Whole Numbers & Order of Operations

Numbers up to 10 million, order of operations with brackets, factors and multiples (HCF/LCM), and mental calculation strategies.

BODMASfactors & multiplesmental math P1 · no calc

BODMAS: Brackets → ÷× → +− Factor: divides exactly into the number Multiple: in the times-table of that number HCF: highest common factor LCM: lowest common multiple

Part A — Order of operations (Paper 1 · no calculator)

Question 1 — Short answer (2 marks) Paper 1 · no calculator BODMAS

Evaluate each expression. Show your working clearly.

(a) 48 ÷ (12 − 4) + 3 × 5

(b) 100 − 6 × (4 + 2²)

(a) 48 ÷ (12 − 4) + 3 × 5

Brackets= 48 ÷ 8 + 3 × 5

÷ and ×= 6 + 15

+= 21

(a) = 21

(b) 100 − 6 × (4 + 2²)

Index first= 100 − 6 × (4 + 4)

Brackets= 100 − 6 × 8

Multiply= 100 − 48

Subtract= 52

(b) = 52

Common trap: In (b), computing 6 × 4 before handling the brackets gives 24 + 2² = 28, then 100 − 28 = 72 — wrong. Indices inside brackets must be resolved before the bracket is evaluated.

Question 2 — Short answer (2 marks) Paper 1 · no calculator BODMAS · word problem

A shopkeeper arranges 6 boxes of oranges on a shelf. Each box holds 12 oranges. He then removes 3 oranges from each of 4 boxes to display them separately. Write a numerical expression and find the total number of oranges remaining in the boxes.

Expression6 × 12 − 4 × 3

Multiply= 72 − 12

Subtract= 60

60 oranges remain in the boxes

Writing the expression first is worth a method mark even if arithmetic slips. Note: no brackets needed here because × is naturally evaluated before − under BODMAS.

Part B — Factors and multiples (Paper 1 · no calculator)

Question 3 — Short answer (2 marks) Paper 1 · no calculator Factors · HCF

(a) List all the common factors of 36 and 48.
(b) What is the highest common factor (HCF) of 36 and 48?

Factors of 361, 2, 3, 4, 6, 9, 12, 18, 36

Factors of 481, 2, 3, 4, 6, 8, 12, 16, 24, 48

(a) Common1, 2, 3, 4, 6, 12

(b) HCF= 12 (the largest one)

(a) 1, 2, 3, 4, 6, 12 | (b) HCF = 12

List factors in pairs starting from 1: 1×36, 2×18, 3×12, 4×9, 6×6. This systematic approach ensures no factor is missed.

Question 4 — Short answer (2 marks) Paper 1 · no calculator Multiples · LCM

Bus A stops at a bus stop every 8 minutes. Bus B stops every 12 minutes. Both buses stop at the same time at 9:00 am. At what time will they next stop at the bus stop together?

Multiples of 88, 16, 24, 32, 40, 48…

Multiples of 1212, 24, 36, 48…

LCM= 24 minutes

Next time= 9:00 am + 24 min = 9:24 am

Next together at 9:24 am

The LCM gives the first time both cycles align. This is the standard "bus / traffic light / bell" type — always identify it as an LCM problem, not HCF.

Part C — Combined (Paper 2 · calculator · word problem)

Question 5 — Long answer (4 marks) Paper 2 · calculator HCF · real-world context Distinction level

Mrs Tan has 84 red beads and 60 blue beads. She wants to pack them into identical bags so that each bag has the same number of red beads and the same number of blue beads, with no beads left over.

(a) What is the greatest number of bags she can make?
(b) How many red and blue beads are in each bag?
(c) If she adds 12 more red beads to her original collection and repacks using the same rule, how many bags can she make now?

Factors of 841, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

Factors of 601, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

(a) HCF= 12 bags

(b) Red= 84 ÷ 12 = 7 red beads per bag

Blue= 60 ÷ 12 = 5 blue beads per bag

Factors of 961, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

HCF(96,60)Common factors: 1, 2, 3, 4, 6, 12 → HCF = 12 bags

(a) 12 bags | (b) 7 red, 5 blue | (c) still 12 bags

Common trap in (c): Assuming adding 12 red beads must change the answer. Re-computing HCF(96, 60) = 12 shows it stays the same — always recalculate, never assume.