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Question 1

(a) Evaluate \( \left( \scriptsize 3\frac{1}{4} \: + \: 1 \frac{1}{2} \right) \: \div \: \normalsize \frac{7}{12} \)

Solution

\( \left(  \frac{13}{4} \: + \: \frac{3}{2} \right) \: \div \:  \frac{7}{12} \)

= \( \left(  \frac{15 \: + \: 6}{4} \right) \: \div \:  \frac{7}{12} \)

= \( \left(  \frac{21}{4} \right) \: \div \:  \frac{7}{12} \)

= \(   \frac{21}{4}  \: \times \:  \frac{12}{7} \)

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

= 9

 

(b) Calculate the perimeter of the triangle below

Solution

The triangle above is an equilateral triangle since all its sides are equal

∴ (5x + 9) = (3x + 15)

collect like terms

5x – 3x = 15 – 9

2x = 6

x = \( \frac{6}{2} \)

x = 3

or

(5x + 9) = (7x + 3)

collect like terms

5x – 7x = 3 – 9

-2x = -6

x = \( \frac{-6}{-2} \)

x = 3

or

(7x + 3) = (3x + 15)

collect like terms

7x – 3x = 15 – 3

4x = 12

x = \( \frac{12}{4} \)

x = 3

Hence if we substitute the value of x into each side we get

5x + 9 =

5 x 3 + 9

= 24cm

or

7x + 3

= 7 x 3 + 3

= 21 + 3 

= 24cm

0r

3x + 15

= 3 x 3 + 15

= 9 + 15

= 24cm

∴ The perimeter of the traingle = 24cm + 24cm + 24cm = 72cm

Question 2

(a) The sum of the interior angles of a regular polygon is 1980º. Find the number of sides of the polygon

Solution

Sum of the interior angles of a polygon

= (n – 2) x 180º = 1980º

Divide both sides by 180º

n – 2 = \( \frac{1980^o}{180^o} \)

n – 2 = 11

n = 11 + 2

n = 13

The regular polygon has 13 sides

 

(b) The table shows the distribution of marks of 7 students in two separate subjects.

Maths 12 8 10 14 4 13 2
Eng 10 15 11 12 9 6 14

 

Solution

(i) Find the mean marks of each of the subjects

Mean mark for Mathematics (x)

= \( \frac{12+8+10+14+4+13+2}{7} \)

= \( \frac{63}{7} \)

= 9

 

(ii) What is the range of marks in Mathematics

Range of marks in Mathematics = 14 – 2

= 12

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