Mathematics (Core)

Paper 1 | Objectives | 50 Questions

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Year: 2004

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#	Question	Ans
1.	1. Evaluate $202_{t h r e e}^{2} - 112_{t h r e e}^{2}$ A. 21120 B. 21121 C. 21112 D. 21011 Show Content Detailed Solution $202_{t h r e e}^{2}$ when converted to base ten $= (202_{3})^{2} 202_{3} = 2 \times 3^{2} + 0 \times 3^{1} + 2 \times 3^{0} = 18 + 0 + 2 = 20_{t e n}; (202_{3})^{2} = (20)_{t e n}^{2} = 400 112_{t h r e e}^{2}$ when converted to base ten $= (112_{3})^{2} 112_{3} = 1 \times 3^{2} + 1 \times 3^{1} + 2 \times 3^{0} = 9 + 3 + 2 = 14_{t e n} (112_{3})^{2} = (14)_{t e n}^{2} = 196_{t e n} E v a l u a t e ⟹ 400 - 196 = 204$ Reconvert to base three $\begin{matrix} 3 & 204 \\ 3 & 69 & R 0 \\ 3 & 22 & R 2 \\ 3 & 7 & R 1 \\ 2 & 2 & R 1 \\ 0 & R 2 ↑ \end{matrix} = 21120_{3}$	A
2.	If $y = 23_{f i v e} + 101_{t h r e e}$ , find y, leaving your answer in base two A. 1110 B. 10111 C. 11101 D. 111100 Show Content Detailed Solution $23_{f i v e} = X_{t e n}; X_{t e n} = 2 \times 5^{1} + 3 \times 5^{0} = 10 + 3 = 13 101_{t h r e e} = P_{t e n}; P_{t e n} = 1 \times 3^{2} + 0 \times 3^{1} + 1 \times 3^{0} = 9 + 0 + 1 = 10_{t e n} Y = 13 + 10 = 23_{t e n}$ ; Converting to base two $\begin{matrix} 2 & 23 \\ 2 & 11 & R 1 \\ 2 & 5 & R 1 \\ 2 & 2 & R 1 \\ 2 & 1 & R 0 \\ 0 & R 1 ↑ \end{matrix} = y = 10111_{2}$	B
3.	Given that sin (5x - 28)^o = cos (3x - 50)^o,0 < x < 90^o, find the value of x A. 14^o B. 21^o C. 32^o D. 39^o Show Content Detailed Solution Sin (5x – 28)^o = cos (3x - 50)^o Since by the trigonometry relation Sin(5x – 28)^o = cos[90 – (5x – 28)]^o Hence cos(3x – 50)^o = cos[90 – (5x – 28)]^o 3x – 50 = 90 - (5x-28) 3x – 50 = 90 – 5x + 28 3x + 5x = 90 + 28 + 50 8x = 168 $x = \frac{168}{8} = 21^{\circ}$	B
4.	Solve for t in the equation $\frac{3}{4} t + \frac{1}{3} (21 - t)$ = 11, A. $\frac{9}{13}$ B. $\frac{9}{5}$ C. 5 D. $9 \frac{3}{5}$ Show Content Detailed Solution $\frac{3}{4} t + \frac{1}{3} (21 - t) = 11; \frac{3 t}{4} + \frac{7}{1} - \frac{t}{3} = \frac{11}{1} \frac{3 \times 3 t + 7 \times 12 - 4 \times t = 11 \times 12}{12} 9 t + 84 - 4 t = 132; 5 t = 132 - 84 5 t = 48; t = \frac{48}{5} = 9 \frac{3}{5}$	D
5.	A school girl spends $\frac{1}{4}$ of her pocket money on books and $\frac{1}{3}$ on dress. What fraction remains? A. $\frac{5}{6}$ B. $\frac{7}{12}$ C. $\frac{5}{12}$ D. $\frac{1}{6}$ Show Content Detailed Solution Let the girls pocket money be rep. by x. The amount spent on books = $\frac{1}{4} o f x = \frac{x}{4}$ The amount spent on dress $= \frac{1}{3} o f x = \frac{x}{3}$ ∴The fraction that remains = $\frac{x}{1} - (\frac{x}{4} + \frac{x}{3}) \frac{3 x + 4 x}{12}; = \frac{12 x - 7 x}{12} = \frac{5 x}{12} = \frac{5}{12}$	C
6.	In the diagram, $R \hat{P} Q = Q \hat{R} Y, P \hat{Q} R = R \hat{Y} Q, \| Q P \| = 3 c m \| Q Y \| = 4 c m a n d \| R Y \| = 5 c m . F i n d \| Q R \|$ A. 2.0cm B. 2.5cm C. 6.4cm D. 10.0cm Show Content Detailed Solution $\frac{P R}{Q R} = \frac{Q R}{Q Y} = \frac{Q P}{Y R}$ (SSS Congruence) $\frac{Q R}{4} = \frac{8}{5}$ $Q R = \frac{4 \times 8}{5}$ = 6.4 cm	C
7.	Find the value of x in the diagram A. 10^o B. 28^o C. 36^o D. 44^o Show Content Detailed Solution The sum of angles at point = 360^o (x+10)^o + (4x+50)^o + 20^o + 3x^o + 2x^o = 360^o 10x + 80^o = 360^o 10x = 360^o - 80^o = 280^o $\frac{280}{10} = 28^{\circ}$	B
8.	There are m boys and 12 girls in a class. What is the probability of selecting at random a girl from the class? A. $\frac{m}{12}$ B. $\frac{12}{m}$ C. $\frac{12}{m + 12}$ D. $\frac{12}{m - 12}$ Show Content Detailed Solution Prob. (a girl) $= \frac{N u m b e r o f g i r l s}{N u m b e r o f b o y s a n d g i r l s i n c l a s s} = \frac{12}{m + 12}$	C
9.	Simplify $7 \frac{1}{2} - (2 \frac{1}{2} + 3) \div 16 \frac{1}{2}$ and correct your answer to the nearest whole number A. 33 B. 8 C. 7 Show Content Detailed Solution $7 \frac{1}{2} - (2 \frac{1}{2} + 3) \div 16 \frac{1}{2}; \frac{15}{2} - (\frac{5}{2} + \frac{3}{1}) \div \frac{33}{2} \frac{15}{2} - (\frac{5 + 6}{2}) \div \frac{33}{2}; \frac{15}{2} - \frac{11}{2} \div \frac{33}{2} \(\frac{11}{2}$ ÷ $\frac{33}{2}$ = $\frac{11}{2}$ * $\frac{2}{33}$ = $\frac{11}{33}$ or $\frac{1}{3}$ (when simplified) $\frac{15}{2}$ - $\frac{1}{3}$ = $\frac{43}{6}$ 7.166 The nearest whole number is 7	C
10.	The angle of elevation of the top of a tower from a point on the ground which is 36m away from the foot of the tower is 30^o. Calculate the height of the tower. A. 62.35m B. 20.78m C. 18.00m D. 10.39m Show Content Detailed Solution Where H is the height of the tower H = ? $T a n 30^{\circ} = \frac{H}{36} \Rightarrow H = 36 \times t a n 30^{\circ} H = 36 \times 0.5774 = 20.79$	B

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Detailed Solution

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