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Unit A5 Section 2
Calculations with Standard Form

When using standard form it is possible to multiply and divide numbers, taking advantage of the form in which they are written.

Note that we can use the rules

10a × 10b = 10a + b
10a ÷ 10b = 10a − b

where a and b are numbers.

Worked Examples

1

Find 4 × 1018 × 3 × 104

To do this calculation, you multiply together the 4 and the 3 and then multiply together the 1018 and the 104 .

4 × 1018 × 3 × 104 = 4 × 3 × (1018 × 104)
= 12 × 1022

This result is not in standard form so the final stage is to convert the result to standard form.

12 × 1022 = 1.2 × 10 × 1022
= 1.2 × 1023

4 × 1018 × 3 × 104 = 1.2 × 1023
2

Find

(a)

3.2 × 104 × 5 × 10−3

Multiply together the 3.2 and the 5 and then multiply together the 104 and the 10−3.

3.2 × 104 × 5 × 10−3 = 3.2 × 5 × 104 × 10−3
= 16.0 × 101

This number is not in standard form so converting gives

16.0 × 101 = 1.6 × 10 × 101
= 1.6 × 102 .
(b)

(6 × 108) ÷ (3 × 104)

Division follows a similar approach to multiplication. First divide 6 by 3 and then divide 108 by 104 .

(6 × 108) ÷ (3 × 104) = (6 ÷ 3) × (108 ÷ 104)
= 2 × 104

This result is in standard form so no further work is required.
(c)

(7.2 × 103) ÷ (6 × 104)

First divide 7.2 by 6 and then divide 103 by 104 .

(7.2 × 103) ÷ (6 × 104) = (7.2 ÷ 6) × (103 ÷ 104)
= 1.2 × 10−1

This result is in standard form.

Challenge!

In astronomy, the distance between stars is measured in light years, which is the distance travelled by light in a year.

One light year = 3 × 105 × 60 × 60 × 24 × 365 km.

This is approximately 9 460 800 000 000 km.

How long would it take for light to travel from the Sun to the Earth if their distance apart is 1.5 × 108 km?

Exercises

Do the following calculations, making sure that your answer is in standard form.
Do not use a calculator.

(a)
3 × 108 × 2 × 104 = × 10
(b)
2 × 105 × 4 × 103 = × 10
(c)
9 × 106 × 1 × 1010 = × 10
(d)
5 × 103 × 4 × 108 = × 10
(e)
6 × 103 × 4 × 1011 = × 10
(f)
3 × 10−2 × 4 × 108 = × 10
(g)
1.2 × 106 × 2.4 × 105 = × 10
(h)
1.1 × 106 × 2 × 10−4 = × 10
(i)
8.1 × 108 × 7.2 × 10−2 = × 10
(j)
5.2 × 103 × 1.3 × 10−7 = × 10
(k)
6.2 × 10−3 × 2.1 × 10−6 = × 10
(l)
1.8 × 10−4 × 2.5 × 10−9 = × 10

Give the answers to the following calculations in standard form.
Do not use a calculator.

(a)
(8 × 106) ÷ (2 × 102) = × 10
(b)
(9 × 105) ÷ (3 × 102) = × 10
(c)
(8 × 104) ÷ (4 × 102) = × 10
(d)
(1.6 × 105) ÷ (2 × 102) = × 10
(e)
(3.6 × 108) ÷ (3 × 102) = × 10
(f)
(4.8 × 1012) ÷ (4 × 103) = × 10
(g)
(8.1 × 104) ÷ (3 × 105) = × 10
(h)
(4.5 × 103) ÷ (9 × 10−5) = × 10
(i)
(1.64 × 108) ÷ (4 × 10−12) = × 10
(j)
(1.32 × 105) ÷ (1.2 × 10−3) = × 10
(k)
(9.6 × 104) ÷ (3.2 × 10−5) = × 10
(l)
(1.21 × 10−4) ÷ (1.1 × 106) = × 10

Do the following using a calculator, giving your answers in standard form.

(a)
(4.2 × 106)2 × 10
(b)
(3.7 × 10−2)2 × 10
(c)
(1.2 × 10−5)3 × 10
(d)
× 10 (to 4 s.f.)
(e)
6.2 × 108 × 1.2 × 1014 × 10
(f)
3.8 × 104 × 4.1 × 10−12 × 10
(g)
(1.84 × 106) ÷ (1.92 × 107) × 10 (to 5 s.f.)
(h)
× 10
(i)
× 10
(j)
× 10 (to 4 s.f.)
(k)
4.8 × 1011 + 3.2 × 1010 × 10
(l)
6.8 × 1012 − 4.7 × 1010 × 10

There are 8.64 × 104 seconds in one day. How many seconds are there in:

(a)
10 days × 10 seconds
(b)
1 week × 10 seconds
(c)
1 year? × 10 seconds

The mass of an electron is 9.1 × 10−31 kg. Find the mass of:

(a)
3 × 1018 electrons × 10 kg
(b)
4 × 1032 electrons × 10 kg
(c)
7 × 108 electrons. × 10 kg

The mean distance of the earth from the sun is 149.6 million kilometres.

(a)

Write the number 149.6 million in standard index form.

× 10

The earth travels a distance, D km, in one day. The value of D is given by the formula

D =

(b)

Calculate the value of D, giving your answer in standard index form.

× 10 km (to 5 s.f.)

The number 10100 is called a googol.

(a)

Write the number, 50 googols, in standard index form.

× 10
A nanometre is 10−9 metres.
(b)

Write 50 nanometres in metres.
Give your answer in standard index form.

× 10 m

Investigation

Han Sin, a Chinese general, devised a method to count the number of soldiers that he had. First, he ordered his soldiers to form groups of 3, followed by groups of 5 and then groups of 7. In each case he noted down the remainder. Using the three remainders, he was able to calculate the exact number of soldiers he had without doing the actual counting.
Do you know how he did it?