Explanation

(a) \(3(x + y) + 45 = 180°\) ( sum of angles on a straight line)

\(3x + 3y + 45 = 180 \implies 3x + 3y = 135\)

\(3x = 135 - 3y \implies x = 45 - y\)

Also,

\(5x + y + y = 180\)

\(5x + 2y = 180..... (*)\)

Put x = 45 - y into (*) above,

\(5(45 - y) + 2y = 180 \implies 225 - 5y + 2y = 180\)

\(225 - 3y = 180 \implies 3y = 45\)

\(y = \frac{45}{3} = 15°\)

\(x = 45 - y\)

\(x = 45 - 15 = 30°\)

(b)

Let the distance BD = x and CD = y.

In \(\Delta ABD\),

\(\tan 53 = \frac{33}{x}\)

\(x = \frac{33}{\tan 53}\)

= \( 24.868 cm\)

In \(\Delta ACD\),

\(\tan 67 = \frac{33}{y}\)

\(y = \frac{33}{\tan 67}\)

\(y = 14.01 cm\)

\(\therefore BC = x + y \)

\(24.868 + 14.01 = 38.878 cm\)

\(\approxeq 38.9 cm\)

Correct Answer: D. 5 significant figures

Explanation

The two trailing zeros in the number are not significant, but the other five are, making it a five-figure number.

Correct Answer: D. 8cm

Explanation

In the diagram, \(\frac{16}{12} = \frac{VX}{6}\) (similar \(\Delta\)s)

VX = \(\frac{16 \times 6}{12}\)

= 8cm

Correct Answer: A. 25

Correct Answer: C. 25\(^o\)

Explanation

L = \(\frac{\theta}{360^o}\) = 2\(\pi\)

\(\frac{55}{36} = \frac{\theta}{360^o} \times 2 \times \frac{22}{7} \times 3.5\)

\(\theta = \frac{360 \times 55 \times 44 \times 0.5}{3 \times 44 \times 0.5}\)

= \(\theta = 25^o\)

Correct Answer: D. 47

Explanation

(3\(\frac{1}{2} + 4\frac{1}{3}) \div (\frac{5}{6} - \frac{2}{3}\)) = (7\(\frac{3 + 2}{6}) \div \frac{(5 - 4)}{6}\)

7\(\frac{5}{6} \div \frac{1}{6} = \frac{47}{6} \times \frac{6}{1}\) = 47

Correct Answer: B. \(125 m\)

Correct Answer: C. -\(\frac{5}{12}\)

Explanation

P * q = \(\frac{q^2 - p^2}{2pq}\).

3 * 2 (where p = 3, q = 2)

i.e 3 *2 = \(\frac{3^2 - 2^2}{2 *3 * 2}\)

= \(\frac{4 - 9}{12}\)

= \(\frac{-5}{12}\)

Correct Answer: D. 9

Explanation

Average speed = \(\frac{Distance}{Time}\)

\(\frac{400km}{hr} = \frac{60km}{Time}\)

Time = \(\frac{60km}{400 km/hr}\)

= \(\frac{60hr}{400}\)

M = \(\frac{60hr}{400} \times \frac{60min}{1hr}\)

= 9 minutes

Explanation

(a) \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\)

Multiply through by the LCM of 4 and 8, to clear fractions :

\(8 \times 4 + 8 \times \frac{3}{4}(x + 2) \leq 8 \times \frac{3}{8}x + 8 \)

\(32 + 6(x + 2) \leq 3x + 8 \)

\(32 + 6x + 12 \leq 3x + 8 \)

\(44 + 6x \leq 3x + 8\)

\(6x - 3x \leq 8 - 44 \implies 3x \leq -36\)

\(x \leq - 12\).

(b)

Area \((A_{1})\), of rectangle PQRS = \(20(20 + x) cm^{2}\)

Area \((A_{2})\), of square = \(x \times x = x^{2} cm^{2}\)

\(\implies 20(20 + x) - x^{2} = 484\)

\(400 + 20x - x^{2} = 484\)

\(x^{2} - 20x + 484 - 400 = x^{2} - 20x + 84 = 0\)

\(x^{2} - 14x - 6x + 84 = 0\)

\(x( x - 14) - 6(x - 14) = 0 \)

\((x - 14)(x - 6) = 0\)

\(\therefore \text{x = 6 cm or 14 cm}\)