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

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1. Answer to Question 1

   1. \(3^{5+4}=\) \(3^9\) and \(7^{8-3}=\) \(7^5\) When we multiply powers of the same base we add the indices, and when we divide we subtract them.
   2. \(2^{3\times4}=\) \(2^{12}\) and \(5^0=\) 1 A power of a power means we multiply the indices. Any non-zero number to the power 0 is 1.
   3. \(2\times4=8\) and \(x^3\times x^5=x^8\), so \(8x^8\) Multiply the numbers and add the indices of \(x\) separately. The answer is \(8x^8\).
   4. \(12\div3=4\) and \(a^{7-2}=a^5\), so \(4a^5\). \((3b^2)^3=3^3b^{6}=\) \(27b^6\) Divide the numbers and subtract the indices. For \((3b^2)^3\), cube the 3 and multiply the index 2 by 3.

2. Answer to Question 2

   1. \(5^0=\) 1 and \(10^{-3}=\dfrac1{1000}=\) 0.001 Any non-zero number to the power 0 equals 1. A negative index means the reciprocal, so \(10^{-3}=\tfrac1{10^3}\).
   2. \(4^{-2}=\dfrac1{4^2}=\) \(\dfrac1{16}\) The negative index gives the reciprocal of \(4^2=16\). So \(4^{-2}=\tfrac1{16}\).
   3. \(\left(\dfrac13\right)^{-2}=3^2=\) 9 Turn the fraction upside down to make the index positive, then square. So the value is 9.
   4. \(\dfrac12+\dfrac14=\) \(\dfrac34\) Write each power as a fraction, then use the common denominator 4: \(\tfrac24+\tfrac14=\tfrac34\).

3. Answer to Question 3

   1. 12 and 5 \(12\times12=144\) and \(5\times5\times5=125\), so the square root of 144 is 12 and the cube root of 125 is 5.
   2. 64 and 225 \(2^6=2\times2\times2\times2\times2\times2=64\) and \(15^2=15\times15=225\).
   3. −3 and −64 A negative number has a negative cube root because \((-3)^3=-27\). An odd power of a negative number is negative, so \((-4)^3=-64\).
   4. 0.3 and \(\dfrac7{10}\) (0.7) \(0.3\times0.3=0.09\), and the square root of a fraction is the square root of the top over the square root of the bottom: \(\tfrac{7}{10}\).

4. Answer to Question 4

   1. \(4.5\times10^6\) and \(8.3\times10^{10}\) Put the decimal point after the first non-zero digit, then count how many places it moves. The point moves 6 and 10 places to the left, so the powers are positive.
   2. \(7.2\times10^{-4}\) and \(6.05\times10^{-5}\) For small numbers the point moves to the right, so the powers are negative: 4 places and 5 places.
   3. 32 000 Multiply 3.2 by \(10^4=10\,000\), which moves the point 4 places to the right. So \(3.2\times10^4=32\,000\).
   4. 0.00705 A power of \(-3\) means divide by 1000, which moves the point 3 places to the left. So \(7.05\times10^{-3}=0.00705\).

5. Answer to Question 5

   1. \(3\times2=6\) and \(10^{4+3}=10^7\): \(6\times10^7\) Multiply the numbers, then add the powers of 10. The answer is \(6\times10^7\).
   2. \(8\div2=4\) and \(10^{6-2}=10^4\): \(4\times10^4\) Divide the numbers, then subtract the powers of 10. The answer is \(4\times10^4\).
   3. \(420\,000+30\,000=450\,000=\) \(4.5\times10^5\) To add numbers in standard form, change them to ordinary numbers (or to the same power of 10) first. The answer is \(4.5\times10^5\).
   4. \(6.3\times5=31.5\) and \(10^{3-5}=10^{-2}\), giving \(31.5\times10^{-2}=\) \(3.15\times10^{-1}\) 31.5 is not between 1 and 10, so we write \(31.5=3.15\times10^1\) and add the powers: \(10^1\times10^{-2}=10^{-1}\).

6. Answer to Question 6

   1. \(2.5\times10^{-1}\), \(4.1\times10^4\), \(3.2\times10^5\), \(9.8\times10^5\) The power of 10 decides the size first; only when the powers are equal do we compare the numbers in front.
   2. \(6.01\times10^{-4}\), \(5.9\times10^{-3}\), \(6\times10^{-3}\), \(0.0061\) (\(0.0061=6.1\times10^{-3}\))
   3. \(2.1\times10^9=21\times10^8\), which is bigger than \(7.3\times10^8\), so \(2.1\times10^9\) is bigger (higher power of 10).
   4. \(2^{10}=1024\), \(10^3=1000\), \(3^7=2187\). In order: \(10^3\), \(2^{10}\), \(3^7\).

7. Answer to Question 7

   1. Time \(=\dfrac{1.5\times10^8}{3\times10^5}=0.5\times10^3=\) 500 s (\(5\times10^2\)). Time \(=\) distance \(\div\) speed. Divide the numbers and subtract the powers of 10. Light takes 500 seconds to reach the Earth.
   2. \(500\div60=8\) remainder 20, so 8 minutes 20 seconds. 8 minutes is 480 seconds, which leaves 20 seconds. So light takes 8 minutes 20 seconds.
   3. \(\dfrac{3.84\times10^5}{3\times10^5}=\) 1.28 s. The powers of 10 are equal, so they cancel. Light takes 1.28 seconds to reach us from the Moon.
   4. \(\dfrac{1.5\times10^8}{3.84\times10^5}=390.6\), so about 391 times further. Divide the larger distance by the smaller one. The Sun is about 391 times further away than the Moon.

8. Answer to Question 8

   1. \(\dfrac{8.1\times10^9}{6.7\times10^7}=120.9\), so about 121 times. Divide the world population by the UK population. The world population is about 121 times bigger.
   2. \(\dfrac{6.7\times10^7}{8.1\times10^9}\times100=\) 0.83%. Divide the UK population by the world population and multiply by 100. About 0.83% of the world population lives in the UK.
   3. \(\dfrac{6.7\times10^7}{2.4\times10^5}=279.2\), so about 279 people per km². Population density is the number of people divided by the area. There are about 279 people per km².
   4. \(6.7\times10^7\times1.1=7.37\times10^7\), so \(7.37\times10^7\). A 10% increase gives the multiplier 1.1. The new population is \(7.37\times10^7\).

9. Answer to Question 9

   1. \(\dfrac{10^{-3}}{1.06\times10^{-10}}=0.943\times10^7=\) \(9.43\times10^6\) Divide the distance by the diameter of one atom. 0.943 is not between 1 and 10, so we write it as \(9.43\times10^{-1}\) and the answer becomes \(9.43\times10^6\) atoms.
   2. \(\dfrac{2\times10^{-4}}{1.06\times10^{-10}}=1.887\times10^6=\) \(1.89\times10^6\) atoms. Divide the diameter of the grain by the diameter of an atom. About \(1.89\times10^6\) atoms fit across the grain.
   3. \(\dfrac{7\times10^{-5}}{1.06\times10^{-10}}=6.6\times10^5=\) \(6.6\times10^5\) times. Divide the width of the hair by the diameter of an atom. The hair is about \(6.6\times10^5\) times wider.
   4. \(1.06\times10^{-10}\text{ m}=0.106\times10^{-9}\) m \(=\) 0.106 nm. One nanometre is \(10^{-9}\) m, so we divide by \(10^{-9}\). The atom is 0.106 nm across.

10. Answer to Question 10

    1. \(\dfrac{10^{12}}{4\times10^6}=0.25\times10^6=\) \(2.5\times10^5\) (250 000) photos. Divide the size of the drive by the size of one photo. About 250 000 photos fit on the drive.
    2. \(\dfrac{10^{12}}{5\times10^6}=\) \(2\times10^5\) (200 000) songs. Divide the size of the drive by the size of one song. About 200 000 songs fit on the drive.
    3. Songs: \(2000\times5\times10^6=10^{10}\). Photos: \(50\,000\times4\times10^6=2\times10^{11}\). Total \(=2.1\times10^{11}\) bytes. Work out the songs and the photos separately, then add: \(1\times10^{10}+20\times10^{10}=21\times10^{10}\). The total is \(2.1\times10^{11}\) bytes.
    4. \(\dfrac{2.1\times10^{11}}{10^{12}}=0.21=\) 21%. Divide the bytes used by the size of the drive. So 21% of the drive is used.

11. Answer to Question 11

    1. \(\dfrac{1}{2.5\times10^{-6}}=\) \(4\times10^5\).
    2. 15 cm \(=0.15\) m: \(0.15\times4\times10^5=\) \(6\times10^4\). Change 15 cm into metres, then multiply by the number of bacteria in 1 m.
    3. 6 hours is 18 doubling periods: \(500\times2^{18}=131\,072\,000=\) \(1.31\times10^8\).
    4. Need \(500\times2^n>10^{12}\), so \(2^n>2\times10^9\). \(2^{31}=2.147\times10^9\) works and \(2^{30}=1.07\times10^9\) does not. So \(n=31\) periods \(=31\times20=620\) min \(=\) 10 h 20 min.

12. Answer to Question 12

    1. \(T=0.1\times2^n\). The thickness starts at 0.1 mm and is multiplied by 2 for each fold, so after \(n\) folds it has been multiplied by \(2^n\).
    2. \(0.1\times2^{10}=0.1\times1024=\) 102.4 mm. Substitute \(n=10\) into the formula. The thickness is 102.4 mm.
    3. \(0.1\times2^{20}=104\,857.6\) mm \(=104.8576\) m \(=\) \(1.05\times10^2\) m. Divide by 1000 to change millimetres into metres. The paper would be about \(1.05\times10^2\) m thick.
    4. Need \(0.1\times2^n>2000\) mm, so \(2^n>20\,000\). \(2^{14}=16\,384\) is too small and \(2^{15}=32\,768\) works. So 15 folds. Change 2 m into 2000 mm so the units match. So 15 folds are needed.

13. Answer to Question 13

    1. \(\dfrac{2.7\times10^{12}}{6.7\times10^7}=4.03\times10^4=\) £\(4.0\times10^4\) per person. Divide the debt by the number of people. The debt is about £\(4.0\times10^4\) per person.
    2. Share for 4 people \(=1.612\times10^5\). \(\dfrac{1.612\times10^5}{6.4\times10^4}=\) 2.5 years. The family's share is 4 times the share per person. Divide it by the yearly income: their share equals about 2.5 years of income.
    3. \(2.7\times10^{12}\times0.025=6.75\times10^{10}\), so £\(6.75\times10^{10}\). 2.5% of the debt is \(0.025\times2.7\times10^{12}\). The debt grows by £\(6.75\times10^{10}\) in the first year.
    4. \(2.7\times10^{12}\times1.025=2.7675\times10^{12}\approx\) £\(2.77\times10^{12}\). A 2.5% increase gives the multiplier 1.025. The new debt is about £\(2.77\times10^{12}\).

14. Answer to Question 14

    1. \(\dfrac{1.8\times10^7}{2.4\times10^3}=0.75\times10^4=\) 7500 kettles. Divide the power of the farm by the power of one kettle. The farm could run 7500 kettles at once.
    2. \(1.8\times10^7\times8\times10^3=14.4\times10^{10}=\) \(1.44\times10^{11}\) Wh. Energy in watt-hours is power times hours. Multiply the numbers and add the powers of 10, then rewrite 14.4 as \(1.44\times10^1\).
    3. A house uses \(3.2\times10^3\times10^3=3.2\times10^6\) Wh. \(\dfrac{1.44\times10^{11}}{3.2\times10^6}=4.5\times10^4=\) 45 000 houses. Change the house's use into watt-hours so that the units match, then divide. About 45 000 houses could be supplied.
    4. \(1.44\times10^{11}\div10^3=\) \(1.44\times10^8\) kWh. There are 1000 watt-hours in a kilowatt-hour, so we divide by \(10^3\). The output is \(1.44\times10^8\) kWh.

15. Answer to Question 15

    1. \(\sqrt{25}=\) 5 and \(\sqrt[3]{27}=\) 3 A power of \(\tfrac12\) is a square root and a power of \(\tfrac13\) is a cube root.
    2. \((\sqrt[3]8)^2=2^2=\) 4 and \((\sqrt[4]{16})^3=2^3=\) 8 The denominator of the index gives the root and the numerator gives the power. Take the root first to keep the numbers small.
    3. \(27^{-\frac23}=\dfrac1{(\sqrt[3]{27})^2}=\dfrac19\). \(\left(\dfrac14\right)^{-\frac32}=4^{\frac32}=(\sqrt4)^3=\) 8. So \(\dfrac19\) and 8. A negative index means the reciprocal, so we turn the base upside down first.
    4. \(\dfrac1{32}=2^{-5}\), so \(x=\) −5. \(9^x=27\) means \(3^{2x}=3^3\), so \(2x=3\) and \(x=\) 1.5. Write both sides as powers of the same base, then set the indices equal.

16. Answer to Question 16

    1. \(2^{n+1}=2\times2^n\), so \(2^n+2\times2^n=3\times2^n\). ✓ Using \(a^{m+n}=a^m\times a^n\), the term \(2^{n+1}\) is twice \(2^n\). One lot of \(2^n\) plus two lots makes three lots.
    2. \(5^{n+2}=25\times5^n\), so \(25\times5^n-5^n=24\times5^n\). ✓ Using the index laws, \(5^{n+2}=5^2\times5^n\). Twenty-five lots minus one lot leaves twenty-four lots of \(5^n\).
    3. \(4^{n+1}=4\times4^n\), which is 4 times a whole number, so it is a multiple of 4. Since \(4^n\) is a whole number for every whole number \(n\), \(4^{n+1}\) is a multiple of 4.
    4. \(\dfrac{6^{n+1}}{2^n}=\dfrac{6\times6^n}{2^n}=6\times\dfrac{6^n}{2^n}=6\times3^n\). ✓ Powers with the same index can be divided base by base: \(\tfrac{6^n}{2^n}=\left(\tfrac62\right)^n=3^n\).

17. Answer to Question 17

    1. \(81=3^4\), so \(x+1=4\) and \(x=\) 3 Write 81 as a power of 3, then the indices must be equal. Check: \(3^{3+1}=3^4=81\).
    2. \(64=2^6\), so \(2x=6\) and \(x=\) 3 Write 64 as a power of 2. Check: \(2^{2\times3}=2^6=64\).
    3. \(\dfrac1{25}=5^{-2}\), so \(x=\) −2 Since \(25=5^2\), its reciprocal is \(5^{-2}\). Check: \(5^{-2}=\tfrac1{25}\).
    4. \(4=2^2\) and \(8=2^3\), so \(2^{2x}=2^3\), \(2x=3\) and \(x=\) 1.5 4 and 8 are both powers of 2, so we write both sides with base 2. Check: \(4^{1.5}=(\sqrt4)^3=8\).

18. Answer to Question 18

    1. \(\dfrac{5\times20}{0.2}=\dfrac{100}{0.2}=\) 500
    2. \(\sqrt{\dfrac{9\times50}{2}}=\sqrt{225}=\) 15
    3. Accurate value \(\dfrac{4.8\times19.7}{0.21}=450.29\). Error \(=\dfrac{500-450.29}{450.29}\times100=\) 11.0%.
    4. It is an overestimate: 4.8 rounds up to 5, 19.7 up to 20, and 0.21 down to 0.2 in the denominator, so every rounding makes the answer bigger.

19. Answer to Question 19

    1. \(\dfrac{1.99\times10^{30}}{5.97\times10^{24}}=0.3333\times10^6=\) \(3.33\times10^5\). Divide the numbers and subtract the powers of 10, then rewrite 0.3333 as \(3.333\times10^{-1}\). The Sun is about \(3.33\times10^5\) times heavier than the Earth.
    2. \(\dfrac{1.9\times10^{27}}{5.97\times10^{24}}=0.3183\times10^3=\) 318 times. Divide the masses in the same way. Jupiter is about 318 times heavier than the Earth.
    3. \(\dfrac{1.9\times10^{27}}{1.99\times10^{30}}=0.955\times10^{-3}=\) \(9.5\times10^{-4}\) (about 0.1%). Divide Jupiter's mass by the Sun's mass. Jupiter has about \(9.5\times10^{-4}\) of the Sun's mass.
    4. \(1.9\times10^{27}+5.97\times10^{24}=1.9\times10^{27}+0.00597\times10^{27}=1.90597\times10^{27}\approx\) \(1.91\times10^{27}\) kg. To add, write both numbers with the same power of 10 first. The combined mass is about \(1.91\times10^{27}\) kg.

20. Answer to Question 20

    1. \(2^{100}=(2^{10})^{10}=1024^{10}\), which is bigger than \(1000^{10}=10^{30}\). So \(2^{100}\) is bigger (\(1.27\times10^{30}\)).
    2. \(2^{100}=1.27\times10^{30}\) has 31 digits.
    3. The pattern repeats every 4. \(2025\div4=506\) remainder 1, so the last digit is the 1st in the cycle: 7.
    4. \(1024\approx1000\) (2.4% bigger). \(2^{40}=(2^{10})^4\approx(10^3)^4=10^{12}\). The exact value is \(1.0995\times10^{12}\).