Rounding and Estimation: Free Response

The place named is the hundredths, the second digit after the point, so the two multiples of one hundredth on either side of the reading come from keeping the hundredths digit as it stands and from raising it by one:

$8.46 \quad \text{and} \quad 8.47.$

Halfway between them is the lower neighbour plus half of one hundredth, which is five thousandths:

$\frac{8.46 + 8.47}{2} = 8.465.$

Now measure the reading against each neighbour by subtracting:

$8.4617 - 8.46 = 0.0017, \qquad 8.47 - 8.4617 = 0.0083.$

The first distance is the smaller one, so the lower neighbour is genuinely the nearer of the two and nothing here is a matter of choice: the direction is forced by which multiple is closer.

$8.4617 \approx 8.46.$

The answer keeps two decimal places, because that is what "to the nearest hundredth" asks for. Writing $8.5$ would be a rounding to a different place, and writing $8.4617$ would be no rounding at all.

Both readings have $6$ in the hundredths place, so both lie between the same two neighbours, $8.46$ and $8.47$, and the mark halfway between those is $8.465$.

Take the first reading. Its thousandths digit is $4$, and subtracting shows where it stands:

$8.465 - 8.4649 = 0.0001.$

The difference is positive, so $8.4649$ has not reached the halfway mark. The lower neighbour is nearer:

$8.4649 \approx 8.46.$

The second reading has thousandths digit $5$, with a further $2$ standing beyond it:

$8.4652 - 8.465 = 0.0002.$

This time the reading is strictly past the halfway mark, so the upper neighbour is nearer:

$8.4652 \approx 8.47.$

That second reading is worth a second look. A $5$ in the deciding place makes it look like a borderline case, but it is not one: the $2$ beyond the $5$ carries it strictly past halfway, so nearness by itself picks the upper neighbour and no tie-breaking rule is consulted. A number lands exactly on the halfway mark only when the deciding digit is $5$ and every digit after it is zero, and that case alone is settled by an agreed convention rather than by distance.

Fix the place being rounded to, and call the digit immediately to its right the deciding digit. The mark halfway between the two neighbours is the lower neighbour with a $5$ in the deciding digit's place and nothing after it.

Now bound the tail. Everything to the right of the deciding digit is worth less than one full unit of the deciding digit's place, however many digits there are. In the classmate's reading the deciding digit is the thousandths $4$, and the tail beyond it is worth at most a hair under one thousandth, even if every one of its digits were a $9$:

$8.4649 < 8.464999 < 8.465,$

and that stays true however many $9$s you write, because nine ten-thousandths and nine hundred-thousandths and so on always total less than one thousandth. So a deciding digit of $4$ or less leaves the number short of the halfway mark whatever follows it, and the lower neighbour is nearer. A deciding digit of $6$ or more puts the number past the mark, and the digits after it only add value, never take it away, so the upper neighbour is nearer. In both of those cases the tail is powerless, and one digit really is enough.

The case the digit does not finish on its own is a deciding digit of exactly $5$. With something nonzero after it the number is strictly past the mark, as $8.4652$ is, and distance still decides. With nothing but zeros after it the number lands exactly on the mark, both neighbours are the same distance away, and nearness picks neither. There is nothing left to measure, so that case is settled by an agreed tie-breaking rule instead, and the rule these lessons agree on is round half up.

So the classmate is half right about one thing and wrong about the main thing. Reading further can tell you whether a deciding digit of $5$ is a genuine tie or not, which does matter. What reading further can never do is reverse a decision that a deciding digit of $4$ or less, or of $6$ or more, has already made.

The hundredths place is the second digit after the point, which holds the second $9$, so the digit that decides the direction is the thousandths digit, $7$. Since $7$ is more than $5$, the reading is strictly past the halfway mark $19.995$ and the upper neighbour is nearer, so the hundredths place goes up. The two neighbours are

$19.99 \quad \text{and} \quad 20.00,$

and the distances confirm the choice:

$19.9973 - 19.99 = 0.0073, \qquad 20.00 - 19.9973 = 0.0027.$

Now follow the carry. Raising nine hundredths by one hundredth gives ten hundredths, which is one tenth, and a single place cannot hold ten. So the hundredths place empties to $0$ and passes one tenth to its left. That tenth arrives at a tenths place already holding $9$, so the same thing happens again: ten tenths is one whole, the tenths place empties to $0$, and one whole moves left. The ones place holds $9$ as well, so ten ones is one ten, the ones place empties to $0$, and the tens digit $1$ becomes $2$:

$19.9973 \approx 20.00.$

The two trailing zeros are not decoration. They record that the rounding was to the nearest hundredth. Writing $20$ names the same value but claims a rounding to the nearest whole number instead.

Rounding to the nearest tenth puts the rounding place at the first digit after the point, so the deciding digit is the hundredths $9$. That is above $5$, so round up, and the carry starts again: ten tenths is one whole, so the tenths place empties and the ones $9$ carries too.

$19.9973 \approx 20.0.$

Rounding to the nearest whole number puts the rounding place at the ones, so the deciding digit is the tenths $9$, again above $5$. Round up, and $19$ becomes $20$:

$19.9973 \approx 20.$

All three answers name the same value, twenty, because the reading was already within three thousandths of it. What differs is how each answer is written: $20.00$ keeps two decimal places, $20.0$ keeps one, and $20$ keeps none. That count is the record of which place was rounded to, so reporting $20$ when the nearest hundredth was asked for throws the record away.

None of the three roundings was a tie. Their deciding digits are $7$, $9$ and $9$, every one of them strictly above $5$, so in each case one neighbour really is nearer than the other and the direction is forced by distance. No agreed rule had to be consulted anywhere in this part.

Start with the check, because it settles the matter without redoing any work. Rounding to the nearest hundredth has only two possible outputs here, the two neighbouring multiples of one hundredth:

$19.99 \quad \text{or} \quad 20.00.$

The recorded $19.10$ is neither of them, so it cannot be this reading rounded to the nearest hundredth, whatever route produced it. A second check says the same thing by size. A rounding to the nearest hundredth never moves a number by more than half a hundredth, which is five thousandths, and the entry sits far further away than that:

$19.9973 - 19.10 = 0.8973.$

That is more than one hundred and seventy times the largest move such a rounding is allowed, so no amount of care in the digits could produce it.

Now the diagnosis. The decision was right: the deciding digit is above $5$, so the hundredths place does go up, and $9$ raised by one is ten. The step that fails is what was done with that ten. A single place holds one digit, so ten cannot be written into the hundredths place at all: ten hundredths is one tenth, and it has to leave that place and travel left, exactly as a column addition carries. Instead the ten was written across the two decimal places the answer had to have, as the digits $1$ and $0$, which both failed to carry the ten out of the hundredths place and overwrote the tenths $9$ that should have received it. That is why the entry keeps two decimal places and still lands nowhere near either neighbour.

Completing the carry properly empties the hundredths to $0$, sends one tenth into a tenths place already holding $9$, which starts the same carry again, and produces

$19.9973 \approx 20.00.$

One more tell is worth carrying away. Rounding up cannot make a number smaller, and $19.10$ is smaller than the reading, so the direction alone shows something has gone wrong before a single digit is examined.

The tenths place is the first digit after the point, so both values lie between the same two multiples of one tenth:

$4.2 \quad \text{and} \quad 4.3.$

Measure the first value against each neighbour:

$4.25 - 4.2 = 0.05, \qquad 4.3 - 4.25 = 0.05.$

The two distances are equal. Under the round half up convention these lessons use, a value that lands exactly on the halfway mark goes to the upper neighbour, so

$4.25 \approx 4.3.$

Now the second value. The three digits after the tenths are $5$, $0$ and $3$, so it stands a little beyond the halfway mark:

$4.2503 - 4.2 = 0.0503, \qquad 4.3 - 4.2503 = 0.0497.$

Here the distances are not equal, and the smaller one belongs to the upper neighbour:

$4.2503 \approx 4.3.$

The two values round to the same tenth, and the four distances are what the next part is for: they show that the two roundings did not arrive at $4.3$ the same way.

A value has a nearest tenth exactly when one of its two distances is smaller than the other, so compare the two distances in each case and let them answer.

For $4.2503$ the distances are $0.0503$ and $0.0497$, and they are not equal:

$0.0497 < 0.0503.$

The smaller distance belongs to the upper neighbour, so $4.3$ genuinely is the nearest tenth to $4.2503$. Nothing else is needed, and no rule about the digit $5$ was consulted to get there.

For $4.25$ the two distances came out the same:

$4.25 - 4.2 = 0.05 = 4.3 - 4.25.$

So neither neighbour is nearer. This value has no nearest tenth, and the phrase "round to the nearest tenth" has nothing to pick out. That is not a gap in the arithmetic; it is a genuine tie, and no amount of further looking will break it, because there is nothing left to measure.

What supplies the answer in that case is an agreement. A tie has to be broken by a rule people settle on in advance, and the rule these lessons use is round half up: a value landing exactly on the halfway mark goes to the upper neighbour. Under that agreement $4.25$ rounds to $4.3$. Notice how different that is from the other case: one of the two values was rounded by measurement, and the other by convention, even though both answers came out as $4.3$.

Take the classmate's reasoning first. It argues from a rule to a fact about distance, and that direction is impossible. The distances at the halfway mark are equal, and a rule cannot make one of two equal distances smaller:

$4.25 - 4.2 = 4.3 - 4.25.$

So the reasoning is unsound, even though the value it lands on is the one this course reports. That is the thing worth noticing: the answer comes from the agreement, and the agreement was made precisely because nearness had nothing to say. A rule can decide what to do when two distances are equal; it cannot report that they are unequal.

Now change the agreement and watch what moves. Round half to even sends a value landing exactly on the halfway mark to whichever neighbour carries an even digit in the rounding place. For $4.25$ the neighbours are $4.2$ and $4.3$, whose tenths digits are $2$ and $3$, so the even one is chosen:

$4.25 \approx 4.2 \quad \text{(round half to even)}.$

For $4.2503$ nothing changes, because it is not a tie. It stands strictly past the halfway mark, so it has a genuinely nearer neighbour, and every tie-breaking rule leaves it alone:

$4.2503 \approx 4.3 \quad \text{(either convention)}.$

That contrast is the whole point. Which neighbour is nearer is a fact about the number, checkable by subtraction, and no convention can overturn it. What to do when neither neighbour is nearer is a choice, and different communities make it differently: round half up is standard for everyday work and is what these lessons use, while round half to even is common in statistics and in calculators, because sending every tie upward nudges a long column of rounded values high. Exactly one of the two values in this question can be moved by that choice, and it is the one whose two distances came out equal.

Round each number to the place named. The price $7.45$ has tenths digit $4$, which is below $5$, so it rounds down to $7$ dollars. The count $18$ has ones digit $8$, which is above $5$, so it rounds up to $20$ boards. Both roundings are forced by distance; neither is a tie.

$7.45 \times 18 \approx 7 \times 20 = 140.$

So the estimate is about $140$ dollars, and it took one multiplication that can be done without paper.

Now the exact cost. Eighteen boards is twenty of them less two of them:

$7.45 \times 20 = 149, \qquad 7.45 \times 2 = 14.90, \qquad 149 - 14.90 = 134.10.$

The boards cost $134.10$ dollars exactly. Compare the two figures:

$140 - 134.10 = 5.90.$

The estimate sits $5.90$ dollars above the exact cost, which is close for a figure arrived at in the head, and the gap is worth remembering rather than forgetting: the next part is about what a gap of that size does to the estimate's usefulness.

Take the two roundings one at a time and ask what each does to the product on its own.

The price was rounded from $7.45$ down to $7$, a drop of $0.45$ on every board. Charging less per board than the boards cost makes the product too small, so this rounding pushes the figure down. The count was rounded from $18$ up to $20$, an increase of two boards. Paying for more boards than the club is buying makes the product too large, so this rounding pushes the figure up.

$7 < 7.45 \quad \text{and} \quad 20 > 18.$

Two pushes in opposite directions do not add up to a direction. Which way the estimate lands depends on which push is bigger, and that is a comparison of amounts rather than something visible in the roundings themselves. Here the price cut of $0.45$, across the twenty boards the estimate pays for, is worth

$0.45 \times 20 = 9.00,$

while the two extra boards at their true price are worth

$2 \times 7.45 = 14.90.$

The upward push is the larger, so this estimate ends up above the true cost, and by exactly the difference of those two amounts, $14.90 - 9.00 = 5.90$ dollars. Nothing in the roundings alone predicted that.

To be certain of the direction, do not let the roundings disagree. Round both numbers up:

$7.45 \times 18 \le 8 \times 20 = 160.$

Each factor has been replaced by something at least as large, and both quantities are positive, so the product can only grow. That makes $160$ dollars a figure the true cost cannot exceed. Rounding the price down and leaving the count alone, $7 \times 18 = 126$ dollars, gives a figure the true cost cannot fall below, by the same argument run the other way.

The estimate is $140$ dollars and the amount set aside is $150$ dollars, so the estimate is under the limit and the order looks affordable. The question is whether looking affordable and being affordable are the same thing here, and they are not, because an estimate carries an error whose size has to be compared with the gap it is being asked to clear:

$150 - 140 = 10.$

Ten dollars of room. Now measure what the roundings did. One of them paid for two boards the club is not buying, which at the true price is

$2 \times 7.45 = 14.90,$

more than the ten dollars of room on its own. The other rounding pulled in the opposite direction, so the two partly cancel and the net shift turns out to be smaller than either, but nothing in the estimate says so in advance. That is the difficulty: a figure assembled from shifts of that size, with no guarantee of direction, cannot settle a ten-dollar question, and the fact that it happens to fall on the affordable side of the limit is not evidence that the true cost does.

What settles it is the exact cost:

$134.10 < 150.$

The order does fit, with $150 - 134.10 = 15.90$ dollars left over. The exact figure was needed for that verdict even though the estimate turned out to point the same way, and that is the ordinary situation rather than bad luck.

The general lesson is about which questions each tool answers. An estimate answers a question about size: is this bill about a hundred dollars or about a thousand, is each board a few dollars or a few hundred. Those answers are separated by far more than the estimate's own error, so the error cannot reach across them. A question that turns on a limit, where the verdict flips if the total moves by ten dollars, is finer than that, so an estimate carrying no guarantee cannot answer it.

That is not the same as saying only exact arithmetic can. A figure the true cost is certain not to exceed does settle a limit question, provided that figure is itself under the limit. The $160$ dollars from part B is not, but a tighter guaranteed figure is: rounding only the price up, and leaving the count alone, gives

$7.45 \times 18 \le 8 \times 18 = 144,$

and $144 < 150$, so the order is certain to fit without the exact cost ever being computed. Rounding only the count up does the same, since $7.45 \times 18 \le 7.45 \times 20 = 149 < 150$. What cannot settle the question is a figure with no guaranteed direction, however close to the limit it lands.

An estimate is only worth making if the rounded number is easier to work with than the original, so round to a number that $8$ divides exactly. The multiples of $8$ on either side of $58.4$ are

$56 \quad \text{and} \quad 64.$

Measure the distance to each:

$58.4 - 56 = 2.4, \qquad 64 - 58.4 = 5.6.$

So $56$ is the nearer one, and that choice is forced by distance rather than by any convention. Divide:

$58.4 \div 8 \approx 56 \div 8 = 7.$

Each bed gets about $7$ kilograms of feed. Rounding $58.4$ to the nearest ten instead would give $60$, which is a perfectly good rounding, but $60 \div 8 = 7.5$ is more work to do in the head than $56 \div 8$ and buys nothing here: both figures put the answer in the neighbourhood of seven kilograms, which is all a check on size needs.

Put the two figures side by side. The estimate says about $7$ kilograms a bed; the record says $0.73$ kilograms, which is about seven tenths of a kilogram. Those are not the same size, and the question is whether the rounding could account for the gap. It could not, and the way to show that is to bound what the rounding was able to do. Rounding to the nearest multiple of $8$ never moves a number by more than half of $8$, which is $4$ kilograms of feed, and that change is shared over $8$ beds:

$4 \div 8 = 0.5.$

So the estimate is certain to be within half a kilogram a bed of the truth. The record sits

$7 - 0.73 = 6.27$

from the estimate, more than twelve times that bound, so the record cannot be explained by the rounding. It is the wrong size, and no careful re-reading of its digits will rescue it.

The shape of the mistake is the commonest one in decimal work: the digits are right and the decimal point is in the wrong place. Do the division exactly to see it. Eight goes into $58$ seven times with $2$ left over, and the remaining $2.4$ shared over $8$ is $0.3$:

$58.4 \div 8 = 7.3.$

Check it by multiplying back, which is the operation division undoes:

$7.3 \times 8 = 58.4.$

That reproduces the feed the gardener started with, so $7.3$ kilograms a bed is right. The recorded $0.73$ carries exactly the digits $7$ and $3$ in that order, with the point one place too far left, which is the same as dividing the true figure by ten:

$0.73 \times 10 = 7.3.$

The estimate did not produce the right answer and was never going to. What it did was refuse a wrong one, which is the job it was doing.

A check works by comparing a disagreement against what the estimate can guarantee, so the first job is to say what it guarantees. The estimate came from replacing $58.4$ with the nearest multiple of $8$. That kind of rounding never moves a number by more than half of $8$, which is $4$ kilograms of feed, and the change is shared over $8$ beds:

$4 \div 8 = 0.5.$

So the estimate of $7$ kilograms a bed is certain to be within half a kilogram of the truth, and a value with a bound is all this check carries.

It is worth being clear about what that throws away on purpose. This particular rounding moved the total down by a known $2.4$ kilograms, so anyone who tracks the direction as well as the size finds that the truth is exactly $2.4 \div 8 = 0.3$ above the estimate, which is the exact answer rather than a check on one. A size check gives that up deliberately: it is meant to cost seconds and to work when the difference has not been kept, so it remembers how far out the estimate may be and not which way.

Now put the suggested figure against the bound. A record of $7.4$ kilograms a bed disagrees with the estimate by

$7.4 - 7 = 0.4,$

and $0.4$ is inside the half-kilogram bound. The truth could be $7.4$, or it could be anything else within half a kilogram of $7$, and the check cannot tell those apart. A disagreement inside the bound refuses nothing and confirms nothing.

A misplaced decimal point is a different animal. Moving the point one place multiplies a figure by ten or divides it by ten, so a figure that should be near $7$ arrives near seven tenths or near seventy:

$0.7 \quad \text{and} \quad 70.$

The recorded $0.73$ disagrees with the estimate by $6.27$, more than twelve times the bound, and no rounding of the total could account for that. This is why an estimate is a strong guard against a misplaced point: the mistake it is hunting is enormous next to what the estimate might be out by, so the figure can be refused outright.

Two statements can now be made, and each needs its own care. First, a figure can be refused when its disagreement with the estimate exceeds the bound, because the truth cannot be further from the estimate than the bound allows. That is a one-way test, and a disagreement inside the bound settles nothing either way. Second, an error is certain to be caught only when it is larger than twice the bound, because even if the estimate is out by the full bound in the unhelpful direction, an error that big still leaves a disagreement above the bound. Between those two sizes the outcome depends on the case: a record of $6.9$ is only $0.1$ from the estimate and slips straight through, while a record of $7.9$ is $0.9$ from it and is refused.

So the kind of mistake this puts within reach is anything that changes the size of an answer, the misplaced decimal point above all. What it usually leaves out is a figure wrong in its final digit, whose disagreement with the estimate is normally smaller than the bound, though $7.9$ shows that a final-digit slip can be large enough to be caught. When you need a finer check, shrink the bound: round less aggressively, so more errors fall outside it, at the cost of an estimate that takes more work. What no estimate becomes is a substitute for the exact arithmetic.