Structure and Calculation

1. 1

   Write $180$ as a product of prime factors in index form.

   • A

     $2^2 \times 3^2 \times 5$

   • B

     $2^2 \times 3 \times 5^2$

   • C

     $2^3 \times 3 \times 5$

   • D

     $2 \times 3^2 \times 10$

   Check answer

   Answer & explanation

   Correct: option A

   $180 = 2 \times 90 = 2 \times 2 \times 45 = 2 \times 2 \times 3 \times 15 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5$. Option D is wrong because 10 is not prime.

2. 2

   What is the LCM of 24 and 36?

   • A

     $12$

   • B

     $72$

   • C

     $144$

   • D

     $864$

   Check answer

   Answer & explanation

   Correct: option B

   $24 = 2^3 \times 3$ and $36 = 2^2 \times 3^2$. LCM: take highest powers — $2^3 \times 3^2 = 8 \times 9 = 72$. The HCF is 12 (shared factors, lower powers), which is a common distractor.

3. 3

   Evaluate $64^{-2/3}$ without a calculator.

   • A

     $-16$

   • B

     $\dfrac{1}{16}$

   • C

     $16$

   • D

     $\dfrac{1}{4}$

   Check answer

   Answer & explanation

   Correct: option B

   $64^{-2/3} = \dfrac{1}{64^{2/3}}$. Then $64^{2/3} = (\sqrt[3]{64})^2 = 4^2 = 16$. So $64^{-2/3} = \dfrac{1}{16}$. The negative index gives a reciprocal — it does not make the answer negative.

4. 4

   Simplify $\sqrt{75}$.

   • A

     $5\sqrt{3}$

   • B

     $3\sqrt{5}$

   • C

     $15\sqrt{3}$

   • D

     $25\sqrt{3}$

   Check answer

   Answer & explanation

   Correct: option A

   $\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}$. The key is finding the largest perfect-square factor of 75, which is 25.

5. 5

   Calculate $(4.5 \times 10^{-3}) \times (6 \times 10^4)$. Give your answer in standard form.

   • A

     $2.7 \times 10^2$

   • B

     $27 \times 10^1$

   • C

     $2.7 \times 10^{-12}$

   • D

     $2.7 \times 10^{-7}$

   Check answer

   Answer & explanation

   Correct: option A

   Multiply the coefficients: $4.5 \times 6 = 27$. Add the indices: $10^{-3} \times 10^4 = 10^1$. So $27 \times 10^1 = 2.7 \times 10^2$. The adjustment step (from $27 \times 10^1$ to $2.7 \times 10^2$) is essential — the coefficient must be $< 10$.

6. 6

   What is the value of $3 + 2^4 \div (8 - 4) \times 2$?

   • A

     $11$

   • B

     $8$

   • C

     $35$

   • D

     $19$

   Check answer

   Answer & explanation

   Correct: option A

   Brackets: $8 - 4 = 4$. Indices: $2^4 = 16$. Division then multiplication (left to right): $16 \div 4 = 4$, then $4 \times 2 = 8$. Addition: $3 + 8 = 11$.

7. 7

   Rationalise the denominator of $\dfrac{10}{\sqrt{5}}$.

   • A

     $2\sqrt{5}$

   • B

     $\sqrt{5}$

   • C

     $2$

   • D

     $10\sqrt{5}$

   Check answer

   Answer & explanation

   Correct: option A

   $\dfrac{10}{\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{10\sqrt{5}}{5} = 2\sqrt{5}$.

Exam-style multi-part questions. Try each part on paper, then reveal the model answer to compare your working.

1. Q15 marks total

   Show all answers

   Scenario

   Two bus routes depart from the same stop. Route A departs every 24 minutes. Route B departs every 36 minutes. They both depart together at 09:00.

   1. (a)

      Find the HCF of 24 and 36. Show your prime factorisation.

      [3]
      Show answer

      Model answer · 3 marks

      Prime factorisations: $24 = 2^3 \times 3 \qquad 36 = 2^2 \times 3^2$

      HCF: shared primes, lower powers. $\text{HCF} = 2^2 \times 3 = 4 \times 3 = 12$

      Marking:

      • 1 mark for each correct prime factorisation.
      • 1 mark for HCF = 12.
   2. (b)

      Find the next time both buses depart together from the stop.

      [2]
      Show answer

      Model answer · 2 marks

      The next time both buses depart together is after the LCM of 24 and 36 minutes.

      $\text{LCM} = 2^3 \times 3^2 = 8 \times 9 = 72\,\text{minutes}$

      $09\!:\!00 + 72\,\text{min} = 10\!:\!12$

      Marking:

      • 1 mark for LCM = 72.
      • 1 mark for converting 72 minutes to the correct time (10:12).

   Examiner note

   AQA examiners report that students often find LCM when HCF is required, and vice versa. Reading the question carefully — 'largest equal group' vs 'next time coinciding' — determines which to use.

2. Q28 marks total

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   Scenario

   This question is about indices and surds.

   1. (a)

      Evaluate $8^{-4/3}$ without a calculator. Give your answer as a fraction.

      [3]
      Show answer

      Model answer · 3 marks

      Step 1: apply the negative index — take the reciprocal. $8^{-4/3} = \frac{1}{8^{4/3}}$

      Step 2: apply the fractional index — root first, then power. $8^{4/3} = \left(\sqrt[3]{8}\right)^4 = 2^4 = 16$

      Step 3: substitute back. $8^{-4/3} = \frac{1}{16}$

      Marking:

      • 1 mark for $\dfrac{1}{8^{4/3}}$.
      • 1 mark for $\sqrt[3]{8} = 2$.
      • 1 mark for final answer $\dfrac{1}{16}$.
   2. (b)

      Expand and simplify $(5 + \sqrt{3})(5 - \sqrt{3})$.

      [2]
      Show answer

      Model answer · 2 marks

      Use the difference of two squares identity $(a+b)(a-b) = a^2 - b^2$: $(5 + \sqrt{3})(5 - \sqrt{3}) = 5^2 - (\sqrt{3})^2 = 25 - 3 = 22$

      Marking:

      • 1 mark for recognising/applying difference of two squares.
      • 1 mark for $22$.
   3. (c)

      Write $\dfrac{3}{\sqrt{7} - 2}$ in the form $a + b\sqrt{7}$ where $a$ and $b$ are integers.

      [3]
      Show answer

      Model answer · 3 marks

      Multiply numerator and denominator by the conjugate $\sqrt{7} + 2$:

      $\frac{3}{\sqrt{7} - 2} \times \frac{\sqrt{7} + 2}{\sqrt{7} + 2} = \frac{3(\sqrt{7} + 2)}{(\sqrt{7})^2 - 4} = \frac{3(\sqrt{7} + 2)}{7 - 4} = \frac{3(\sqrt{7} + 2)}{3}$

      $= \sqrt{7} + 2$

      So $a = 2$, $b = 1$.

      Marking:

      • 1 mark for multiplying by the conjugate $\sqrt{7} + 2$.
      • 1 mark for correct denominator = 3.
      • 1 mark for final answer $2 + \sqrt{7}$.
3. Q35 marks total

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   Scenario

   The mass of a proton is $1.67 \times 10^{-27}$ kg. The mass of an electron is $9.11 \times 10^{-31}$ kg.

   1. (a)

      Calculate the total mass of one proton and one electron. Give your answer in standard form.

      [3]
      Show answer

      Model answer · 3 marks

      Convert to the same power of 10: $1.67 \times 10^{-27} = 16700 \times 10^{-31}$

      Add: $16700 \times 10^{-31} + 9.11 \times 10^{-31} = 16709.11 \times 10^{-31}$

      Convert to standard form: $= 1.670911 \times 10^{-27} \approx 1.67 \times 10^{-27}\,\text{kg}$

      Marking:

      • 1 mark for correct strategy (same power of 10).
      • 1 mark for correct addition.
      • 1 mark for answer in standard form (accept $1.671 \times 10^{-27}$ or similar appropriate rounding).
   2. (b)

      How many times heavier is a proton than an electron? Give your answer to 3 significant figures.

      [2]
      Show answer

      Model answer · 2 marks

      $\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}} = \frac{1.67}{9.11} \times 10^{-27 - (-31)} = 0.1833\ldots \times 10^{4}$

      $= 1.833\ldots \times 10^{3} \approx 1830\,\text{(to 3 s.f.)}$

      Marking:

      • 1 mark for correct division of coefficients and subtraction of powers.
      • 1 mark for $1830$ (or $1.83 \times 10^3$) to 3 s.f.

   Examiner note

   Part (a) is a common source of lost marks: students subtract the powers instead of aligning them and adding. Always write both numbers with the same power of 10 before adding or subtracting.