Indices and standard form – Year 10 GCSE maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \((4x^3)^2=16x^6\) (1), times \(3x\) is \(48x^7\) (1) Square both the 4 and the power: \(4^2=16\) and \(x^{3\times2}=x^6\). Then multiply by \(3x\), adding the indices. The answer is \(48x^7\).
   2. \(\left(\dfrac52\right)^2=\) \(\dfrac{25}{4}\) (1) and \(9^{\frac12}=\) 3 (1) A negative index means the reciprocal, so turn the fraction upside down and square it. A power of \(\tfrac12\) is a square root, so \(9^{\frac12}=3\).

2. Answer to Question 2 (4 marks)

   1. \(8.6\times10^{-5}\) (1) The decimal point moves 5 places to the right to give 8.6, so the power is \(-5\).
   2. \(3\times4=12\) and \(10^{5-2}=10^3\) (1), so \(12\times10^3=\) \(1.2\times10^4\) (1) 12 is not between 1 and 10, so we write \(12=1.2\times10^1\). The answer is \(1.2\times10^4\).
   3. \(7.2\div9=0.8\) and \(10^4\), so \(8\times10^3\) (1) 0.8 is not between 1 and 10, so \(0.8\times10^4=8\times10^3\).

   Check (c) by reversing the division: \(8000\times900=7\,200\,000\), which is \(7.2\times10^6\) as given.

3. Answer to Question 3 (4 marks)

   1. \(\dfrac{3\times10^8}{3.4\times10^2}=\) \(8.82\times10^5\) (1 for division, 1 for the answer) Divide the speed of light by the speed of sound. Light is about \(8.82\times10^5\) times faster than sound.
   2. \(\dfrac{5.1\times10^3}{3.4\times10^2}=\) 15 seconds (1 for the method, 1 for the answer) The light arrives almost at once, so the delay is the time the sound takes: time \(=\) distance \(\div\) speed. The thunder is heard 15 seconds after the flash.

4. Answer to Question 4 (4 marks)

   1. \(\dfrac{60\times0.4}{0.2}=\) 120 (1 for rounding to 1 s.f., 1 for the answer) Round 58.7 to 60, 0.41 to 0.4 and 0.19 to 0.2. The estimate is 120.
   2. Accurate value \(=\dfrac{58.7\times0.41}{0.19}=126.7\) (1), so the estimate is smaller: 0.41 and 58.7 were rounded down (0.4) or only slightly up while the denominator 0.19 was rounded up to 0.2 (1).

   In detail: 58.7 rises by about 2%, 0.41 falls by about 2% and 0.19 rises by about 5%, so the fraction becomes smaller overall.

5. Answer to Question 5 (4 marks)

   1. \(64^{\frac13}=4\), then \(4^{-2}=\) \(\dfrac1{16}\) (1 for each stage) The denominator 3 means a cube root and the minus sign means the reciprocal. So \(64^{-\frac23}=\tfrac1{16}\).
   2. \(\dfrac1{16}=2^{-4}\) (1), so \(x+3=-4\) and \(x=\) −7 (1) Write \(\tfrac1{16}\) as a power of 2, then set the indices equal. Check: \(2^{-7+3}=2^{-4}=\tfrac1{16}\). So \(x=-7\).

   In (a), working the other way round, \(64^2=4096\) and \(\sqrt[3]{4096}=16\), gives the same result but is harder without a calculator. A common error is to give \(-16\) or \(-\tfrac{128}{3}\).

Total : 20 marks