WAEC SSCE - Further Mathematics - 2010…

Question 1 Report

Find the equation of the tangent to the curve $y = 4x^{2} - 12x + 7$ at point (2, -1).

y + 4x - 9 = 0

y - 4x - 9 = 0

y - 4x + 9 = 0 Correct answer

y + 4x + 9 = 0

Answer Details

View Answer

Question 2 Report

If $y = 2(2x + \sqrt{x})^{2}$, find $\frac{\mathrm d y}{\mathrm d x}$.

$2\sqrt{x}(2x + \sqrt{2})$

$4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})$ Correct answer

$4(2x + \sqrt{x})(2 + \sqrt{x})$

$8(2x + \sqrt{x})(2 + \sqrt{x})$

Answer Details

To differentiate the given function with respect to x, we will use the chain rule and the power rule of differentiation. Let's start by using the power rule of differentiation to find the derivative of $(2x + \sqrt{x})^{2}$ with respect to x: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})\] Next, we use the chain rule to differentiate $\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})$ with respect to x: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x}) = 2 + \frac{1}{2\sqrt{x}}\] Substituting this into the previous equation, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(2x + \sqrt{x}) \cdot \left(2 + \frac{1}{2\sqrt{x}}\right)\] Expanding this expression, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 4(2x + \sqrt{x}) + 2\sqrt{x}(2x + \sqrt{x})\] Simplifying further, we get: \[\frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 8x + 6\sqrt{x}\] Finally, using the constant multiple rule of differentiation, we can find the derivative of the given function: \[\frac{\mathrm d y}{\mathrm d x} = 2 \cdot \frac{\mathrm d}{\mathrm d x} (2x + \sqrt{x})^{2} = 2(8x + 6\sqrt{x}) = 16x + 12\sqrt{x}\] Therefore, the correct answer is: $4(2x + \sqrt{x})(2 + \frac{1}{2\sqrt{x}})$.

View Answer

Question 3 Report

The probabilities that a husband and wife will be alive in 15 years time are m and n respectively. Find the probability that only one of them will be alive at that time.

m + n

m + n - 2mn Correct answer

1 - mn

Answer Details

The probability that only one of the husband and wife will be alive in 15 years can happen in two ways: either the husband is alive and the wife is not, or the wife is alive and the husband is not. The probability of the first case is (m)(1-n) = m - mn, and the probability of the second case is (n)(1-m) = n - mn. Therefore, the total probability that only one of them will be alive is (m - mn) + (n - mn) = m + n - 2mn. Hence, the correct answer is: m + n - 2mn.

Read lesson note on Probability (WAEC)

Probability

View Answer

Question 4 Report

If $\frac{5}{\sqrt{2}} - \frac{\sqrt{8}}{8} = m\sqrt{2}$, where m is a constant. Find m.

$1\frac{1}{2}$

$1\frac{1}{4}$

$2\frac{1}{4}$ Correct answer

$2\frac{1}{2}$

Answer Details

Read lesson note on Surds (WAEC)

Surds

View Answer

Question 5 Report

Two particles are fired together along a smooth horizontal surface with velocities 4 m/s and 5 m/s. If they move at 60° to each other, find the distance between them in 2 seconds.

$2\sqrt{61}$

$\sqrt{42}$

$2\sqrt{21}$ Correct answer

$2\sqrt{10}$

Answer Details

Read lesson note on Trigonometry (WAEC)

Trigonometry

View Answer

Question 6 Report

Find the value of $\cos(60° + 45°)$ leaving your answer in surd form.

$\frac{6 + \sqrt{2}}{4}$

$\frac{3 + \sqrt{6}}{4}$

$\frac{\sqrt{2} - \sqrt{6}}{4}$ Correct answer

$\frac{3 - \sqrt{6}}{4}$

Answer Details

Using the sum to product formula, we have: \begin{align*} \cos(60° + 45°) &= \cos 60° \cos 45° - \sin 60° \sin 45° \\ &= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} - \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \\ &= \frac{\sqrt{2} - \sqrt{6}}{4} \end{align*} Therefore, the correct answer is option C: $\frac{\sqrt{2} - \sqrt{6}}{4}$.

Read lesson note on Surds (WAEC)

Surds

View Answer

Question 7 Report

The distance s metres of a particle from a fixed point at time t seconds is given by $s = 7 + pt^{3} + t^{2}$, where p is a constant. If the acceleration at t = 3 secs is $8 ms^{-2}$, find the value of p.

$\frac{1}{3}$ Correct answer

$\frac{4}{9}$

$\frac{5}{9}$

$1$

Answer Details

Given, distance of a particle from a fixed point at time t seconds is given by, $$s = 7 + pt^{3} + t^{2}$$ We can find the acceleration of the particle by differentiating the distance equation twice with respect to time. $$\frac{d}{dt} s = \frac{d}{dt} (7 + pt^{3} + t^{2})$$ $$\Rightarrow \frac{ds}{dt} = 3pt^2 + 2t$$ $$\frac{d^{2}}{dt^{2}} s = \frac{d}{dt} (3pt^2 + 2t)$$ $$\Rightarrow \frac{d^{2}s}{dt^{2}} = 6pt + 2$$ We are given that the acceleration at t = 3 sec is $8 ms^{-2}$, so we can substitute the values in the second derivative of the distance equation to get, $$6p(3) + 2 = 8$$ $$\Rightarrow 18p = 6$$ $$\Rightarrow p = \frac{6}{18} = \frac{1}{3}$$ Hence, the value of p is $\frac{1}{3}$. Therefore, is correct.

Read lesson note on Polynomial Functions (WAEC)

Polynomial Functions

View Answer

Question 8 Report

If $\sin x = -\sin 70°, 0° < x < 360°$, determine the two possible values of x.

110°, 250°

110°, 290°

200°, 250°

250°, 290° Correct answer

Answer Details

Read lesson note on Trigonometry (WAEC)

Trigonometry

View Answer

Question 9 Report

If (x - 3) is a factor of $2x^{2} - 2x + p$, find the value of constant p.

-12 Correct answer

-6

Answer Details

If (x-3) is a factor of the quadratic expression, then the expression can be factored into the form: $$(x - 3)(ax + b)$$ where a and b are constants. Multiplying out the brackets gives: $$2x^{2} - 2x + p = (x - 3)(ax + b) = ax^{2} + (b - 3a)x - 3b$$ Since the coefficients of the quadratic expression are equal to those of the factored expression, we can equate the corresponding coefficients to get a system of equations: $$a = 2$$ $$b - 3a = -2$$ $$-3b = p$$ Solving these equations simultaneously, we get: $$a = 2, b = 3a - 2 = 4, p = -3b = -12$$ Therefore, the constant p is -12.

Read lesson note on Polynomial Functions (WAEC)

Polynomial Functions

View Answer

Question 10 Report

The coefficient of the 7th term in the binomial expansion of $(2 - \frac{x}{3})^{10}$ in ascending powers of x is

$\frac{560}{243}$

$\frac{841}{243}$

$\frac{1120}{243}$ Correct answer

$\frac{4481}{243}$

Answer Details

Read lesson note on Binomial Theorem (WAEC)

Binomial Theorem

View Answer

Question 11 Report

Calculate, correct to one decimal place, the length of the line joining points X(3, 5) and Y(5, 1).

4.0

4.2

4.5 Correct answer

5.0

Answer Details

Read lesson note on Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

View Answer

Question 12 Report

What is the angle between $a = (3i - 4j)$ and $b = (6i + 4j)$?

13°

87° Correct answer

100°

110°

Answer Details

Read lesson note on Trigonometry (WAEC)

Trigonometry

View Answer

Question 13 Report

P, Q, R, S are points in a plane such that PQ = 8i - 5j, QR = 5i + 7j, RS = 7i + 3j and PS = xi + yj. Find (x, y).

(-6, -15)

(-6, 5)

(20, 5) Correct answer

(20, 15)

Answer Details

Given that PQ = 8i - 5j, QR = 5i + 7j, RS = 7i + 3j, and PS = xi + yj. Since PQ + QR + RS = PS, we have: (8i - 5j) + (5i + 7j) + (7i + 3j) = xi + yj Simplifying the left-hand side, we get: 20i + 5j = xi + yj Equating the corresponding components, we have: 20 = x and 5 = y Therefore, the coordinates of point S are (20, 5). Hence, the correct answer is (20, 5).

Read lesson note on Vectors (WAEC)

Vectors

View Answer

Question 14 Report

The roots of a quadratic equation are $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$. Find its equation.

$x^{2} - 6x - 9 = 0$

$x^{2} - 6x + 6 = 0$ Correct answer

$x^{2} + 6x - 9 = 0$

$x^{2} + 6x + 6 = 0$

Answer Details

The roots of a quadratic equation of the form $ax^{2} + bx + c = 0$ are given by the formula: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ If the roots are $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$, then we have: $$x = 3 - \sqrt{3} \quad \text{or} \quad x = 3 + \sqrt{3}$$ Therefore, we can write the equation as: $$(x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0$$ Expanding the brackets, we get: $$x^{2} - (3 - \sqrt{3} + 3 + \sqrt{3})x + (3 - \sqrt{3})(3 + \sqrt{3}) = 0$$ Simplifying, we get: $$x^{2} - 6x + 6 = 0$$ Hence, the equation of the quadratic with roots $(3 - \sqrt{3})$ and $(3 + \sqrt{3})$ is $x^{2} - 6x + 6 = 0$. Therefore, the answer is.

Read lesson note on Surds (WAEC)

Surds

View Answer

Question 15 Report

Find the domain of $f(x) = \frac{x}{3 - x}, x \in R$, the set of real numbers.

${x : x \in R, x \neq 3}$ Correct answer

${x : x \in R, x \neq 1}$

${x : x \in R, x \neq 0}$

${x : x \in R, x\neq -3}$

Answer Details

The function $f(x) = \frac{x}{3-x}$ is defined for all real numbers except when the denominator is zero, i.e., when $3-x=0$. Therefore, the domain of the function is all real numbers except $x=3$, or in interval notation: ${x : x \in R, x \neq 3}$. So the correct option is: ${x : x \in R, x \neq 3}$.

Read lesson note on Rational Functions (WAEC)

Rational Functions

View Answer

Question 16 Report

Given that $P = {x : \text{x is a factor of 6}}$ is the domain of $g(x) = x^{2} + 3x - 5$, find the range of x.

{-1, 5, 13}

{5, 13, 49}

{1, 2, 3, 6}

{-1, 5, 13, 49} Correct answer

Answer Details

Read lesson note on Polynomial Functions (WAEC)

Read lesson note on Functions (WAEC)

Polynomial Functions

Functions

View Answer

Question 17 Report

If $\overrightarrow{OA} = 3i + 4j$ and $\overrightarrow{OB} = 5i - 6j $ where O is the origin and M is the midpoint of AB, find OM.

-2i - 10j

-2i + 2j

4i - j Correct answer

4i + j

Answer Details

We can find the midpoint M of AB by using the midpoint formula: \[\overrightarrow{OM} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}\] Substituting the values we have: \[\overrightarrow{OM} = \frac{(3i + 4j) + (5i - 6j)}{2} = \frac{8i - 2j}{2} = 4i - j\] Therefore, the answer is 4i - j.

Read lesson note on Vectors (WAEC)

Vectors

View Answer

Question 18 Report

A stone is thrown vertically upwards and its height at any time t seconds is $h = 45t - 9t^{2}$. Find the maximum height reached.

45. 25 m

45.50 m

56.00 m

56.25 m Correct answer

Answer Details

The height of the stone at any time t seconds is given by $h = 45t - 9t^{2}$. To find the maximum height reached, we need to find the vertex of the parabolic function $h$, since the vertex represents the highest point of the parabola. The vertex of the parabola is given by the formula $\frac{-b}{2a}$, where $a$ and $b$ are the coefficients of the quadratic function. In this case, $a = -9$ and $b = 45$. Substituting these values into the formula for the vertex, we get: \[\frac{-b}{2a} = \frac{-45}{2(-9)} = \frac{45}{18} = 2.5.\] Therefore, the stone reaches its maximum height after 2.5 seconds. To find the maximum height, we substitute $t = 2.5$ into the equation for $h$: \[h = 45(2.5) - 9(2.5)^2 = 56.25\text{ m}.\] Therefore, the maximum height reached by the stone is 56.25 meters. The answer is (D) 56.25 m.

Read lesson note on Differentiation (WAEC)

Differentiation

View Answer

Question 19 Report

Yomi was asked to label four seats S, R, P, Q. What is the probability he labelled them in alphabetical order?

$\frac{1}{24}$ Correct answer

$\frac{1}{6}$

$\frac{2}{13}$

$\frac{1}{4}$

Answer Details

There are 4 seats to be labeled, so Yomi can label the first seat with any of the 4 letters. After labeling the first seat, there are 3 letters left to choose from for the second seat, 2 letters for the third seat, and only 1 letter left for the last seat. Thus, there are a total of 4 x 3 x 2 x 1 = 24 possible ways to label the seats. Out of these 24 possible ways, only one arrangement is in alphabetical order (i.e., S, P, Q, R). Therefore, the probability that Yomi labeled the seats in alphabetical order is 1/24. Therefore, the correct option is (a) $\frac{1}{24}$.

Read lesson note on Probability (WAEC)

Read lesson note on Co-ordinate Geometry (WAEC)

Probability

Co-ordinate Geometry

View Answer

Question 20 Report

Calculate, correct to one decimal place, the acute angle between the lines 3x - 4y + 5 = 0 and 2x + 3y - 1 = 0.

70.6° Correct answer

50.2°

39.8°

19.4°

Answer Details

Read lesson note on Matrices And Linear Transformation (WAEC)

Read lesson note on Co-ordinate Geometry (WAEC)

Matrices And Linear Transformation

Co-ordinate Geometry

View Answer

Question 21 Report

Two forces (2i - 5j)N and (-3i + 4j)N act on a body of mass 5kg. Find in $ms^{-2}$, the magnitude of the acceleration of the body.

$\frac{\sqrt{2}}{5}$ Correct answer

$5\sqrt{2}$

$2\sqrt{5}$

$\frac{5\sqrt{2}}{2}$

Answer Details

To find the magnitude of acceleration, we need to use Newton's Second Law, which states that the force acting on an object is equal to its mass multiplied by its acceleration. In this case, we have two forces: (2i - 5j)N and (-3i + 4j)N. To find the net force, we add these two vectors by adding their corresponding components: (2i - 5j)N + (-3i + 4j)N = -i - j N So the net force is (-i - j)N. Now, using Newton's Second Law, we can find the acceleration: F = ma where F is the net force, m is the mass, and a is the acceleration. In our case, the net force is (-i - j)N and the mass is 5kg, so we have: (-i - j)N = 5kg * a Solving for a, we get: a = (-i - j)N / 5kg To find the magnitude of a, we take the square root of the sum of the squares of its components: |a| = sqrt((-1)^2 + (-1)^2) N/kg = sqrt(2) N/kg Converting to $ms^{-2}$ by dividing by 5, we get: |a| = sqrt(2)/5 $ms^{-2}$ Therefore, the answer is $\frac{\sqrt{2}}{5}$.

Read lesson note on Vectors (WAEC)

Vectors

View Answer

Question 22 Report

Find the least value of n for which $^{3n}C_{2} > 0, n \in R$.

$\frac{1}{3}$

$\frac{1}{6}$

$\frac{2}{3}$ Correct answer

Answer Details

Read lesson note on Permutation And Combinations (WAEC)

Permutation And Combinations

View Answer

Question 23 Report

Find the maximum value of $2 + \sin (\theta + 25)$.

3 Correct answer

Answer Details

The value of $\sin(\theta + 25)$ varies between -1 and 1. Therefore, the maximum value of $2 + \sin(\theta + 25)$ occurs when $\sin(\theta + 25) = 1$. This occurs when $\theta + 25 = \frac{\pi}{2} + 2n\pi$, where $n$ is an integer. Thus, $\theta = \frac{\pi}{2} - 25 + 2n\pi$. Substituting this value of $\theta$ into the expression for $2 + \sin(\theta + 25)$, we get: \[2 + \sin\left(\frac{\pi}{2} + 2n\pi\right) = 2 + 1 = 3\] Therefore, the maximum value of $2 + \sin(\theta + 25)$ is 3.

Read lesson note on Trigonometry (WAEC)

Trigonometry

View Answer

Question 24 Report

A bicycle wheel of diameter 70 cm covered a distance of 350 cm in 2 seconds. How many radians per second did it turn?

5 Correct answer

Answer Details

To find the radians per second that the bicycle wheel turns, we need to find the angle in radians that the wheel turns in 1 second. The distance covered by the bicycle wheel in 1 second can be found by dividing the distance covered in 2 seconds by 2: \[\frac{350 \, \text{cm}}{2 \, \text{s}} = 175 \, \text{cm/s}\] To convert this to radians per second, we need to know the circumference of the wheel in centimeters. The circumference of the wheel is given by: \[\pi \times \text{diameter} = \pi \times 70 \, \text{cm} = 220 \pi \, \text{cm}\] So, in one revolution, the wheel covers a distance of 220π cm. The angle in radians that the wheel turns in 1 second can be found by dividing the distance covered in 1 second by the circumference of the wheel: \[\frac{175 \, \text{cm/s}}{220 \pi \, \text{cm/rev}} = \frac{5}{2\pi} \, \text{rev/s}\] To convert revolutions per second to radians per second, we multiply by 2π: \[\frac{5}{2\pi} \times 2\pi = 5 \, \text{rad/s}\] Therefore, the bicycle wheel turns at a rate of 5 radians per second. So the answer is (A) 5.

Read lesson note on Vectors (WAEC)

Vectors

View Answer

Question 25 Report

Evaluate $\int_{1}^{2} \frac{4}{x^{3}} \mathrm {d} x$

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

$-\frac{15}{16}$

$\frac{15}{16}$

$1\frac{1}{2}$ Correct answer

Answer Details

Read lesson note on Integration (WAEC)

Integration

View Answer

Question 26 Report

A company took delivery of 12 vehicles made up of 7 buses and 5 saloon cars for two of its departments; Personnel and General Administration. If the Personnel department is to have at least 3 saloon cars, in how many ways can these vehicles be distributed equally between the departments?

350

455

462 Correct answer

571

Answer Details

Read lesson note on Permutation And Combinations (WAEC)

Permutation And Combinations

View Answer

Question 27 Report

For what values of x is $\frac{x^{2} - 9x + 18}{x^{2} + 2x - 35}$ undefined?

6 or 3

-18 or -9

-7 or 5 Correct answer

-5 or 7

Answer Details

The expression is undefined when the denominator is equal to zero because division by zero is undefined. Therefore, we can find the values of x that make the denominator zero by setting it equal to zero and solving for x: $$x^{2} + 2x - 35 = 0$$ We can factor this quadratic equation as: $$(x + 7)(x - 5) = 0$$ So, the denominator is equal to zero when x = -7 or x = 5. Therefore, the expression is undefined for these values of x. So the correct answer is: - -7 or 5

Read lesson note on Rational Functions (WAEC)

Rational Functions

View Answer

Question 28 Report

If $16^{3x} = \frac{1}{4}(32^{x - 1})$, find the value of x.

$-1$ Correct answer

$\frac{-1}{3}$

$\frac{-3}{7}$

$\frac{-5}{19}$

Answer Details

Read lesson note on Surds (WAEC)

Read lesson note on Sets (WAEC)

Surds

Sets

View Answer

Question 29 Report

If $h(x) = x^{3} - \frac{1}{x^{3}}$, evaluate $h(a) - h(\frac{1}{a})$.

-1

$2a^{3} - \frac{2}{a^{3}}$ Correct answer

$\frac{2}{a^{3}} - 2a^{3}$

Answer Details

We can start by computing the values of $h(a)$ and $h(\frac{1}{a})$, and then find their difference. First, we have $h(a) = a^{3} - \frac{1}{a^{3}}$. Next, we have $h(\frac{1}{a}) = (\frac{1}{a})^{3} - \frac{1}{(\frac{1}{a})^{3}} = \frac{1}{a^{3}} - a^{3}$. Therefore, \begin{align*} h(a) - h(\frac{1}{a}) &= \left(a^{3} - \frac{1}{a^{3}}\right) - \left(\frac{1}{a^{3}} - a^{3}\right) \\ &= a^{3} - \frac{1}{a^{3}} - \frac{1}{a^{3}} + a^{3} \\ &= 2a^{3} - \frac{2}{a^{3}}. \end{align*} So the answer is $2a^{3} - \frac{2}{a^{3}}$. Therefore, is the correct answer.

Read lesson note on Polynomial Functions (WAEC)

Polynomial Functions

View Answer

Question 30 Report

Given that $\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4$ and y = 6 when x = 3, find the equation for y.

$x^{3} - 4x - 9$ Correct answer

$x^{3} - 4x + 9$

$x^{3} + 4x - 9$

$x^{3} + 4x + 9$

Answer Details

View Answer

Question 31 Report

Simplify $(1 + 2\sqrt{3})^{2} - (1 - 2\sqrt{3})^{2}$

$8\sqrt{3}$ Correct answer

$2 - 4\sqrt{3}$

Answer Details

Read lesson note on Surds (WAEC)

Surds

View Answer

Question 32 Report

Two forces $F_{1} = (7i + 8j)N$ and $F_{2} = (3i + 4j)N$ act on a particle. Find the magnitude and direction of $F_{1} - F_{2}$.

$(4\sqrt{2} N, 000°)$

$(4\sqrt{2} N, 045°)$ Correct answer

$(4\sqrt{2} N, 090°)$

$(4\sqrt{2} N, 180°)$

Answer Details

To find the magnitude and direction of $F_{1} - F_{2}$, we need to subtract the components of $F_{2}$ from $F_{1}$. \begin{align*} F_{1} - F_{2} &= (7i + 8j) - (3i + 4j)\\ &= 4i + 4j\\ \end{align*} The magnitude of $F_{1} - F_{2}$ is given by: \begin{align*} |F_{1} - F_{2}| &= \sqrt{(4)^2 + (4)^2}\\ &= 4\sqrt{2} N\\ \end{align*} The direction of $F_{1} - F_{2}$ is given by: \begin{align*} \theta &= \tan^{-1}\left(\frac{4j}{4i}\right)\\ &= \tan^{-1}(1)\\ &= 45°\\ \end{align*} Therefore, the magnitude and direction of $F_{1} - F_{2}$ are $(4\sqrt{2} N, 045°)$. Answer is correct.

Read lesson note on Statics (WAEC)

Statics

View Answer

Question 33 Report

The initial velocity of an object is $u = \begin{pmatrix} -5 \\ 3 \end{pmatrix} ms^{-1}$. If the acceleration of the object is $a = \begin{pmatrix} 3 \\ -4 \end{pmatrix} ms^{-2}$ and it moved for 3 seconds, find the final velocity.

$\begin{pmatrix} -14 \\ 15 \end{pmatrix} ms^{-1}$

$\begin{pmatrix} -2 \\ 1 \end{pmatrix} ms^{-1}$

$\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}$ Correct answer

$\begin{pmatrix} 14 \\ -9 \end{pmatrix} ms^{-1}$

Answer Details

The final velocity of an object can be found using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we have: v = (-5, 3) + 3(3, -4) v = (-5, 3) + (9, -12) v = (4, -9) Therefore, the final velocity of the object is $\begin{pmatrix} 4 \\ -9 \end{pmatrix} ms^{-1}$. The correct option is (C).

Read lesson note on Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

View Answer

Question 34 Report

The mean age of 15 pupils in a class is 14.2 years. One new pupil joined the class and the mean changed to 14.1 years. Calculate the age of the new pupil.

12.4 years

12.6 years Correct answer

13.2 years

14.1 years

Answer Details

Let's use the formula for the mean (average) of a set of numbers: \[\text{mean} = \frac{\text{sum of all numbers}}{\text{number of numbers}}\] Let the age of the new pupil be x. Before the new pupil joined the class, the sum of ages of the 15 pupils would be: \[15 \times 14.2 = 213\] After the new pupil joined the class, the sum of ages of the 16 pupils would be: \[15 \times 14.2 + x = 213 + x\] We know that the new mean is 14.1, so we can set up an equation: \[\frac{15 \times 14.2 + x}{16} = 14.1\] Multiplying both sides by 16, we get: \[15 \times 14.2 + x = 16 \times 14.1\] Simplifying the left side, we get: \[213 + x = 225.6\] Subtracting 213 from both sides, we get: \[x = 12.6\] Therefore, the age of the new pupil is 12.6 years. Answer: (b) 12.6 years.

Read lesson note on Statistics (WAEC)

Statistics

View Answer

Question 35 Report

Find the direction cosines of the vector $4i - 3j$.

$\frac{9}{10}, \frac{27}{10}$

$\frac{17}{27}, -\frac{17}{27}$

$\frac{4}{5}, -\frac{3}{5}$ Correct answer

$\frac{4}{7}, \frac{-3}{7}$

Answer Details

Direction cosines are the cosines of the angles that a vector makes with the positive x, y and z axes. Let's denote the vector by $\mathbf{v} = 4\mathbf{i} - 3\mathbf{j} + 0\mathbf{k}$. Then the direction cosines of $\mathbf{v}$ are: \begin{align*} \cos\alpha &= \frac{4}{5} \\ \cos\beta &= \frac{-3}{5} \\ \cos\gamma &= 0 \end{align*} Therefore, the correct option is: - $\frac{4}{5}, -\frac{3}{5}$

Read lesson note on Vectors (WAEC)

Vectors

View Answer

Question 36 Report

In a class of 50 pupils, 35 like Science and 30 like History. What is the probability of selecting a pupil who likes both Science and History?

0.10

0.30 Correct answer

0.60

0.70

Answer Details

We can use the formula: P(A and B) = P(A) + P(B) - P(A or B) where A is the event of liking Science, B is the event of liking History, P(A and B) is the probability of selecting a pupil who likes both Science and History, and P(A or B) is the probability of selecting a pupil who likes either Science or History or both. We know that there are 50 pupils in the class, 35 like Science and 30 like History. This means that the maximum number of pupils who could like both Science and History is 30 (since 30 already like History). Therefore, we have: P(A and B) <= 30/50 = 0.6 Also, we know that P(A or B) = P(A) + P(B) - P(A and B) (as shown above). Therefore: P(A and B) = P(A) + P(B) - P(A or B) = 35/50 + 30/50 - P(A or B) = 65/50 - P(A or B) We also know that P(A or B) <= 1, since it is the probability of selecting a pupil who likes either Science or History or both. Therefore: P(A and B) >= 65/50 - 1 = 15/50 = 0.3 Thus, the probability of selecting a pupil who likes both Science and History is between 0.3 and 0.6. The only option that falls within this range is 0.30, so the answer is: - 0.30

Read lesson note on Probability (WAEC)

Probability

View Answer

Question 37 Report

If $\begin{vmatrix} 3 & x \\ 2 & x - 2 \end{vmatrix} = -2$, find the value of x.

-8 Correct answer

-4

Answer Details

Read lesson note on Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

View Answer

Question 38 Report

Find the axis of symmetry of the curve $y = x^{2} - 4x - 12$.

x = -2

y = -2

x = 2 Correct answer

y = 2

Answer Details

To find the axis of symmetry of a quadratic equation in the form of $y = ax^2 + bx + c$, we need to use the formula: $$x = \frac{-b}{2a}$$ Comparing the given equation $y = x^2 - 4x - 12$ with the standard form of the quadratic equation, we have: $$a = 1, b = -4, c = -12$$ Substituting the values in the formula, we get: $$x = \frac{-(-4)}{2(1)} = 2$$ Therefore, the axis of symmetry of the curve is the vertical line passing through the point $(2,0)$ and the equation of this line is: $$x = 2$$ Hence, the correct answer is "x = 2".

View Answer

Question 39 Report

Simplify $\frac{\log_{5} 8}{\log_{5} \sqrt{8}}$.

-2

$\frac{-1}{2}$

$\frac{1}{2}$

2 Correct answer

Answer Details

We can use the properties of logarithms to simplify this expression. Recall that $\log_{a} b^{c} = c\log_{a} b$ and $\log_{a} \sqrt{b} = \frac{1}{2}\log_{a} b$. So, \[\frac{\log_{5} 8}{\log_{5} \sqrt{8}} = \frac{\log_{5} 8}{\frac{1}{2}\log_{5} 8} = \frac{2\log_{5} 8}{\log_{5} 8} = \boxed{2}.\] Therefore, the answer is 2.

Read lesson note on Indices And Logarithmic Functions (WAEC)

Indices And Logarithmic Functions

View Answer

Question 40 Report

The third of geometric progression (G.P) is 10 and the sixth term is 80. Find the common ratio.

2 Correct answer

Answer Details

We can use the formula for the nth term of a geometric progression to solve this problem. Let a be the first term and r be the common ratio, then the third term is ar^2 and the sixth term is ar^5. We are given that ar^2 = 10 and ar^5 = 80. Dividing the second equation by the first, we get: (ar^5)/(ar^2) = 80/10 Simplifying and canceling a, we get: r^3 = 8 Taking the cube root of both sides, we get: r = 2 Therefore, the common ratio is 2, and the answer is (a) 2.

Read lesson note on Sets (WAEC)

Sets

View Answer






Press & Hold to Drag Around
Click Here to Close

WAEC SSCE - Further Mathematics - 2010 (Objective)