A man is standing in the corridor of a 10-storey building and looking down at a tall tree in front of the building. He sees the top of the tree at angle of depression of 30o. If the tree is 200m tall and the man’s eyes are 300m above the ground, calculate the angle of depression of the foot tree as seen by the man
- A. 30o
- B. 60o
- C. 45o
- D. 25o
If \(\sqrt{3^{\frac{1}{x}}}\) = \(\sqrt{9}\) then the value of x is:
- A. \(\frac{3}{4}\)
- B. \(\frac{4}{3}\)
- C. \(\frac{1}{3}\)
- D. \(\frac{2}{3}\)
- E. \(\frac{1}{2}\)
A father is now three times as old as his son. Twelve years ago he was six times as old as his son. How old are the son and the father?
- A. 20 and 45
- B. 100 and 150
- C. 45 and 65
- D. 35 and 75
- E. 20 and 60
If y = x2 – 2x – 3, Find the least value of y and corresponding value of x
- A. x = 3, y = 3
- B. x = 1, y = -3
- C. x = 4, y = 1
- D. x = 1, y = -4
- E. x = 2, y = -3
The following table relates the number of objects f corresponding to a certain size x. What is the average size of an object?
\(\begin{array}{c|c} f & 1 & 2 & 3 & 4 & 5 \\ \hline x & 1 & 2 & 4 & 8 & 16\end{array}\)
- A. \(\frac{31}{15}\)
- B. \(\frac{31}{5}\)
- C. \(\frac{129}{5}\)
- D. \(\frac{43}{5}\)
- E. \(\frac{16}{5}\)
The size of a quantity first doubles and then increases by a further 16%. After a short time its size decreases by 16%. What is the net increases in size of the quantity?
- A. \(\frac{59300}{625}\)
- B. \(\frac{50900}{625}\)
- C. 200%
- D. +100%
What is log7(49a) – log10(0.01)?
- A. \(\frac{49^a}{100}\)
- B. \(\frac{a}{2}\) + 2
- C. 72a + 2
- D. 2a + 2
- E. \(\frac{2a}{2}\)
What is the greatest straight line distance between two vertices (corners) of a cube whose sides are 2239cm long?
- A. \(\sqrt{2239cm}\)
- B. \(\sqrt{2}\) x 2239cm
- C. \(\frac{\sqrt{3}}{2}\) 2239cm
- D. \(\sqrt{3}\) x 2239cm
- E. 4478cm
An arc of circle of radius 2cm subtends an angle of 60º at the centre. Find the area of the sector
- A. \(\frac{2 \pi}{3}\)cm2
- B. \(\frac{\pi}{2}\)cm2
- C. \(\frac{\pi}{3}\)cm2
- D. \(\pi\)cm2
A cylinder of height h and radius r is open at one end. Its surface area is
- A. 2\(\pi\)rh
- B. \(\pi\)r2h
- C. 2\(\pi\)rh + \(\pi\)r2
- D. 2\(\pi\)rh + 2\(\pi\)r2
Simplify \(\frac{1 – x^2}{x – x^2}\), where x \(\neq\) 0
- A. \(\frac{1}{x}\)
- B. \(\frac{1 - x}{x}\)
- C. \(\frac{1 + x}{x}\)
- D. \(\frac{1}{x - 1}\)
- E. \(\frac{-x - 1}{1}\)
For the set of numbers 2, 3, 5, 6, 7, 7, 8
- A. The median is greater than the mode
- B. the mean is greater than the mode
- C. the mean is greater than the median
- D. the median is equal to the mean
- E. the mean is less than the median
12 men complete a job in 9 days. How many men working t the same rate would be required to complete the job in 6 days?
- A. 24
- B. 18
- C. 12
- D. 9
- E. 8
Find the value of log\(_{10}\)\(\frac{1}{40}\), given that log10\(_4\) = 0.6021
- A. 1.3979
- B. 2.3979
- C. -1.6021
- D. 2.6021
Find the value of (4\(^{\frac{1}{2}}\))\(^6\)
- A. 1
- B. 2
- C. 4
- D. 6
- E. 64
If x3 – 12x – 16 = 0 has x = -2 as a solution then the equaion has
- A. x = -4 as a solution also
- B. 3 roots all different
- C. 3 roots with two equal and the third different
- D. 3 roots all equal
- E. only one root
Solve the simultaneous linear equations: 2x + 5y = 11, 7x + 4y = 2
- A. x = -8, y = 1
- B. x = -2, y = 4
- C. x = \(\frac{-34}{27}\), y = \(\frac{73}{27}\)
- D. x = 7, y = -9
On each market day Mrs. Bassey walks to the market from her home at a steady speed. This journey normally takes her 2 hours to complete. She finds, however, that by increasing her usual speed by 1 km/hr she can save 20 minutes. Find her usual speed in km/hr
- A. 1\(\frac{2}{3}\)
- B. 2
- C. 5
- D. 6
- E. 10
Which of the following values of the variable x, (a)x = 0, (b)x = -3, (c)x = 9, satisfy the inequalities 0 < \(\frac{x + 3}{x – 1}\) < 2?
- A. (a), (b), (c)
- B. (c)
- C. None of the choices
- D. All of the above
Make x the subject of the equation a(b + c) + \(\frac{5}{d}\) – 2 = 0
- A. c = \(\frac{2d - 5 - b}{ad}\)
- B. c = \(\frac{2d - 5 - abd}{ad}\)
- C. c = \(\frac{2d - 5 - ab}{ad}\)
- D. c = \(\frac{5 - 2d - b}{ad}\)
PQRS is a cyclic quadrilateral with PQ as diameter of the circle. If < PQS = 15o find < QRS
- A. 75o
- B. 37\(\frac{1}{2}\)o
- C. 127\(\frac{1}{2}\)o
- D. 105o