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JSS2: MATHEMATICS - 1ST TERM

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  1. Properties of Whole Numbers I | Week 1
    4 Topics
    |
    1 Quiz
  2. Properties of Whole Numbers II | Week 2
    4 Topics
    |
    1 Quiz
  3. Properties of Whole Numbers III | Week 3
    5 Topics
    |
    1 Quiz
  4. Indices | Week 4
    2 Topics
    |
    1 Quiz
  5. Laws of Indices | Week 5
    5 Topics
    |
    1 Quiz
  6. Whole Numbers & Decimal Numbers | Week 6
    4 Topics
    |
    1 Quiz
  7. Standard Form | Week 7
    3 Topics
    |
    1 Quiz
  8. Significant Figures (S.F) | Week 8
    4 Topics
    |
    1 Quiz
  9. Fractions, Ratios, Proportions & Percentages I | Week 9
    6 Topics
    |
    1 Quiz
  10. Fractions, Ratios, Proportions & Percentages II | Week 10
    4 Topics
    |
    1 Quiz
  11. Fractions, Ratios, Proportions & Percentages III | Week 11
    3 Topics
    |
    1 Quiz
  12. Approximation & Estimation | Week 12
    1 Topic
    |
    1 Quiz
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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

View Answers

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.

View Answers

Evaluation Questions (1 & 2)

Answers:

1. ₦2. million,  ₦1.9 million

 

2.
a. 90
b. 219
c. 0.74
d. 2380
e. 0.098

Evaluation Questions (3 & 4)

Answers:

3
a. 100
b. 60
c. 540
d. 4000
e. 9

4. 100  000