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Exam questions · Maths · Algebra

Substitution and Rearranging Formulae

  • 7 exam questions
  • 19 marks
  • 10 quick checks
  1. 1 Work out [2 marks]

    Work out the value of \(5p - 2q\) when \(p = 3\) and \(q = -4\).

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    Model answer

    \(5 \times 3 - 2 \times (-4) = 15 + 8 = 23\).

    Mark scheme

    • \(15\) or \(-2 \times (-4) = +8\) — M1
    • 23 — A1
  2. 2 Work out [2 marks]

    Work out the value of \(3x^2\) when \(x = -4\).

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    Model answer

    The power comes first: \((-4)^2 = 16\). Then \(3 \times 16 = 48\).

    Mark scheme

    • \((-4)^2 = 16\) — M1
    • 48 — A1
  3. 3 Work out [4 marks]

    A formula for the distance \(s\) is \(s = \tfrac{1}{2}(u + v)t\). (a) Work out \(s\) when \(u = 3\), \(v = 11\) and \(t = 6\). [2 marks] (b) Make \(v\) the subject of the formula. [2 marks]

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    Model answer

    (a) \(s = \tfrac{1}{2} \times (3 + 11) \times 6 = \tfrac{1}{2} \times 14 \times 6 = 42\). (b) Multiply both sides by 2: \(2s = (u + v)t\). Divide by \(t\): \(\dfrac{2s}{t} = u + v\). Subtract \(u\): \(v = \dfrac{2s}{t} - u\).

    Mark scheme

    • (a) \(\tfrac{1}{2} \times 14 \times 6\) or \(7 \times 6\) — M1
    • (a) 42 — A1
    • (b) \(\dfrac{2s}{t} = u + v\) — M1
    • (b) \(v = \dfrac{2s}{t} - u\) — A1
  4. 4 Make [2 marks]

    Make \(x\) the subject of \(y = 4x + 7\).

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    Model answer

    Subtract 7 from both sides to get \(y - 7 = 4x\), then divide by 4 to get \(x = \dfrac{y - 7}{4}\).

    Mark scheme

    • \(y - 7 = 4x\) — M1
    • \(x = \dfrac{y - 7}{4}\) — A1
  5. 5 Make [2 marks]

    Make \(r\) the subject of the formula \(A = \pi r^2\).

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    Model answer

    Divide both sides by \(\pi\) to get \(\dfrac{A}{\pi} = r^2\). Take the square root of both sides to get \(r = \sqrt{\dfrac{A}{\pi}}\).

    Mark scheme

    • \(\dfrac{A}{\pi} = r^2\) — M1
    • \(r = \sqrt{\dfrac{A}{\pi}}\) — A1
  6. 6 Make [3 marks]

    Make \(a\) the subject of the formula \(v^2 = u^2 + 2as\).

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    Model answer

    Subtract \(u^2\) from both sides to get \(v^2 - u^2 = 2as\). Divide both sides by \(2s\) to get \(a = \dfrac{v^2 - u^2}{2s}\).

    Mark scheme

    • \(v^2 - u^2 = 2as\) — M1
    • Divides both sides by \(2s\) — M1
    • \(a = \dfrac{v^2 - u^2}{2s}\) — A1
  7. 7 Work out [4 marks]

    The cost, \(C\) pounds, of hiring a bike for \(d\) days is given by \(C = 12 + 8d\). (a) Work out the cost of hiring the bike for 5 days. [1 mark] (b) Paul paid \(\pounds 84\). For how many days did he hire the bike? [2 marks] (c) Make \(d\) the subject of the formula. [1 mark]

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    Model answer

    (a) \(12 + 8 \times 5 = 12 + 40 = \pounds 52\). (b) \(12 + 8d = 84\), so \(8d = 72\) and \(d = 9\) days. (c) \(C - 12 = 8d\), so \(d = \dfrac{C - 12}{8}\).

    Mark scheme

    • (a) \(\pounds 52\) — B1
    • (b) \(8d = 72\) or \(84 - 12 = 72\) — M1
    • (b) 9 days — A1
    • (c) \(d = \dfrac{C - 12}{8}\) — B1

Quick check

  1. 1

    Make \(a\) the subject of \(P = 2(a + b)\).

    1. A\(a = P - 2b\)
    2. B\(a = 2P - b\)
    3. C\(a = \dfrac{P}{2} - b\)
    4. D\(a = \dfrac{P}{2} + b\)
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    C: \(a = \dfrac{P}{2} - b\)

    Divide both sides by 2 to get \(\dfrac{P}{2} = a + b\), then subtract \(b\).

  2. 2

    Use \(A = \tfrac{1}{2}(a + b)h\) to work out \(A\) when \(a = 7\), \(b = 11\) and \(h = 5\).

    1. A45
    2. B18
    3. C35
    4. D90
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    A: 45

    \(\tfrac{1}{2} \times 18 \times 5 = 45\).

  3. 3

    Work out the value of \(2a + b\) when \(a = 4\) and \(b = -3\).

    1. A\(-5\)
    2. B5
    3. C11
    4. D1
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    B: 5

    \(2 \times 4 + (-3) = 8 - 3 = 5\).

  4. 4

    Work out the value of \(x^2\) when \(x = -5\).

    1. A25
    2. B\(-10\)
    3. C\(-25\)
    4. D10
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    A: 25

    \((-5) \times (-5) = 25\), because a negative multiplied by a negative is positive.

  5. 5

    Work out the value of \(3x^2\) when \(x = -2\).

    1. A\(-36\)
    2. B36
    3. C12
    4. D\(-12\)
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    C: 12

    The power is done first: \((-2)^2 = 4\), then \(3 \times 4 = 12\).

  6. 6

    Make \(x\) the subject of \(y = x + 7\).

    1. A\(x = 7 - y\)
    2. B\(x = y + 7\)
    3. C\(x = 7y\)
    4. D\(x = y - 7\)
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    D: \(x = y - 7\)

    Subtract 7 from both sides to get \(x = y - 7\).

  7. 7

    Make \(x\) the subject of \(y = 4x - 1\).

    1. A\(x = \dfrac{y + 1}{4}\)
    2. B\(x = 4(y + 1)\)
    3. C\(x = \dfrac{y - 1}{4}\)
    4. D\(x = \dfrac{y}{4} + 4\)
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    A: \(x = \dfrac{y + 1}{4}\)

    Add 1 to both sides to get \(y + 1 = 4x\), then divide by 4.

  8. 8

    Make \(t\) the subject of \(v = u + at\).

    1. A\(t = v - u - a\)
    2. B\(t = \dfrac{v - u}{a}\)
    3. C\(t = \dfrac{v + u}{a}\)
    4. D\(t = a(v - u)\)
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    B: \(t = \dfrac{v - u}{a}\)

    Subtract \(u\) from both sides to get \(v - u = at\), then divide by \(a\).

  9. 9

    What is the value of \((2x)^2\) when \(x = 3\)?

    1. A12
    2. B36
    3. C6
    4. D18
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    B: 36

    \(2x = 6\), and \(6^2 = 36\). Note that \(2x^2\) would be \(2 \times 9 = 18\).

  10. 10

    Make \(r\) the subject of \(A = \pi r^2\).

    1. A\(r = \dfrac{A^2}{\pi}\)
    2. B\(r = \dfrac{\sqrt{A}}{\pi}\)
    3. C\(r = \dfrac{A}{\pi}\)
    4. D\(r = \sqrt{\dfrac{A}{\pi}}\)
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    D: \(r = \sqrt{\dfrac{A}{\pi}}\)

    Divide by \(\pi\) to get \(r^2 = \dfrac{A}{\pi}\), then take the square root of both sides.