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Exam questions · Maths · Algebra

Substitution and Rearranging Formulae

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

    Work out the value of \(3a + 2b\) when \(a = 4\) and \(b = -5\).

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    Model answer

    \(3 \times 4 + 2 \times (-5) = 12 - 10 = 2\).

    Mark scheme

    • \(3 \times 4 = 12\) or \(2 \times (-5) = -10\) — M1
    • 2 — A1
  2. 2 Work out [3 marks]

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

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    Model answer

    \(2 \times (-2)^2 - 3 \times (-2) = 2 \times 4 + 6 = 14\).

    Mark scheme

    • \((-2)^2 = 4\) — M1
    • \(2 \times 4\) and \(-3 \times (-2) = +6\) — M1
    • 14 — A1
  3. 3 Work out [4 marks]

    The formula \(v = u + at\) is used in science. (a) Work out the value of \(v\) when \(u = 5\), \(a = 3\) and \(t = 4\). (2 marks) (b) Make \(t\) the subject of \(v = u + at\). (2 marks)

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    Model answer

    (a) \(v = 5 + 3 \times 4 = 5 + 12 = 17\). (b) Subtract \(u\) from both sides to get \(v - u = at\), then divide by \(a\) to get \(t = \dfrac{v - u}{a}\).

    Mark scheme

    • (a) \(5 + 3 \times 4\) or \(5 + 12\) — M1
    • (a) 17 — A1
    • (b) \(v - u = at\) — M1
    • (b) \(t = \dfrac{v - u}{a}\) — A1
  4. 4 Make [2 marks]

    Make \(x\) the subject of \(y = 5x - 3\).

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    Model answer

    Add 3 to both sides to get \(y + 3 = 5x\), then divide by 5 to get \(x = \dfrac{y + 3}{5}\).

    Mark scheme

    • \(y + 3 = 5x\) — M1
    • \(x = \dfrac{y + 3}{5}\) — A1
  5. 5 Make [3 marks]

    Make \(m\) the subject of the formula \(E = \tfrac{1}{2}mv^2\).

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    Model answer

    Multiply both sides by 2 to get \(2E = mv^2\). Divide both sides by \(v^2\) to get \(m = \dfrac{2E}{v^2}\).

    Mark scheme

    • \(2E = mv^2\) — M1
    • Divides both sides by \(v^2\) — M1
    • \(m = \dfrac{2E}{v^2}\) — A1
  6. 6 Work out [5 marks]

    A plumber charges a call-out fee of \(\pounds 35\) plus \(\pounds 22\) for each hour of work. The total cost is \(C\) pounds for \(h\) hours. (a) Write down a formula for \(C\) in terms of \(h\). (2 marks) (b) Work out the total cost for 4 hours. (1 mark) (c) A job costs \(\pounds 167\). Work out how many hours the plumber worked. (2 marks)

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    Model answer

    (a) \(C = 35 + 22h\). (b) \(35 + 22 \times 4 = 35 + 88 = \pounds 123\). (c) \(35 + 22h = 167\), so \(22h = 132\) and \(h = 6\) hours.

    Mark scheme

    • (a) \(22h\) or 35 as the fixed amount identified — M1
    • (a) \(C = 35 + 22h\) — A1
    • (b) \(\pounds 123\) — B1
    • (c) \(22h = 132\) or \(167 - 35 = 132\) — M1
    • (c) 6 hours — A1
  7. 7 Make [4 marks]

    Make \(x\) the subject of \(y = \dfrac{x + 3}{x - 2}\).

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    Model answer

    Multiply both sides by \((x - 2)\): \(y(x - 2) = x + 3\). Expand: \(xy - 2y = x + 3\). Collect the \(x\) terms: \(xy - x = 3 + 2y\). Factorise: \(x(y - 1) = 2y + 3\). So \(x = \dfrac{2y + 3}{y - 1}\).

    Mark scheme

    • \(y(x - 2) = x + 3\) — M1
    • \(xy - 2y = x + 3\) — M1
    • \(xy - x = 2y + 3\) and \(x(y - 1)\) — M1
    • \(x = \dfrac{2y + 3}{y - 1}\) — A1

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.