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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