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Order of Operations: 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 Two powers

Evaluate $7^2-4^2$ .
1. Hint 1
  A power represents repeated multiplication.
2. Hint 2
  Work out $7\times7$ and $4\times4$ before subtracting.
Answer

$33$ .

Full solution

Each exponent applies to the number immediately before it.
Evaluate both powers.
$\begin{aligned}7^2&=49\\4^2&=16\end{aligned}$

Subtract the second result from the first.
$49-16=33$

Check the subtraction by adding back: $33+16=49$ .

Answer

$33$ .

Key idea

Evaluate the powers before subtracting their values.
Problem 1: Mark complete
Problem 2 A shared tier

Evaluate $48\div6\times3\div2$ .
1. Hint 1
  Multiplication and division share the same priority.
2. Hint 2
  Begin with $48\div6$ , then multiply that result by $3$ before dividing by $2$ .
Answer

$12$ .

Full solution

Work from left to right through the multiplication and division.
$\begin{aligned}48\div6&=8\\8\times3&=24\\24\div2&=12\end{aligned}$

The final division checks with
$12\times2=24.$

Answer

$12$ .

Key idea

When multiplication and division share an ungrouped chain, work from left to right.
Problem 2: Mark complete
Problem 3 A running total

Evaluate $31-9+4-6$ .
1. Hint 1
  Addition and subtraction share one tier, so neither kind automatically comes first.
2. Hint 2
  Start by subtracting $9$ from $31$ , then continue in the written order.
Answer

$20$ .

Full solution

Keep a running total, moving left to right.
$\begin{aligned}31-9&=22\\22+4&=26\\26-6&=20\end{aligned}$

Check the final subtraction: $20+6=26$ .

Answer

$20$ .

Key idea

Addition and subtraction in an ungrouped chain act on the running total from left to right.
Problem 3: Mark complete
Problem 4 Groups inside groups

Evaluate $64\div[2\times(9-5)]+3^2$ .
1. Hint 1
  A group inside another group must be settled before the outer group can be used.
2. Hint 2
  First subtract inside the parentheses, then multiply that result by $2$ .
3. Hint 3
  Evaluate the power and the division before adding their results.
Answer

$17$ .

Full solution

The parentheses form the innermost group.
Then finish the brackets.
$\begin{aligned}9-5&=4\\2\times4&=8\end{aligned}$

The expression is now $64\div8+3^2$ .
Evaluate the power, then the division, then the addition.
$\begin{aligned}3^2&=9\\64\div8&=8\\8+9&=17\end{aligned}$

The division checks with
$8\times8=64.$

Answer

$17$ .

Key idea

Work outward through nested groups, then apply the usual priorities to the remaining expression.
Problem 4: Mark complete
Problem 5 Sharing the remaining cards

A club has 36 loose cards and opens 4 packs of 6 cards each. It sets aside 12 of all these cards and shares the rest equally among 6 friends. Write one expression, using a fraction bar, that shows the whole calculation for the number of cards each friend receives, then evaluate it.
1. Hint 1
  The amount shared is the total number of cards after the set-aside cards have been removed.
2. Hint 2
  Combine the loose cards with the cards in the packs, then subtract the 12 cards.
3. Hint 3
  Put the amount shared above the fraction bar and the number of friends below it.
Answer

$\frac{36+4\times6-12}{6}$ , or any fraction over $6$ whose numerator combines the same three amounts, such as $\frac{4\times6+36-12}{6}$ ; $8$ cards per friend.

Full solution

The packs contribute $4\times6$ cards.
Combine that with the loose cards and remove the set-aside cards.
The expression for one friend is $\frac{36+4\times6-12}{6}$ .

The bar groups the entire amount being shared.
Work out that amount first.
$\begin{aligned}4\times6&=24\\36+24&=60\\60-12&=48\end{aligned}$

Divide the remaining cards among the friends.
$\frac{48}{6}=8$

Check: the friends receive $6\times8=48$ cards altogether, and $48+12=60$ accounts for all the cards.

Answer

$\frac{36+4\times6-12}{6}$ , or any fraction over $6$ whose numerator combines the same three amounts, such as $\frac{4\times6+36-12}{6}$ ; $8$ cards per friend.

Key idea

A fraction bar can group the complete amount to be shared before dividing it into equal shares.
Problem 5: Mark complete
Problem 6 Writing a calculation

A calculation starts with 7, adds 2, squares the result, and then divides by 3. Write one expression that follows these instructions and evaluate it.
1. Hint 1
  The expression needs to preserve the sequence described in words.
2. Hint 2
  Place parentheses around the addition so the power applies to its entire result.
3. Hint 3
  After evaluating the power, divide by 3.
Answer

$(7+2)^2\div3$ , or $\frac{(7+2)^2}{3}$ ; $27$ .

Full solution

The addition happens before squaring, so it belongs inside parentheses.
The expression is $(7+2)^2\div3$ .

Under the standard order the parentheses come first, then the power, then the division, which matches the instructions.
$\begin{aligned}7+2&=9\\9^2&=81\\81\div3&=27\end{aligned}$

Check the final division:
$27\times3=81.$

Answer

$(7+2)^2\div3$ , or $\frac{(7+2)^2}{3}$ ; $27$ .

Key idea

Parentheses make a power apply to the result of a whole addition.
Problem 6: Mark complete
Problem 7 Building a fraction

A fraction has the expression $6\times7-18$ as its numerator and the expression $3+9\div3$ as its denominator. Write the fraction with both expressions in place and find its value.
1. Hint 1
  The fraction bar treats the full numerator and the full denominator as separate groups.
2. Hint 2
  Above the bar, multiply before subtracting; below the bar, divide before adding.
3. Hint 3
  Divide the completed numerator by the completed denominator.
Answer

$\frac{6\times7-18}{3+9\div3}$ ; $4$ .

Full solution

The required fraction is $\frac{6\times7-18}{3+9\div3}$ .
First evaluate its numerator.
$\begin{aligned}6\times7&=42\\42-18&=24\end{aligned}$

Evaluate the denominator with division before addition.
$\begin{aligned}9\div3&=3\\3+3&=6\end{aligned}$

The denominator is $6$ , which is nonzero, so the division is defined.
$\frac{24}{6}=4$

Check by multiplication:
$4\times6=24.$

Answer

$\frac{6\times7-18}{3+9\div3}$ ; $4$ .

Key idea

The usual operation priorities apply separately within the numerator and denominator of a fraction.
Problem 7: Mark complete
Problem 8 A student's first step

A student starts evaluating $42-18\div3+2$ by replacing $3+2$ with $5$ , giving $42-18\div5$ . Identify the error and find the correct value of the original expression.
1. Hint 1
  Check whether the chosen first operation belongs to the highest tier present.
2. Hint 2
  The addition is outside the division; compute $18\div3$ before doing either addition or subtraction.
Answer

Error: grouping $3+2$ as the divisor. Correct value: $38$ .

Full solution

The student has treated the divisor as $(3+2)$ , but the expression divides $18$ by $3$ .
Adding first introduces grouping that was not written.

Division has priority.
Then subtraction and addition proceed from left to right.
$\begin{aligned}18\div3&=6\\42-6&=36\\36+2&=38\end{aligned}$

Check the division:
$6\times3=18.$

Answer

Error: grouping $3+2$ as the divisor. Correct value: $38$ .

Key idea

An addition next to a divisor does not join that divisor unless grouping symbols include it.
Problem 8: Mark complete
Problem 9 Parentheses around products

An expression is written as $(2\times7)+(3\times4)$ . Can both pairs of parentheses be removed without changing the value? Give the value and explain which agreed rule supports your decision.
1. Hint 1
  Ask which operations the expression would perform first with no parentheses.
2. Hint 2
  Compare the reading of $2\times7+3\times4$ with the two products shown in the original.
Answer

Yes; $26$ . Multiplication has priority over addition.

Full solution

Without the parentheses, $2\times7+3\times4$ still asks for both products before the addition.
The parentheses do not change that reading.

Evaluate the two products and then add them.
$\begin{aligned}2\times7&=14\\3\times4&=12\\14+12&=26\end{aligned}$

Ranking multiplication above addition lets each product stand inside a sum without its own parentheses.

Answer

Yes; $26$ . Multiplication has priority over addition.

Key idea

Multiplication taking priority over addition lets products in a sum be written without parentheses around them.
Problem 9: Mark complete
Problem 10 Two proposed readings

Mira and Ben read $36\div3+3$ without first agreeing on an order. Mira divides before adding. Ben adds $3+3$ before dividing. Find each result and explain why the written expression needs an agreed reading. Which result does the standard order assign?
1. Hint 1
  An expression communicates one value reliably when its writer and reader follow the same order.
2. Hint 2
  Mira is reading $(36\div3)+3$ , while Ben is reading $36\div(3+3)$ .
3. Hint 3
  Evaluate each grouped reading, then compare each with the standard priority of division and addition.
Answer

Mira: $15$ . Ben: $6$ . The two readings give different values. Standard order: $15$ .

Full solution

Mira divides first and then adds.
$\begin{aligned}36\div3&=12\\12+3&=15\end{aligned}$

Ben adds first and uses that sum as the divisor.
$\begin{aligned}3+3&=6\\36\div6&=6\end{aligned}$

The two readings give different numbers.
A shared convention tells the writer and reader which calculation the ungrouped expression names.
Under the standard order, division precedes addition, so the value is $15$ .
To request Ben's calculation, write $36\div(3+3)$ .

Answer

Mira: $15$ . Ben: $6$ . The two readings give different values. Standard order: $15$ .

Key idea

A shared order of operations gives a written expression the same meaning for its writer and its reader.
Problem 10: Mark complete

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