Lesson Progress
0% Complete

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

1st movement:

Time = (t + 3)

Speed = 45km/h,

Let xkm distance = 45t + 3      …………(i)

2nd 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 1st 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 = 36km

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

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

Responses

Your email address will not be published. Required fields are marked *

error: