Example 4

**(i)** Tunde is four times as old as his son. In four years’ time, he will be three times as old. How old are they now?

**Solution:**

⇒ Let the son’s age = x

⇒ Thus, Tunde’s age = 4x

⇒ In four year’s time Son’s age = (x + 4)

⇒ In four year’s time the son will be three times as old = 3(x + 4)

⇒ In four year’s time Tunde’s age = 4x + 4

⇒ Therefore, (4x + 4) = 3(x + 4)

**Expand the brackets**

⇒ 4x + 4 = 3x + 12

**Collect like terms**

⇒ 4x – 3x = 12 – 4

⇒ x = 8

∴ Son’s age = 8years and Tunde’s age = 4 × 8 = 32 years

**(ii).** The sum of two numbers is 40. When \( \scriptsize 3\frac{1}{4} \) times the larger number is subtracted from \( \scriptsize 5 \frac{1}{2} \) times the smaller, the difference is -25. Find the two numbers.

**Solution:**

Let the two numbers be x and y

**∴** x + y = 40……………..(i)

Let “x” be the larger number and “y” be the smaller number

⇒ \( \frac{11y}{2} \: – \: \frac{13x}{4} = \: \scriptsize -25\)

Clear the fractions by **multiplying both sides by 4**

⇒ \( \scriptsize 4 \: \times \: \normalsize \frac{11y}{2} \: – \: \scriptsize 4 \: \times \: \normalsize \frac{13x}{4} \scriptsize = 4 \: \times \: – 25 \)

22y – 13x = -100……………..(ii)

Recall x + y = 40

∴ x = 40 – y ……(iii)

**Substitute for x in equation (ii)**

⇒ 22y – 13(40 – y) = -100

⇒ 22y – 520 + 13y = -100

**Collect like terms**

⇒ 22y + 13y = 520 – 100

⇒ 35y = 420

**Divide both sides by 35**

⇒ \( \frac {35y}{35} = \frac {420}{35} \)

∴ y = 12

**Substitute for y in equation (iii)**

⇒ x = 40 – 12

⇒ x = 28

Therefore, the two numbers are 12 and 28.

**(iii)** The distance between two villages P and Q is xkm. A train running between P and Q arrives 3 minutes behind the scheduled time when it travels at an average speed of 45km/h, but arrives 9 minutes ahead of the scheduled time when it travels at an average speed of 60km/h, calculate:

**a.** The scheduled time for the journey

**b.** The distance x (WAEC)

**Solution:**

**a.** Recall D = Speed x Time

**1 ^{st} movement:**

Time = (t + 3)

Speed = 45km/h,

Let xkm distance = 45t + 3 …………(i)

**2 ^{nd} movement:**

Time = (t – 9)

Speed = 60km/h

Let xkm distance = 60t – 9 ………..(ii)

**Equating equations (i) and (ii), we have **

45(t + 3) = 60(t – 9)

**Divide both sides by 15**

⇒ 3(t + 3) = 4(t- 9)

**Expand the brackets**

⇒ 3t + 9 = 4t – 36

**Collect like terms**

⇒ 4t – 3t = 36 + 9

t = 45 minutes

⇒ \( \frac {45}{60} \scriptsize hr = \normalsize \frac {3}{4}\scriptsize hr \)

**b.** Recall: Distance = Speed x Time

**Using the 1 ^{st} movement**

Speed = 45km/h

Time = t + 3 = 45 + 3

= 48 minutes

= \( \frac {48}{60} \scriptsize hr = \normalsize \frac {4}{5} \scriptsize hr \)

= \( \scriptsize 45 \: \times \: \normalsize \frac{4}{5} \)

= 9 x 4 = 36km

Distance = 36*km*

**Hint: **To solve a word problem leading to an equation follow these steps:

- Decide the unknown quantity and represent it by a letter such as x
- State clearly the units used when necessary and convert to the same unit.
- Form an equation to represent the facts provided by the problems
- Solve the equation using a suitable method and use the solution to answer the questions in words.
- Check the solution (optional but useful)

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