Powers, roots and standard form – Year 8 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. The square root asks which number multiplied by itself gives 196. Since \(14\times14=196\), \(\sqrt{196}=\)14 (1 mark). The cube root asks which number cubed gives 125: \(5\times5\times5=125\), so \(\sqrt[3]{125}=\)5 (1 mark).
   2. When multiplying powers of the same base, add the indices; when dividing, subtract them. So \(3^{8}\times3^{2}=3^{10}\), then \(3^{10}\div3^{4}=\)\(3^{6}\). (1 mark)
   3. A negative index means the reciprocal: \(2^{-3}=\dfrac{1}{2^{3}}=\)\(\dfrac18\), which is 0.125. (1 mark)

   A common mistake in (c) is writing \(-8\) or \(-6\). A negative index never makes the answer negative.

2. Answer to Question 2 (4 marks)

   1. Use a factor tree or repeated division: \(90=2\times45=2\times3\times15=2\times3\times3\times5\) (1 mark). In index form: \(2\times3^{2}\times5\) (1 mark).
   2. In the same way, \(84=2^{2}\times3\times7\). The HCF uses the primes common to both, each with the lower power: one 2 and one 3. HCF \(=2\times3=\)6. (1 mark)
   3. The LCM uses every prime that appears, each with the higher power: \(2^{2}\times3^{2}\times5\times7=4\times9\times35=\)1 260. (1 mark)

   Check: HCF × LCM \(=6\times1260=7\,560\), and \(90\times84=7\,560\). This product rule holds for any two numbers.

3. Answer to Question 3 (4 marks)

   1. Move the decimal point so that the number is between 1 and 10: 0.000 072 becomes 7.2. The point moved 5 places to the right, so the power is negative: \(7.2\times10^{-5}\). (1 mark)
   2. Multiply the numbers and add the powers: \(4\times3=12\) and \(10^{5}\times10^{4}=10^{9}\). Since 12 is not between 1 and 10, write \(12=1.2\times10\): \(1.2\times10^{10}\). (1 mark)
   3. Divide the numbers and subtract the powers: \(9\div3=3\) and \(10^{8-(-2)}=10^{10}\): \(3\times10^{10}\). (1 mark)
   4. Powers differ, so change to ordinary numbers first: \(6\,000+400=6\,400=\)\(6.4\times10^{3}\). (1 mark)

4. Answer to Question 4 (4 marks)

   1. The two numbers have the same power of 10, so subtract the front numbers: \((2.3-1.5)\times10^{8}=0.8\times10^{8}\) (1 mark). The front number must be between 1 and 10, so \(0.8\times10^{8}=\)\(8\times10^{7}\) km (1 mark).
   2. Time = distance ÷ speed \(=\dfrac{2.3\times10^{8}}{3\times10^{5}}=\dfrac{2.3}{3}\times10^{3}=0.7666\ldots\times10^{3}=766.66\ldots\) s (1 mark). To 3 significant figures this is 767 s (1 mark).

   767 seconds is about 12.8 minutes. In other words, the sunlight seen on Mars left the Sun almost 13 minutes earlier. Check (a): 80 000 000 km is 80 million km, and \(230-150=80\) million.

5. Answer to Question 5 (4 marks)

   1. Doubling once each hour for 6 hours multiplies by \(2^{6}=64\) (1 mark). The number is \(500\times64=32\,000\), which in standard form is \(3.2\times10^{4}\) (1 mark).
   2. We need \(500\times2^{h}\gt1\,000\,000\), so \(2^{h}\gt1\,000\,000\div500=2\,000\) (1 mark). Powers of 2: \(2^{10}=1\,024\), which is too small, and \(2^{11}=2\,048\), which is enough. The culture first exceeds one million after 11 hours (1 mark).

   Check: after 10 hours there are \(500\times1024=512\,000\) bacteria, and after 11 hours \(1\,024\,000\). Growth like this, multiplying by the same factor each hour, is called exponential growth.

Total : 20 marks