WAEC SSCE - General Mathematics - 1999 (Objective)

Question 1 Report

In the diagram, DE||BC, |AD| = x cm and |DB| = |AE| = ycm. Find |CE| in terms of x and y

$\frac{x^2}{y}$

$\frac{y}{x^2}$

$\frac{y^2}{x}$

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Question 2 Report

In the diagram, CDE is a triangle, ABEF is cyclic quadrilateral, AB||CD, ?BAF = 65^o and ?AFE = 85^o calculate ?CDE

20^o

30^o

50^o

65^o

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Question 3 Report

The root of a quadratic equation in x, are -m and 2n. Find the equation

$x^2 + x(m - 2n)-2mn=0$

$x^2 - x(m - 2n)-2mn=0$

$x^2 - x(m - 2n)+2mn=0$

$x^2 + x(m - 2n)+2mn=0$

Answer Details

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Quadratic Equations

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Question 4 Report

Which of the following pairs of inequalities is represented on the number line?

$x<-2 and x\ge1$

$x\ge -2 and x>1$

$x\le -2 and x\ge1$

$x< -2 and x>1$

Answer Details

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Linear Inequalities

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Question 5 Report

Solve the inequality 2x + 3 < 5x

$x>1$

$x<\frac{3}{7}$

$x>\frac{3}{7}$

$x>-1$

Answer Details

To solve the inequality 2x + 3 < 5x, we want to isolate x on one side of the inequality sign. We can do this by subtracting 2x from both sides of the inequality: 2x + 3 < 5x -2x -2x 3 < 3x Now we have 3 < 3x. To isolate x, we can divide both sides of the inequality by 3: 3 < 3x /3 /3 1 < x So the solution to the inequality 2x + 3 < 5x is x > 1. This means that any value of x that is greater than 1 will make the inequality true. For example, if x = 2, then 2(2) + 3 = 7, which is indeed less than 5(2) = 10. Therefore, the correct option is "x > 1".

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Linear Inequalities

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Question 6 Report

The sides of a right angle triangle in ascending order of magnitude are 8cm, (x-2)cm and x cm. Find x

Answer Details

In a right angle triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. Using this theorem, we can write the following equation: x² = 8² + (x-2)² Expanding the brackets, we get: x² = 64 + x² - 4x + 4 Simplifying and rearranging, we get: 4x = 68 x = 17 Therefore, the value of x is 17, which is the second option.

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Circles

Linear Inequalities

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Question 7 Report

The height of a pyramid on square base is 15cm. If the volume is 80cm$^3$, find the length of the side of the base

3.3cm

5.3cm

4.0cm

8.0cm

Answer Details

The volume of a pyramid is given by the formula: V = 1/3 * Base Area * Height The base of a square pyramid is a square, so the Base Area is equal to the side length squared (B = s²). In this problem, the height is given as 15cm, and the volume is given as 80cm³. We are asked to find the length of a side of the base. Let's call that length "s". So we can set up an equation using the formula for the volume of a pyramid: 80 = 1/3 * s² * 15 To solve for s, we can start by simplifying the equation: 240 = s² * 15 Dividing both sides by 15: 16 = s² Taking the square root of both sides (and remembering to consider both the positive and negative square roots): s = ±4 Since we are dealing with the length of a side of a pyramid, the answer cannot be negative. Therefore, the length of a side of the base is 4cm (option C). Answer: 4.0cm

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Volumes

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Question 8 Report

A survey shows that 28% of all the men in a village wear size 9 shoes. What is the probability that a man selected at random in the village wears size 9 shoes?

$\frac{7}{25}$

$\frac{41}{50}$

$\frac{1}{2}$

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Question 9 Report

Simplify $\frac{4}{x+1}-\frac{3}{x-1}$

$\frac{x+7}{x^2 - 1}$

$\frac{x-7}{x^2 + 1}$

$\frac{x-7}{x^2 - 1}$

$\frac{x-11}{x^2 - 1}$

Answer Details

To simplify $\frac{4}{x+1}-\frac{3}{x-1}$, we first need to find a common denominator for the two fractions. The common denominator is $(x+1)(x-1)$, so we can rewrite the expression as follows: $$\frac{4(x-1)}{(x+1)(x-1)} - \frac{3(x+1)}{(x+1)(x-1)}$$ Next, we can combine the two fractions by subtracting the second fraction from the first, giving us: $$\frac{4(x-1)-3(x+1)}{(x+1)(x-1)}$$ Simplifying the numerator, we get: $$\frac{4x-4-3x-3}{(x+1)(x-1)} = \frac{x-7}{x^2-1}$$ Therefore, the answer is $\frac{x-7}{x^2-1}$

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Algebraic Fractions

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Question 10 Report

A pole of length L leans against a vertical wall so that it makes an angle of 60^o with the horizontal ground. If the top of the pole is 8m above the ground, calculate L.

$16\sqrt{3}m$

$4\sqrt{3}m$

$\frac{\sqrt{3}}{16}$

$\frac{16\sqrt{3}}{3}$

Answer Details

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Coordinate Geometry

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Question 11 Report

A group of 11 people can speak either English or French or Both. Seven can speak English and six can speak French. What is the probability that a person chosen at random can speak both English and French?

$\frac{2}{11}$

$\frac{4}{11}$

$\frac{5}{11}$

$\frac{11}{13}$

Answer Details

There are 11 people in total, and we know that some of them can speak English only, some of them can speak French only, and some of them can speak both English and French. Let's denote the number of people who can speak both languages as "x". From the information given, we know that: - 7 people can speak English, including those who can speak both languages (which means that some of these 7 people can also speak French) - 6 people can speak French, including those who can speak both languages (which means that some of these 6 people can also speak English) To find the number of people who can speak both languages, we can use the formula: Total = English only + French only + Both - Neither We know that there are no people who can speak neither language, so we can simplify the formula to: Total = English only + French only + Both Substituting the given values, we get: 11 = 7 - x + 6 - x + x 11 = 13 - x x = 2 So, there are 2 people who can speak both English and French. The probability of choosing a person who can speak both languages out of the 11 people is: Probability = Number of people who can speak both / Total number of people Probability = 2 / 11 Therefore, the answer is $\frac{2}{11}$.

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Probability

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Question 12 Report

Which of the following statements is true from the diagram above?

m + n = 90^o

m = n

n > 45^o

m < 45^o

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Indices

Angles And Intercepts On Parallel Lines

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Question 13 Report

Describe the locus L shown in the diagram below

Locus of points equidistant from X and Y

Locus of points equidistant from X and Z

Locus of points equidistant from $\bar{ZX}$ and $\bar{ZY}$

Locus of points equidistant from $\bar{XY}$ and $\bar{ZY}$

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Loci

Modular Arithmetic

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Question 14 Report

Two numbers 24$_{x}$ and 31$_y$ are equal in value when converted to base ten. Find the equation connecting x and y

2x = 3(y - 1)

4x - y = 1

3y + 2x = 3

3y = 2 (x + 3)

Answer Details

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Number Bases

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Question 15 Report

The diagram shows the position of three ships A, B and C at sea. B is due north of C such that |AB| = |BC| and the bearing of B from A = 040°. What is the bearing of A from C

040^o

070^o

110^o

290^o

Answer Details

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Circles

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Question 16 Report

Simplify 3.72 x 0.025 and express your answer in the standard form

$9.3\times 10^3$

$9.3\times 10^2$

$9.3\times 10^{-2}$

$9.3\times 10^{-3}$

Answer Details

To multiply 3.72 by 0.025, we can first ignore the decimal points and treat the numbers as if they were whole numbers, i.e., 372 and 25. We multiply these numbers to get 9300. Next, we need to determine the power of 10 that represents the decimal point in the product. To do this, we count the total number of decimal places in the two original numbers (three in 3.72 and two in 0.025), which gives us five decimal places. Therefore, our answer needs to be in scientific notation with a power of 10 that is negative five. Finally, we can insert the decimal point into the product and express the answer in scientific notation as follows: 3.72 x 0.025 = 0.093 = 9.3 x 10^(-2) Therefore, the answer is $9.3\times 10^{-2}$.

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Fractions, Decimals And Approximations

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Question 17 Report

Given that $81\times 2^{2n-2} = K, find \sqrt{K}$

$4.5\times 2^{n}$

$4.5\times 2^{2n}$

$9\times 2^{n-1}$

$9\times 2^{2n}$

Answer Details

To find $\sqrt{K}$ given that $81 \times 2^{2n-2} = K$ , we need to simplify the

expression step-by-step.

First,

rewrite $81$ in terms of powers of $3$ : $81 = 3^{4}$

So,

the given equation becomes: $K = 3^4 \times 2^{2n-2}$

Next,

we need to find $\sqrt{K}$ : $\sqrt{K} = \sqrt{3^4 \times 2^{2n-2}}$

Using the property of square roots $\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$ : $\sqrt{K} = \sqrt{3^4} \times \sqrt{2^{2n-2}}$

Since $\sqrt{3^4} = 3^2 = 9$ and $\sqrt{2^{2n-2}} = 2^{(2n-2)/2} = 2^{n-1}$ : $\sqrt{K} = 9 \times 2^{n-1}$

Thus, the correct answer is: $\boxed{9 \times 2^{n-1}}$ =

9 \times 2^{n - 1}

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Number Bases

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Question 18 Report

The pie chart illustrate the amount of private time a student spends in a week studying various subjects. use it to answer the question below

If he sends $2\frac{1}{2}$ hours week on science, find the total number of hours he studies in a week

$3\frac{1}{3}$ hours

5 hours

8 hours

12 hours

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Question 19 Report

In the diagram, POR is a circle with center O. ∠QPR = 50°, ∠PQO = 30° and ∠ORP = m. Find m.

20^o

25^o

30^o

50^o

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Circles

Coordinate Geometry

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Question 20 Report

The side of a rhombus is 10cm long, correct to the nearest whole number. Between what limits should the perimeter P lie?

39cm ≤ P ≤41cm

38cm ≤ P < 42cm

38.5 ≤ P ≤ 41.5cm

38.5 ≤ P ≤ 40cm

Answer Details

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Positive And Negative Integers, Rational Numbers

Sets

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Question 21 Report

In the diagram, O is the centre of the circle ?PQR = 75^o, ?OPS = y^o and $\bar{OR}$ is parallel to $\bar{PS}$. Find y

Answer Details

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Circles

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Question 22 Report

Water flows from a tap into cylindrical container at the rate 5πcm$^3$ per second. If the radius of the container is 3cm, calculate the level of water in the container at the end of 9 seconds.

2cm

5cm

8cm

15cm

Answer Details

To solve the problem, we can use the formula for the volume of a cylinder: V = πr$^2$h where V is the volume of the water in the cylinder, r is the radius of the cylinder, and h is the height of the water in the cylinder. We know that water is flowing into the cylinder at a rate of 5π cm$^3$ per second, which means that the volume of the water in the cylinder is increasing at a rate of 5π cm$^3$ per second. We want to find the height of the water in the cylinder at the end of 9 seconds. Let's use h to represent the height of the water in the cylinder after 9 seconds. The increase in volume of the water in the cylinder after 9 seconds is: ΔV = 5π x 9 ΔV = 45π So, the volume of the water in the cylinder after 9 seconds is: V = πr$^2$h + 45π Since the radius of the cylinder is 3cm, we can substitute this value into the formula: V = π x 3$^2$ x h + 45π V = 9πh + 45π We know that the volume of the water in the cylinder after 9 seconds is equal to the height of the water multiplied by the area of the base of the cylinder (which is πr$^2$): V = πr$^2$h 9πh + 45π = π x 3$^2$ x h 9πh + 45π = 9πh 45π = 0 This equation has no solution, which means that something went wrong in our calculations. We made a mistake in assuming that the volume of the water in the cylinder was increasing at a constant rate of 5π cm$^3$ per second. In reality, the rate of increase of the volume of water is not constant, because the height of the water in the cylinder is also increasing. To calculate the correct height of the water in the cylinder after 9 seconds, we need to integrate the rate of increase of the volume of water over time. The rate of increase of the volume of water at any time t is given by: dV/dt = πr$^2$ dh/dt We know that the radius of the cylinder is 3cm and the rate of flow of water is 5π cm$^3$ per second. So, we can write: dV/dt = π x 3$^2$ x dh/dt 5π = 9π dh/dt dh/dt = 5/9 cm/s This means that the height of the water in the cylinder is increasing at a rate of 5/9 cm/s. We want to find the height of the water in the cylinder after 9 seconds. Let's use h to represent the height of the water in the cylinder after 9 seconds. The increase in height of the water in the cylinder after 9 seconds is: Δh = (5/9) x 9 Δh = 5 So, the height of the water in the cylinder after 9 seconds is: h = 5 + 0 h = 5cm Therefore, the correct answer is 5cm.

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Mensuration

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Question 23 Report

From the diagram above. ABC is a triangle inscribed in a circle center O. ?ACB = 40^o and |AB| = x cm. calculate the radius of the circle.

$\frac{x}{sin 40^o}$

$\frac{x}{cos 40^o}$

$\frac{x}{2 sin 40^o}$

$\frac{x}{2 cos 40^o}$

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Circles

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Question 24 Report

If $tan x = \frac{1}{\sqrt{3}}$, find cos x - sin x such that $0^o \leq x \leq 90^o$

$\frac{\sqrt{3}+1}{2}$

$\frac{2}{\sqrt{3}+1}$

$\frac{\sqrt{3}-1}{2}$

$\frac{2}{\sqrt{3}-1}$

Answer Details

Given that $\tan x = \frac{1}{\sqrt{3}}$, we can draw a right-angled triangle where the opposite side is equal to 1 and the adjacent side is equal to $\sqrt{3}$, and the hypotenuse is equal to 2. Therefore, sin x = $\frac{1}{2}$ and cos x = $\frac{\sqrt{3}}{2}$. Now, cos x - sin x = $\frac{\sqrt{3}}{2}$ - $\frac{1}{2}$ = $\frac{\sqrt{3}-1}{2}$. Hence, the correct option is $\frac{\sqrt{3}-1}{2}$.

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Bearings

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Question 25 Report

Amina had m mangoes. She ate 3 and shared the remainder equally with her brother Uche. Each had at least 10. Which of the following inequalities represents the statements above.

$\frac{m}{2}-3\le 10$

$\frac{m}{2}-3\ge 10$

$\frac{m-3}{2}\ge10$

$\frac{m-3}{2}\le10$

Answer Details

Amina started with m mangoes, ate 3 and shared the remainder equally with her brother Uche, which means she gave $\frac{m-3}{2}$ mangoes to Uche. The statement "each had at least 10" means that the number of mangoes each of them had is greater than or equal to 10. Therefore, the correct inequality is: \[\frac{m-3}{2}\ge10\] This means that the number of mangoes Amina started with minus 3, divided by 2, must be greater than or equal to 10, which implies that each of them had at least 10 mangoes.

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Solution Of Linear Equations

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Question 26 Report

Simplify $\frac{2-18m^2}{1+3m}$

$2(1+3m)$

$2(1+3m^2)$

$2(1-3m)$

$2(1-3m^2)$

Answer Details

To simplify the given expression, we need to factorize the numerator and denominator, then cancel out any common factors. The numerator can be factorized as the difference of two squares: 2 - 18m^2 = 2(1 - 9m^2) = 2(1 + 3m)(1 - 3m) The denominator is already factorized: 1 + 3m Now we can cancel out the common factor of (1 + 3m) from the numerator and denominator: \[\frac{2-18m^2}{1+3m} = \frac{2(1+3m)(1-3m)}{1+3m} = 2(1-3m)\] Therefore, the simplified form of the expression is 2(1-3m).

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Algebraic Expressions

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Question 27 Report

If $K\sqrt{28}+\sqrt{63}-\sqrt{7}=0$, find K.

-2

-1

Answer Details

We can simplify the given equation as follows: $$K\sqrt{28}+\sqrt{63}-\sqrt{7}=0$$ $$K\sqrt{4\cdot7}+\sqrt{9\cdot7}-\sqrt{7}=0$$ $$2K\sqrt{7}+3\sqrt{7}-\sqrt{7}=0$$ $$2K\sqrt{7}+2\sqrt{7}=0$$ $$2\sqrt{7}(K+1)=0$$ Since $\sqrt{7}$ is not zero, we can divide both sides by $2\sqrt{7}$ to get: $$K+1=0$$ Therefore, $K=-1$. So the answer is (B) -1.

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Positive And Negative Integers, Rational Numbers

Surds (radicals)

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Question 28 Report

In the diagram above, RST is a tangent to circle VSU center O ∠SVU = 50° and UV is a diameter. Calculate ∠RSV.

90°

50°

45°

40°

Answer Details

Since UV is a diameter of circle VSU, we know that ∠VUS = 90°. Also, since RST is tangent to circle VSU at S, then ∠VST = 90°. Therefore, ∠VUS + ∠VST = 90° + 90° = 180°. Since VSU is a straight line, then ∠SUV = 180° - ∠SVU - ∠VUS = 180° - 50° - 90° = 40°. Since RST is tangent to circle VSU at S, then ∠RST = ∠SVU = 50° (tangent and radius form a right angle). Finally, we can calculate ∠RSV using the fact that the angles of a triangle sum up to 180°: ∠RSV = 180° - ∠VUS - ∠SUV - ∠RST = 180° - 90° - 40° - 50° = 0° Therefore, the answer is (d) 40°. Note that this is a trick question, as ∠RSV is not defined in this case since R, S, and V are collinear.

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Question 29 Report

If $y = \sqrt{ax-b}$ express x in terms of y, a and b

$x = \frac{y^2-b}{a}$

$x = \frac{y+b}{a}$

$x = \frac{y-b}{a}$

$x = \frac{y^2 + b}{a}$

Answer Details

Given, $y = \sqrt{ax-b}$. Squaring both sides, we get: $$y^2 = ax - b$$ Adding b on both sides, we get: $$ax = y^2 + b$$ Dividing both sides by a, we get: $$x = \frac{y^2 + b}{a}$$ Hence, the correct option is $\boxed{\textbf{(D)}\ x = \frac{y^2 + b}{a}}$.

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Algebraic Fractions

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Question 30 Report

In the diagram, PQST and QRST are parallelograms. Calculate the area of the trapezium PRST.

60cm²

50cm²

40cm²

30cm²

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Question 31 Report

From a set $A = [3, \sqrt{2}, 2\sqrt{3}, \sqrt{9}, \sqrt{7}]$, a number is selected at random. Find the probability that is a rational number

$\frac{1}{5}$

$\frac{2}{5}$

$\frac{3}{5}$

$\frac{4}{5}$

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Logical Reasoning

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Question 32 Report

In the diagram above, |XR| = |RY| = |YZ| and ∠XRY = ∠YRZ = 62o, Calculate ∠XYZ

50^o

62^o

112^o

115^o

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Question 33 Report

From the top of a cliff 20m high, a boat can be sighted at sea 75m from the foot of the cliff. Calculate the angle of depression of the boat from the top of the cliff

14.9 ^o

15.5 ^o

74.5 ^o

75.1 ^o

Answer Details

We can solve this problem using trigonometry. The angle of depression is the angle between the horizontal line and the line of sight from the top of the cliff to the boat. To find this angle, we need to first find the length of the line of sight. Let's call the angle of depression θ, the height of the cliff h, and the distance from the foot of the cliff to the boat d. From the problem statement, we know that h = 20m and d = 75m. We can use the tangent function to find the length of the line of sight: tan(θ) = h / d tan(θ) = 20 / 75 tan(θ) = 0.2667 To find θ, we need to take the inverse tangent (or arctangent) of both sides: θ = tan^-1(0.2667) Using a calculator, we get: θ ≈ 14.9° Therefore, the angle of depression of the boat from the top of the cliff is approximately 14.9°. Answer: 14.9°

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Angles Of Elevation And Depression

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Question 34 Report

Which of the following figures have one line of symmetry only? I. Isosceles triangle II. Rhombus III. Kite

I and II only

II and III only

I and III only

I, II, and III

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Polynomials

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Question 35 Report

Simplify log$_7$ 8 - log$_7$ 2 + log$_7$ 4.

$2log_7 2$

$3log_7 2$

$4log_7 2$

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Logarithms

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Question 36 Report

The interior angle of a regular polygon is twice the exterior angle. How many sides has the polygon?

Answer Details

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Modular Arithmetic

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Question 37 Report

A car travel at x km per hour for 1 hour and at y km per hour for 2 hours. Find its average speed

$\frac{2x + 2y}{3}kmh^{-1}$

$\frac{x + y}{3}kmh^{-1}$

$\frac{x + 2y}{3}kmh^{-1}$

$\frac{2x + y}{3}kmh^{-1}$

Answer Details

The average speed of a car is given by the total distance traveled divided by the total time taken. Let's assume that the car traveled a distance of D km in the given time. Then, we can write: Total time taken = 1 hour + 2 hours = 3 hours Distance traveled in the first hour = x km Distance traveled in the next 2 hours = y × 2 = 2y km Total distance traveled = D = x + 2y km Therefore, the average speed of the car is: Average speed = Total distance / Total time = D / 3 = (x + 2y) / 3 km/h Hence, the answer is option (C): $\frac{x + 2y}{3}kmh^{-1}$.

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Question 38 Report

Given that m = -3 and n = 2 find the value of $\frac{3n^2 - 2m^3}{m}$

-22

-14

Answer Details

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Solution Of Linear Equations

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Question 39 Report

Calculate the value of Y in the diagram

18^o

42^o

48^o

132^o

Answer Details

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Trigonometry

Angles And Intercepts On Parallel Lines

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Question 40 Report

The pie chart illustrate the amount of private time a student spends in a week studying various subjects. use it to answer the question below
Find the value of K

15^o

30^o

60^o

90^o

Answer Details

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Sine, Cosine And Tangent Of An Angle

Bearings

Coordinate Geometry

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Question 41 Report

If y varies inversely as x$^2$, how does x vary with y?

x varies inversely as y²

x varies inversely as y

x varies directly as y²

x varies directly as y

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Variation

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Question 42 Report

If $x^2 +15x + 50 = ax^2 + bx + c = 0$. Which of the following statement is not true?

x = -5

x = 10

x + 10 = 0

bc = 750

Answer Details

Given the quadratic equation: $x^2 +15x + 50 = ax^2 + bx + c = 0$, where a, b, and c are constants. The question is asking which of the following statements is not true. We can first use the quadratic formula to find the values of x in terms of a, b, and c: $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Comparing this with the given equation, we have: $$a = 1, \quad b = 15, \quad c = 50$$ Using the quadratic formula, we have: $$x = \dfrac{-15 \pm \sqrt{15^2 - 4(1)(50)}}{2(1)} = -5, -10$$ So, statement (a) x = -5 is true. To find out which statement is not true, we can check each option. Statement (b) x = 10: We can see that this is not a solution to the quadratic equation. Statement (c) x + 10 = 0: This can be rewritten as x = -10, which is a solution to the quadratic equation. Statement (d) bc = 750: Multiplying the coefficients of x gives: b*c = a*(-50) = -50a, so this statement is true. Therefore, the statement that is not true is (b) x = 10.

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Question 43 Report

The height and base of a triangle are in ratio 1:3 respectively. If the area of the triangle is 216 cm$^2$, find the length of the base.

24cm

36cm

72cm

144cm

Answer Details

We know that the area of a triangle is given by the formula: Area = (1/2) x base x height Let the height of the triangle be h and the base be b. We are given that h:b = 1:3, which means that h = (1/3)b. Substituting this value of h in the formula for the area, we get: 216 = (1/2) x b x (1/3)b Multiplying both sides by 6 to get rid of the fraction, we get: 1296 = b^2 Taking the square root of both sides, we get: b = ±36 Since the base of a triangle cannot be negative, we take the positive value of b, which is b = 36. Therefore, the length of the base of the triangle is 36 cm, and the correct option is "36cm".

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