Word Problems Leading to Simple Linear Equations

Worked Example 9.2.1:

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

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

Cross  multiply

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

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

Collect 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 × 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 × 10

= 40 years

Age of daughter = 10 years

Work to do: