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Rounding and Estimation: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 The scale labels

A scale can label a reading with either $27.8$ or $27.9$ . Which label is nearer to $27.86$ ?
1. Hint 1
  Compare the distances from the reading to the two allowed labels.
2. Hint 2
  Subtract the smaller number from the larger number for each distance.
Answer

$27.9$ .

Full solution

The distance from the lower label is
$27.86-27.8=0.06$

The distance from the upper label is
$27.9-27.86=0.04$

Since $0.04<0.06$ , the label $27.9$ is nearer.
To the nearest tenth, $27.86\approx27.9$ .

Answer

$27.9$ .

Key idea

Rounding selects the allowed value with the smaller distance from the original number.
Problem 1: Mark complete
Problem 2 The logger entry

A logger rounds each reading to the nearest thousandth and then records how many thousandths the rounded value contains. What does it record for a reading of $0.003672$ ?
1. Hint 1
  First decide which thousandth is nearer to the reading.
2. Hint 2
  After rounding, express that value as a count of parts each worth one thousandth.
Answer

$4$ .

Full solution

The reading is between $0.003$ and $0.004$ , above their midpoint $0.0035$ .
It therefore rounds to
$0.003672\approx0.004$

That rounded value is four thousandths.
$0.004=\frac{4}{1000}$

The logger records the count $4$ , rather than the decimal value of the reading.

Answer

$4$ .

Key idea

After rounding to a place, the rounded value is a whole number of units of that place, so $0.004$ is four thousandths.
Problem 2: Mark complete
Problem 3 The combined charge

A service makes two charges of $0.486$ dollars each. The combined charge is then rounded to the nearest cent. What combined amount is billed?
1. Hint 1
  Only the combined charge is rounded, so first work with the exact charges.
2. Hint 2
  Add the charges, then round the sum to two decimal places, since a cent is one hundredth of a dollar.
Answer

$0.97$ dollars.

Full solution

First add the exact charges.
$0.486+0.486=0.972$

A cent is one hundredth of a dollar.
The thousandths digit is $2$ , so the hundredths digit stays as it is.
$0.972\approx0.97$

The billed amount is $0.97$ dollars.
Rounding each charge first would give $0.49+0.49=0.98$ dollars, one cent more, so rounding before adding changes the bill.

Answer

$0.97$ dollars.

Key idea

When a total must be rounded, compute the exact total before rounding it once.
Problem 3: Mark complete
Problem 4 The unfinished record

A reading is written $17.9\square$ , where the box holds one digit. Which digits in the box make the reading round to $18.0$ to the nearest tenth? Use round half up for an exact tie.
1. Hint 1
  The missing digit is immediately after the rounding place.
2. Hint 2
  Find the midpoint between the two possible rounded readings, including how an exact tie is treated.
Answer

$5$ , $6$ , $7$ , $8$ and $9$ .

Full solution

The two neighboring tenths are $17.9$ and $18.0$ , with midpoint $17.95$ .
The missing digit is in the hundredths column.
At the midpoint, round half up gives
$17.95\approx18.0$

Each larger digit also rounds upward.
Digits below $5$ stay below the midpoint, for example
$17.94\approx17.9$

Thus the allowed digits are $5$ , $6$ , $7$ , $8$ and $9$ .
Rounding up turns the nine tenths into a whole, so the recorded result is $18.0$ .

Answer

$5$ , $6$ , $7$ , $8$ and $9$ .

Key idea

Find the midpoint and include its tie convention when recovering digits from a rounded result.
Problem 4: Mark complete
Problem 5 The trip log

A trip meter reads $12.384$ km before a journey and $17.08$ km afterward. What distance should be entered in a log that reports journeys to the nearest hundredth of a km?
1. Hint 1
  The journey distance is the change in the meter reading.
2. Hint 2
  Pad the final reading to three decimal places before subtracting, then watch for a $9$ that carries when the hundredths digit rounds up.
Answer

$4.70$ km.

Full solution

Pad the final reading to thousandths and subtract the starting reading.
$17.080-12.384=4.696$

The thousandths digit is $6$ , so the hundredths digit rounds upward.
Raising nine hundredths carries into the tenths.
$4.696\approx4.70$

The log should record $4.70$ km.
Before rounding, adding $4.696$ km to the starting reading returns the final reading.

Answer

$4.70$ km.

Key idea

When the digit in the rounding place is a $9$ that rounds up, it carries into the place on its left, so $4.696$ rounds to $4.70$ .
Problem 5: Mark complete
Problem 6 The two measurements

Two measurements are $6.48$ meters and $2.51$ meters. Estimate their difference by rounding each to the nearest whole number. Without finding the exact difference, say whether the estimate is above or below it and why; then find the exact difference to check.
1. Hint 1
  The estimate uses rounded measurements, while the exact difference uses the original values.
2. Hint 2
  Note which way each measurement moved when it was rounded, and how each move changes a difference.
3. Hint 3
  Subtract the original measurements last, to check the prediction you made from those moves.
Answer

Estimate: $3$ meters, below the exact difference of $3.97$ meters.

Full solution

To the nearest whole number, $6.48\approx6$ and $2.51\approx3$ .
The estimate is
$6-3=3$

Rounding lowered the first measurement, and a smaller starting number gives a smaller difference.
Rounding raised the second measurement, and subtracting a larger number also gives a smaller difference.
Both changes shrink the difference, so the estimate is below the exact difference.

The exact difference is
$6.48-2.51=3.97$

The estimate is below it, as predicted, by $0.97$ meters.
$3.97-3=0.97$

Answer

Estimate: $3$ meters, below the exact difference of $3.97$ meters.

Key idea

When a difference is estimated, rounding the first number down and the second number up both push the estimate below the exact difference.
Problem 6: Mark complete
Problem 7 The carrier limit

Two containers weigh $18.4$ kg and $13.7$ kg. A carrier can hold at most $32$ kg. Estimate their combined weight by rounding each weight to the nearest whole number. Does the estimate settle whether the carrier can hold both? Decide whether it can.
1. Hint 1
  Compare the estimate with the limit, then ask how far the two roundings could have moved the estimate away from the true total.
2. Hint 2
  Add the rounded weights and the original weights separately.
Answer

Estimate: $32$ kg, which does not settle it; exact total: $32.1$ kg; the carrier cannot hold both.

Full solution

The rounded weights are $18$ kg and $14$ kg, giving
$18+14=32$

The estimate lands exactly on the $32$ kg limit.
Each weight moved by less than half a kilogram when it was rounded, so the true total could lie on either side of the limit, and the estimate does not settle the question.

The exact total is
$18.4+13.7=32.1$

The exact total is $0.1$ kg over the $32$ kg limit, so the carrier cannot hold both.
Rounding moved $18.4$ down by $0.4$ but $13.7$ up by $0.3$ , so the estimate came out $0.1$ kg too low and seemed to fit.

Answer

Estimate: $32$ kg, which does not settle it; exact total: $32.1$ kg; the carrier cannot hold both.

Key idea

Use exact amounts when an estimate is too close to a stated limit to settle a decision.
Problem 7: Mark complete
Problem 8 The covered digits

A terminating decimal starts with $6.47$ , and its remaining digits are covered. Using round half up, which of these can be settled now: the number to the nearest whole number, to the nearest tenth, and to the nearest hundredth? Give each value that can be settled and explain.
1. Hint 1
  The visible digits fix a range the number must lie in; compare that range with the midpoint between the two neighbors at each place asked.
2. Hint 2
  The covered digits add less than one hundredth to the visible starting value.
3. Hint 3
  A rounding is settled when the whole range sits on one side of its midpoint, and still open when the range reaches both sides.
Answer

Nearest whole number: $6$ ; nearest tenth: $6.5$ ; nearest hundredth: cannot be settled ( $6.47$ or $6.48$ ).

Full solution

Whatever the covered digits are, they add less than one hundredth to $6.47$ , so the number is at least $6.47$ and below $6.48$ .

For the nearest whole number, the neighbors are $6$ and $7$ , with midpoint $6.5$ .
The whole range lies below $6.5$ , so the number rounds to $6$ .

For the nearest tenth, the neighbors are $6.4$ and $6.5$ , with midpoint $6.45$ .
The whole range lies past $6.45$ , so the number rounds to $6.5$ .

For the nearest hundredth, the neighbors are $6.47$ and $6.48$ , with midpoint $6.475$ .
The range reaches both sides of it: $6.471$ would round to $6.47$ , and $6.479$ would round to $6.48$ .

That rounding depends on the covered thousandths digit, so it cannot be settled yet.

Answer

Nearest whole number: $6$ ; nearest tenth: $6.5$ ; nearest hundredth: cannot be settled ( $6.47$ or $6.48$ ).

Key idea

For a nonnegative terminating decimal under round half up, a rounding can be settled once the digit just right of the rounding place is visible.
Problem 8: Mark complete
Problem 9 The matching results

Under round half up, both $4.250$ and $4.251$ round to $4.3$ to the nearest tenth. A student says that switching to round half down would change both results. Is the student correct? Explain.
1. Hint 1
  The same rounded result does not require the same relationship to the midpoint.
2. Hint 2
  Compare each original value with the midpoint between the neighboring tenths.
Answer

No; under round half down $4.250$ becomes $4.2$ , while $4.251$ still rounds to $4.3$ .

Full solution

The midpoint between $4.2$ and $4.3$ is $4.25$ .
The first value is exactly there.
Its distance to the lower tenth is
$4.250-4.2=0.050$

Its distance to the upper tenth is the same.
$4.3-4.250=0.050$

So $4.250$ is a tie, and the tie rule decides it: round half up gives $4.3$ , while round half down gives $4.2$ .

The second value is past the midpoint.
Its distance to the lower tenth is
$4.251-4.2=0.051$

Its distance to the upper tenth is
$4.3-4.251=0.049$

Since $4.3$ is strictly nearer, $4.251$ rounds to $4.3$ under either tie rule.
Switching rules changes only the result for $4.250$ , so the student is not correct.

Answer

No; under round half down $4.250$ becomes $4.2$ , while $4.251$ still rounds to $4.3$ .

Key idea

Switching the tie convention changes a rounded result only when the value sits exactly on the midpoint.
Problem 9: Mark complete
Problem 10 The filling log

A machine fills $18$ bins with $4.35$ liters each. Its log printed the total's digits as $783$ but lost the decimal point. Use an estimate, with the amount in each bin rounded to the nearest whole number, to place the point, and explain, from the most each bin's amount could have moved when it was rounded, why no other placement of the point is possible.
1. Hint 1
  The total amount is the number of bins multiplied by the amount that goes into each bin.
2. Hint 2
  Compare the rounded product with the values the digits can make as the point moves, and keep the one of the right size.
3. Hint 3
  Each bin's amount moved by less than half a liter when it was rounded. How far could eighteen such moves shift the total?
Answer

Estimate: $72$ liters; total: $78.3$ liters.

Full solution

The amount in each bin rounds to $4$ liters, and $18$ is already a whole number.
The estimate is
$18\times4=72$

Placing the point in the digits $783$ gives values such as $7.83$ , $78.3$ and $783$ .
Only $78.3$ is near $72$ , so the total has two digits before its point.

Each bin's amount moved by less than half a liter when it was rounded, so the eighteen moves together shift the total by less than
$18\times0.5=9$

So the true total lies within $9$ liters of $72$ , between $63$ and $81$ .

$78.3$ lies in that range, while every other placement is at most $7.83$ or at least $783$ , far outside it, so no other placement of the point is possible.

As a check, multiply the digits as whole numbers.
$18\times435=7830$

The factor $4.35$ has two decimal places, so the product is $78.30$ , which is $78.3$ liters and matches the printed digits $783$ .

Answer

Estimate: $72$ liters; total: $78.3$ liters.

Key idea

Knowing how far rounding could move each input bounds how far an estimate can miss, and that bound can rule out a misplaced decimal point.
Problem 10: Mark complete

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