Foundations of Arithmetic: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

In the number $7{,}269$ , what is the value of the digit $2$ ?

Answer choices for question 1 A $2$ B $20$ C $200$ D $2{,}000$
2

The equation $(6 + 11) + 9 = 6 + (11 + 9)$ illustrates which property?

Answer choices for question 2 A Commutative property of addition B Associative property of addition C Distributive property D Additive identity
3

Evaluate $50 - 8 \times 4 + 2$ .

Answer choices for question 3 A $20$ B $16$ C $170$ D $2$
4

Which expression has the same value as $7 \times 26$ ?

Answer choices for question 4 A $7 \times 20 + 6$ B $7 \times 20 + 7 \times 6$ C $7 \times 20 + 20 \times 6$ D $7 \times 2 + 7 \times 6$
5

Using each of the digits $4$ , $0$ , $7$ and $9$ exactly once, what is the largest four-digit number that can be written?

Answer choices for question 5 A $9{,}704$ B $9{,}470$ C $4{,}079$ D $9{,}740$
6

Which of these shows $28 + 42$ written as a product, collected completely?

Answer choices for question 6 A $14 \times (2 + 3)$ B $7 \times (4 + 6)$ C $2 \times (14 + 21)$ D $14 \times (2 + 42)$
7

Evaluate $2 \times [\,30 - (4^2 + 5)\,]$ .

Answer choices for question 7 A $34$ B $38$ C $18$ D $9$
8

Which numeral equals $9{,}000 + 600 + 70$ ?

Answer choices for question 8 A $967$ B $9{,}607$ C $9{,}760$ D $9{,}670$
9

Exactly one of these four equations is true. Which one?

Answer choices for question 9 A $(60 \div 6) \div 2 = 60 \div (6 \div 2)$ B $(9 + 14) + 6 = 9 + (14 + 6)$ C $44 - 12 - 5 = 44 - (12 - 5)$ D $8 \times 0 = 8$
10

Which single product equals $9 \times 17 + 9 \times 3$ ?

Answer choices for question 10 A $9 \times 20$ B $9 \times 51$ C $18 \times 20$ D $81 \times 20$
11

Which one-line expression names the same number as $\dfrac{24 + 48}{8 - 2}$ ?

Answer choices for question 11 A $24 + 48 \div 8 - 2$ B $(24 + 48) \div 8 - 2$ C $(24 + 48) \div (8 - 2)$ D $24 + 48 \div (8 - 2)$
12

The digits $5$ , $2$ , $8$ and $0$ are each used exactly once to write a four-digit number in which the $8$ contributes $8{,}000$ . What is the smallest such number?

Answer choices for question 12 A $8{,}052$ B $8{,}250$ C $2{,}058$ D $8{,}025$
13

To work out $8 \times 46 \times 5$ , a student first writes it as $8 \times 5 \times 46$ , and then as $(8 \times 5) \times 46$ . Which property licenses the second step?

Answer choices for question 13 A Associative property of multiplication B Commutative property of multiplication C Distributive property D Multiplicative identity
14

Which expression has the same value as $5 \times 96$ ?

Answer choices for question 14 A $5 \times 100 - 4$ B $5 \times 100 - 5 \times 4$ C $5 \times 100 + 5 \times 4$ D $5 \times 90 - 5 \times 4$
15

Which list is ordered from least to greatest?

Answer choices for question 15 A $40{,}087 < 4{,}807 < 48{,}070 < 4{,}870$ B $4{,}870 < 4{,}807 < 40{,}087 < 48{,}070$ C $4{,}807 < 4{,}870 < 40{,}087 < 48{,}070$ D $48{,}070 < 40{,}087 < 4{,}870 < 4{,}807$
16

Evaluate $\dfrac{13 \times 37 + 13 \times 23}{13}$ .

Answer choices for question 16 A $336$ B $780$ C $851$ D $60$
17

A number is written as $(6 \times 1{,}000) + (0 \times 100) + (8 \times 10) + (3 \times 1)$ . Which statement about that number is true?

Answer choices for question 17 A It is $6{,}083$ , and the term $(0 \times 100)$ holds the hundreds column open even though it adds nothing to the sum. B It is $683$ , because a term worth $0$ names no column and can be left out. C It is $6{,}183$ , because multiplying $100$ by $0$ leaves the $100$ untouched. D It is $0$ , because a sum containing a product with a factor of $0$ collapses to $0$ .
18

A student computes $15 \times 24$ by splitting both factors: $15 \times 24 = (10 + 5) \times (20 + 4) = 10 \times 20 + 5 \times 4 = 220$ . What is $15 \times 24$ ?

Answer choices for question 18 A $220$ B $300$ C $360$ D $260$
19

One of these four rewritings changes the value of the expression it starts from. Which one?

Answer choices for question 19 A $19 + 26 + 4$ rewritten as $19 + (26 + 4)$ B $60 - 15 - 5$ rewritten as $60 - (15 - 5)$ C $5 \times 18 \times 2$ rewritten as $(5 \times 2) \times 18$ D $14 + 9 + 6$ rewritten as $6 + 14 + 9$
20

Evaluate $4 \times (25 + 11) - 3 \times (25 - 13)$ .

Answer choices for question 20 A $1{,}692$ B $23$ C $82$ D $108$

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Where you stand

Keep the PLACE a digit occupies separate from the VALUE it contributes, and move between a numeral and its expanded form, including what a zero digit is doing. Review: Place Value and the Number System
Compare and order whole numbers by place, and build the largest or the smallest number a given set of digits allows. Review: Place Value and the Number System
Evaluate a numerical expression under the agreed order, from the exponent tier downward, including nested grouping and a fraction bar acting as a grouping symbol. Review: Order of Operations
Apply the commutative, associative, identity and zero properties, and recognize that subtraction and division are neither commutative nor associative. Review: Properties of Addition and Multiplication
Spread a factor across a sum or a difference, including splitting one factor to turn an awkward product into two easy ones. Review: The Distributive Property
Collect the greatest common factor out of a sum, and judge whether the collection is finished. Review: The Distributive Property

Core practice

10 problems from across the chapter. 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 Changed counter digit

A counter displays $507{,}260$ . Its $7$ is replaced by $0$ , with every other digit left in the same position. How much does the displayed number decrease?
1. Hint 1
  A digit contributes an amount determined by its position.
2. Hint 2
  Locate the $7$ by counting places from the right, then multiply $7$ by that place value.
Answer

$7{,}000$ .

Full solution

The $7$ is in the thousands place.
Its contribution is
$7\times1{,}000=7{,}000$

Replacing that digit with $0$ removes its contribution and preserves the other places.
The new display is $500{,}260$ .
$507{,}260-500{,}260=7{,}000$

Check by restoring the removed amount:
$500{,}260+7{,}000=507{,}260$

Answer

$7{,}000$ .

Key idea

Replacing one digit by zero removes that digit's value when all other digits keep their positions.
Problem 1: Mark complete
Problem 2 Power and quotient

Evaluate $6^2\div9\times2$ .
1. Hint 1
  Identify the highest operation tier before starting.
2. Hint 2
  Evaluate the power, then work through division and multiplication from left to right.
Answer

$8$ .

Full solution

The exponent counts two factors of $6$ .
$6^2=36$

Division and multiplication share a tier, so divide before multiplying here.
$\begin{aligned}36\div9&=4\\4\times2&=8\end{aligned}$

As a check, four groups of $9$ make $36$ , and two copies of that quotient make $8$ .

Answer

$8$ .

Key idea

Evaluate powers before multiplication and division, then work left to right within their shared tier.
Problem 2: Mark complete
Problem 3 Ribbon for a club

Ribbon is sold in 30-meter rolls. A club orders 9 rolls, and the supplier cuts 2 meters from every roll before delivery. How many meters of ribbon does the club receive?
1. Hint 1
  Every roll loses the same length before delivery.
2. Hint 2
  Find what 9 full rolls would hold, then remove the 2 meters cut from each of the 9 rolls.
Answer

252 meters.

Full solution

Each delivered roll holds $30-2$ meters, so the club receives $9(30-2)$ meters.
The cut comes off every roll, so the factor $9$ multiplies the $2$ as well as the $30$ .
$\begin{aligned}9(30-2)&=9\times30-9\times2\\9\times30-9\times2&=270-18\\270-18&=252\end{aligned}$

Check by finding one delivered roll first: $30-2=28$ , and
$9\times28=252$

Answer

252 meters.

Key idea

A cut taken from every roll is multiplied by the number of rolls, just as the full length is.
Problem 3: Mark complete
Problem 4 Swapped counter places

A machine stores a count equal to $5{,}000+600+9$ . A fault swaps its hundreds and tens digits while leaving all other digits in place. Write the faulty count in expanded form and determine whether it is smaller or larger than the original.
1. Hint 1
  An absent contribution still needs a digit in the numeral.
2. Hint 2
  Write the original numeral with its zero digit, then exchange the hundreds and tens digits.
3. Hint 3
  Compare the two numerals from the left until their digits differ.
Answer

$5{,}000+60+9$ ; smaller, with $5{,}069<5{,}609$ .

Full solution

The original has 5 thousands, 6 hundreds, no tens and 9 ones, so it is $5{,}609$ .
The swap puts $0$ in the hundreds place and $6$ in the tens place.
$5{,}069=5{,}000+60+9$

The thousands digits agree.
The hundreds digits are the first to differ: the faulty count has $0$ where the original has $6$ .
Therefore
$5{,}069<5{,}609$

The fault changes the contribution of the $6$ from $600$ to $60$ .
That decrease agrees with the comparison.

Answer

$5{,}000+60+9$ ; smaller, with $5{,}069<5{,}609$ .

Key idea

Swapping a digit with a zero changes its contribution, and the first differing place settles the comparison.
Problem 4: Mark complete
Problem 5 A four-step number machine

A number machine starts with $841$ . It adds $0$ , multiplies the result by $1$ , multiplies that result by $0$ , and finally adds $23$ . Find its output, and decide whether a different whole-number starting value would change it.
1. Hint 1
  Consider what each operation does to the current value, especially when the multiplier changes.
2. Hint 2
  Adding $0$ and multiplying by $1$ preserve the current value. Follow the operation that comes next.
Answer

$23$ ; every whole-number starting value gives the same output.

Full solution

Adding zero and multiplying by one leave the starting value unchanged.
Multiplying by zero then removes its effect.
$\begin{aligned}841+0&=841\\841\times1&=841\\841\times0&=0\\0+23&=23\end{aligned}$

For any whole-number input, the first two steps preserve it and the third step produces $0$ .
The last step therefore produces $23$ for every such input.
Check with a different starting value, $0$ : all three initial steps leave $0$ , followed by the same final addition.

Answer

$23$ ; every whole-number starting value gives the same output.

Key idea

Multiplication by zero makes the final value independent of the starting input when the later operations are fixed.
Problem 5: Mark complete
Problem 6 A stacked expression

Evaluate $\dfrac{2(1+6)^2}{28\div2}$ .
1. Hint 1
  The fraction bar groups the entire numerator and the entire denominator.
2. Hint 2
  In the numerator, settle the parentheses before the power and the outside multiplication.
3. Hint 3
  After evaluating the denominator, divide the completed numerator by it.
Answer

$7$ .

Full solution

Start with the innermost group, then evaluate its square and the multiplication.
$\begin{aligned}1+6&=7\\7^2&=49\\2\times49&=98\end{aligned}$

The denominator is a separate group.
$28\div2=14$

The denominator is $14$ , which is nonzero, so divide the completed numerator by it.
$\frac{98}{14}=7$

Check that the quotient times the denominator recovers the numerator:
$7\times14=98$

Answer

$7$ .

Key idea

A fraction bar keeps its numerator and denominator grouped while the operation tiers govern each group.
Problem 6: Mark complete
Problem 7 Updating identical packets

A volunteer has 18 pencils and 30 erasers. She makes the greatest possible number of identical packets, using every item. Each packet must contain the same positive number of pencils and the same positive number of erasers.

She then puts 2 additional erasers into every packet. How many packets are there, and how many items are in all the updated packets combined?
1. Hint 1
  The packet count must divide both original item counts exactly.
2. Hint 2
  Find the greatest possible packet count first, then find the contents of one packet.
3. Hint 3
  Update the eraser count in one packet before counting each item type across all packets.
Answer

6 packets; 60 items.

Full solution

The packet count divides both $18$ and $30$ .
The divisors of $18$ are $1$ , $2$ , $3$ , $6$ , $9$ and $18$ .
Of these, $9$ and $18$ do not divide $30$ , but $6$ does.
Thus the greatest possible count is 6 packets.

Find the original contents of each packet.
$\begin{aligned}18\div6&=3\\30\div6&=5\end{aligned}$
The original total can be written as
$18+30=6(3+5)$

Each updated packet has 3 pencils and $5+2$ erasers.
$5+2=7$
Count both parts across all 6 packets.
$\begin{aligned}6(3+7)&=6\times3+6\times7\\6\times3&=18\\6\times7&=42\\18+42&=60\end{aligned}$

Check the change against the original total.
There were $18+30$ items and 2 erasers were added to each of 6 packets.
$\begin{aligned}18+30&=48\\6\times2&=12\\48+12&=60\end{aligned}$

Answer

6 packets; 60 items.

Key idea

The greatest shared divisor determines the packet count, and updating each packet changes the totals for every copy.
Problem 7: Mark complete
Problem 8 A proposed rewrite

A student replaces $4[7+(20\div5)]$ with $4\times7+20\div5$ . Find the value of each expression and decide whether the replacement preserves the value. Explain your decision.
1. Hint 1
  Check which quantities are multiplied by the outside $4$ in each expression.
2. Hint 2
  In the first expression, evaluate the inner division and then the entire bracket.
3. Hint 3
  In the second expression, complete its multiplication and division before its addition.
Answer

$44$ and $32$ , respectively; the replacement does not preserve the value.

Full solution

The bracket groups both $7$ and the quotient for multiplication by $4$ .
$\begin{aligned}20\div5&=4\\7+4&=11\\4\times11&=44\end{aligned}$

In the replacement, the $4$ multiplies just the $7$ .
$\begin{aligned}4\times7&=28\\20\div5&=4\\28+4&=32\end{aligned}$

The replacement leaves the quotient unmultiplied by $4$ .
To preserve the first expression, both parts must receive that factor.
$\begin{aligned}4(7+4)&=4\times7+4\times4\\28+16&=44\end{aligned}$
The two values, $44$ and $32$ , disagree, so the replacement does not preserve the value.

Answer

$44$ and $32$ , respectively; the replacement does not preserve the value.

Key idea

Dropping the brackets around a group that a factor multiplies requires that factor to multiply every term of the group.
Problem 8: Mark complete
Problem 9 Testing a general claim

A learner notices that swapping the inputs in $9-9$ leaves its value unchanged. The learner also notices that $(36\div1)\div1$ and $36\div(1\div1)$ have the same value.

The learner concludes that subtraction is commutative and division is associative. For each conclusion, say what was special about the learner's example, and give a counterexample to each conclusion that fails. For the division counterexample, use positive whole numbers with exact whole-number quotients in both groupings.
1. Hint 1
  A property must hold for every allowed choice of numbers, so a successful example does not establish it.
2. Hint 2
  To test the subtraction claim, try reversing two unequal whole numbers.
3. Hint 3
  To test the division claim, choose a divisor greater than $1$ inside the group and compare the two groupings.
Answer

The subtraction example used two equal numbers, and the division example divided by $1$ . Neither conclusion is valid; for example, $11-4\ne4-11$ and $(32\div4)\div2\ne32\div(4\div2)$ .

Full solution

The subtraction example uses equal inputs, so swapping them does not change the written calculation.
Unequal inputs reveal the failure.
$11-4=7$
But $4-11$ takes more than the available $4$ and goes below zero, so it is not $7$ .
This is a counterexample to commutativity.

The division example uses $1$ , which leaves the values unchanged in both groupings.
Instead compare these exact divisions.
Group the first pair:
$\begin{aligned}32\div4&=8\\8\div2&=4\end{aligned}$
Group the last pair:
$\begin{aligned}4\div2&=2\\32\div2&=16\end{aligned}$

Every divisor in the example is positive and nonzero, and every quotient is a whole number.
The two groupings give $4$ and $16$ , so division fails associativity.
One failing example is enough to reject each general conclusion.

Answer

The subtraction example used two equal numbers, and the division example divided by $1$ . Neither conclusion is valid; for example, $11-4\ne4-11$ and $(32\div4)\div2\ne32\div(4\div2)$ .

Key idea

A successful example can illustrate a property, but a single counterexample disproves a claim that it holds in every case.
Problem 9: Mark complete
Problem 10 Ravi's two lines

Ravi writes $32+56=8(4+7)$ . He then replaces $8(4+7)$ with $16(2+7)$ , saying that the $4$ inside can still be divided by $2$ .

Decide whether each line preserves the original sum and whether the first line has the greatest common factor of $32$ and $56$ outside the parentheses. Explain your decisions.
1. Hint 1
  An outside factor must reproduce both original terms, not just the first one.
2. Hint 2
  Check what each outside factor gives when multiplied by each inside number.
3. Hint 3
  A further whole-number factor must divide both $4$ and $7$ .
Answer

The first line preserves the sum and has the greatest common factor, $8$ , outside; the replacement does not preserve the sum.

Full solution

The first line reproduces both original terms.
$\begin{aligned}8\times4&=32\\8\times7&=56\\32+56&=88\end{aligned}$

The possible divisors of $4$ greater than $1$ are $2$ and $4$ , and neither divides $7$ .
Thus $4$ and $7$ share no factor greater than $1$ , so $8$ is already the greatest shared factor removed from the original terms.

The replacement still reproduces $32$ but changes the other term.
$\begin{aligned}16\times2&=32\\16\times7&=112\\32+112&=144\end{aligned}$

Doubling the outside factor would require halving both inside terms to preserve their contributions.
Ravi halves just one, so his replacement changes the sum.
Check the valid expression directly:
$\begin{aligned}4+7&=11\\8\times11&=88\end{aligned}$

Answer

The first line preserves the sum and has the greatest common factor, $8$ , outside; the replacement does not preserve the sum.

Key idea

A common factor is fully collected when the remaining whole-number terms share no factor greater than one.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Pre-Algebra. You can skip it. → Continue to Chapter 2: Integers and the Number Line →

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