Maths1,443Updated 13 Sept 202626 pages

GCSE Maths Paper 1: Practice and Revision Questions

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Nasim Adatiya@asimdatiya_mzhrgjqei
Welcome to your GCSE Maths revision guide for Edexcel Higher Paper 1! This guide covers essential topics that frequently appear on Paper 1. We'll break down the key concepts, solving techniques, and example questions to help you tackle these important areas with confidence.
GCSE Maths paper 1  – page 1

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Suggested Revision Topics

Getting ready for your Edexcel Higher GCSE Paper 1? These topics appear frequently and deserve your attention:

Mathematical operations like fractions, indices, and standard form form the foundation of many questions. Make sure you're confident with these basics.

Algebraic skills including sequences, factorising, equation rearrangement and function problems are highly testable areas. Pay special attention to quadratics!

Geometry and statistics topics like ratio, surface area, probability trees, and data analysis are also common test areas.

Top Tip: While this list is a good starting point, remember to revise all topics thoroughly. Examiners can include any part of the syllabus, so don't leave gaps in your knowledge.

Continue through this guide for focused revision on each of these key topics, with worked examples to help you master them.

GCSE Maths paper 1  – page 2

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Fractions - 4 Operations

Fractions questions come up regularly and testing all four operations (add, subtract, multiply, divide). Let's see how to handle them:

For multiplication, simply multiply the numerators together and multiply the denominators together: 23×34=2×33×4=612=12\frac{2}{3} \times \frac{3}{4} = \frac{2\times3}{3\times4} = \frac{6}{12} = \frac{1}{2}

For division, multiply by the reciprocal of the second fraction: ab÷cd=ab×dc\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}

For addition and subtraction, you need a common denominator first. For 2314\frac{2}{3} - \frac{1}{4}, convert to 812312=512\frac{8}{12} - \frac{3}{12} = \frac{5}{12}

With mixed numbers, convert to improper fractions first: 217=1572\frac{1}{7} = \frac{15}{7} before performing operations.

Remember: Always simplify your final answer to its lowest terms or convert back to a mixed number if required by the question!

Test yourself: Can you work out 4152234\frac{1}{5} - 2\frac{2}{3} as a mixed number?

GCSE Maths paper 1  – page 3

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Laws of Indices

Indices are crucial for higher maths. The laws of indices give us shortcuts for working with powers.

When multiplying powers with the same base, add the indices: m3×m4=m3+4=m7m^3 \times m^4 = m^{3+4} = m^7

When dividing powers with the same base, subtract the indices: 32q4r44q3r=8q43r41=8q1r3\frac{32q^4r^4}{4q^3r} = 8q^{4-3}r^{4-1} = 8q^1r^3

For a power of a power, multiply the indices: (p2)5=p2×5=p10(p^2)^5 = p^{2\times5} = p^{10}

For negative indices, use xn=1xnx^{-n} = \frac{1}{x^n}, so 23=123=182^{-3} = \frac{1}{2^3} = \frac{1}{8}

When calculating complex expressions like 37×3233\frac{3^7 \times 3^{-2}}{3^3}, apply the rules step by step: 37×3233=37233=3533=353=32=9\frac{3^7 \times 3^{-2}}{3^3} = \frac{3^{7-2}}{3^3} = \frac{3^5}{3^3} = 3^{5-3} = 3^2 = 9

Watch out! You can only apply these rules when the bases are the same. Terms like c5+c2c^5 + c^2 cannot be simplified using the laws of indices because they involve addition.

Test yourself: Can you simplify (d4)3(d^4)^3 using the laws of indices?

GCSE Maths paper 1  – page 4

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Estimation and Approximation

Estimation helps you check if your calculator answer is reasonable. The key is to round each number to 1 significant figure first.

To estimate 790×28949\frac{790 \times 289}{49}, round to get 800×30050=24000050=4800\frac{800 \times 300}{50} = \frac{240000}{50} = 4800

This estimation technique is particularly useful for:

  • Checking whether a calculator answer is reasonable
  • Determining which of two possible answers is correct
  • Quickly working out approximate solutions

For real-world problems, estimates often involve rounding and using π3\pi \approx 3 for circle calculations. For example, to estimate how many boxes of grass seed are needed for a circular garden with radius 10m: Area ≈ 3×102=300m23 \times 10^2 = 300m^2 Number of boxes ≈ 300÷466.5300 ÷ 46 ≈ 6.5 boxes (round up to 7)

Important: Examiners often ask if your estimate is an underestimate or overestimate. Think about whether rounding up or down affects your final answer - rounding values up usually leads to overestimates.

When estimating, showing your working is essential for full marks!

GCSE Maths paper 1  – page 5

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Standard Form (with Calculations)

Standard form writes numbers as a×10na \times 10^n where 1a<101 ≤ a < 10 and nn is an integer. This is especially useful for very large or very small numbers.

To convert to standard form:

  • Move the decimal point so there's just one digit before it
  • Count how many places you moved it
  • If you moved right, nn is negative; if left, nn is positive

For example: 0.00007547=7.547×1050.00007547 = 7.547 \times 10^{-5}

To convert from standard form to an ordinary number:

  • If nn is positive, move the decimal point nn places right
  • If nn is negative, move the decimal point nn places left

For example: 3.42×104=342003.42 \times 10^4 = 34200

For calculations with standard form:

  • Use your calculator's [EXP][EXP] or [×10x][\times 10^x] button
  • For multiplication: (a×10m)×(b×10n)=(a×b)×10m+n(a \times 10^m) \times (b \times 10^n) = (a \times b) \times 10^{m+n}
  • For division: a×10mb×10n=ab×10mn\frac{a \times 10^m}{b \times 10^n} = \frac{a}{b} \times 10^{m-n}

Tip: When comparing numbers in standard form, first compare the powers of 10, then the numbers. 6.212×1086.212 \times 10^8 is smaller than 4.73×1094.73 \times 10^9 because 10810^8 is smaller than 10910^9.

Practice converting between forms and doing calculations to master this topic!

GCSE Maths paper 1  – page 6

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Product of Prime Factors

Prime factorisation means writing a number as a product of its prime factors. This is essential for finding LCM and HCF.

To find the prime factorisation:

  1. Divide the number by the smallest prime number that goes into it
  2. Keep dividing until you can't go further
  3. Write as a product or in index form

For example: 36=2×2×3×3=22×3236 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2

For the Lowest Common Multiple (LCM):

  1. Find the prime factorisation of each number
  2. Include each prime factor the maximum number of times it appears in either number
  3. Multiply these prime factors together

For the Highest Common Factor (HCF):

  1. Find the prime factorisation of each number
  2. Include each prime factor the minimum number of times it appears in both numbers
  3. Multiply these prime factors together

Visual tip: Draw Venn diagrams for LCM and HCF calculations. The LCM includes all factors (the whole diagram), while the HCF includes only the overlapping factors.

For example, finding the LCM of 40 and 56: 40 = 23×52^3 \times 5 56 = 23×72^3 \times 7 LCM = 23×5×7=2802^3 \times 5 \times 7 = 280

Master this technique for solving various number problems!

GCSE Maths paper 1  – page 7

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Averages from Frequency Tables

Frequency tables organize data sets, and you'll need to calculate averages from them.

For the median from a frequency table:

  1. Find the middle position (n+12\frac{n+1}{2} where nn is the total frequency)
  2. Create a running total column
  3. Find which group the median lies in

For example, for 25 women's dress sizes, the median position is the 13th value, which falls in the size 12 group.

For the mean from a grouped frequency table:

  1. Use the midpoint of each group (xx)
  2. Multiply each midpoint by its frequency (fxfx)
  3. Add all the fxfx values and divide by the total frequency

For example, calculating the mean race time: Mean=fxf=33518=18.6\text{Mean} = \frac{\sum fx}{\sum f} = \frac{335}{18} = 18.6 seconds

Exam tip: For grouped data, you can only estimate the mean and median since you don't know the exact values within each group. Examiners may ask about this limitation.

The mode is simply the group with the highest frequency, and the range is the difference between the upper bound of the highest group and the lower bound of the lowest group.

Practice these calculations to master interpreting data in different formats!

GCSE Maths paper 1  – page 8

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Linear Sequences and nth Term

Sequences questions test your ability to spot patterns and express them algebraically.

To find the nth term of a linear sequence:

  1. Find the common difference dd between terms
  2. Multiply d by n for the "growing part"
  3. Find what you need to add or subtract to get the actual terms

For example, for the sequence 3, 8, 13, 18, 23:

  • Common difference = 5
  • Growing part = 5n
  • Adjustment = 3 - 5 = -2
  • So the nth term = 5n - 2

For sequences with patterns (like the counter example):

  1. Count how many items in each pattern
  2. See how this relates to the pattern number
  3. Write as an expression in terms of n

For example, if pattern 1 has 4 counters, pattern 2 has 7, and pattern 3 has 10:

  • Difference = +3 each time
  • Formula = 3n + 1

Remember: For arithmetic sequences, the nth term is always in the form an + b, where a is the common difference.

To check your answer, substitute different values of n to make sure you get the correct terms in the sequence!

GCSE Maths paper 1  – page 9

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Expand and Factorise (Single Bracket)

Expanding and factorising are fundamental algebraic skills.

To expand a single bracket:

  1. Multiply each term inside the bracket by the term outside
  2. Simplify if needed

For example: 342x4 - 2x = 12 - 6x

To factorise an expression:

  1. Find the highest common factor (HCF) of all terms
  2. Express the expression as HCF × (remaining terms)

For example: 4p + 6 = 22p+32p + 3

For expressions with powers, look for common factors including variables: 9x² + 6x = 3x3x+23x + 2

Tip: When factorising, always check your answer by expanding it again. You should get back to the original expression.

These skills are building blocks for more complex algebra, including solving equations and working with quadratic expressions. Make sure you're confident with them before moving on to harder topics!

Test yourself: Can you factorise 15y - 10?

GCSE Maths paper 1  – page 10

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Factorising Quadratics (and Solving)

Factorising quadratics is a key skill for solving quadratic equations.

For quadratics in the form x² + bx + c:

  1. Find two numbers that multiply to give c and add to give b
  2. Put these numbers as the second terms in two brackets

For example: x² + 4x + 3 = x + 3$$x + 1

  • We need numbers that multiply to give 3 and add to give 4
  • 3 × 1 = 3 and 3 + 1 = 4, so we use 3 and 1

For solving quadratic equations:

  1. Make sure the equation equals zero
  2. Factorise the quadratic
  3. Set each bracket equal to zero and solve for x

For example: x² - 7x - 18 = 0

  • Factorise: x + 2$$x - 9 = 0
  • Either x + 2 = 0 or x - 9 = 0
  • So x = -2 or x = 9

Special case: For perfect square quadratics like x² + 6x + 9, the answer is x+3x + 3², because x+3x + 3² = x² + 6x + 9.

This method only works for quadratics that can be factorised. For others, you'll need the quadratic formula or completing the square method.

Practice factorising different types of quadratics to build your confidence!

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