QuanticaPrealgebraOpen in the courseA step counter reads \(4{,}732\), and you round it to \(4{,}700\). The halfway mark between \(4{,}700\) and \(4{,}800\) is \(4{,}750\), and \(4{,}732\) falls below it, so the lower neighbor wins. That single comparison is all rounding ever is. This lesson turns it into a one-digit check and then handles ties, carries, and negative values.
Problem
Round \(4{,}732\) to the nearest hundred. The midpoint between \(4{,}700\) and \(4{,}800\) is \(4{,}750\). Which neighbor is \(4{,}732\) closer to?
Show a hint
- Mark the two neighbors \(4{,}700\) and \(4{,}800\) and find the point exactly between them. Then ask which side of that point \(4{,}732\) falls on.
- The midpoint is \(4{,}750\). Since \(4{,}732\) is below \(4{,}750\), it is closer to the neighbor on the left, which is \(4{,}700\).
Show the full solution
\(4{,}732\) sits between \(4{,}700\) and \(4{,}800\), and the midpoint is \(4{,}750\). Since \(4{,}732\) is below that, it leans toward the lower neighbor. The gap down is \(32\) and the gap up is \(68\), so the nearest hundred is \(\boxed{4700}\). Only a value at or above \(4{,}750\) would tick up to \(4{,}800\).
Problem
A festival shows \(68{,}431\) attendees. Round to the nearest thousand. The deciding digit is the hundreds digit. What is \(68{,}431\) rounded to the nearest thousand?
Show a hint
- The place you are rounding to is the thousands place, the \(8\) in \(68{,}431\). The digit that decides what happens to it is the next one to the right, the hundreds digit. Circle that digit and ask whether it is big enough to push the thousands place up.
- The hundreds digit is \(4\). Since \(4\) is \(4\) or less, the thousands digit stays as it is, it does not tick up. Now drop everything to the right of the thousands place to zero and read off the result.
Show the full solution
The thousands digit is the \(8\), so the answer is either \(68{,}000\) or \(69{,}000\). The deciding digit is the one just to its right, the hundreds digit \(4\). Since \(4\) is under \(5\), the thousands digit stays put and everything after it drops to zero. $$68{,}431 \approx \boxed{68{,}000}$$ The midpoint check agrees, since \(68{,}431\) is below \(68{,}500\).
Problem
Round \(2{,}950\) to the nearest hundred. The value lands exactly on the midpoint \(2{,}950\), so the tie rule sends it up. The hundreds digit is \(9\), so a carry rolls forward. What is the result?
Show a hint
- Find the two nearest hundreds that hug \(2{,}950\), then find the midpoint between them. Which way does the tie rule send you?
- You are rounding up from \(2{,}900\) toward \(3{,}000\). The hundreds digit is a 9 and cannot go to 10, so the carry pushes into the thousands. The result is the next clean hundred above \(2{,}950\).
Show the full solution
The hundreds on either side of \(2{,}950\) are \(2{,}900\) and \(3{,}000\), and the midpoint is \(2{,}950\) itself. A tie rounds to the larger value, so we go up to \(\boxed{3000}\). The hundreds digit is a \(9\), so it cannot climb. It rolls to \(0\) and carries \(1\) into the thousands, turning the \(2\) into a \(3\).
Problem
A 3D printer reads \(7.83\) m of filament remaining. Round to the nearest tenth. The hundredths digit \(3\) decides. What is \(7.83\) rounded to the nearest tenth?
Show a hint
- Your two neighbors are \(7.8\) and \(7.9\), and the line between them is at \(7.85\). The only digit that decides which neighbor you get is the hundredths digit, the one right after the tenths place.
- The hundredths digit of \(7.83\) is \(3\). Since \(7.83\) is below the midpoint \(7.85\), it stays with the lower neighbor. Keep the tenths digit as it is and drop the rest.
Show the full solution
The tenths on either side of \(7.83\) are \(7.8\) and \(7.9\), and the midpoint is \(7.85\). The hundredths digit is \(3\), so the reading sits below the midpoint and stays with the lower neighbor. Keep the tenths digit and drop the rest, giving \(\boxed{7.8}\).
Problem
A tide gauge reads \(3.617\) m. The harbor chart records to the nearest hundredth. The thousandths digit \(7\) decides. Round \(3.617\) to the nearest hundredth.
Show a hint
- You only care about which side of the midpoint \(3.615\) the reading falls on. Compare \(3.617\) to \(3.615\). Is it below the midpoint, at the midpoint, or above it?
- The digit that decides everything is the one just past the hundredths place, the thousandths digit. Here that digit is \(7\). Anything \(5\) or higher tips you up to the larger neighbor, so keep two digits after the point and bump the hundredths up.
Show the full solution
The hundredths on either side of \(3.617\) are \(3.61\) and \(3.62\), and the midpoint is \(3.615\). The thousandths digit is \(7\), which is \(5\) or more, so round up. The harbor chart records \(\boxed{3.62}\). The distances agree, since \(3.617\) is \(0.007\) above \(3.61\) but only \(0.003\) below \(3.62\).
Problem
A lab scale reads \(1.995\) grams. Round to the nearest hundredth. The thousandths digit \(5\) is a tie, so round up. The hundredths digit is \(9\), triggering a carry. What is the result?
Show a hint
- The deciding digit is the thousandths digit, the \(5\) in \(1.99\underline{5}\). A \(5\) is the exact halfway case, so round the hundredths place up to the larger value. The only catch is that the digit you are bumping is a \(9\), so think about where it goes when it cannot just become a \(10\).
- Bumping the hundredths \(9\) up writes a \(0\) there and carries \(1\) into the tenths place. But the tenths digit is also \(9\), so it becomes \(0\) and carries \(1\) into the ones place, turning \(1\) into \(2\). So \(1.99\) climbs to \(2.00\). Keep both decimal places.
Show the full solution
Keeping two decimal places makes the thousandths digit the decider, and here it is \(5\) with nothing behind it, an exact tie. So the hundredths place rounds up. That digit is a \(9\), so it becomes \(0\) and carries \(1\) into the tenths, which is also \(9\), so that becomes \(0\) and carries into the ones, lifting the \(1\) to a \(2\). $$1.995 \approx \boxed{2.00}$$ Keep both zeros, since the notebook still records two decimal places.
Problem
A freezer reads \(-83\) degrees. Round to the nearest ten. The neighbors are \(-90\) and \(-80\). Measure each gap and pick the closer one. What is the result?
Show a hint
- Find the gap from \(-83\) to each neighbor. How far is it left to \(-90\), and how far right to \(-80\)? The shorter trip wins.
- From \(-83\) it is \(7\) steps left to reach \(-90\), but only \(3\) steps right to reach \(-80\). The nearer multiple of ten is \(-80\).
Show the full solution
The multiples of ten around \(-83\) are \(-90\) and \(-80\). The gap down to \(-90\) is \(7\), and the gap up to \(-80\) is only \(3\), so the technician records \(\boxed{-80}\) degrees. Negatives use the same midpoint test as positives. The midpoint here is \(-85\), and \(-83\) sits to the right of it.
Problem
A thermostat reads \(-2.35\) degrees. Round to the nearest tenth. \(-2.35\) is the exact midpoint between \(-2.4\) and \(-2.3\). The tie rule rounds to the larger value. What does the display show?
Show a hint
- Forget the minus sign for a second and just picture the number line. Mark \(-2.4\) on the left and \(-2.3\) on the right, then drop \(-2.35\) in between. Is it closer to one of them, or stuck dead in the middle?
- It is a perfect tie, so use the rule, round to the larger value. "Larger" means farther to the right on the number line. Between \(-2.4\) and \(-2.3\), the one on the right is \(-2.3\), so that is your answer.
Show the full solution
The tenths around \(-2.35\) are \(-2.4\) and \(-2.3\), and \(-2.35\) is \(0.05\) from each, so it is an exact tie. A tie rounds to the larger value, and the larger of two negatives is the one farther right. The display shows \(\boxed{-2.3}\). It feels backwards because \(2.3\) is less than \(2.4\), but with the minus signs attached, \(-2.3\) is the bigger number.
Problem
Three crates weigh \(487\), \(213\), and \(356\) kg. Round each to the nearest hundred, then add the rounded values. What is your estimated total?
Show a hint
- Handle one crate at a time. Look at the part of each weight after the hundreds and decide whether it pushes the number up to the next hundred or leaves it at the hundred below.
- Round each one. \(487\) is past the halfway mark so it becomes \(500\), \(213\) is below halfway so it becomes \(200\), and \(356\) is past halfway so it becomes \(400\). Now add \(500 + 200 + 400\).
Show the full solution
For \(487\), the leftover past the hundred is \(87\), above \(50\), so it rounds up to \(500\). For \(213\) the leftover is \(13\), so it stays at \(200\). For \(356\) the leftover is \(56\), so it rounds up to \(400\). Adding gives \(500 + 200 + 400 = \boxed{1100}\). The exact total is \(1{,}056\), so the estimate tells you instantly whether the van is near its limit.
Problem
A field-trip invoice lists \(29\) students at \(6.10\) dollars each, printing \(1{,}769\) dollars. Round each to a friendly value and multiply. What is your quick estimate of \(29 \times 6.10\)?
Show a hint
- You are not finding the exact total here, you are building a quick comparison number. Replace each part with a friendly nearby value first. What is \(29\) rounded to the nearest ten, and what is \(6.10\) rounded to the nearest whole dollar?
- Rounding \(29\) to the nearest ten gives \(30\), since \(29\) is much closer to \(30\) than to \(20\). Rounding \(6.10\) to the nearest whole dollar gives \(6\), since \(6.10\) is just past \(6\). Now multiply \(30 \times 6\).
Show the full solution
\(29\) rounds to \(30\) at the nearest ten, and \(6.10\) rounds to \(6\) at the nearest dollar, since \(10\) cents is nowhere near the \(6.50\) midpoint. Multiplying the friendly numbers gives \(30 \times 6 = \boxed{180}\). The printed \(1{,}769\) is about ten times that, the signature of a decimal point that slipped one spot. The exact total is \(176.90\).
Problem
A gauge is rounded to the nearest tenth and reports \(4.3\) mm. The original reading was stored to two decimal places. What is the largest two-decimal value that still rounds to \(4.3\)?
Show a hint
- The reported \(4.3\) covers everything from \(4.25\) up to but not including \(4.35\). You only need the values in that stretch that have exactly two decimal places, so start at the top end and step downward.
- Check \(4.35\) first. It sits exactly halfway between \(4.3\) and \(4.4\), so by the larger-value rule it rounds up to \(4.4\), not \(4.3\). That means \(4.35\) is out. Step down one hundredth to \(4.34\) and round it to the nearest tenth to see where it lands.
Show the full solution
A reported \(4.3\) covers everything from \(4.25\) up to just under \(4.35\), because \(4.35\) is a tie that rounds up to \(4.4\) instead. The largest two-decimal value in that stretch is \(\boxed{4.34}\), whose hundredths digit \(4\) sends it down to \(4.3\). The band is half-open, so the lower end is included and the upper end is not.
Problem
A price rounds to \(7\) at the nearest dollar, to \(6.8\) at the nearest tenth, and its hundredths digit is \(0\). Exactly one price to the cent satisfies all three. What is it?
Show a hint
- Turn each clue into an interval. Rounding to the nearest whole dollar gives \(7\), so the price runs from \(6.5\) up to just under \(7.5\). Rounding to the nearest tenth gives \(6.8\), so the price runs from \(6.75\) up to just under \(6.85\). The price must sit inside both ranges at once, which means inside the tighter one.
- The overlap of the two ranges is \(6.75\) up to just under \(6.85\). Now list the prices in that band whose hundredths digit is \(0\). The candidates ending in zero cents are \(6.80\). Check the only one that lands in the band.
Show the full solution
Rounding to \(7\) at the nearest dollar means \(6.5 \le p < 7.5\). Rounding to \(6.8\) at the nearest tenth means \(6.75 \le p < 6.85\), which sits entirely inside the first range. The prices to the cent in that band run \(6.75\) through \(6.84\), and the only one with a hundredths digit of \(0\) is \(\boxed{6.80}\). Turning each rounding clue into a range and overlapping them is the standard way to pin a value down.
Practice these ideas
Practice
A scoreboard reads \(7{,}284\) fans. The press box reports to the nearest hundred. What is \(7{,}284\) rounded to the nearest hundred?
Show the solution
The hundreds around \(7{,}284\) are \(7{,}200\) and \(7{,}300\), and the midpoint is \(7{,}250\). Since \(7{,}284\) is above the midpoint, it climbs, and the press box reports \(\boxed{7300}\) fans.
Practice
A fundraiser total of \(3{,}500\) dollars is to be printed rounded to the nearest thousand. \(3{,}500\) is the exact midpoint. Using the half-way-up rule, what goes on the banner?
Show the solution
\(3{,}500\) is \(500\) from \(3{,}000\) and \(500\) from \(4{,}000\), so neither thousand is closer. That is an exact tie, and our rule sends a tie to the larger value, so the banner reads \(\boxed{4000}\).
Practice
A scale reads \(6.47\) kg. A recipe records to the nearest tenth. Round \(6.47\) to the nearest tenth.
Show the solution
The tenths around \(6.47\) are \(6.4\) and \(6.5\), and the midpoint is \(6.45\). The hundredths digit is \(7\), which is \(5\) or more, so the tenths digit goes up by one and \(6.47\) rounds to \(\boxed{6.5}\).
Practice
A lap is timed at \(0.382\) seconds. The results sheet keeps two decimal places. Round \(0.382\) to the nearest hundredth.
Show the solution
The hundredths around \(0.382\) are \(0.38\) and \(0.39\), and the midpoint is \(0.385\). The thousandths digit is \(2\), below \(5\), so the hundredths digit stays and everything after it is dropped, giving \(\boxed{0.38}\). The gaps confirm it, \(0.002\) down versus \(0.008\) up.
Practice
A bus ride takes exactly \(9.5\) minutes. The schedule lists whole minutes only. \(9.5\) is exactly halfway between \(9\) and \(10\). What whole number does the schedule show?
Show the solution
\(9.5\) is \(0.5\) from \(9\) and \(0.5\) from \(10\), so it is an exact tie. A tie rounds to the larger value, so the schedule lists the trip as \(\boxed{10}\) minutes.
Practice
A fitness band reads \(0.65\) battery and rounds to one decimal place. \(0.65\) is exactly halfway between \(0.6\) and \(0.7\). What does the band display?
Show the solution
\(0.65\) is \(0.05\) from \(0.6\) and \(0.05\) from \(0.7\), so it is a perfect tie. A tie rounds to the larger value, so the fitness band shows \(\boxed{0.7}\).
Practice
A caliper reads \(0.396\) cm and the spec sheet keeps two decimal places. The thousandths digit \(6\) rounds the hundredths \(9\) up, triggering a carry. Round \(0.396\) to the nearest hundredth.
Show the solution
The hundredths digit of \(0.396\) is the \(9\), and the thousandths digit \(6\) is at least \(5\), so that \(9\) rounds up. A \(9\) cannot become a single-digit \(10\), so it turns into \(0\) and carries \(1\) into the tenths, taking \(3\) to \(4\). That gives \(\boxed{0.40}\). Keep the trailing zero so the answer still shows the hundredths place the spec sheet asked for.
Practice
A fuel gauge reads \(8.97\) liters and the trip computer shows one decimal place. The tenths digit is \(9\), so rounding it up triggers a carry into the ones. Round \(8.97\) to the nearest tenth.
Show the solution
The hundredths digit of \(8.97\) is \(7\), which is at least \(5\), so the tenths place rounds up. The tenths digit is already \(9\), so it rolls over to \(0\) and carries \(1\) into the ones, taking the \(8\) to \(9\). The trip computer shows \(\boxed{9.0}\). Write the trailing zero to show the value was rounded to the nearest tenth.
Practice
A wind-chill of \(-46\) degrees is reported to the nearest ten. The two neighboring tens are \(-50\) and \(-40\), with midpoint \(-45\). Which side is \(-46\) on? Round \(-46\) to the nearest ten.
Show the solution
The multiples of ten around \(-46\) are \(-50\) and \(-40\). The gap to \(-50\) is \(4\) and the gap to \(-40\) is \(6\), so \(-46\) rounds to \(\boxed{-50}\). The midpoint is \(-45\), and \(-46\) sits to the left of it, so there is no tie to break.
Practice
An elevation app shows \(-3.5\) m. The value is exactly halfway between \(-4\) and \(-3\). The larger of those two is closer to zero. Round \(-3.5\) to the nearest whole number.
Show the solution
\(-3.5\) is exactly halfway between \(-4\) and \(-3\), so closeness cannot decide and the half-way rule takes over. Larger means farther right on the number line, and \(-3\) is right of \(-4\), so the elevation rounds to \(\boxed{-3}\) meters. Between two negatives, the larger one is always the one closer to zero.
Practice
A freezer probe reads \(-1.25\) degrees. The app rounds to the nearest tenth. \(-1.25\) is an exact tie between \(-1.3\) and \(-1.2\). Round to the larger value. What does the app display?
Show the solution
\(-1.25\) is \(0.05\) from \(-1.3\) and \(0.05\) from \(-1.2\), so it is an exact tie. A tie goes to the larger value, and \(-1.2\) lies to the right of \(-1.3\) on the number line, so the app displays \(\boxed{-1.2}\).
Practice
A concert order lists \(41\) tickets at \(19.80\) dollars each, printing a total of \(81.18\) dollars. Round each to the nearest friendly value and multiply. What is your quick estimate?
Show the solution
\(41\) has a ones digit of \(1\), so it rounds down to \(40\). \(19.80\) has \(80\) cents, past the \(50\) mark, so it rounds up to \(20\) dollars. Multiplying gives \(40 \times 20 = \boxed{800}\). The printed \(81.18\) is nowhere near that ballpark, so the receipt is almost certainly missing a digit.
Practice
Three boxes weigh \(596\), \(207\), and \(398\) grams. Round each to the nearest hundred, then add. What is the estimated total?
Show the solution
Round each weight to the nearest hundred one at a time. For \(596\), the tens digit is \(9\), which is well past halfway, so it rounds up to \(600\). For \(207\), the tens digit is \(0\), so it rounds down to \(200\). For \(398\), the tens digit is \(9\), so it rounds up to \(400\). Now add the friendly numbers, \(600 + 200 + 400 = 1200\). The estimate clears the \(1000\) gram mark with room to spare. \(\boxed{1200}\)
Practice
A rain gauge stores readings to the nearest hundredth but its display shows the nearest tenth, reading \(5.3\) mm. What is the smallest two-decimal stored value that rounds to \(5.3\)?
Show the solution
The tenths around the stored reading are \(5.2\) and \(5.3\), and the midpoint is \(5.25\). By the half-way rule that midpoint rounds up to \(5.3\), while \(5.24\) would drop to \(5.2\). The smallest qualifying value is \(\boxed{5.25}\).