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Question 1a

Solve \( \frac{3}{4}  = \frac{2}{3(x \: + \: 1) } \: – \: \frac{1}{4}\)

Solution

Multiply each term by the L.C.M =  12(x + 1)

\( \normalsize \frac{3}{4} \scriptsize\: \times \:  12(x+1)  = \normalsize\frac{2}{3(x \: + \: 1) } \scriptsize\: \times \:  12(x+1) \: – \: \normalsize\frac{1}{4} \scriptsize\: \times \:  12(x+1) \)

= \( \scriptsize 3 \: \times \: 3(x+1) = 2 \: \times \: 4\: – \: 3(x +1) \)

= \( \scriptsize 9(x+1) = 8 \: – \: 3(x +1) \)

Remove the brackets

= \( \scriptsize 9x \: + \: 9 = 8 \: – \: 3x \: – \: 3 \)

Collect like terms

9x + 3x = 8 – 3 – 9

12x = -4

Divide both sides by 12

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

x = \( \frac{-1}{3} \)

Question 1b

The linear scale factor of two similar shapes is 4:7. If the area of the bigger shape is 490 cm², what is the area of the small shape?

Solution

(Scale factor)2 = Area factor 

\(  \left(\frac{4}{7} \right)^2 = \frac{x}{490} \)

x  = area of smaller shape

\(  \frac{16}{49} = \frac{x}{490} \)

cross multiply

49x = 16 x 490

Divide both sides by 49

\( \frac{49x}{49} = \frac{16 \: \times \: 490}{49} \)

x = 16 x 10

x = 160cm²

Area of smaller shape = 160cm²

Question 2a

A motorcycle uses 15 litres of fuel for a journey of 75km.

(i) What distance will 40 litres cover by the same motorcycle

Solution

15 litres = 75km

1litre = \( \frac{75}{15} \)

= 5km

40 litres = (5 x 40)km

= 200km

(ii) How many litres will be needed for a distance of 350km

Solution

1 litre = 5km

x litre = 350km

Using ratio

\( \frac{1}{x} = \frac{5}{350}\)

cross multiply

5x = 350

Divide both sides by 5

\( \frac{5x}{5} = \frac{350}{5} \)

x = 70 litres

Question 2b

Construct a ΔPQR with

|QR| = 7cm

|PR| = 8cm

<PQR = 60°

Solution

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