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

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1. Answer to Question 1

   1. \(2\times2\times2\times2\times2=\) 32
   2. \(3\times3\times3\times3=\) 81
   3. \(13^2=169\), so 13
   4. \(4^3=64\), so 4
   5. 1 000 000

   \(8^2=64\) and \(9^2=81\), so \(\sqrt{75}\) is between 8 and 9. As 75 is closer to 81 than to 64, an estimate is about 8.7 (calculator: 8.66).

   Knowing the squares up to \(15^{2}\) and the cubes up to \(5^{3}\) by heart makes roots like these instant.

2. Answer to Question 2

   1. Multiplying powers of the same base: add the indices. \(2^{3+4}=2^{7}=\)128.
   2. Dividing: subtract the indices. \(5^{9-6}=5^{3}=\)125.
   3. A power of a power: multiply the indices. \(3^{2\times3}=3^{6}=\)729.
   4. Top first: \(7^{5+2}=7^{7}\); then divide: \(7^{7-4}=7^{3}=\)343.
   5. Any non-zero number to the power 0 is 1, because \(4^{2}\div4^{2}=4^{0}\) and also equals 1.

3. Answer to Question 3

   1. A negative index means the reciprocal: \(2^{-3}=\dfrac{1}{2^{3}}=\)\(\dfrac{1}{8}\).
   2. \(10^{-2}=\dfrac{1}{10^{2}}=\dfrac{1}{100}=\)0.01.
   3. \(5^{-1}=\)\(\dfrac{1}{5}\), the reciprocal of 5.
   4. The reciprocal of \(\dfrac{1}{2}\) is 2, so \(\left(\dfrac{1}{2}\right)^{-2}=2^{2}=\)4.
   5. Add the indices: \(3^{-2+4}=3^{2}=\)9.

   A negative index never makes the answer negative: it tells you to take the reciprocal of the positive power.

4. Answer to Question 4

   Divide repeatedly by primes (2, 3, 5, …) or draw a factor tree, then collect equal primes into powers.

   1. \(60=2^2\times3\times5\)
   2. \(72=2^3\times3^2\)
   3. \(180=2^2\times3^2\times5\)
   4. HCF: take the lowest powers of common primes: \(2^2\times3=\) 12
   5. LCM: take the highest powers: \(2^3\times3^2\times5=\) 360

   Check: \(12\times360=4320=60\times72\); the product of the HCF and LCM equals the product of the two numbers.

5. Answer to Question 5

   Standard form is \(A\times10^{n}\) with \(1\le A<10\). Count how many places the decimal point moves: to the left for large numbers (positive n), to the right for small numbers (negative n).

   1. \(4.5\times10^{4}\)
   2. \(7.2\times10^{6}\)
   3. \(6\times10^{-4}\)
   4. \(3.45\times10^{-5}\)
   5. \(5\times10^{2}\)

   \(0.8\times10^5\) and \(12\times10^3\) are not in standard form because the first number must be at least 1 and less than 10. The correct forms are \(8\times10^4\) and \(1.2\times10^4\).

6. Answer to Question 6

   Multiply or divide the front numbers, use the index laws for the powers of 10, then adjust so that the front number is between 1 and 10.

   1. \(3\times2=6\) and \(10^{4+3}=10^7\): \(6\times10^{7}\)
   2. \(8\div4=2\) and \(10^{7-2}=10^5\): \(2\times10^{5}\)
   3. \(6\times5=30\), so \(30\times10^9=\) \(3\times10^{10}\) (adjust to get a number between 1 and 10)
   4. \(4500+600=5100=\) \(5.1\times10^{3}\)

7. Answer to Question 7

   1. UK: \(6.7\times10^{7}\). London: \(9\times10^{6}\).
   2. \(6.7\times10^7-9\times10^6=6.7\times10^7-0.9\times10^7=\) \(5.8\times10^{7}\)
   3. \(\dfrac{9\times10^6}{6.7\times10^7}=\) 0.13
   4. \(\dfrac{6.7\times10^7}{2.4\times10^5}=279.2\), so about 279 people per km²

   To subtract in standard form, first write both numbers with the same power of 10, as with \(9\times10^{6}=0.9\times10^{7}\).

8. Answer to Question 8

   1. Sun: \(1.5\times10^{8}\) km. Moon: \(3.84\times10^{5}\) km.
   2. \(\dfrac{1.5\times10^8}{3.84\times10^5}=390.6\), so about 391 times
   3. \(\dfrac{1.5\times10^8}{3\times10^5}=5\times10^2=500\) seconds, which is \(500\div60=\) 8.3 minutes
   4. \(\dfrac{3.84\times10^5}{3\times10^5}=\) 1.28 seconds, about 1.3 s

   Dividing numbers in standard form is quick: divide the front numbers and subtract the powers of 10.

9. Answer to Question 9

   1. 10:00: \(2^3=\) 8. 10:20: \(2^4=\) 16.
   2. From 9:00 to 12:00 is 180 minutes, which is \(9\) doublings. \(2^9=\) 512.
   3. \(2^{20}\) needs 20 doublings, which is \(20\times20=400\) minutes \(=\) 6 hours 40 minutes.
   4. \(512=\) \(5.12\times10^{2}\)

   Each 20 minutes multiplies the number by 2, so after n doublings there are \(2^{n}\) bacteria; three doublings happen every hour.

10. Answer to Question 10

    1. 1 KB is \(2^{10}=\)1 024 bytes, not exactly 1 000.
    2. 1 MB \(=2^{10}\) KB \(=2^{10}\times2^{10}=2^{20}=\)1 048 576 bytes.
    3. 64 \(=2^{6}\), and each of GB, MB, KB brings another \(2^{10}\): \(2^{6+10+10+10}=\)\(2^{36}\) bytes.
    4. Divide by subtracting indices: \(2^{36}\div2^{22}=2^{14}=\)16 384 photos.

11. Answer to Question 11

    1. 3 folds: \(2^3=\) 8 layers, thickness 0.8 mm.
    2. \(2^{10}=\) 1024 layers
    3. \(1024\times0.1=\) 102.4 mm = 10.24 cm
    4. 1.8 m = 1800 mm, and \(1800\div0.1=18000\) layers. \(2^{14}=16384\) is too small and \(2^{15}=32768\) is enough, so 15 folds.

    The thickness after n folds is \(0.1\times2^{n}\) mm, so doubling quickly makes a sheet of paper enormously thick.

12. Answer to Question 12

    1. \(4.25\times9.5\times10^{12}=40.375\times10^{12}=\) \(4.04\times10^{13}\) km
    2. \(\dfrac{4.0375\times10^{13}}{5\times10^4}=\) \(8.075\times10^{8}\) hours
    3. \(8.075\times10^8\div(365\times24)=\) \(92180\) years, about 92 000 years
    4. The numbers are very large, so writing all the zeros would be slow and error-prone. Standard form shows the size of the number at a glance.

    Dividing by \(365\times24=8760\) hours in a year changes hours into years.

13. Answer to Question 13

    1. Move the decimal point 5 places to the right to get 7: \(7\times10^{-5}\) m.
    2. The more negative the power, the smaller the number: virus \(1\times10^{-7}\), bacterium \(2\times10^{-6}\), hair \(7\times10^{-5}\). Order: virus, bacterium, hair.
    3. \(\dfrac{7\times10^{-5}}{1\times10^{-7}}=7\times10^{2}=\)700 times.
    4. \(\dfrac{7\times10^{-5}}{2\times10^{-6}}=3.5\times10^{1}=\)35 bacteria.

    Dividing in standard form: divide the front numbers and subtract the powers of 10.

14. Answer to Question 14

    1. The first square has \(2^0\), so the 6th has \(2^5=\) 32
    2. \(2^{63}\)
    3. \(2^{63}=9.22\times10^{18}\), so \(9.22\times10^{18}\)
    4. \(9.22\times10^{18}\div50=1.844\times10^{17}\) g \(=\) \(1.84\times10^{11}\) tonnes

    Square n holds \(2^{n-1}\) grains because the first square starts at \(2^{0}=1\); this is why square 64 holds \(2^{63}\).

15. Answer to Question 15

    1. UK: \(6.7\times10^7\div2.4\times10^5\approx\) 279. Iceland: \(3.8\times10^5\div10^5=\) 4. Singapore: \(5.9\times10^6\div720\approx\) 8194. Canada: \(3.9\times10^7\div10^7=\) 4.
    2. Singapore is the most crowded. Iceland and Canada are the least crowded (about 4 per km²).
    3. \(8194\div279\approx\) 29 times
    4. \(6.7\times10^7\times1.012=6.7804\times10^7\), so \(6.78\times10^{7}\) (to 3 s.f.).

    Population density is population divided by area; the powers of 10 show quickly that Singapore’s small area makes it far more crowded.

16. Answer to Question 16

    1. \(12=2^2\times3\) and \(18=2\times3^2\)
    2. LCM \(=2^2\times3^2=\) 36
    3. 9:36 am
    4. Every 36 minutes. 3 hours = 180 minutes, and \(180\div36=5\), so they ring together at 9:00, 9:36, 10:12, 10:48, 11:24 and 12:00: 6 times including both ends (5 gaps).
    5. HCF \(=2\times3=\) 6. The largest number of minutes that divides both 12 and 18 exactly, so the bells' timings can both be written as multiples of 6 minutes.

17. Answer to Question 17

    1. \(2^{6+3-4}=2^5=\) 32
    2. Ali is correct. Bea forgot to subtract 4. Cal changed the base to 1. Dev multiplied the base by the index law (turned 2 and 2 into 4).
    3. \(2^9=512\) and \(2^4=16\), so \(512\div16=32\). It matches.
    4. \(5^{8-3-2}=5^3=\) 125

    The index laws only work when the bases are the same, and the base never changes: only the indices are added or subtracted.

18. Answer to Question 18

    1. \((2.5\times4)\times10^{-3+6}=10\times10^{3}=\)\(1\times10^{4}\) kg.
    2. Divide by 1 000: \(10^{4}\div10^{3}=\)10 tonnes.
    3. \(\dfrac{1\times10^{4}}{8\times10^{3}}=1.25\). One lorry is not enough, so round up to 2 lorry loads.
    4. \(5\times1\times10^{4}=\)\(5\times10^{4}\) kg, which is 50 tonnes.

    Multiplying in standard form: multiply the front numbers and add the powers of 10, then adjust the front number to lie between 1 and 10.

19. Answer to Question 19

    1. Squares: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Cubes: 1, 8, 27, 64.
    2. A number that is both is a sixth power: \(1^6=1\) and \(2^6=64\). So 1 and 64.
    3. 10 squares have tokens and 4 have bonus tokens (with 1 and 64 in both lists).
    4. \(\sqrt{64}=8\), so she ends on square \(64+8=\) 72.

20. Answer to Question 20

    1. \(365\times24\times60\times60=31\,536\,000\), which is \(3.15\times10^{7}\) s to 3 s.f.
    2. \(80\times3.1536\times10^{7}=2.52\times10^{9}\), so \(2.5\times10^{9}\) s to 2 s.f.
    3. Minutes in 80 years: \(60\times24\times365\times80=4.2\times10^{7}\); times 70 beats gives \(2.94\times10^{9}\), so \(2.9\times10^{9}\) beats.
    4. \(\dfrac{2.52\times10^{9}}{3}=\)\(8.4\times10^{8}\) s asleep, about 27 years.

    Rounding each answer to 2 significant figures is sensible because the 80 years and the 70 beats per minute are only estimates.