Where must one pair of brackets go so that 8 + 4 × 6 - 2 ÷ 2 = 28?

Question 21

HardOriginal question

Where must one pair of brackets go so that $8 + 4 \times 6 - 2 \div 2 = 28$ ?

(8 + 4 \times 6 - 2) \div 2

Correct answerYour answer

(8 + 4) \times 6 - 2 \div 2

Correct answerYour answer

8 + 4 \times (6 - 2 \div 2)

Correct answerYour answer

8 + 4 \times (6 - 2) \div 2

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$(8 + 24 - 2) \div 2 = 30 \div 2 = 15$ .
$12 \times 6 - 1 = 71$ .
$8 + 4 \times 4 \div 2 = 8 + 8 = 16$ .
$8 + 4 \times (6 - 1) = 8 + 4 \times 5 = 28$ . This is the placement that works.

Worth noting

Test each placement fully with BODMAS. Inside the bracket $6 - 2 \div 2$ , the division still comes first, giving 5, not 2.

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