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Decimal Place Value: 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 counter reading

A counter starts at zero. Each click adds one tenth to its reading. What decimal does it show after $14$ clicks?
1. Hint 1
  Count the reading as a number of tenths, including any whole amounts they make.
2. Hint 2
  Ten tenths make one whole, so bundle them before placing the remaining tenths.
Answer

$1.4$ .

Full solution

Fourteen clicks add fourteen tenths.
Ten of those tenths form a whole and four tenths remain.
$\frac{14}{10}=1+\frac{4}{10}$

The counter therefore shows $1.4$ .
It has passed one whole but has not reached two wholes.

Answer

$1.4$ .

Key idea

Ten tenths form a whole, so a count of tenths may need digits on both sides of the point.
Problem 1: Mark complete
Problem 2 The missing digit

Find the digit that completes $1.0\square9=\frac{1049}{1000}$ .
1. Hint 1
  The fraction counts all of the number’s thousandths, including those in its whole part.
2. Hint 2
  Separate one thousand thousandths from the numerator, then place the remaining digits in their columns.
Answer

$4$ .

Full solution

One thousand of the thousandths make one whole.
The rest form the decimal part.
$\frac{1049}{1000}=1+\frac{49}{1000}$

Forty-nine thousandths occupy the hundredths and thousandths columns, with a zero holding the tenths column open.
$\frac{1049}{1000}=1.049$

The missing digit is therefore $4$ .
It contributes four hundredths, and the final nine contributes nine thousandths.

Answer

$4$ .

Key idea

A fraction over one thousand determines the decimal’s whole part and three places after the point.
Problem 2: Mark complete
Problem 3 The marked gap

The gap from $2.6$ to $2.7$ on a number line is divided into ten equal parts. What decimal labels the fourth small mark to the right of $2.6$ ?
1. Hint 1
  Decide what fraction of one whole the entire gap represents.
2. Hint 2
  A tenth split into ten equal parts gives a smaller place; name it, then count the steps from $2.6$ .
Answer

$2.64$

Full solution

The gap is one tenth of a whole, and splitting a tenth into ten equal parts makes each part one hundredth.
Four steps beyond $2.6$ therefore add four hundredths.
$\frac{26}{10}+\frac{4}{100}=\frac{264}{100}$

The mark is labeled $2.64$ .
It lies between $2.60$ and $2.70$ , as required.

Answer

$2.64$

Key idea

Dividing one tenth into ten equal parts produces hundredths.
Problem 3: Mark complete
Problem 4 The label pair

Label A reads "eleven and forty thousandths." Label B shows $11.004$ . Write label A as a decimal and state which label names the larger number.
1. Hint 1
  Use the named final place to translate between the words and decimal notation.
2. Hint 2
  Give both decimals three places, then look for the first column where they differ.
Answer

A: $11.040$ , or $11.04$ ; A is larger.

Full solution

Forty thousandths means $40$ thousandths, so the final digit of $40$ lands in the thousandths column.
The $4$ then sits in the hundredths column, and a zero holds the tenths column open.
$\frac{40}{1000}=0.040$

Label A is therefore $11.040$ .
Its final zero is a trailing zero, so $11.04$ names the same number, but the zero in the tenths column cannot be dropped.

Both labels have $11$ ones and $0$ tenths.
Label A has $4$ hundredths, while label B has $0$ hundredths.
$11.040>11.004$

The first column where the labels differ settles the comparison, so label A names the larger number.

Answer

A: $11.040$ , or $11.04$ ; A is larger.

Key idea

Read the place named by the words before comparing the resulting decimals.
Problem 4: Mark complete
Problem 5 The recorded amount

A recorder describes an amount as $12$ tenths and $15$ thousandths. Write the amount as a decimal and as a sum of its nonzero place values, using fractions for the parts smaller than one.
1. Hint 1
  Both counts are ten or more of their unit, so each one spills over into a larger place.
2. Hint 2
  Bundle ten tenths into one whole and ten thousandths into one hundredth, then place what remains in each column.
Answer

$1.215$ ; $1+\frac{2}{10}+\frac{1}{100}+\frac{5}{1000}$ .

Full solution

Twelve tenths contain one whole and two tenths.
$\frac{12}{10}=1+\frac{2}{10}$

Fifteen thousandths contain one hundredth and five thousandths, because ten thousandths make one hundredth.
$\frac{15}{1000}=\frac{1}{100}+\frac{5}{1000}$

Placing every piece in its column gives $1$ one, $2$ tenths, $1$ hundredth and $5$ thousandths, so the amount is $1.215$ .
In expanded form,
$1.215=1+\frac{2}{10}+\frac{1}{100}+\frac{5}{1000}$

As a check, count everything in thousandths.
Twelve tenths are $1200$ thousandths.
$1200+15=1215$

That is $1215$ thousandths, which is $1.215$ again.

Answer

$1.215$ ; $1+\frac{2}{10}+\frac{1}{100}+\frac{5}{1000}$ .

Key idea

Bundle ten parts into the next larger place before writing the decimal digits.
Problem 5: Mark complete
Problem 6 Pat's fraction

Pat writes $7.09=\frac{79}{100}$ , reading the digits $7$ and $9$ over one hundred. Is Pat correct? Write $7.09$ as one fraction with denominator $100$ .
1. Hint 1
  Before writing $7.09$ as a fraction, compare the size of Pat's fraction with the size of $7.09$ .
2. Hint 2
  Find how many hundredths the whole part contributes, then add the hundredths after the point.
Answer

No; $7.09=\frac{709}{100}$ .

Full solution

Pat's fraction is $79$ hundredths, which is less than one whole.
$\frac{79}{100}=0.79$

That cannot equal $7.09$ , which is more than seven.
Pat dropped the placeholder zero and treated the $7$ ones as $7$ tenths.

Each whole is one hundred hundredths, so the $7$ ones contribute $700$ hundredths.
The digits after the point contribute $9$ hundredths.
$700+9=709$

The number is therefore $709$ hundredths.
$7.09=\frac{709}{100}$

Answer

No; $7.09=\frac{709}{100}$ .

Key idea

One fraction over $100$ for a decimal counts the hundredths in its whole part as well as those after the point.
Problem 6: Mark complete
Problem 7 The four readings

Four readings are $2.508$ , $\frac{258}{100}$ , " $2$ ones, $5$ tenths and $7$ hundredths" and $2.5080$ . List them from least to greatest, keeping all four readings and marking any equal readings with $=$ .
1. Hint 1
  Turn the fraction and the place description into decimals, so that all four readings share one form.
2. Hint 2
  Extra zeros at the far right can make all four readings the same length without changing them.
3. Hint 3
  Compare tenths first, then hundredths, then any further columns that are needed.
Answer

$2.508=2.5080<$ " $2$ ones, $5$ tenths and $7$ hundredths" $<\frac{258}{100}$ ; in decimals, $2.508=2.5080<2.57<2.58$ .

Full solution

The place description puts $5$ in the tenths column and $7$ in the hundredths column after the $2$ ones, so it names $2.57$ .
$2+\frac{5}{10}+\frac{7}{100}=2.57$

The fraction counts $258$ hundredths.
Two hundred of them make two wholes, leaving $58$ hundredths.
$\frac{258}{100}=2.58$

Pad all four readings to four places: $2.5080$ , $2.5800$ , $2.5700$ and $2.5080$ .

All four readings have $5$ tenths, so the hundredths decide.
The two versions of $2.508$ have $0$ hundredths, the place description has $7$ and the fraction has $8$ .

The two versions of $2.508$ agree in every column, so they have the same value.
From least to greatest, $2.508$ and $2.5080$ come first and are equal, then the place description, and the fraction $\frac{258}{100}$ comes last.

Answer

$2.508=2.5080<$ " $2$ ones, $5$ tenths and $7$ hundredths" $<\frac{258}{100}$ ; in decimals, $2.508=2.5080<2.57<2.58$ .

Key idea

Trailing zeros preserve equality while the first differing place settles an order.
Problem 7: Mark complete
Problem 8 The token exchange

Each small token is worth one thousandth of a credit, and each large token is worth one hundredth of a credit. Kai has $3$ large tokens and $5$ small tokens. He trades each large token for ten small tokens and says that, because he now has more tokens, they are worth more. Is Kai correct? Give the value of his tokens before and after the trade as fractions of a credit.
1. Hint 1
  Decide what one large token is worth in small tokens before comparing anything.
2. Hint 2
  Write both totals as fractions of a credit with denominator $1000$ , counting each large token as ten thousandths.
Answer

No; his tokens are worth $\frac{35}{1000}$ of a credit, or $\frac{7}{200}$ , both before and after the trade.

Full solution

Before the trade, the $3$ large tokens are worth $3$ hundredths, which is $30$ thousandths, and the $5$ small tokens are worth $5$ thousandths.
$\frac{30}{1000}+\frac{5}{1000}=\frac{35}{1000}$

One large token is worth ten small ones, since ten thousandths make one hundredth.
$\frac{10}{1000}=\frac{1}{100}$

Trading the $3$ large tokens gives $30$ small tokens, so Kai ends with $35$ small tokens.
$35\times\frac{1}{1000}=\frac{35}{1000}$

The value is the same before and after the trade.
Kai has more tokens, but the $30$ that replaced his large tokens are each worth one tenth of a large token, so the total is unchanged and he is not correct.
In lowest terms the total is $\frac{7}{200}$ , since dividing $35$ and $1000$ by $5$ gives $7$ and $200$ .

Answer

No; his tokens are worth $\frac{35}{1000}$ of a credit, or $\frac{7}{200}$ , both before and after the trade.

Key idea

Trading one hundredth for ten thousandths changes how many pieces there are but not the total, because each new piece is worth one tenth as much.
Problem 8: Mark complete
Problem 9 The extra zero

A display shows $8.05$ . One extra zero is to be inserted after the decimal point, in one of three positions: immediately after the point, between the $0$ and the $5$ , or after the $5$ . Give the number each choice produces, and say which choices keep the value.
1. Hint 1
  Write $8.05$ in expanded form first, so you know what the $5$ is worth before any zero is added.
2. Hint 2
  After each insertion, write the digits in their columns and read what the $5$ is worth.
Answer

Immediately after the point: $8.005$ ; between the $0$ and the $5$ : $8.005$ ; after the $5$ : $8.050$ . Only the zero after the $5$ keeps the value; the other two give the smaller $8.005$ .

Full solution

In $8.05$ the $5$ sits in the hundredths column, so it is worth five hundredths.
$8.05=8+\frac{5}{100}$

A zero inserted immediately after the point, or between the $0$ and the $5$ , leaves two zeros before the $5$ .
Both choices give $8.005$ , where the $5$ has moved into the thousandths column and is worth only five thousandths.

A zero inserted after the $5$ gives $8.050$ .
It is a trailing zero, and the $5$ stays in the hundredths column.

So $8.050$ has the same value as $8.05$ , while the other two choices give a smaller number.
$8.005<8.050$

Answer

Immediately after the point: $8.005$ ; between the $0$ and the $5$ : $8.005$ ; after the $5$ : $8.050$ . Only the zero after the $5$ keeps the value; the other two give the smaller $8.005$ .

Key idea

After the decimal point, a zero inserted before a nonzero digit pushes that digit one place to the right, while a zero added after the last digit leaves the value unchanged.
Problem 9: Mark complete
Problem 10 The narrow gap

Mia says that every terminating decimal strictly between $3.21$ and $3.22$ needs at least three places after the point once all trailing zeros are removed. Is she correct? Explain.
1. Hint 1
  Ask whether a number ending at the hundredths column could fit between the two endpoints.
2. Hint 2
  Write the endpoints in thousandths to see where an extra place could fit.
Answer

Yes.

Full solution

A decimal ending at the hundredths column counts whole hundredths.
The endpoints are $321$ hundredths and $322$ hundredths, with no whole count of hundredths between them.
A shorter decimal can also be written in hundredths by padding it with zeros, so it does not create another possibility.

An additional place does allow values in the gap.
For example, $3.215$ has more thousandths than $3.210$ .
$3.210<3.215$
It has fewer hundredths than $3.220$ .
$3.215<3.220$

Thus a decimal in the gap must retain at least three decimal places after its trailing zeros are removed.

Answer

Yes.

Key idea

A terminating decimal strictly between two neighboring hundredths needs a digit beyond the hundredths place.
Problem 10: Mark complete

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