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Operations with Decimals: 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 Subtracting thousandths

Find $9.3-4.658$ , then check your result by addition.
1. Hint 1
  Give both numbers the same number of decimal places before stacking them with the points in one column.
2. Hint 2
  The thousandths column needs a borrow, and the top number’s hundredths digit is zero, so the borrow must come from the tenths.
Answer

$4.642$ .

Full solution

Pad $9.3$ with two trailing zeros, which does not change its value, so both numbers reach the thousandths.
Stack $9.300$ over $4.658$ with the points in one column.

The thousandths read $0-8$ , and the top hundredths digit is also $0$ , so borrow one tenth: it becomes $9$ hundredths and $10$ thousandths, and $2$ tenths remain.
The thousandths give $10-8=2$ , and the hundredths give $9-5=4$ .
The tenths read $2-6$ , so borrow one whole, making $12$ tenths and $8$ ones.
The tenths give $12-6=6$ , and the ones give $8-4=4$ .
$9.300-4.658=4.642$

Check by adding the result to the number subtracted.
$4.642+4.658=9.300$

Answer

$4.642$ .

Key idea

When a borrow meets a zero digit, borrow from the nearest nonzero place to its left, and each zero it passes through becomes $9$ .
Problem 1: Mark complete
Problem 2 The three-place dividend

Compute $0.594\div0.36$ , and check your quotient by multiplication.
1. Hint 1
  The divisor has two decimal places; decide what would make it a whole number.
2. Hint 2
  Slide both points two places, then divide $59.4$ by $36$ , carrying remainders into tenths and hundredths.
Answer

$1.65$

Full solution

The divisor $0.36$ has two decimal places, so slide both points two places right to make it whole.
$0.594\div0.36=59.4\div36$

Thirty-six fits once into $59$ , leaving $23$ .
The next digit makes $234$ tenths; six groups of $36$ leave $18$ tenths.
Padding a zero after the $4$ turns those into $180$ hundredths, which divide into five groups.
$59.4\div36=1.65$

Check by multiplying the quotient by the divisor, which should give back the dividend.
$1.65\times0.36=0.594$

Answer

$1.65$

Key idea

Sliding both points by the divisor’s decimal places makes the divisor whole; a dividend with more places than the divisor keeps a point, and the long division carries on past it, padding zeros as needed.
Problem 2: Mark complete
Problem 3 The material tub

An empty tub weighs $0.84$ kg. Filled with material, it weighs $3.075$ kg. After $0.65$ kg of material is removed, how much material remains in the tub?
1. Hint 1
  Separate the material from the weight of the tub.
2. Hint 2
  Subtract the empty weight from the filled weight, then subtract the amount removed.
Answer

$1.585$ kg.

Full solution

Pad the empty weight to three places and subtract it from the filled weight.
$3.075-0.840=2.235$

The tub initially holds $2.235$ kg of material.
Subtract the $0.65$ kg removed, again matching decimal places.
$2.235-0.650=1.585$

The remaining material weighs $1.585$ kg.
Adding $0.65$ kg and the tub weight of $0.84$ kg returns the original $3.075$ kg.

Answer

$1.585$ kg.

Key idea

Separate the container’s weight from its contents, padding every amount to the same number of decimal places before subtracting.
Problem 3: Mark complete
Problem 4 The enlarged factors

Two positive decimal factors have an unknown product. The first factor is multiplied by $100$ , and the second by $10$ . Their new product is $40$ . What was the original product?
1. Hint 1
  Track how scaling each factor affects their product.
2. Hint 2
  Undo the combined effect of both multiplications on the new product.
Answer

$0.04$ .

Full solution

The first change makes the product one hundred times as large, and the second change multiplies it by ten again.
Together the two changes multiply the product by
$100\times10=1000$

Divide the new product by $1000$ to recover the original product.
$40\div1000=0.04$

Checking the change forward gives $0.04\times1000=40$ , the stated new product.

Answer

$0.04$ .

Key idea

When each factor is multiplied by a number, the product is multiplied by the product of those numbers.
Problem 4: Mark complete
Problem 5 The card stacks

A stack of $50$ identical cards is $3.9$ cm thick, with no gaps between cards. How thick is one card, and how thick would a stack of $250$ of these cards be?
1. Hint 1
  The thickness of equal cards adds to the thickness of the stack.
2. Hint 2
  Divide the given thickness by the number of cards in the given stack, then multiply the thickness of one card by the number of cards in the new stack.
Answer

One card: $0.078$ cm; $250$ cards: $19.5$ cm.

Full solution

Divide the thickness of the stack by the $50$ cards in it.
Fifty does not fit into $3$ ones, so the ones digit is $0$ , and the $3$ ones join the $9$ tenths to make $39$ tenths, which $50$ does not fit into either, so the tenths digit is $0$ .
Padding a zero makes $390$ hundredths, which hold seven groups of $50$ with $40$ hundredths left over.
Those become $400$ thousandths, which hold exactly eight groups.
$3.9\div50=0.078$

Multiply that thickness by $250$ .
The digit strings give $78\times250=19500$ , and three decimal places give
$0.078\times250=19.500$

Dropping the trailing zeros, the new stack is $19.5$ cm thick.

The new stack is five times the given stack of $50$ cards, and multiplying the given thickness by $5$ confirms the result.
$3.9\times5=19.5$

Answer

One card: $0.078$ cm; $250$ cards: $19.5$ cm.

Key idea

Dividing a decimal total by a whole-number count gives one item’s share, and multiplying that share by a new count gives the new total.
Problem 5: Mark complete
Problem 6 The roller’s progress

A roller advances a sheet $0.375$ meters for each turn, and a fraction of a turn advances it that same fraction of $0.375$ meters. The sheet is $8$ meters long, and the roller stops after $15.6$ turns. How many meters of the sheet have not yet been advanced?
1. Hint 1
  The advance for a number of turns, whole or not, is that number times the advance for one turn.
2. Hint 2
  Find the advance by multiplication, counting the decimal places of both factors, then subtract it from the length of the sheet.
Answer

$2.15$ meters.

Full solution

The advance after $15.6$ turns multiplies two decimals, $0.375\times15.6$ .
The digit strings give $375\times156=58500$ , and the factors have $3+1=4$ decimal places together.
$0.375\times15.6=5.8500$

Dropping the trailing zeros, the roller has advanced $5.85$ meters.
Pad the sheet’s length to hundredths and subtract, borrowing from the ones through the zero tenths.
$8.00-5.85=2.15$

Check by adding the two parts, which should give back the whole sheet.
$5.85+2.15=8.00$

Answer

$2.15$ meters.

Key idea

Count the decimal places of both factors to place the point in the product, then pad the whole to the same places before subtracting the part from it.
Problem 6: Mark complete
Problem 7 The cartridge supply

A supply tank holds $3.12$ liters of liquid. Another $0.48$ liters is added. Each empty cartridge receives exactly $0.22$ liters. How many cartridges can be filled completely from the tank, and how much liquid is left over?
1. Hint 1
  Work out how much liquid is available after the addition.
2. Hint 2
  Slide both points so the divisor is whole, and find how many whole times it fits into the dividend.
3. Hint 3
  Multiply the number of full cartridges by $0.22$ , then subtract that amount from the available liquid.
Answer

$16$ cartridges, with $0.08$ liters left over.

Full solution

Add the amounts of liquid.
$3.12+0.48=3.60$

The number of full cartridges is the whole number of times $0.22$ fits into $3.60$ .
Shift both points two places right.
$3.60\div0.22=360\div22$

Since $22\times16=352$ and $22\times17=374$ , which is more than $360$ , twenty-two fits into $360$ sixteen whole times.
So $16$ cartridges can be filled completely.

Those cartridges use $16\times0.22=3.52$ liters.
Subtract to find what is left.
$3.60-3.52=0.08$

The $0.08$ liters left is less than $0.22$ liters, so no further cartridge can be filled.

Answer

$16$ cartridges, with $0.08$ liters left over.

Key idea

When decimal portions do not fit exactly, count the whole portions, then multiply back and subtract to find what is left over.
Problem 7: Mark complete
Problem 8 Nina’s total

Nina writes $0.74+2.8=3.54$ . She says that the $7$ and the $8$ belong in the same column of her addition. Is her result correct, and is her column choice justified? Explain.
1. Hint 1
  Identify what each of the two named digits is worth.
2. Hint 2
  Write both numbers to hundredths, then combine matching columns and account for any carry.
Answer

Both the result and the column choice are correct.

Full solution

The $7$ in $0.74$ and the $8$ in $2.8$ both count tenths.
Padding the second number gives $2.80$ .
Stacking with the points in one column puts tenths over tenths and hundredths over hundredths.
$0.74+2.80=3.54$

The tenths total fifteen tenths: five tenths remain and one whole is carried.
The hundredths total four, and the carried whole raises the whole part from two to three.
Both parts of Nina’s work are justified.

Answer

Both the result and the column choice are correct.

Key idea

Aligning decimal points puts digits that count the same place value into the same column.
Problem 8: Mark complete
Problem 9 Mara’s note

Mara claims that multiplying two numbers each written with exactly one digit after the point must leave exactly two digits after the point, even after all trailing zeros are removed. Is the claim true? Give a numerical example that supports your decision.
1. Hint 1
  The place-counting rule positions the point before optional trailing zeros are removed.
2. Hint 2
  Try factors whose whole-number product ends in zeros, then apply the place-counting rule.
Answer

No; any pair whose whole-number product ends in $0$ disproves it, for example $3.2\times2.5=8$ or $0.5\times0.2=0.1$ .

Full solution

Take $3.2$ and $2.5$ , each written with one decimal place.
Their whole-number product is
$32\times25=800$

Two decimal places put the product at
$3.2\times2.5=8.00$

The two trailing zeros may be removed.
$8.00=8$

The exact product has no decimal places after that removal, so it disproves Mara’s claim.
The place-counting rule correctly located the point before the zeros were dropped.
Any pair whose whole-number product ends in $0$ also breaks the claim; for example $5\times2=10$ gives $0.5\times0.2=0.10$ , which is $0.1$ .

Answer

No; any pair whose whole-number product ends in $0$ disproves it, for example $3.2\times2.5=8$ or $0.5\times0.2=0.1$ .

Key idea

Place-counting fixes the decimal point before any trailing zeros are removed.
Problem 9: Mark complete
Problem 10 The paired calculations

To evaluate $0.084\div0.012$ , Jo uses $84\div12$ and Ren uses $840\div120$ . Do both calculations give the correct value? Find that value and explain.
1. Hint 1
  A quotient is preserved when both of its numbers are multiplied by the same nonzero amount.
2. Hint 2
  Decide how many places each proposed calculation slides the point in the dividend and in the divisor.
Answer

Yes; both give $7$ .

Full solution

Jo multiplies both numbers by $1000$ .
Ren multiplies both numbers by $10000$ .
Each calculation scales its dividend and divisor by the same nonzero amount, so each preserves the original quotient.

Both divisions give the same value.
$84\div12=7$

For Ren’s calculation,
$840\div120=7$

Checking in the original division gives
$0.012\times7=0.084.$

Answer

Yes; both give $7$ .

Key idea

Both points may slide any number of places to the right, as long as the dividend and divisor slide the same number of places.
Problem 10: Mark complete

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