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JSS3: MATHEMATICS - 1ST TERM

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  1. Binary Number System I | Week 1
    5 Topics
    |
    1 Quiz
  2. Binary Number System II | Week 2
    6 Topics
    |
    1 Quiz
  3. Word Problems I | Week 3
    4 Topics
    |
    1 Quiz
  4. Word Problems with Fractions II | Week 4
    1 Topic
    |
    1 Quiz
  5. Factorisation I | Week 5
    4 Topics
    |
    1 Quiz
  6. Factorisation II | Week 6
    3 Topics
    |
    1 Quiz
  7. Factorisation III | Week 7
    3 Topics
    |
    1 Quiz
  8. Substitution & Change of Subject of Formulae | Week 8
    2 Topics
    |
    1 Quiz
  9. Simple Equations Involving Fractions | Week 9
    3 Topics
    |
    1 Quiz
  10. Word Problems | Week 10
    1 Topic
    |
    1 Quiz



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Lesson 10, Topic 1
In Progress

Word Problems Leading to Simple Linear Equations

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Example 

i. If three-quarters of a certain number is 10 more than one-third of the number what is the number.

Solution

Let the number be x

\( \frac{3}{4}\scriptsize x = 10 \: +\: \normalsize \frac{1}{3}\scriptsize x \)

Take L.C.M

\( \frac{3x}{4} \: + \: \frac{1x}{3} \)

Cross multiply

\( \scriptsize 3x \: \times \: 3 = 4(30 \: + \: x) \)

\( \scriptsize 9x = 120 \: + \: 4x \)

Take like terms

\( \scriptsize 9x \: – \:4x = 120 \)

\( \scriptsize 5x = 120 \)

Divide both sides by 5

\( \frac{ 5x}{5} = \frac{120}{5} \)

\( \frac{ \not{5}x}{\not{5}} = \frac{120}{5} \)

\( \scriptsize x = 24 \)

ii.   A man spends one-third of his monthly salary earning on projects and one-quarter of the remainder on household needs. If he has N15000 left how much does he receive in a month?

Solution

Let the income be x

\( \frac{1}{3} \scriptsize x \: on\: projects \)

The remainder = \( \scriptsize \: 1 \: – \: \normalsize \frac{1}{3} \)

\( \scriptsize \: The\: remainder \: = \normalsize \frac{2}{3} \)

\( \frac{1}{4} \scriptsize \: of \: the\; remainder \: on \: household \: needs\)

= \( \frac{1}{4} \scriptsize \: of \: \normalsize \frac{2}{3} \)

= \( \frac{2}{12} \)

= \( \frac{1}{6} \)

In total He spent = Amount spent on projects + Amount spent on household needs

In total He spent = \(\frac{1}{3} \: + \: \frac{1}{6} \)

Find the L.C.M

\(\frac{2 \: + \: 1}{6} \)

= \(\frac{3}{6} \)

= \(\frac{1}{2} \)

Therefore he has \(\scriptsize 1 \: – \: \frac{1}{2} \) left

Which is equals to \(\frac{1}{2} \)

This means he has half his income, x, left which is also N15,000

\( \frac{1}{2}x = \scriptsize 15000 \)

\( \scriptsize x = 2\: \times \: 15000 \)

\( \scriptsize x = 30000 \)

iii.   Think of a number, treble it, then divide by 4, the result is 12 more than the original number, find the number.

Solution

Let the number be x

treble it = 3x

then divide by 4 = \( \frac{3x}{4} \)

the result is 12 more than the number = 12 + x

Thereore,

\( \frac{3}{4} \scriptsize x = 12 \: + \: x \)

Cross  multiply

\( \scriptsize 3x = 4(12 \: + \: x) \)

\( \scriptsize 3x = 48 \: + \: 4x \)

Take like terms

\( \scriptsize 3x \: – \: 4x = 48 \)

\( \scriptsize \: – \: x = 48 \)

Divide both sides by  -1

\( \frac{-x}{-1} = \frac{48}{-1}\)

\( \scriptsize x = \: – 48 \)

iv. When 12  is subtracted from a number, and the answer is divided into 10,  the result obtained is one-quarter of the original number. Find the number.

Solution 

Let the number be x

\(\frac{x \: -\: 12}{10} = \frac{1}{4}\scriptsize x \)

Cross multiply

\( \scriptsize 4(x \: -\: 12) = 10 \: \times \: x \)

\( \scriptsize 4x \: -\: 48 = 10x \)

Bring like terms together

\( \scriptsize 4x \: -\: 10x = 48\)

\( \scriptsize \: -\: 6x = 48 \)

Divide both sides by -6

\( \frac{-6x}{-6} = \frac{48}{-6}\)

\( \scriptsize x = \: -8\)

v. A mother is four times as old as her daughter. If their age difference is three decades, find the ages of the mother and her daughter.

Solution

A decade is a period of 10 years.

Let x be the age of the daughter in years

Age of mother is therefore 4x (four times as old as her daughter)

Their age difference in years is 4x x, i.e 3x

3 decades = 3 x 10 = 30

Therefore, 3x = 30

Divide both sides by 3

\( \frac{3x}{3} = \frac{30}{3}\)

\( \frac{\not{3}x}{\not{3}} = \frac{30}{3}\)

x = 10 years

Age of mother = 4x

= 4 x 10

= 40 years

Age of daughter = 10 years

Work to do:

1. When 9 is added to four times a certain number, the result is 37. Find the number.

2. When 3 is subtracted from eight times a number the result is 53.  What is the number?

3. Three-quarter of a certain number is 18. Find the number.

4. The sum of two numbers is 13. If the two numbers differ by 3, find the smaller of the two numbers.  

5. When a certain number is added to 2/9 of itself the result is 94 greater than the original number. Find the original number.

Answers:

1. 7
2. 7
3. 24
4. 5
5. 32

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