bound by an upper bound.

Operation	How wide the answer is	For a = 8.6, b = 3.2
a + b	the two input widths ADDED	0.1 + 0.1 = 0.2
a − b	the two input widths ADDED, exactly as for a sum	0.1 + 0.1 = 0.2
a × b	grows with the SIZE of a and b	28.1125 − 26.9325 = 1.18
a ÷ b	grows with the size of a, and blows up as b nears zero	2.746… − 2.630… ≈ 0.115

EXAMPLEA car covers 250 km to the nearest 10 km in 3.5 hours to the nearest 0.1 hour. The fastest possible average speed is the longest distance in the shortest time: 255 ÷ 3.45 = 73.9 km/h. The slowest is 245 ÷ 3.55 = 69.0 km/h. Dividing 255 by 3.55 would have used two upper bounds and produced neither.

Practice · Products and quotients: the flip · 10 questions

P31Write down the upper bound of a × b, where a = 8.6 m and b = 3.2 m, each given to 1 decimal place.Fluency

Answer28.1125 m²

P32Write down the lower bound of a × b.Fluency

Answer26.9325 m²

P33State which bound of b gives the GREATEST value of a ÷ b.Fluency

Answerthe lower bound of b

P34Write down the calculation that gives the upper bound of a ÷ b.Fluency

Answer8.65 ÷ 3.15

P35Determine the upper bound of a ÷ b, to 3 significant figures.Applied

Answer2.75Dividing by a smaller number gives a bigger answer, so the greatest quotient uses the largest a and the SMALLEST b: 8.65 ÷ 3.15 = 2.746031…, which is 2.75 to 3 significant figures. Using 8.65 ÷ 3.25 instead gives 2.66 — a real value of the quotient, but not its upper bound.

P36Determine the lower bound of a ÷ b, to 3 significant figures.Applied

Answer2.63The least quotient uses the smallest a and the LARGEST b: 8.55 ÷ 3.25 = 2.630769…, which is 2.63 to 3 significant figures. Notice that each bound of the answer uses one bound from the top and the opposite bound from the bottom: that is the flip.

P37A car covers 250 km, to the nearest 10 km, in 3.5 hours, to the nearest 0.1 hour. Determine the bounds of its average speed, to 3 significant figures.Applied

Answer69.0 km/h to 73.9 km/hThe distance lies between 245 km and 255 km and the time between 3.45 h and 3.55 h. Greatest speed = 255 ÷ 3.45 = 73.9130… = 73.9 km/h: the longest distance in the shortest time. Least speed = 245 ÷ 3.55 = 69.0140… = 69.0 km/h. The published figure, 250 ÷ 3.5, is about 71.4 km/h, comfortably inside.

P38A rectangle measures 12.4 cm by 7.8 cm, each to 1 decimal place. Determine the bounds of its area exactly.Applied

Answer95.7125 cm² ≤ A < 97.7325 cm²The sides lie between 12.35 cm and 12.45 cm, and between 7.75 cm and 7.85 cm. For a product of two positive quantities the bounds do not cross: least × least = 12.35 × 7.75 = 95.7125 cm² and greatest × greatest = 12.45 × 7.85 = 97.7325 cm². Both bounds are exact and neither is rounded.

P39Density is mass ÷ volume. A block has mass 48 g, to the nearest gram, and volume 15 cm³, to the nearest cm³. Determine the bounds of its density, to 3 significant figures.Challenge

Answer3.06 g/cm³ to 3.34 g/cm³The mass lies between 47.5 g and 48.5 g; the volume between 14.5 cm³ and 15.5 cm³. Greatest density = 48.5 ÷ 14.5 = 3.344827… = 3.34 g/cm³: heaviest possible mass squeezed into the smallest possible volume. Least density = 47.5 ÷ 15.5 = 3.064516… = 3.06 g/cm³. The published 48 ÷ 15 = 3.2 g/cm³ is only the middle of a wide interval.

P40Explain why the upper bound of a ÷ b uses the LOWER bound of b, and determine what a student gets by dividing upper bound by upper bound.Challenge

Answer2.66, which is a possible value but not the upper boundDividing a fixed amount among fewer people gives each more; dividing by a smaller number gives a bigger answer. So the greatest quotient pairs the greatest numerator with the least denominator. A student who divides upper by upper gets 8.65 ÷ 3.25 = 2.661538… = 2.66, which is genuinely one of the values a ÷ b could take — it is just not the largest, and quoting it as the upper bound understates the answer by about 0.085.

Bounds through a formula

Several rounded inputs, one answer, and a decision about each input in turn

A formula with several rounded inputs is not a new problem, it is the same problem repeated: go through the inputs one at a time and decide which bound helps. Anything multiplied on the top takes its lower bound for a lower bound; anything on the bottom takes its upper bound; anything subtracted crosses over; and an exact constant contributes nothing at all. Then do the arithmetic on the exact bounds and round the answer once, at the end.

Both inputs sit on the top of the fraction, so both of them push E the same way. The lower bound of E therefore takes the lower bound of each, and the upper bound takes the upper bound of each. Every figure in the E column is rounded once, at the end, to 3 significant figures.

Worked example 5 · The lower bound of E = Ad² ÷ 700, from two rounded inputs

Write both inputs as intervals first

3 550 000 ≤ A < 3 650 000; 14.5 ≤ d < 15.5

Decide what each input does to E

both are on the top, so both increase E

So the LOWER bound of E takes the lower bound of each

A = 3 550 000 and d = 14.5

Square the depth first, exactly

14.5² = 210.25

Multiply and divide by the exact constant

3 550 000 × 210.25 ÷ 700 = 1 066 267.857…

Round the ANSWER once, at the end

lower bound of E = 1 070 000 (3 s.f.)

A formula with several rounded inputs is not harder, it is just longer: decide, input by input, which bound helps. Anything multiplied on the top takes its lower bound for a lower bound; anything on the bottom takes its upper bound. The 700 is an exact constant and contributes nothing. Do the arithmetic on the exact bounds — 14.5 and 3 550 000, never rounded first — and round only the final answer. The same reasoning gives an upper bound of 1 250 000, so E is settled only to about 1.07 to 1.25 million: the published 1 160 000 has one honest significant figure in it.

Input	As given	The interval	What it does to E
A, the area	3 600 000 m² to 2 s.f.	3 550 000 ≤ A < 3 650 000	on the top, so lower for lower
d, the depth	15 m to the nearest metre	14.5 ≤ d < 15.5	on the top and squared, so lower for lower
700	an exact constant	exactly 700	adds no uncertainty at all
E = Ad² ÷ 700	published as 1 160 000	1 070 000 to 1 250 000 (3 s.f.)	one honest significant figure

EXAMPLEWith A = 3 600 000 m² to 2 s.f. and d = 15 m to the nearest metre, E = Ad² ÷ 700 lies between 1 070 000 and 1 250 000 to 3 significant figures. The published value, 1 160 000, has one honest significant figure in it. The depth is the shakier input: its half-width is 3.33% of itself against the area’s 1.39%, and it is then squared.

Practice · Bounds through a formula · 10 questions

P41In E = Ad² ÷ 700, the area A is 3 600 000 m² to 2 significant figures. Write down the lower bound of A.Fluency

Answer3 550 000 m²

P42The depth d is 15 m, to the nearest metre. Write down the lower bound of d.Fluency

Answer14.5 m

P43State which bound of A and which bound of d are needed for the lower bound of E.Fluency

Answerthe lower bound of each

P44Calculate E from the published figures, to 3 significant figures.Fluency

Answer1 160 000

P45Calculate the lower bound of E, to 3 significant figures.Applied

Answer1 070 000Every input appears on the top of the fraction and every input increases E, so the lower bound of E needs the lower bound of each: A = 3 550 000 and d = 14.5. Then 14.5² = 210.25 and 3 550 000 × 210.25 ÷ 700 = 1 066 267.857…, which is 1 070 000 to 3 significant figures. The bounds 3 550 000 and 14.5 are used exactly — they are never rounded first.

P46Calculate the upper bound of E, to 3 significant figures.Applied

Answer1 250 000Take the upper bound of each input: A = 3 650 000 and d = 15.5. Then 15.5² = 240.25 and 3 650 000 × 240.25 ÷ 700 = 1 252 732.142…, which is 1 250 000 to 3 significant figures. The 700 is an exact constant, so it contributes no uncertainty at all.

P47Determine the width of the interval for E, to 3 significant figures.Applied

Answer186 000The width is the upper bound minus the lower bound: 1 252 732.142… − 1 066 267.857… = 186 464.285…, which is 186 000 to 3 significant figures. The subtraction is done on the exact bounds, not on the rounded ones — using 1 250 000 − 1 070 000 = 180 000 would be rounding the bounds and then rounding again.

P48A box measures 2.4 m by 1.5 m by 0.8 m, each to 1 decimal place. Determine the bounds of its volume, to 3 significant figures.Applied

Answer2.56 m³ to 3.23 m³Every edge lies half a unit either side: 2.35 to 2.45, 1.45 to 1.55, and 0.75 to 0.85. All three multiply, so least × least × least = 2.35 × 1.45 × 0.75 = 2.555625 m³ = 2.56 m³, and greatest gives 2.45 × 1.55 × 0.85 = 3.227875 m³ = 3.23 m³. Three roundings compound: the interval is a quarter of the answer wide.

P49Determine the greatest possible percentage error in the published value of E, to 3 significant figures.Challenge

Answer8.26%The published E is 1 157 142.857…. Upwards, the upper bound is 1 252 732.142…, a gap of 95 589.285…, which is 8.2608…% of the published value. Downwards, the gap to the lower bound is 90 875, which is 7.8534…%. The greatest possible error is therefore 8.26% to 3 significant figures — and the interval is NOT symmetric about the published figure, because d is squared.

P50Determine which of the two inputs E is more sensitive to, and justify your answer with figures.Challenge

Answerd, the depthCompare each input’s half-width as a percentage of itself. For A: 50 000 ÷ 3 600 000 × 100 = 1.39%. For d: 0.5 ÷ 15 × 100 = 3.33%. So d is already the shakier measurement — and it is then squared, which roughly doubles its effect to about 6.67%. Measuring the depth to the nearest 0.1 m would shrink the interval for E far more than an extra significant figure on the area would.

Can you be certain? Working with the worst case

Compare the upper bound of what must fit with the lower bound of what it must fit into

This is what bounds are for. A decision that has to be right — will it fit, is the load legal, will the water last — cannot be taken on the published figures, because nobody gets to choose which value inside the interval turns up. Work with the worst case. To be certain something fits, compare the upper bound of what must fit with the lower bound of what it fits into. If UB(thing) ≤ LB(hole), you are certain; otherwise you are not, and “probably fine” is not an answer.

The shelf’s interval reaches past the gap’s lower bound, so the two overlap and the fit is not guaranteed. Compare the upper bound of what must fit with the lower bound of what it fits into — every time.

Worked example 6 · “It is 90 cm and the gap is 90.4 cm” — can you be sure it fits?

The shelf, to the nearest centimetre

89.5 ≤ s < 90.5

The gap, to 1 decimal place

90.35 ≤ g < 90.45

The dangerous case: widest shelf, narrowest gap

s up to 90.5 against g = 90.35

Compare UB(shelf) with LB(gap)

90.5 > 90.35

So the answer is

no — it is not certain to fit

By how much it could be too wide

up to 90.5 − 90.35 = 0.15 cm

The nominal figures say 90 cm into 90.4 cm, a comfortable 0.4 cm of room. The bounds say the shelf could be almost 90.5 cm and the gap could be 90.35 cm, so it could be 0.15 cm too wide. A guarantee is always the same comparison: the upper bound of what has to fit against the lower bound of what it fits into. Everything else — the published figures, the averages, the “it should be fine” — is about the typical case, and nobody gets to choose the typical case. If UB(thing) ≤ LB(hole) you are certain; otherwise you are not, and the honest answer is “measure again”.

The decision	What to compare	The test	Worked case
Will it fit into the gap?	UB(thing) against LB(gap)	certain if UB(thing) ≤ LB(gap)	the shelf: 90.5 against 90.35 — NOT certain
Is the load under the limit?	UB(load) against LB(limit)	certain if UB(load) ≤ LB(limit)	596 against 595 — NOT certain
Will it last long enough?	LB(supply) ÷ UB(rate)	round the answer DOWN	4 950 ÷ 47.5 = 104.2, so 104 days
How many will certainly fit?	n × UB(each) against LB(limit)	round the answer DOWN	595 ÷ 74.5 = 7.98, so 7 crates

EXAMPLEA hoist is rated at 600 kg to the nearest 10 kg and eight crates each weigh 74 kg to the nearest kilogram. The nominal arithmetic says 8 × 74 = 592 kg against 600 kg, comfortable. The bounds say the load can reach 8 × 74.5 = 596 kg while the rating can be as low as 595 kg. It is not certain, and the honest answer is seven crates.

Practice · Can you be certain? Working with the worst case · 10 questions

P51A shelf is 90 cm long, to the nearest centimetre. Write down its upper bound.Fluency

Answer90.5 cm

P52The gap it must fit into is 90.4 cm, to 1 decimal place. Write down the lower bound of the gap.Fluency

Answer90.35 cm

P53State whether the shelf is certain to fit.Fluency

Answerno

P54State which bound of the shelf and which bound of the gap must be compared to be certain.Fluency

Answerthe upper bound of the shelf against the lower bound of the gap

P55Determine by how much the shelf could be too wide.Applied

Answerby up to 0.15 cmThe shelf can be almost 90.5 cm and the gap can be as narrow as 90.35 cm, so in the worst case the shelf is 90.5 − 90.35 = 0.15 cm too wide. It is “up to” 0.15 cm because 90.5 cm is the upper bound and is not itself attainable. The nominal figures, 90 cm into 90.4 cm, look like a comfortable 0.4 cm of room.

P56A box is 34 cm wide, to the nearest centimetre. A doorway is 35 cm wide, to the nearest centimetre. Determine whether the box is certain to pass through, and justify your answer.Applied

Answeryes, it is certainThe box is less than 34.5 cm and the doorway is at least 34.5 cm. The two bounds are equal, and that is enough: the box’s upper bound is not attainable, so the box is always strictly narrower than 34.5 cm, while the doorway is always at least 34.5 cm. Every possible box fits every possible doorway — but only just, and only because the interval is half-open at the top.

P57A hoist is rated at 600 kg, to the nearest 10 kg. Eight crates each weigh 72 kg, to the nearest kilogram. Determine whether the load is certainly safe.Applied

Answeryes: at most 580 kg against a limit of at least 595 kgEach crate is under 72.5 kg, so eight of them come to less than 8 × 72.5 = 580 kg. The rating is at least 600 − 5 = 595 kg. Since 580 ≤ 595, every possible load is under every possible limit, so the lift is certainly safe with 15 kg to spare in the worst case.

P58The same hoist is loaded with eight crates each weighing 74 kg, to the nearest kilogram. Determine whether the load is certainly safe, and comment on what the nominal figures suggest.Applied

Answerno: up to 596 kg against a limit of as little as 595 kgEach crate can be almost 74.5 kg, so the load can be almost 8 × 74.5 = 596 kg, while the rating can be as low as 595 kg. Since 596 > 595, there are possible loads that exceed possible limits, so it is not certain. The nominal arithmetic, 8 × 74 = 592 kg against 600 kg, suggests 8 kg of comfort that the measurements do not actually guarantee.

P59Determine the greatest whole number of those 74 kg crates that the hoist is certain to carry.Challenge

Answer7 cratesThe worst case is n crates of almost 74.5 kg against a limit of 595 kg, so we need n × 74.5 ≤ 595, that is n ≤ 7.986…. The greatest whole number is 7, and 7 × 74.5 = 521.5 kg, safely under 595 kg. Eight crates reach 596 kg, which is over. Notice that the answer is rounded DOWN, whatever the decimal: rounding 7.986 up to 8 would put people under a load that can exceed the rating.

P60A tank holds 5 000 litres, to the nearest 100 litres. A household draws 45 litres a day, to the nearest 5 litres. Determine the greatest number of whole days the tank is certain to last.Challenge

Answer104 daysThe worst case is the least water with the fastest use: at least 4 950 litres, drawn at up to 47.5 litres a day. 4 950 ÷ 47.5 = 104.2105… days, so 104 whole days are guaranteed. The published figures, 5 000 ÷ 45 = 111.1 days, promise a week more than the measurements can support — and rounding 104.2 up to 105 would be a promise the tank cannot keep.

The bounds routine

Four moves, whatever the question turns out to be about.

Write both bounds down before you calculate anythingFind the rounding unit, halve it, and write LB ≤ x < UB for every rounded input. A question you cannot start is nearly always a question whose intervals have not been written out.

Decide, input by input, which bound helpsFor a bigger answer: bigger on the top, SMALLER on the bottom, and smaller for anything being subtracted. Like with like for + and ×; crossed over for − and ÷. If you are unsure, work out all four corners and take the biggest and the smallest.

Keep the bounds exact all the way through14.5 and 3 550 000 are exact consequences of what was written down, not estimates. Never round a bound, and never round a step. Exact constants such as the 700 in Ad² ÷ 700 add no uncertainty at all.

✓

Round the answer once, and then answer the questionGive a derived bound to 3 significant figures, money in pula to 2 decimal places. Then say what it means: for a guarantee, compare the upper bound of what must fit with the lower bound of what it fits into, and round a “how many” answer DOWN.

The mistakes that cost the most marks

1Using the upper bound of every input to get the upper bound of a quotient

Wrong“the biggest a ÷ b must be 8.65 ÷ 3.25 = 2.66”

Rightthe divisor takes its LOWER bound: 8.65 ÷ 3.15 = 2.75 to 3 significant figures

This is the most expensive single error in the whole Extended bounds row, and it comes from a reasonable-sounding rule that happens to be wrong: “upper bound means upper bounds everywhere”. It is right for a sum and right for a product, and it is wrong for a difference and wrong for a quotient. The reason is not a convention to memorise, it is arithmetic you already trust: dividing a fixed amount among fewer people gives each person more. So the largest quotient puts the largest possible numerator over the smallest possible denominator, and the smallest quotient does the opposite. Here 8.65 ÷ 3.25 = 2.66 is a perfectly real value of a ÷ b — it is just not the largest one, and quoting it as the upper bound understates the answer by about 0.085, which in a tolerance decision is the difference between certain and not certain. When in doubt, work out all four corners and take the biggest and the smallest; it costs twenty seconds and it cannot go wrong.

2Subtracting like for like when finding the bounds of a difference

Wrong“lower bound = 8.55 − 3.15 and upper bound = 8.65 − 3.25, so a − b = 5.4 both ways”

Rightthe bounds cross: min = 8.55 − 3.25 = 5.3 and max = 8.65 − 3.15 = 5.5

The giveaway that this method is broken is built into its own answer: both “bounds” come out as 5.4, so it claims that a difference of two measurements, neither of which is known exactly, is itself known exactly. That cannot be true, and the interval it produces has zero width. What goes wrong is the direction of the second input. In a + b, making b bigger makes the answer bigger, so like goes with like. In a − b, making b bigger makes the answer SMALLER, so the largest answer needs the smallest b. Written out: max = UB(a) − LB(b) and min = LB(a) − UB(b). Two further facts are worth carrying away. First, the width of a difference is the two input widths ADDED, exactly as for a sum — uncertainty never cancels. Second, a difference of two nearly equal rounded numbers is the most dangerous calculation in the topic, because the answer shrinks while its interval does not.

3Rounding the bounds themselves

Wrong“the lower bound of 3 600 000 m² to 2 s.f. is about 3.6 million, and 14.5 is about 15”

Rightbounds are EXACT: 3 550 000 and 14.5 go into the calculation untouched, and only the final answer is rounded

A bound is not a measurement and it is not an estimate. It is an exact consequence of what somebody wrote down: if an area is given to 2 significant figures as 3 600 000 m², then its lower bound is exactly 3 550 000 m², not approximately, and 14.5 m is exactly the lower bound of a depth given as 15 m to the nearest metre. Rounding a bound does two kinds of damage. It destroys the guarantee — a lower bound that has been rounded up is no longer a lower bound, and a decision built on it is no longer safe. And it corrupts the arithmetic before it starts, so the error grows through every step that follows. There is exactly one place in a bounds question where rounding is allowed: the final derived answer, once, at the end, to 3 significant figures. The width of an interval is also computed from the exact bounds — here 1 252 732.142… − 1 066 267.857… = 186 464.285…, so 186 000 to 3 s.f., and never 1 250 000 − 1 070 000.

Tasks to submit

Criterion B · Investigating patterns

How wide is the answer?

Two quantities are measured and rounded, and then something is done with them. The answer is no longer a number either — it is an interval, and that interval has a width. How does the width of the answer depend on what you did to the two quantities? Adding, subtracting, multiplying and dividing all take the same two intervals in and all give an interval out, so it is a fair race. Two of the four behave identically and the other two do not behave like each other at all. The only tools you need are the bound rules from this unit and a calculator.

One column is given. Work the other three out yourself, keeping every value exact, and then look hard at the bottom row.

Part 1 · structured

Operation	Lower bound	Upper bound	Width
a + b	16	18	2
a − b
a × b
a ÷ b

Predict, before you calculate anything, which of the four operations will give the WIDEST interval and which the narrowest, and say whether you expect a + b and a − b to come out the same or different. Write the prediction down; you will be asked to compare it with what you find.
Complete the table for a = 12 and b = 5, each given to the nearest whole number, so that each is trapped between bounds one unit apart. For each of a + b, a − b, a × b and a ÷ b give the lower bound, the upper bound and the width. Keep the sums and differences exact; give the quotient to 3 significant figures but work out its width from the EXACT bounds.
Identify the pattern. What do the widths of a + b and a − b have in common, and what do they have in common with the two rounding units you started from? Then look at the product and the quotient: say, with numbers, how their widths compare with the sizes of a and b themselves.
Determine a general rule for the width of the answer when two quantities with interval widths w(a) and w(b) are added, and another for when they are subtracted. Write each rule as a formula in w(a) and w(b), not in words alone. Then determine a rule for the width of a × b as well, in terms of a, b, w(a) and w(b).
Test and verify your rules on a pair that is not in the table: a = 7.4 and b = 2.5, each given to 1 decimal place. Then test them again on a = 12 and b = 5 given to the nearest 0.5. Say how close each rule came, and whether the sum rule and the difference rule still agree with each other. A rule tested only on the numbers it was built from has not been tested.
Justify your rules. Explain why a difference must widen by exactly as much as a sum, when subtracting looks as though it ought to cancel some of the uncertainty out. Then explain why the product’s width depends on the SIZE of the two quantities while the sum’s does not, and say what that means for a calculation that multiplies two large rounded numbers together.

Part 2 · open

Now you choose what to investigate. Here are three openings — take one, or propose your own. You decide which quantities to try, how many you need, how to record them and how you will test whatever you find.

Chain a third quantity on. Investigate what happens to the width when three rounded quantities are added, and when three are multiplied. Determine a rule for n quantities added, test it, and say whether the same kind of rule can work for n quantities multiplied.
Squeeze the divisor. Investigate the width of a ÷ b as b gets smaller while its rounding unit stays the same — try b = 5, 2, 1 and 0.5, each to the nearest 0.1. Determine what happens to the width as the divisor’s lower bound approaches zero, justify what you find, and state the value of b at which you would refuse to quote an answer at all.
Break the difference. Investigate what happens when a and b are nearly equal — for example 8.6 and 8.4, each to 1 decimal place. Compare the width of a − b with the answer itself, determine how the two are related, and justify why a difference of two nearly equal rounded numbers is the most dangerous calculation in this unit.

Whichever you choose, take it through the same ladder: predict, identify the pattern, determine a general rule, test and verify it on cases you have not already used, and justify why it must hold and where it stops holding.

Submit: your completed table, your prediction written before the calculations, your rules for the width of a sum, a difference and a product, your verification on the 7.4 and 2.5 pair and on the nearest-0.5 pair, and your reasoning, as a scan or a typed document. Marked out of 8 against Criterion B. Completing the table earns the lower bands; the general rules, the verification on pairs that are not in the table and the justification are what reach levels 5–8. Part 2 must show your own choice of method. Keep every bound exact and round only a quotient, once, to 3 significant figures.

Remember, you should communicate in an organized and coherent manner.

Criterion D · Applying maths in real-life contexts

Will it fit, and will it be legal?

A builders’ yard in Gaborone has a delivery to send out this morning: a moulded water tank, eight bags of cement and a bundle of roofing sheets, on one flat-bed trailer, to a customer about 90 km away. The yard manager has a data sheet, and the data sheet tells you far more than you need. Two questions have to be answered before the trailer leaves, and both of them are yes-or-no: will the tank fit across the bed, and is the load certainly inside the trailer’s rating? Every figure on the sheet is a rounded measurement, so neither question can be answered from the nominal numbers alone.

The data sheet (illustrative figures, written for this task)

Tank base diameter: 1.9 m, to the nearest 0.1 m
Tank height: 2.3 m, to the nearest 0.1 m
Tank stated capacity: 5 000 litres, to the nearest 100 litres
Empty tank mass: 78 kg, to the nearest kilogram
Trailer bed: 2.0 m wide and 3.6 m long, each to the nearest 0.1 m
Trailer load limit: 750 kg, to the nearest 50 kg
Also on the trailer: 8 bags of cement at 50 kg each, to the nearest kilogram, and roofing sheets totalling 240 kg, to the nearest 10 kg
Yard gateway: 2.4 m wide, to the nearest 0.1 m
Delivery fee: P450.00, a fixed charge (exact)
Diesel: P22.40 a litre on the day (exact)
Fuel use: 14 litres per 100 km, to 2 significant figures
Distance to the customer: about 90 km each way, to the nearest 10 km
Customer’s roof catchment: 96 m², to 2 significant figures
Last season’s rainfall: 380 mm, to 2 significant figures
Straps carried: 6, rated at 400 kg each
Yard hours: 07:00–17:00 on weekdays, 07:00–12:00 on Saturday

These figures are illustrative. The yard, the customer, the load, the prices and the distances were invented for this task. No real business, person, product or price list is named or implied, nothing here is a quotation, and there is no source to look up. What is authentic is the shape of the problem: tolerance decisions really are made on rounded measurements, and the nominal figures really do mislead.

The two intervals touch at 1.95 m and do not overlap. Because the tank’s upper bound is open and the bed’s lower bound is closed, the tank is always strictly narrower than the bed — certain to fit, and certain only because the interval is half-open.

The nominal load is 718 kg and the nominal rating is 750 kg, so the published figures promise 32 kg of room. The load can reach 727.5 kg and the rating can be as low as 725 kg, so the bars overlap and the trip is not certain to be legal.

Identify which figures on the data sheet are rounded and to what, and write down the interval each one stands for. Then justify, item by item, which entries cannot affect either decision — deal with the yard gateway, the roof catchment, the rainfall and the yard hours by name — and state which two figures on the sheet are exact. (Strand i — identifying the relevant elements)
Select the strategies you will use and justify each choice. For the fit, say which bound of the tank and which bound of the bed decide it, and why. For the load, say which bound of every item in the load and which bound of the rating decide it, and why. Explain why the comparison is never between two nominal figures. (Strand ii — selecting appropriate strategies)
Apply them. Calculate (a) the intervals for the tank’s base and the trailer bed, and decide the fit; (b) the interval for the total mass on the trailer, and compare it with the interval for the rating; (c) the greatest number of cement bags that is certainly safe; and (d) the cost of a second delivery run, in pula to 2 decimal places, as an interval — fuel plus the fixed fee. (Strand iii — applying the strategy)
Justify the accuracy of your two decisions. For each one, say which bound is the DANGEROUS bound and what would go wrong if you used the other. Explain exactly why the nominal figures mislead here: 718 kg against 750 kg looks like 32 kg of room, so determine what the worst case really is and how it can be worse than the published sum of the published parts. Discuss the fit decision, where the two bounds meet at 1.95 m: does that make the answer yes, no, or undecidable, and what does the half-open interval have to do with it? Finally, justify why none of the bounds may be rounded on the way through. (Strand iv — accuracy of the solution)
Justify a recommendation to the yard manager, who has to decide before the trailer leaves. State clearly what you would send today and what you would hold back, justify it with your intervals rather than with the nominal figures, and put a price on being wrong by comparing the cost of a second run with the consequence of an overloaded trailer. Then discuss the limitations of your own answer: which single extra measurement, taken more precisely, would change the decision most, and what have you assumed that the data sheet does not actually say? (Strand v — does it make sense in context)

Submit: your interval list for every relevant figure, your two decisions each with the comparison that settles it, the greatest safe number of bags, the cost interval for a second run, and written answers for strands i, iv and v. Marked out of 8 against Criterion D at Extended level, where the command term is justify rather than describe: a solution that computes the intervals and stops cannot reach the upper bands. Give money in pula to 2 decimal places and everything else to 3 significant figures, round rather than truncate, and never round a bound.

When you are ready

Take the mastery quiz →
Download this unit as a PDF

The mastery quiz is 10 questions drawn at random from a bank of 25. You need 70% to pass and you have three attempts. Type numbers as digits only, with no units and no spaces, and pula to 2 decimal places with no P. Work through the practice questions first — especially Concept 4, where the divisor takes the opposite bound, and Concept 6, where the worst case decides.

Printable version
Download Unit 18 as a PDF — the same lesson, laid out for printing and for working offline.

MYP Mathematics · The Complete Course Unit 18 · Trapped Between Bounds

MYP Mathematics · The Complete Course