OCR GCSE Maths · OCR 6

Algebra

50 maths questions with worked answers with answers for OCR 6 Algebra in OCR GCSE Maths (spec J560). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below. Not official OCR material.

All 50 questions and answers

Question 1 · Vocabulary · OCR 6.01a
State whether 4(x + 2) = 4x + 8 is an equation, a formula or an identity.
Answer:
It is true for every value of x.
It is an identity.
Question 2 · Vocabulary · OCR 6.01a
State whether 3x + 1 = 10 is an equation or an identity.
Answer:
It is true only for x = 3.
It is an equation.
Question 3 · Vocabulary · OCR 6.01a
In A = πr², what is A called?
Answer:
A is the subject of the formula.
It is the subject.
Question 4 · Notation · OCR 6.01b
Simplify 3a × 4b.
Answer:
Multiply the numbers and write the letters together.
12ab.
Question 5 · Notation · OCR 6.01b
Simplify 5x + 3y − 2x + y.
Answer:
Collect like terms: 5x − 2x = 3x and 3y + y = 4y.
3x + 4y.
Question 6 · Notation · OCR 6.01b
Simplify 12p²q ÷ 4pq.
Answer:
12 ÷ 4 = 3, p² ÷ p = p, q ÷ q = 1.
3p.
Question 7 · Brackets · OCR 6.01d
Expand 4(x + 3).
Answer:
Multiply each term in the bracket by 4.
4(x + 3) = 4x + 12.
Question 8 · Brackets · OCR 6.01d
Expand −3(2y − 5).
Answer:
Multiply each term in the bracket by −3.
−3(2y − 5) = −6y + 15.
Question 9 · Brackets · OCR 6.01d
Simplify 2(x + 4) + 3(x − 1).
Answer:
Expand: 2x + 8 + 3x − 3.
Collect like terms: 5x + 5.
Question 10 · Factorising · OCR 6.01e
Factorise 15y − 20.
Answer:
HCF of 15 and 20 is 5.
5(3y − 4).
Question 11 · Factorising · OCR 6.01e
Factorise fully 8a²b + 12ab².
Answer:
HCF is 4ab.
4ab(2a + 3b).
Question 12 · Factorising · OCR 6.01e
Factorise x² + 7x + 12.
Answer:
Find two numbers that multiply to 12 and add to 7: 3 and 4.
(x + 3)(x + 4).
Question 13 · Turning points · OCR 6.01f · Higher only
Write x² − 6x + 11 in the form (x − a)² + b.
Answer:
(x − 3)² = x² − 6x + 9.
(x − 3)² + 2.
Question 14 · Turning points · OCR 6.01f · Higher only
Use completing the square to find the turning point of y = x² + 8x + 3.
Answer:
(x + 4)² − 16 + 3 = (x + 4)² − 13.
Turning point (−4, −13), a minimum.
Question 15 · Quadratic graphs · OCR 6.01f
The roots of a quadratic are x = 1 and x = 5. Find the x-coordinate of the turning point.
Answer:
The turning point lies halfway between the roots.
x = 1 + 52 = 3.
Question 16 · Algebraic fractions · OCR 6.01g · Higher only
Simplify x² − 9x² + 5x + 6.
Answer:
Factorise: (x + 3)(x − 3)(x + 3)(x + 2).
x − 3x + 2.
Question 17 · Algebraic fractions · OCR 6.01g · Higher only
Write 2x + 3x + 1 as a single fraction.
Answer:
Common denominator x(x + 1): 2(x + 1) + 3xx(x + 1).
5x + 2x(x + 1).
Question 18 · Algebraic fractions · OCR 6.01g · Higher only
Simplify 4x + 8x² − 4.
Answer:
4(x + 2)(x + 2)(x − 2).
4x − 2.
Question 19 · Interpreting · OCR 6.02a
A solution to a quadratic for the length of a side is x = −6. Why is it rejected?
Answer:
A length cannot be negative.
Only the positive solution makes sense.
Question 20 · Modelling · OCR 6.02a
Three identical tickets cost £27. Write and solve an equation for price p.
Answer:
Let p be the price of one ticket: 3p = 27.
Divide by 3: p = £9.
Question 21 · Modelling · OCR 6.02a
A rectangle has width x and length x + 3. Its perimeter is 22. Find x.
Answer:
Perimeter: 2x + 2(x + 3) = 22.
Expand and simplify: 4x + 6 = 22.
4x = 16, so x = 4.
Question 22 · Substitution · OCR 6.02b
Find the value of 3a² − 2b when a = 4 and b = 5.
Answer:
3 × 4² = 3 × 16 = 48 and 2 × 5 = 10.
48 − 10 = 38.
Question 23 · Substitution · OCR 6.02b
Find the value of p + q2 when p = −3 and q = 11.
Answer:
p + q = 8.
8 ÷ 2 = 4.
Question 24 · Substitution · OCR 6.02b
Use v = u + at to find v when u = 5, a = 3 and t = 7.
Answer:
v = 5 + 3 × 7.
v = 5 + 21 = 26.
Question 25 · Rearranging · OCR 6.02c
Make x the subject of y = 3x − 7.
Answer:
Add 7: y + 7 = 3x.
x = y + 73.
Question 26 · Rearranging · OCR 6.02c
Make r the subject of C = 2πr.
Answer:
Divide both sides by 2π.
r = C2π.
Question 27 · Rearranging · OCR 6.02c · Higher only
Make t the subject of s = at + bt.
Answer:
st = at + b, so st − at = b.
t(s − a) = b, so t = bs − a.
Question 28 · Formulae · OCR 6.02d
If C = 5n + 2, calculate C when n = 6.
Answer:
Substitute n = 6: C = 5 × 6 + 2.
C = 30 + 2 = 32.
Question 29 · Linear equations · OCR 6.03a
Solve 5x + 3 = 28.
Answer:
5x = 25.
x = 5.
Question 30 · Linear equations · OCR 6.03a
Solve 4(x − 2) = 20.
Answer:
x − 2 = 5.
x = 7.
Question 31 · Linear equations · OCR 6.03a
Solve 7x − 4 = 3x + 16.
Answer:
4x = 20.
x = 5.
Question 32 · Quadratic formula · OCR 6.03b · Higher only
Solve x² + 4x − 3 = 0, giving answers to 2 d.p.
Answer:
x = −4 ± √(16 + 12)2 = −4 ± √282.
x = 0.65 or x = −4.65.
Question 33 · Quadratic formula · OCR 6.03b · Higher only
Solve 3x² − 5x − 1 = 0 to 2 d.p.
Answer:
x = 5 ± √(25 + 12)6 = 5 ± √376.
x = 1.85 or x = −0.18.
Question 34 · Simultaneous equations · OCR 6.03c
Solve x + y = 10 and x − y = 4.
Answer:
Add: 2x = 14, so x = 7.
y = 3.
Question 35 · Simultaneous equations · OCR 6.03c
Solve 3x + 2y = 16 and x + 2y = 8.
Answer:
Subtract: 2x = 8, so x = 4.
4 + 2y = 8, so y = 2.
Question 36 · Simultaneous equations · OCR 6.03c
Solve 2x + 3y = 13 and 5x − y = 7.
Answer:
Multiply the second by 3: 15x − 3y = 21. Add: 17x = 34, so x = 2.
10 − y = 7, so y = 3.
Question 37 · Quadratic graphs · OCR 6.03d
Describe the shape of the graph of y = −x² + 4.
Answer:
The coefficient of x² is negative.
An n-shaped parabola with a maximum at (0, 4).
Question 38 · Discriminant · OCR 6.03d · Higher only
How many real solutions does x² + 2x + 5 = 0 have?
Answer:
b² − 4ac = 4 − 20 = −16.
It is negative, so there are no real solutions.
Question 39 · Iteration · OCR 6.03e · Higher only
Use xₙ₊₁ = √(2xₙ + 5) with x₀ = 3 to find x₁ and x₂ to 3 d.p.
Answer:
x₁ = √11 = 3.317.
x₂ = √(2 × 3.3166 + 5) = √11.633 = 3.411.
Question 40 · Iteration · OCR 6.03e · Higher only
Show that x³ − 4x − 1 = 0 has a root between 2 and 3.
Answer:
When x = 2: 8 − 8 − 1 = −1. When x = 3: 27 − 12 − 1 = 14.
The sign changes, so there is a root between 2 and 3.
Question 41 · Iteration · OCR 6.03e · Higher only
Show that x² − 3x − 1 = 0 can be rearranged to x = 3 + 1x.
Answer:
x² = 3x + 1.
Divide by x: x = 3 + 1x.
Question 42 · Inequalities · OCR 6.04a
Solve 3x − 5 > 10.
Answer:
3x > 15.
x > 5.
Question 43 · Inequalities · OCR 6.04a
Solve 2(x + 3) ≤ 14.
Answer:
x + 3 ≤ 7.
x ≤ 4.
Question 44 · Inequalities · OCR 6.04a
List the integers n such that −2 < n ≤ 3.
Answer:
−2 is not included; 3 is included.
−1, 0, 1, 2, 3.
Question 45 · Functions · OCR 6.05a
f(x) = 4x − 1. Find f(3).
Answer:
4 × 3 − 1.
f(3) = 11.
Question 46 · Functions · OCR 6.05a · Higher only
f(x) = 2x + 5 and g(x) = x². Find fg(3).
Answer:
g(3) = 9.
f(9) = 2 × 9 + 5 = 23.
Question 47 · Term-to-term · OCR 6.06a
The sequence 4, 9, 14, 19, … continues. Write the next two terms.
Answer:
The rule is add 5.
24 and 29.
Question 48 · Term-to-term · OCR 6.06a
The first term is 3 and the rule is "multiply by 2 then subtract 1". Find the first four terms.
Answer:
3, 3 × 2 − 1 = 5, 5 × 2 − 1 = 9, 9 × 2 − 1 = 17.
3, 5, 9, 17.
Question 49 · Special sequences · OCR 6.06b
Write the first five square numbers.
Answer:
1², 2², 3², 4², 5².
1, 4, 9, 16, 25.
Question 50 · Fibonacci · OCR 6.06b
Write the next two terms of 1, 1, 2, 3, 5, 8, ….
Answer:
Each term is the sum of the previous two.
13 and 21.

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