Topic Content:
- Meaning of Approximation
- Using Approximation to Estimate Answers
- Worked Examples
- End of Lesson Evaluation Questions
What are Approximations?
Approximations are not exact, but close enough value to the original value that can be used in a calculation.
For example;
1. If the bus ride takes 57 minutes, one can say it is “a one-hour bus ride”
2. If a cord measures 3.14 cm, one can say it is 3 cm (which is not the actual value).
The exact number may be greater than or less than the approximation.
Worked Example 12.1.1:
Round off 285.563 correct to
a. 1s.f
b. 2s.f
c. 1d.p
d. 2 d.p
Solution
a. 285.563 = 300 to 1s.f
b. 285.563 = 290 to 2s.f.
c. 285.563 = 285.6 to 1d.p
d. 285.563 = 285.56 to 2 d.p
Worked Example 12.1.2:
Calculate the following and round your answer to the given degree of accuracy.
a. 679.209 + 20.95 (2 d.p)
b. 7.63 × 0.6213 (3s.f)
c. 792.65 – 37.809 (2 d.p)
Solution
a. 679.209 + 20.95 (2 d.p) = 700. 159 = 700.16 to 2 d.p
b. 7.63 × 0.6213 (3s.f) = 7.63 × 0.6213 = 4.740519 = 4.74 to 3s.f
c. 792.65 – 37.809 (2 d.p) = 754.841 = 754.84 to 2 d.p
Worked Example 12.1.3:
As a newspaper editor, state two ways the following numbers will appear in your paper.
a. 59 279 fans attended a football match
b. 7907 new jobs created
Solution
a. 59 279 fans attended a football match can be written as 60 000 to 1s.f.
i.e. 60 thousand fans.
b. 7907 = 8000 new jobs to 1s.f
8 thousand
or
7900 jobs to 2s.f
Using Approximation to Estimate Answers:
Approximations can be used to estimate answers to calculations. This will give one a rough idea of what the real answer should be.
Worked Example 12.1.4:
Use reasonable approximation to estimate the cost of 402 exercise books at ₦1.85 each.
Solution
Cost of 402 exercise books = 402 × ₦1.85
402 – Round off to 1 s.f = 400
₦1.85 – Round off to 1 s.f = 2
Cost of 402 exercise books = 400 × 2 = ₦800
Worked Example 12.1.5:
Find the approximate values for:
a. \( \frac{407 \: \times \: 196}{35} \)
b. 607 × 68
c. \( \scriptsize 9 \frac{4}{5} \: + \: \scriptsize 8 \frac{1}{7}\)
d. \( \scriptsize 7 \frac{1}{6} \: + \: \scriptsize 2 \frac{3}{5}\)
e. 1218 × 12
Solution
a. \( \frac{407 \: \times \: 196}{35} \)
\( \frac{407 \rightarrow 1 s.f \: \times \: 196 \rightarrow 1 s.f}{35 \rightarrow 1 s.f} \)= \( \frac{400 \: \times \: 200}{40} \\ \scriptsize = 10 \: \times \: 200 \\ \scriptsize = 2000 \)
b. \(\scriptsize 607 \: \times \: 68 \)
\(\scriptsize 607 \rightarrow 1 s.f \: \times \: 68\rightarrow 1 s.f \\ \scriptsize = 600 \: \times \: 70 \\ \scriptsize = 42,000 \)c. \( \scriptsize 9 \frac{4}{5} \: + \: \scriptsize 8 \frac{1}{7}\)
Compare the denominator and numerators, if the fraction is greater than \( \frac{1}{2} \) or 0.5 we round up, if it’s less than \( \frac{1}{2} \) or 0.5 we round down.
Let’s look at the first fraction
\( \frac{4}{5} \)To compare with half we multiply the denominator and numerator by 2 and also multiply \( \frac{1}{2} \) by 5.
\( \frac{4 \; \times \; 2}{5 \; \times \; 2} = \frac{8}{10} \) \( \frac{1 \; \times \; 5}{2 \; \times \; 5} = \frac{5}{10} \) \( \frac{8}{10}\scriptsize \; is \; greater \; than \; \normalsize \frac{5}{10} \)Therefore, \( \frac{4}{5}\scriptsize \; is \; greater \; than \; \normalsize \frac{1}{2} \)
so we round up.
\( \scriptsize 9 \frac{4}{5} = 9 + 1 = 10 \)For the second fraction;
\( \frac{1}{7}\scriptsize\; is \; less \; than \; \normalsize \frac{1}{2} \)so we round down
\( \scriptsize 8 \frac{1}{7}\) = 8 + 0 = 8
\( \scriptsize \therefore 9 \frac{4}{5} \; + \; \scriptsize 8 \frac{1}{7} = 10 + 8 = 18\)d. \( \scriptsize 7 \frac{1}{6} \: + \: \scriptsize 2 \frac{3}{5}\)
Comparing the fractions.
\(\frac{1}{6} \rightarrow\) less than ½ so we round down
\(\frac{3}{5} \rightarrow\) more than ½ so we round-up
(7 + 0) – (2 + 1)
= 7 – 3
= 4
e. \(\scriptsize 1218 \: \times \: 12 \)
\(\scriptsize 1218 \rightarrow 1 s.f \: \times \: 12\rightarrow 1 s.f \\ = \scriptsize 1000 \: \times \: 10 \\ \scriptsize = 10000 \)Or
\(\scriptsize 1218 \rightarrow 2 s.f \: \times \: 12\rightarrow 2 s.f \\ = \scriptsize 1200 \: \times \: 12 \\ \scriptsize = 14,400 \)Evaluation Questions:
1. How do you think ₦1 885 096 will be written in your newspaper as a newspaper editor?
2. Calculate the following and give your answer correct to the degree of accuracy given.
a. 7.606 + 81.35 + 098 to 1 s.f
b. 406 × 0.54 to 3 s.f
c. 0.046 × 16 correct to 2 s.f
d. 297 × 8 to 3 s.f
e. 14.05 × 0.007 correct to 3 s.f
3. Solve first and then Round off the answer to 1s.f.
a. 82 + 19
b. 5.09 + 57.8
c. \( \frac{92 \: \times \: 25 \: \times \: 81}{48 \: \times \: 8} \)
d. \( \frac{1905 \: \times \: 58 }{33} \)
e. \( \scriptsize 5 \frac{3}{8} \: \times \: 1 \frac{3}{4}\)
4. The cost of shipping 82 vehicles was ₦43 674, estimate the cost of shipping 198 vehicles and then calculate the actual cost.
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