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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 Progress
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Example 1

What must be added to the positive difference between \( \scriptsize 4 \frac{3}{10} \: and \: 7 \frac{2}{5}\) to obtain 25?

Solution

Let the number be y

\( \scriptsize y \: + \: \left ( 7 \frac{2}{5}\: – \; 4 \frac{3}{10} \right) \scriptsize = 25 \)

\( \scriptsize y + \left (\normalsize \frac{37}{5}\: – \: \frac{43}{10} \right) \scriptsize = 25 \)

\( \scriptsize y \: + \: \normalsize \frac{\left(74 \: – \: 43 \right) }{10} \scriptsize = 25 \)

\( \scriptsize y \: + \: \normalsize \frac{31}{10} \scriptsize = 25 \)

Multiply both sides by 10

\( \scriptsize 10 \: \times \: y \:+ \:10 \: \times \: \normalsize \frac{31}{10} \scriptsize = 25 \: \times \: 10 \)

\(\scriptsize = 10y \: + \: 31 = 250 \)

\(\scriptsize 10y = 250\: – \: 31 \)

\(\scriptsize 10y = 219 \)

Divide both sides by 10

\( \frac{10y}{10} = \frac{219}{10} \)

\( \scriptsize y = 21.9 \: or \: 21 \frac{9}{10} \)

Example 2

The sum of three numbers is 30. The middle number is 5 less than the largest number, while the smallest number is one-third of the largest number. Find the three numbers.

Solution 

Let the numbers be x, y, z

 sum = 30

Let the largest number be z

y = \( \scriptsize z \: – \: 5 \)

x = \( \frac{1}{3} \scriptsize z \)

Therefore:

\( \scriptsize x \: + \: y\: + \: z = 30 \) ………………. 1

\( \frac{1}{3} \scriptsize z \: + \: z \: – \: 5 \: + \: z = 30\)………………. 2

Add 5 to both sides in equation 2

\( \frac{1}{3} \scriptsize z +z \; – \; 5 + 5 + z = 30 + 5\)

\( \frac{1}{3} + \frac{2z}{1}\scriptsize = 35\)

Find L.C.M

\( \frac{z + 6z}{3}\scriptsize = 35\)

\( \frac{7z}{3}\scriptsize = 35\)

\( \scriptsize 7z = 35 \: \times \: 3\)

z = \( \frac{35 \: \times \: 3}{7} \)

z = \( \scriptsize 5 \: \times \: 3 \)

z = 15

y = \( \scriptsize z \: – \: 5 \)

Substitute value of z into the equation

y = \( \scriptsize 15 \: – \: 5 = 10\)

x = \( \frac{1}{3} \scriptsize z \)

Substitute value of z into the equation

=\( \frac{1}{3} \scriptsize \: \times \: 15 = 5\)

The numbers are 5, 10, 15.

Example 3

A man is four times as old as his son. Five years ago the man was seven times as old as the son.

a. How old are they now?

b. How old were they five years ago? If the sum of their ages is 50.

Solution

a.

Let the man’s age be x.

Let the son’s age by y.

Therefore x + y =  50 ……………….. (1)

Man is four times as old as his son:

x  =  4y  ……………….. (2)

Five years ago the man was seven times as old as the son:

x = 5 – 7y……………….. (3)

Substitute eq (2)  into eq (1)

4y + y =  50

5y = 50

Divide both sides by 5

\( \frac{5y}{5} =\frac{50}{5} \)

y = 10

y = 10 years

Substituting y into equation 2  

= 4y = 4 x 10 =  40 years 

b. x –  5 =  7y

40 – 5  = 7y

35 = 7y

7y = 35

Divide both sides by 7

\( \frac{7y}{7} =\frac{35}{7} \)

y = 5

y = 5 years

x  =  40 – 5   =  35yrs 

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