OCR GCSE Mathematics: Number Operations and Integers — Revision Notes

Condensed for the final weeks. For the full explanation, use the Number Operations and Integers study guide.

Tier: four techniques below are in the Higher tier column of the J560 specification and are marked (Higher): converting recurring decimals to fractions (2.02a), fractional indices (3.01a-b), surds and rationalising denominators (3.03a-b), and calculating the bounds of a calculation (4.01c). Everything else is Foundation content, assessed at both tiers.

Integers and primes

BIDMAS — Brackets, Indices, Division and Multiplication, Addition and Subtraction. Division and multiplication rank equally and are worked left to right, as do addition and subtraction.

Primes have exactly two factors, so 1 is not prime and 2 is the only even prime.

Prime factorisation gives HCF and LCM:

• HCF — product of the lowest power of each common prime.
• LCM — product of the highest power of every prime present.

Negative numbers: two like signs multiply or divide to give a positive; two unlike signs give a negative.

Useful check: HCF × LCM = the product of the two original numbers.

Worked example. Two bells ring at intervals of 84 s and 120 s, together at 09:00. When do they next ring together?

This is an LCM problem, not HCF — the bells coincide on a common multiple.

84  = 2^2 x 3 x 7
120 = 2^3 x 3 x 5
LCM = 2^3 x 3 x 5 x 7 = 840 s = 14 minutes -> 09:14

Choosing between HCF and LCM is the real difficulty: LCM answers “when do things next coincide”, HCF answers “what is the largest equal group I can make”.

Fractions

Multiply across; divide by multiplying by the reciprocal; add and subtract with a common denominator. Convert mixed numbers to improper fractions before multiplying or dividing.

Recurring decimals to fractions (Higher): let x equal the decimal, multiply by a power of 10 so the recurring part aligns, subtract, and solve.

Percentages

increase by 30%:  x 1.3        decrease by 30%:  x 0.7
percentage change = (change / ORIGINAL) x 100
reverse percentage: DIVIDE by the multiplier
compound growth: P x (multiplier)^n

The two errors that dominate:

1. Dividing by the new value instead of the original in percentage change.
2. Subtracting rather than dividing in reverse percentages. A price of £78 after a 30% rise was £78 ÷ 1.3 = £60, not £78 × 0.7 = £54.60.

Note also that a 30% rise followed by a 30% fall does not return to the start: 1.3 × 0.7 = 0.91, a 9% net loss.

Estimating by rounding every value to 1 significant figure before calculating is a quick way to sanity-check whether a final answer is plausible.

Ratio

Divide by the total number of parts, then multiply. Read carefully whether the question gives the total, one share, or the difference between shares — which is given determines how you find the value of one part.

Indices and standard form

a^m x a^n = a^(m+n)     a^m / a^n = a^(m-n)     (a^m)^n = a^(mn)
a^0 = 1                 a^-n = 1/a^n
(Higher only)           a^(m/n) = (n-th root)^m

A negative index means reciprocal, not a negative answer: 3⁻² = 1/9 (negative integer indices are Foundation). (Higher) For a negative fractional index, flip first: (9/16)^(−1/2) = (16/9)^(1/2) = 4/3.

standard form:  A x 10^n     with  1 <= A < 10

After multiplying or dividing, re-check that A lies between 1 and 10 and adjust the power.

Surds (Higher)

sqrt(a) x sqrt(b) = sqrt(ab)        sqrt(72) = 6 sqrt(2)

Rationalise by the surd, or by the conjugate for two terms.

√a + √b ≠ √(a+b) — check with √9 + √16 = 7, not √25 = 5.

Bounds

For a value rounded to the nearest unit u, bounds are ± u/2. Writing the error interval for a single rounded or truncated value is Foundation (see Error intervals below).

(Higher) Calculating the upper and lower bounds of a calculation (4.01c):

Want	Add	Subtract	Multiply	Divide
Maximum	UB + UB	UB − LB	UB × UB	UB ÷ LB
Minimum	LB + LB	LB − UB	LB × LB	LB ÷ UB

Subtraction and division cross over. That single fact is most of the topic.

Error intervals

A value rounded to a given precision could have come from a range either side of it — a length recorded as 6.3 cm to 1 decimal place satisfies 6.25 ≤ length < 6.35. The lower bound uses ≤ and the upper uses <, because 6.35 would itself round up to 6.4.

Exam traps

• Treating 1 as prime.
• Working left to right instead of applying BIDMAS.
• Dividing by the new value in percentage change.
• Subtracting instead of dividing in reverse percentages.
• Reading a negative index as a negative answer.
• (Higher) Using UB ÷ UB for a maximum quotient.
• Rounding partway through.

Self-test

1. Find the HCF and LCM of 48 and 72 by prime factorisation.
2. A price is £78 after a 30% increase. What was it before?
3. (Higher) Evaluate (9/16)^(−1/2).
4. (Higher) Rationalise 3/(4 + √2).
5. (Higher) How do you find the maximum value of a − b from bounds?
6. Two bells ring at intervals of 84 s and 120 s, together at 09:00. When do they next ring together?
7. Write the error interval for a length recorded as 6.3 cm to 1 decimal place.

Answers: 1. 48 = 2⁴×3, 72 = 2³×3²; HCF = 2³×3 = 24, LCM = 2⁴×3² = 144. 2. 78 ÷ 1.3 = £60. 3. Flip to (16/9)^(1/2) = 4/3. 4. Multiply top and bottom by (4 − √2) to get 3(4 − √2)/14. 5. Upper bound of a minus the lower bound of b. 6. LCM of 84 and 120 = 2³×3×5×7 = 840 s = 14 minutes, so 09:14. 7. 6.25 ≤ length < 6.35.