QuanticaPrealgebraOpen in the courseEach place-value column is worth ten times the one on its right, and those columns only ever build whole numbers. But a length between \(14\) and \(15\) has no whole-number name. So keep the columns running to the right, each one worth a tenth of the one before it.
Problem
In \(5.83\) the digit \(8\) sits in the first place past the decimal point, the tenths place. What is the value of the \(8\) alone? Write your answer as a fraction \(\tfrac{a}{b}\).
Show a hint
- The value of a digit is the digit times the worth of its place. The 8 is sitting in the tenths place, and one single tenth is worth \(10^{-1} = \tfrac{1}{10}\). So you have 8 of those tenths.
- Take the digit \(8\) and multiply it by the place value \(\tfrac{1}{10}\). That is \(8 \times \tfrac{1}{10}\), which stacks 8 copies of one tenth on top of each other. Write the result as a single fraction.
Show the full solution
The \(8\) sits one step past the point, in the tenths place, and one unit of that place is worth \(10^{-1}=\tfrac{1}{10}\). Eight of those is \(8 \times \tfrac{1}{10} = \boxed{\tfrac{8}{10}}\). A digit's value is always the digit times the worth of its place, so the \(5\) and the \(3\) do not affect this one.
Problem
In \(5.83\) the digit \(3\) sits one step right of the tenths place, in the hundredths place. What is the \(3\) worth all by itself? Write your answer as a fraction \(a/b\).
Show a hint
- The 3 is one step to the right of the tenths place. Stepping right means the place value gets divided by 10 again, so this new place is worth \(\tfrac{1}{10}\) of \(\tfrac{1}{10}\).
- One tenth of one tenth is \(\tfrac{1}{10}\times\tfrac{1}{10}=\tfrac{1}{100}\), which is \(10^{-2}\). The digit there is a \(3\), so its value is \(3\) of those, meaning \(3\times\tfrac{1}{100}\).
Show the full solution
Step one place right from the tenths and the place value divides by \(10\) again, so the \(3\) sits in the hundredths place, worth \(\tfrac{1}{10}\times\tfrac{1}{10}=\tfrac{1}{100}\). Three of those is \(3\times\tfrac{1}{100}=\boxed{3/100}\). That is \(10^{-2}\), and it should come out smaller than the \(8\) tenths next door, since places shrink as you move right.
Problem
Write the decimal with 5 in the ones place, 7 in tenths, nothing in hundredths, and 3 in thousandths as a single decimal.
Show a hint
- Set up the columns left to right just as they are named. One column for the ones, then the decimal point, then tenths, then hundredths, then thousandths. Drop each given digit into its own column and ask what belongs in the hundredths column when the part is \(\frac{0}{100}\).
- The digits in order are 5, then a point, then 7, then 0, then 3. That zero is not optional. It is the placeholder that keeps the 3 sitting in the thousandths column instead of letting it drift left. Read the columns straight across to get the decimal.
Show the full solution
Drop each digit into its own column, left to right. Ones \(5\), then the point, then tenths \(7\), hundredths \(0\), thousandths \(3\). Reading straight across gives \(\boxed{5.703}\). That middle zero is doing real work. Without it you would write \(5.73\), which puts the \(3\) in the hundredths place, ten times bigger than intended.
Problem
Write \(0.27\) as a single fraction over a power of ten in the form \(\frac{a}{b}\). Use a common denominator of \(100\).
Show a hint
- You have two pieces, \(\frac{2}{10}\) and \(\frac{7}{100}\). To add them you need the same denominator. Which power of ten do both pieces fit into, \(10\) or \(100\)?
- Rewrite \(\frac{2}{10}\) over \(100\) by multiplying top and bottom by \(10\), which gives \(\frac{20}{100}\). Now add the hundredths together, \(\frac{20}{100}+\frac{7}{100}\), and keep the denominator the same.
Show the full solution
The \(2\) is in tenths and the \(7\) is in hundredths, so \(0.27=\frac{2}{10}+\frac{7}{100}\). Rewrite the first piece over \(100\) as \(\frac{20}{100}\) and add, \(\frac{20}{100}+\frac{7}{100}=\boxed{\frac{27}{100}}\). Two decimal places always give a denominator of \(10^{2}=100\), with the digits after the point read as the numerator.
Problem
A jeweller records a mass as \(\frac{9}{10}\) of a gram. Write it as a decimal.
Show a hint
- A denominator of \(10\) means tenths, and the tenths place is the first slot to the right of the decimal point. So the only digit you need to write past the point is the \(9\).
- Put the \(9\) in the tenths place, just after the point, and write a \(0\) in the ones place in front so the answer reads \(0.9\) instead of a lonely \(.9\).
Show the full solution
A denominator of \(10\) means tenths, and tenths is the first place past the point, so the \(9\) goes there. There are no whole grams, so a \(0\) fills the ones place, giving \(\boxed{0.9}\). That leading zero is only there so the answer does not arrive as a bare \(.9\) that is easy to misread.
Problem
Are \(0.3\) and \(0.30\) equal as numbers? Answer yes or no.
Show a hint
- A trailing zero lives in a brand new place to the right. Ask yourself what that place is actually holding. The hundredths place in \(0.30\) holds a \(0\), so it contributes \(\frac{0}{100}\), and adding nothing to a number leaves it alone.
- Turn each reading into a fraction and see if they match. \(0.3\) means \(\frac{3}{10}\). For \(0.30\), write \(\frac{30}{100}\) and simplify by dividing top and bottom by \(10\) to get \(\frac{3}{10}\). Same fraction means same number.
Show the full solution
Read each one as a fraction. \(0.3\) is \(\frac{3}{10}\). In \(0.30\) the \(3\) is still in tenths and the new \(0\) sits in hundredths, adding \(\frac{0}{100}=0\), so \(0.30=\frac{3}{10}\) as well. Same fraction, same number, so \(\boxed{\text{yes}}\). Tacking on a trailing zero is the same move as multiplying top and bottom by \(10\), which never changes a value.
A whole number has an exact spot on the number line, but \(7.4\) falls between \(7\) and \(8\). Zoom in on that gap. Chop it into ten tenths, then chop each tenth into ten hundredths. The same divide-by-ten step that built the columns lets you land on any decimal.
Problem
The unit from \(0\) to \(1\) is split into \(10\) equal steps. \(0.7\) means seven tenths. Which numbered mark does \(0.7\) land on?
Show a hint
- The marks are \(0.1, 0.2, 0.3\) and so on. Walk along them one tenth at a time and count how many steps it takes to arrive at \(0.7\).
- The number \(0.7\) means \(7\) tenths, that is \(7 \times \tfrac{1}{10}\), so it sits \(7\) one-tenth steps from \(0\).
Show the full solution
\(0.7\) means seven tenths, that is \(7 \times \tfrac{1}{10}\), so it sits seven one-tenth steps from \(0\) and lands on mark \(\boxed{7}\). The digit in the tenths place gives the count directly, so you never have to walk the marks.
Problem
Stack \(0.7\) and \(0.68\) with decimal points aligned. Read from left to right one place at a time. Which is larger?
Show a hint
- Line up the decimal points and look only at the tenths place to start. One number has \(7\) tenths there, the other has \(6\) tenths. The hundredths and anything smaller come later and only matter if the tenths tie.
- The tenths place already disagrees, so that is where it ends. \(7\) tenths beats \(6\) tenths, and even the largest possible hundredths digit in \(0.68\) cannot add up to a full extra tenth. So the number with \(7\) in the tenths place wins.
Show the full solution
Pad \(0.7\) to \(0.70\) so both numbers show the same places, then read from the left. The tenths place has a \(7\) against a \(6\), and that first disagreement settles it, so \(\boxed{0.7}\) is larger. The most a hundredths digit can be worth is \(9\) hundredths, still less than one whole tenth, so nothing to the right can close a gap opened in the tenths.
Problem
Which number is larger, \(0.5\) or \(0.4999\)? Stack them with points aligned and compare place by place from the left.
Show a hint
- Write them one above the other with the decimal points aligned, \(0.5000\) over \(0.4999\). Now scan left to right and find the first place where the two digits are different. That place, and nothing after it, settles the question.
- The first differing column is the tenths. There \(0.5\) shows a \(5\) and \(0.4999\) shows a \(4\). Five tenths already beats four tenths, so the larger number is decided before you ever reach those \(9\)s. All three \(9\)s live in the hundredths, thousandths, and ten-thousandths, places far too tiny to add up to a whole extra tenth.
Show the full solution
Pad to the same length and compare \(0.5000\) with \(0.4999\) from the left. The tenths place has a \(5\) against a \(4\), so \(\boxed{0.5}\) is larger. Over a common denominator the gap is \(\frac{5000}{10000}-\frac{4999}{10000}=\frac{1}{10000}\). Those three \(9\)s buy finer detail, not more size.
Problem
Order \(0.4\), \(0.39\), \(0.401\), \(0.4009\), and \(0.41\) from smallest to largest by aligning decimal points. Which number is third in the list?
Show a hint
- First order them. Line up the decimal points and read the tenths place. Four splits begin with \(4\) tenths and one with \(3\) tenths, so the \(3\)-tenths one is smallest. Then pad the rest with trailing zeros to the same length and compare.
- Padded to four places the four tied splits are \(0.4000, 0.4010, 0.4009, 0.4100\), so compare \(4000, 4010, 4009, 4100\). The full order is \(0.39, 0.4, 0.4009, 0.401, 0.41\). Pick the one in the middle.
Show the full solution
Only \(0.39\) opens with \(3\) tenths, so it is the smallest. Pad the other four to \(0.4000, 0.4010, 0.4009, 0.4100\) and compare them like the whole numbers \(4000, 4010, 4009, 4100\). The full order is \(0.39, 0.4, 0.4009, 0.401, 0.41\), so third is \(\boxed{0.4009}\). It shows the most digits yet is not the biggest, because those extra digits sit in the thousandths and ten-thousandths where they are worth almost nothing.
Problem
Multiplying by \(10\) shifts every digit one place left, so the decimal point appears to hop right. Multiplying by \(100\) shifts two places. What is \(3.7 \times 100\)?
Show a hint
- Each factor of \(10\) shifts every digit one place toward higher value. You have \(100\), which is two tens, so that is two places. Watch where the \(3\) and the \(7\) land.
- Holding the point still, multiplying by \(100\) makes it appear to move two places to the right. Starting from \(3.7\), one place gives \(37\), and a second place gives \(370\).
Show the full solution
Multiplying by \(100\) shifts every digit two places toward higher value. The \(3\) goes from ones to hundreds, worth \(300\), and the \(7\) goes from tenths to tens, worth \(70\). Together that is \(300+70=\boxed{370}\). Watching the point instead of the digits gives the same thing, two places right, \(3.7\) to \(37\) to \(370\).
Problem
A tag should read \(4.50\) dollars but printed \(45.0\) dollars instead, shifting the decimal point one place right. By how many times is the printed price larger than the correct price?
Show a hint
- Look at where the digit 4 lives in each price. In 4.50 it is in the ones place, worth 4. In 45.0 it is in the tens place, worth 40. Every digit has jumped up exactly one place.
- When each digit moves one place to the left, its value gets ten times bigger. So the whole price is ten times bigger. You can check by asking what you multiply 4.50 by to reach 45.0.
Show the full solution
In \(4.50\) the \(4\) is in the ones place, worth \(4\). In \(45.0\) that same \(4\) is in the tens place, worth \(40\). Every digit moved one place left, so every digit is worth ten times as much and so is the whole price, making the printed tag \(\boxed{10}\) times larger. Moving the point one place right always multiplies by \(10\), and \(4.50 \times 10 = 45.0\) confirms it.
Practice these ideas
Practice
Four splits in seconds: Runner A \(0.205\), B \(0.21\), C \(0.2\), D \(0.2049\). Ordered smallest to largest, which split is the second smallest?
Show the solution
Pad every split to four places, \(0.2000, 0.2049, 0.2050, 0.2100\), and read left to right. All four tie at \(2\) tenths. In the hundredths only \(0.2100\) has a \(1\), so it is the largest. The other three split at the thousandths, where the digits are \(0\), \(4\), \(5\). The order is \(0.2, 0.2049, 0.205, 0.21\), so the second smallest is \(\boxed{0.2049}\).
Practice
In \(4.36\), swap the tenths and hundredths digits. What new decimal do you get?
Show the solution
The swap touches only the two digits after the point. The \(6\) moves into tenths and the \(3\) moves into hundredths, while the \(4\) stays where it is, giving \(\boxed{4.63}\). Swapping changes the number because the tenths place is worth ten times the hundredths place.
Practice
Write \(0.013\) as a single fraction \(a/b\) with a power-of-ten denominator.
Show the solution
The last digit \(3\) sits in the thousandths place, so the denominator is \(1{,}000\), and the digits after the point read as \(13\) on top. That gives \(\boxed{13/1000}\). The leading zero adds nothing to the numerator, but it does push the \(3\) out to thousandths, which is what sets the denominator.
Practice
Write \(\frac{7}{100}\) as a decimal.
Show the solution
A denominator of \(100=10^{2}\) means hundredths, the second place after the point. Put the \(7\) there and a \(0\) in the tenths place to hold it open, giving \(\boxed{0.07}\). Drop that inner zero and you get \(0.7\), which is ten times too big.
Practice
A cartridge holds \(100\) drops, each weighing \(0.306\) grams. What is the total weight in grams? Compute \(0.306 \times 100\).
Show the solution
Multiplying by \(100\) slides every digit two columns left. The \(3\) runs from tenths to ones to tens, worth \(30\), and the \(6\) runs from thousandths to hundredths to tenths, worth \(0.6\). Adding gives \(\boxed{30.6}\) grams. Each factor of ten is one slide, so \(100\) is two.
Practice
A weather station records \(8.2\) mm of rain, then converts to meters by dividing by \(1{,}000\). What is the rainfall in meters?
Show the solution
Dividing by \(1{,}000 = 10^{3}\) slides every digit three columns right. The \(8\) runs from ones down to thousandths and the \(2\) lands in ten-thousandths, with zeros filling the empty tenths and hundredths. So \(8.2 \div 1{,}000 = \boxed{0.0082}\). Shrinking by a factor of a thousand should land far below \(1\), and it does.
Practice
A point on the number line is at \(14.7\). What is the smaller of the two whole numbers it sits between?
Show the solution
\(14.7\) is \(14\) whole units plus \(7\) tenths, so it sits past \(14\) but short of \(15\). The smaller of the two is \(\boxed{14}\). The digits left of the point always name the largest whole number the value has fully reached.
Practice
The value \(0.73\) falls between two neighboring tenths. Which of those two tenths is the larger one?
Show the solution
The tenths digit of \(0.73\) is \(7\), and the \(3\) hundredths carries it just past \(0.7\). It is \(73\) hundredths, still short of the \(80\) hundredths that make \(0.8\), so it lies between \(0.7\) and \(0.8\) and the larger tenth is \(\boxed{0.8}\).
Practice
Which number is larger, \(0.6\) or \(0.5999\)?
Show the solution
Pad the short one to \(0.6000\) and compare with \(0.5999\) from the left. The tenths place has a \(6\) against a \(5\), so \(\boxed{0.6}\) is larger. More digits does not mean more value. As fractions, \(\frac{6000}{10000}\) beats \(\frac{5999}{10000}\) by a single ten-thousandth.
Practice
Which number is larger, \(0.07\) or \(0.7\)?
Show the solution
In \(0.7\) the \(7\) is in the tenths place, worth \(\frac{7}{10}\). In \(0.07\) the zero takes the tenths place and pushes the \(7\) out to hundredths, worth \(\frac{7}{100}\). Since \(\frac{7}{10}=\frac{70}{100}\), the larger number is \(\boxed{0.7}\), ten times as big. An inner zero moves a digit, so it changes the value.
Practice
Four forms: \(0.6\), \(0.60\), \(0.600\), and \(0.06\). How many of these name the exact same amount as \(0.6\)?
Show the solution
\(0.6\) is \(\frac{6}{10}\). Then \(0.60=\frac{60}{100}=\frac{6}{10}\) and \(0.600=\frac{600}{1000}=\frac{6}{10}\), so counting \(0.6\) itself that is \(\boxed{3}\) forms naming the same amount. \(0.06\) is the odd one out, since its zero sits before the \(6\) and drops it to hundredths, ten times smaller.
Practice
A receipt should read \(8.30\) dollars but printed \(83.0\) dollars. How many times larger is the printed amount?
Show the solution
In \(8.30\) the \(8\) sits in the ones column, and in \(83.0\) it has slid into the tens column, so it counts for ten times as much and every other digit does too. The printed amount is \(\boxed{10}\) times larger. Checking directly, \(83.0 \div 8.3 = 10\), which is exactly what moving the point one place right does.
Practice
An app should have multiplied \(4.5\) grams by \(10\) but divided by \(10\) instead, showing \(0.45\) grams. How many times larger is the correct amount, \(45\) grams, than \(0.45\) grams?
Show the solution
Getting from \(0.45\) up to \(45\) moves the point two places right, and each place is a factor of \(10\), so the correct amount is \(10 \times 10 = \boxed{100}\) times larger. Dividing directly agrees, \(\frac{45}{0.45}=\frac{4500}{45}=100\). The gap is two steps because the app divided by \(10\) where it should have multiplied by \(10\).
Practice
Using tiles \(0\), \(3\), \(8\) in the three blank slots of \(0.\underline{\phantom{0}}\,\underline{\phantom{0}}\,\underline{\phantom{0}}\), build the smallest positive decimal greater than zero.
Show the solution
The tenths place counts for the most, so park the \(0\) there. Of the two digits left, the smaller one goes in the stronger place, so \(3\) in hundredths and \(8\) in thousandths, giving \(\boxed{0.038}\). Swapping them gives the bigger \(0.083\), and any nonzero tile in tenths jumps you to at least \(0.3\).