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JSS3: MATHEMATICS - 1ST TERM

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

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