The length has been rounded to the nearest 10 cm, and half of 10 is 5.

\(260 + 5 = 265\)

A length of exactly 265 cm would round to 270 cm instead, so the true length always stays below 265 cm. That is what the upper bound is: the number the length never quite reaches.

One decimal place means the mass has been rounded to the nearest 0.1 kg, and half of 0.1 is 0.05 — not 0.5.

\(1.2 - 0.05 = 1.15\)

A mass of exactly 1.15 kg does round to 1.2 kg, so it is possible: the lower bound is included.

The capacity has been rounded to the nearest 50 ml, and half of 50 is 25.

\(850 + 25 = 875\)

A capacity of exactly 875 ml would round to 900 ml instead, so the capacity stays below 875 ml.

The half is never automatically 0.5. It is always half of whatever unit the measurement was rounded to.

Rounding to the nearest 5 g means the true mass, \(x\), satisfies \(62.5 \leq x < 67.5\).

The lower end is included: a mass of exactly 62.5 g does round to 65 g. The upper end is not: a mass of exactly 67.5 g would round to 70 g instead, so \(x\) can get arbitrarily close to 67.5 but never reach it.

\(66.75\) sits inside that interval, so the answer is Yes.

The interval is the same one as part (a): \(62.5 \leq x < 67.5\), lower end included, upper end never reached.

\(62.5\) sits exactly at the lower bound, which the interval does include, so the answer is Yes.

Each one is correct to the nearest 10 cm, so its true length is within 5 cm of 210 cm: from 205 up to 215 cm.

The total is biggest when every one of the 6 items is at its own upper bound — the uncertainty adds up 6 times over, not just once:

\(6 \times 215 = 1290\) cm

Now take every item at its lower bound instead:

\(6 \times 205 = 1230\) cm

So the total length is at least 1230 cm and stays below 1290 cm.

The square root of \(k\) lies from \(4.5\) up to (not including) \(5.5\), so

\(4.5^2 \le k < 5.5^2\)

\(20.25 \le k < 30.25\)

The whole numbers in this interval run from \(21\) to \(30\).

The smallest possible value of \(k\) is \(21\).

\(4.5^2 = 20.25\)

\(16\) is less than \(20.25\), so its square root is less than \(4.5\).

That root rounds to \(4\), not \(5\), so \(k\) cannot be \(16\).

The area has to reach 7689 m\(^2\) whatever the true sides are, so the argument must use the smallest area the playground could have.

Correct to the nearest 20 m, each side could be up to 20 m ÷ 2 = 10 m below the value given.

Lower bound of the first side: \(120 - 10 = 110\) m

Lower bound of the second side: \(80 - 10 = 70\) m

Both sides are positive, so the area is smallest when both sides are smallest:

\(110 \times 70 = 7700\) m\(^2\)

Compare with the minimum: 7700 m\(^2\) is greater than 7689 m\(^2\).

Even the smallest possible area is bigger than 7689 m\(^2\), so the playground can be approved.

Working with the sides as printed gives \(120 \times 80 = 9600\) m\(^2\), and the upper bounds give 11700 m\(^2\); neither shows that the area cannot fall short, because the true area could be smaller.

Rounding to 1 decimal place puts each mass within 0.05 kg of the value written down.

Upper bound of A: \(8.4 + 0.05 = 8.45\) kg

Upper bound of B: \(3.1 + 0.05 = 3.15\) kg

The total is biggest when both parcels are as heavy as they can be:

\(8.45 + 3.15 = 11.6\) kg

This is the one to slow down on. For a total you take both upper bounds, but a difference is not like that.

\(\text{A} - \text{B}\) is biggest when A is as heavy as it can be and B is as light as it can be — a lighter B leaves a bigger gap.

\(8.45 - 3.05 = 5.4\) kg

Taking both upper bounds would give \(8.45 - 3.15 = 5.3\), which is just the difference of the rounded numbers and is not a bound at all.

Now the other way round: the gap is smallest when A is as light as it can be and B is as heavy as it can be.

\(8.35 - 3.15 = 5.2\) kg

So the difference in mass is at least 5.2 kg and less than 5.4 kg.

The rule worth remembering: for a sum, both bounds go the same way; for a difference, they swap.

Half of 1 cm is 0.5 cm, so each measurement is within 0.5 cm of the number given.

Upper bound of the length: \(30 + 0.5 = 30.5\) cm

Upper bound of the width: \(25 + 0.5 = 25.5\) cm

Both sides are being multiplied and both are positive, so the area is biggest when both are biggest:

\(30.5 \times 25.5 = 777.75\) cm²

Now take both measurements as small as they can be:

\(29.5 \times 24.5 = 722.75\) cm²

So the area is at least 722.75 cm² and stays below 777.75 cm².

Notice how far apart those are. Multiplying 30 by 25 gives 750, but half a centimetre of uncertainty on each side is worth several square centimetres of uncertainty in the area — which is why an area worked out from rounded lengths cannot be quoted to the same accuracy as the lengths were.

"Definitely" means the answer has to hold even in the worst case, so both quantities need their upper bounds: the most that could already be in the van, and the most the new crate could weigh.

Load already in, correct to the nearest 50 kg, could be up to \(1250 + 25 = 1275\) kg.

Crate, correct to the nearest 10 kg, could be up to \(90 + 5 = 95\) kg.

Worst-case total: \(1275 + 95 = 1370\) kg.

Comparing this to the 1386 kg limit: \(1370\) is within 1386, so the crate can definitely be added safely — Yes.

Comparing the rounded values as written down, \(1250 + 90 = 1340\) kg, would always look safe, but that ignores how much more each rounded value could really be — it is not a bound at all.

The sign gives the limit: the lift can safely carry at most 143 kg.

Total mass, correct to the nearest 10 kg: it is at least \(210 - 10/2 = 205\) kg and at most \(210 + 10/2 = 215\) kg.

Mass of the person leaving, correct to the nearest kg: it is at least \(72.5\) kg and at most \(73.5\) kg.

The remaining mass is a difference, so it is greatest when the most is left behind, that is the greatest total less the smallest mass leaving:

\(215 - 72.5 = 142.5\) kg

It is least when the smallest total loses the greatest mass leaving:

\(205 - 73.5 = 131.5\) kg

The remaining mass is between \(131.5\) kg and \(142.5\) kg, and the limit is 143 kg. The greatest possible remaining mass is not over the limit.

So the answer is: Yes.

The distance is rounded to the nearest 10 m, so its bounds are 460 ± 5. The time is rounded to the nearest 0.1 s, so its bounds are 76.8 ± 0.05.

Distance: from 455 m up to 465 m. Time: from 76.75 s up to 76.85 s.

A speed is a distance divided by a time, so it is biggest when the distance is biggest and the time is smallest — dividing by a smaller number gives a larger answer.

\(465 \div 76.75 \approx 6.06\) m/s (to 3 significant figures)

The other way round: the speed is smallest when the distance is smallest and the time is largest.

\(455 \div 76.85 \approx 5.92\) m/s (to 3 significant figures)

Both bounds of the time are used, but each one goes with the opposite bound of the distance. That crossing over is the whole of this question.