Order of operations — BODMAS - PEMDAS with nested brackets: Exercises

Level 1 — Recognition

Goal: spot which tier acts first. No long computation.

L1.1 In the expression $7 + 6 \times 2$ , which single operation must you do first, and why?

Recall Solution

WHAT: The two operations present are $+$ and $\times$ . WHY: $\times$ lives one tier above $+$ (multiplication is repeated addition, so it must be unpacked before adding). So multiply first.

$6 \times 2 = 12$
$7 + 12 = 19$ Answer: $19$ — the multiplication $6 \times 2$ happens first.

L1.2 True or false: in $10 - 4 + 3$ you should add before you subtract (because A comes before S in BODMAS). Give the correct value.

Recall Solution

False. $+$ and $-$ share one tier, so we go strictly left→right.

$10 - 4 = 6$
$6 + 3 = 9$ Answer: $9$ . (Doing $4 + 3$ first would wrongly give $10 - 7 = 3$ .)

L1.3 Which do you evaluate first in $\frac{9 + 3}{6 - 2}$ : the top, the bottom, or the division? Just name the rule.

Recall Solution

The fraction bar is a hidden bracket: it groups the whole top and the whole bottom. So you finish each grouped part before dividing.

Top: $9 + 3 = 12$ . Bottom: $6 - 2 = 4$ . Then $12 \div 4 = 3$ . Rule named: a fraction bar brackets numerator and denominator separately. (See Fractions — numerator and denominator as grouping.)

Level 2 — Application

Goal: run the full evaluation loop on clean expressions.

L2.1 Evaluate $48 \div 6 \times 2$ .

Recall Solution

$\div$ and $\times$ share a tier → left→right.

$48 \div 6 = 8$
$8 \times 2 = 16$ Answer: $16$ . (Doing $6 \times 2$ first would wrongly give $48 \div 12 = 4$ .)

L2.2 Evaluate $5 + 2 \times 3^{2}$ .

Recall Solution

WHAT/WHY: Orders (powers) beat $\times$ , which beats $+$ . Climb down the ladder.

Power: $3^{2} = 9$
Multiply: $2 \times 9 = 18$
Add: $5 + 18 = 23$ Answer: $23$ .

L2.3 Evaluate $2 \times (7 - 3) + 12 \div 4$ .

Recall Solution

Bracket first: $7 - 3 = 4$ . Now $2 \times 4 + 12 \div 4$ .
Same-tier $\times \div$ left→right: $2 \times 4 = 8$ , then $12 \div 4 = 3$ .
Add: $8 + 3 = 11$ . Answer: $11$ .

L2.4 Evaluate $\frac{20 - 2 \times 4}{3}$ .

Recall Solution

The bar brackets the top: $(20 - 2 \times 4)$ .

Inside top, $\times$ before $-$ : $2 \times 4 = 8$ , so top $= 20 - 8 = 12$ .
Divide: $12 \div 3 = 4$ . Answer: $4$ .

Level 3 — Analysis

Goal: nested groupings, signs, and hidden brackets.

L3.1 Evaluate $18 - [4 + (10 - 2 \times 3)]$ .

Recall Solution

Innermost first.

Inside $(10 - 2 \times 3)$ : $\times$ before $-$ → $2 \times 3 = 6$ , then $10 - 6 = 4$ .
Now $18 - [4 + 4] = 18 - 8$ .
$= 10$ . Answer: $10$ .

L3.2 Evaluate $- 4^{2} + (- 4)^{2}$ .

Recall Solution

WHY the two terms differ: in $- 4^{2}$ the power binds tighter than the unary minus, so it means $- (4^{2})$ . In $(- 4)^{2}$ the bracket forces the sign into the base first. (See Negative numbers and the unary minus sign.)

$- 4^{2} = - (4 \times 4) = - 16$
$(- 4)^{2} = (- 4) \times (- 4) = + 16$
Sum: $- 16 + 16 = 0$ . Answer: $0$ .

L3.3 Evaluate $\frac{6 + 3}{3} + \frac{6}{3} + 3$ . Notice how the bar changes things.

Recall Solution

Each bar brackets its own numerator.

$\frac{6 + 3}{3} = \frac{9}{3} = 3$
$\frac{6}{3} = 2$
Now $3 + 2 + 3 = 8$ . Answer: $8$ . (If someone wrote it flat as $6 + 3 \div 3 + 6 \div 3 + 3$ they'd get $6 + 1 + 2 + 3 = 12$ — a different expression, because the bar's grouping vanished.)

L3.4 Evaluate $2^{3} \times {5 - [8 \div (1 + 3)]}$ .

Recall Solution

Deepest bracket outward.

$(1 + 3) = 4$
$[8 \div 4] = 2$
${5 - 2} = 3$
Power: $2^{3} = 8$
$8 \times 3 = 24$ . Answer: $24$ .

Level 4 — Synthesis

Goal: combine powers, nested brackets, fractions and signs in one problem.

L4.1 Evaluate $\frac{( 3 + 1 ) ^{2} - 4}{6} + 2 \times (5 - 3^{2})$ .

Recall Solution

Two grouped chunks — handle each fully, then combine.

Left fraction: top $(3 + 1)^{2} - 4$ . Bracket $(3 + 1) = 4$ , power $4^{2} = 16$ , then $16 - 4 = 12$ . Divide: $12 \div 6 = 2$ .
Right term: inside $(5 - 3^{2})$ : power $3^{2} = 9$ , then $5 - 9 = - 4$ . Multiply: $2 \times (- 4) = - 8$ .
Combine: $2 + (- 8) = - 6$ . Answer: $- 6$ .

L4.2 Evaluate $10 - 2 [3 + (2^{2} - 1)] + 4 \div 2$ .

Recall Solution

Innermost $(2^{2} - 1)$ : power $2^{2} = 4$ , then $4 - 1 = 3$ .
Outer bracket $[3 + 3] = 6$ .
Now $10 - 2 \times 6 + 4 \div 2$ .
Same-tier $\times \div$ left→right: $2 \times 6 = 12$ , then $4 \div 2 = 2$ . Line: $10 - 12 + 2$ .
$+ / -$ left→right: $10 - 12 = - 2$ , then $- 2 + 2 = 0$ . Answer: $0$ .

L4.3 Substitute $a = 3, b = - 2$ into $a^{2} - b^{2} + 2 ab$ and evaluate (this links to Algebraic expressions — evaluating and substitution).

Recall Solution

WHY bracket every substitution: when a variable is negative, dropping the bracket lets a stray minus attach to the power. Substitute with brackets.

$a^{2} = 3^{2} = 9$
$b^{2} = (- 2)^{2} = 4$
$2 ab = 2 \times 3 \times (- 2) = - 12$
Now $9 - 4 + (- 12) = 5 - 12 = - 7$ . Answer: $- 7$ .

Level 5 — Mastery

Goal: the notorious traps and self-checking under pressure.

L5.1 Evaluate the internet-famous $48 \div 2 (1 + 2)$ using the standard convention, and state the rule that removes the ambiguity.

Recall Solution

Rule: $2 (1 + 2)$ means $2 \times (1 + 2)$ — a multiplication of the same tier as $\div$ . So after the bracket, go left→right. (See Calculator vs mental arithmetic — parsing expressions.)

Bracket: $(1 + 2) = 3$ . Line: $48 \div 2 \times 3$ .
Left→right: $48 \div 2 = 24$ , then $24 \times 3 = 72$ . Answer: $72$ . (When authoring, add brackets — write $\frac{48}{2} (3)$ or $\frac{48}{2 \times 3}$ — to remove all doubt.)

L5.2 Evaluate $\frac{- 2 ^{2} + ( - 2 ) ^{3}}{( - 1 ) ^{2} - 3 ^{0}}$ . Watch every sign and the fraction bar.

Recall Solution

WHY careful: unary minus vs power on top; a zero exponent on the bottom.

Top: $- 2^{2} = - (2^{2}) = - 4$ . $(- 2)^{3} = - 8$ . Sum $= - 4 + (- 8) = - 12$ .
Bottom: $(- 1)^{2} = 1$ . $3^{0} = 1$ (any nonzero base to the power $0$ is $1$ ; see Exponents and powers — laws of indices). So bottom $= 1 - 1 = 0$ .
The bottom is $0$ → division by zero is undefined. Answer: undefined (division by zero). The lesson: always finish the denominator before declaring an answer.

L5.3 Evaluate $3 + 4 \times [2^{3} - (6 \div (1 + 2))]^{2} - 5$ .

Recall Solution

Deepest bracket outward, tracking tiers.

Deepest $(1 + 2) = 3$ .
$(6 \div 3) = 2$ .
Outer bracket $[2^{3} - 2]$ : power $2^{3} = 8$ , then $8 - 2 = 6$ .
The whole bracket is raised: $6^{2} = 36$ .
Now $3 + 4 \times 36 - 5$ . Multiply first: $4 \times 36 = 144$ .
Line: $3 + 144 - 5$ . Left→right: $3 + 144 = 147$ , then $147 - 5 = 142$ . Answer: $142$ .

L5.4 A calculator returns $2$ for the input 8 ÷ 2 ^ 2, but a student expected $16$ . Who is right, and where did the student's reading go wrong?

Recall Solution

The calculator is right: the answer is $2$ .

WHY: Orders (the power) beat $\div$ . So $2^{2} = 4$ first, then $8 \div 4 = 2$ .
The student's error: they read left→right ( $8 \div 2 = 4$ , then $4^{2} = 16$ ), applying the same-tier tie-breaker across tiers. Power is a higher tier and always jumps the queue. Answer: $2$ .

Master check (fill the blanks)

Recall Cover and self-test

$7 + 6 \times 2$ ::: $19$ $10 - 4 + 3$ ::: $9$ $48 \div 6 \times 2$ ::: $16$ $5 + 2 \times 3^{2}$ ::: $23$ $- 4^{2} + (- 4)^{2}$ ::: $0$ $2^{3} \times {5 - [8 \div (1 + 3)]}$ ::: $24$ $48 \div 2 (1 + 2)$ ::: $72$ $8 \div 2^{2}$ ::: $2$ $\frac{- 2 ^{2} + ( - 2 ) ^{3}}{( - 1 ) ^{2} - 3 ^{0}}$ ::: undefined (denominator $= 0$ )

Connections

Order of Operations — BODMAS / PEMDAS with Nested Brackets (parent)
Arithmetic operations — addition, subtraction, multiplication, division
Exponents and powers — laws of indices
Fractions — numerator and denominator as grouping
Negative numbers and the unary minus sign
Algebraic expressions — evaluating and substitution
Calculator vs mental arithmetic — parsing expressions

Exercises — Order of operations — BODMAS - PEMDAS with nested brackets

Level 1 — Recognition

Level 2 — Application

Level 3 — Analysis

Level 4 — Synthesis

Level 5 — Mastery

Master check (fill the blanks)

Connections

Difficulty Ladder