WAEC SSCE - Further Mathematics - 2012…

Question 1 Report

From the diagram above, $h[g(3)]$ is

$\beta$ Correct answer

${s, \beta}$

${s, w, \beta}$

Answer Details

Read lesson note on Functions (WAEC)

Functions

View Answer

Question 2 Report

The marks obtained by 10 students in a test are as follows: 3, 7, 6, 2, 8, 5, 9, 1, 4 and 10. Find the mean mark.

4.50

5.50 Correct answer

6.50

6.75

Answer Details

To find the mean mark, we add up all the marks and divide by the number of students. Adding up all the marks, we get: 3 + 7 + 6 + 2 + 8 + 5 + 9 + 1 + 4 + 10 = 55 There are 10 students, so we divide the sum of the marks by 10: 55 / 10 = 5.50 Therefore, the mean mark is 5.50. Hence, the correct option is: 5.50.

Read lesson note on Statistics (WAEC)

Statistics

View Answer

Question 3 Report

$g \circ h$ is

one- to- one

onto Correct answer

a relation

a series

Answer Details

Read lesson note on Binary Operations (WAEC)

Binary Operations

View Answer

Question 4 Report

Evaluate $\log_{10}(\frac{1}{3} + \frac{1}{4}) + 2\log_{10} 2 + \log_{10} (\frac{3}{7})$

-3

0 Correct answer

$\frac{5}{6}$

Answer Details

To simplify this expression, we can first use the identity: $$\log_{a}(b) + \log_{a}(c) = \log_{a}(bc)$$ Using this identity, we can simplify the given expression as follows: \begin{align*} \log_{10}\left(\frac{1}{3}+\frac{1}{4}\right) + 2\log_{10}(2) + \log_{10}\left(\frac{3}{7}\right) &= \log_{10}\left(\frac{7}{12}\right) + \log_{10}(2^2) + \log_{10}\left(\frac{3}{7}\right) \\ &= \log_{10}\left(\frac{7}{12} \cdot 2^2 \cdot \frac{3}{7}\right) \\ &= \log_{10}(1) \\ &= 0 \end{align*} Therefore, the answer is 0.

Read lesson note on Indices And Logarithmic Functions (WAEC)

Indices And Logarithmic Functions

View Answer

Question 5 Report

If P(x - 3) + Q(x + 1) = 2x + 3, find the value of (P + Q).

2 Correct answer

Answer Details

To find the value of (P + Q), we need to first expand the left-hand side of the equation using distributive property. P(x - 3) + Q(x + 1) = Px - 3P + Qx + Q Then we can simplify it by combining the like terms. Px + Qx - 3P + Q + 2x + 3 = 0 Now, we can group the like terms together: (P + Q)x - 3P + Q + 3 = 2x + 3 Since the coefficients of x on both sides of the equation are equal, we can equate the corresponding coefficients of x: (P + Q) = 2 Therefore, the value of (P + Q) is 2. So the correct answer is: - 2

Read lesson note on Functions (WAEC)

Functions

View Answer

Question 6 Report

The equation of a circle is $3x^{2} + 3y^{2} + 24x - 12y = 15$. Find its radius.

5 Correct answer

Answer Details

To find the radius of the circle, we need to use the standard form of the equation of a circle, which is $(x - a)^{2} + (y - b)^{2} = r^{2}$, where $(a, b)$ is the center of the circle and $r$ is the radius. To convert the given equation to standard form, we can complete the square for both $x$ and $y$: \begin{align*} 3x^{2} + 3y^{2} + 24x - 12y &= 15 \\ 3(x^{2} + 8x) + 3(y^{2} - 4y) &= 15 \\ 3(x^{2} + 8x + 16) + 3(y^{2} - 4y + 4) &= 15 + 3(16) + 3(4) \\ 3(x + 4)^{2} + 3(y - 2)^{2} &= 72 \\ (x + 4)^{2} + (y - 2)^{2} &= 8^{2} \end{align*} Comparing this with the standard form, we see that the center of the circle is $(-4, 2)$ and the radius is $8$. Therefore, the answer is (d) 5.

Read lesson note on Co-ordinate Geometry (WAEC)

Read lesson note on Polynomial Functions (WAEC)

Co-ordinate Geometry

Polynomial Functions

View Answer

Question 7 Report

A fair die is tossed twice. Find the probability of obtaining a 3 and a 5.

$\frac{5}{12}$

$\frac{2}{3}$

$\frac{1}{18}$

$\frac{1}{36}$ Correct answer

Answer Details

When a die is tossed, there are six possible outcomes, each with equal probability. Therefore, the probability of getting a 3 on the first toss is 1/6, and the probability of getting a 5 on the second toss is also 1/6. Since we want both events to occur, we need to multiply their probabilities: \[\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\] So, the probability of obtaining a 3 and a 5 is 1/36. Therefore, the correct option is: - $\frac{1}{36}$

Read lesson note on Probability (WAEC)

Probability

View Answer

Question 8 Report

The distance s in metres covered by a particle in t seconds is $s = \frac{3}{2}t^{2} - 3t$. Find its acceleration.

$1 ms^{-2}$

$2 ms^{-2}$

$3 ms^{-2}$ Correct answer

$4 ms^{-2}$

Answer Details

To find the acceleration, we need to differentiate the distance formula with respect to time (t): \begin{align*} s &= \frac{3}{2}t^{2} - 3t \\ \frac{d}{dt}s &= \frac{d}{dt}\left(\frac{3}{2}t^{2}\right) - \frac{d}{dt}(3t) \\ \frac{d}{dt}s &= 3t - 3 \\ \end{align*} Therefore, the acceleration is the second derivative of the distance formula with respect to time: \begin{align*} \frac{d^{2}}{dt^{2}}s &= \frac{d}{dt}(3t - 3) \\ \frac{d^{2}}{dt^{2}}s &= 3 \\ \end{align*} Thus, the acceleration of the particle is a constant value of 3 $ms^{-2}$. Therefore, the answer is $3 ms^{-2}$.

Read lesson note on Differentiation (WAEC)

Differentiation

View Answer

Question 9 Report

If (x + 1) is a factor of the polynomial $x^{3} + px^{2} + x + 6$. Find the value of p.

-8

-4 Correct answer

Answer Details

If (x + 1) is a factor of the polynomial, it means that if we substitute -1 in the polynomial, it should give us zero. Therefore: $(-1)^3 + p(-1)^2 -1 + 6 = 0$ Simplifying the above equation, we get: $-1 + p + 5 = 0$ $p = -4$ Hence, the value of p is -4. So the correct answer is.

Read lesson note on Sets (WAEC)

Read lesson note on Polynomial Functions (WAEC)

Sets

Polynomial Functions

View Answer

Question 10 Report

Differentiate $\frac{x}{x + 1}$ with respect to x.

$\frac{1}{x + 1}$

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

$\frac{1 - x}{x + 1}$

$\frac{1 - x}{(x + 1)^{2}}$

Answer Details

To differentiate $\frac{x}{x + 1}$ with respect to x, we can use the quotient rule of differentiation, which states that for functions u(x) and v(x), the derivative of $\frac{u(x)}{v(x)}$ is given by: \[\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}\] Applying this rule to the given function, we have: \[u(x) = x\] \[v(x) = x + 1\] So, we need to find u'(x) and v'(x): \[u'(x) = 1\] \[v'(x) = 1\] Substituting these values into the quotient rule formula, we get: \[\frac{d}{dx} \left( \frac{x}{x + 1} \right) = \frac{1(x + 1) - x(1)}{(x + 1)^2}\] Simplifying the numerator and denominator, we get: \[\frac{d}{dx} \left( \frac{x}{x + 1} \right) = \frac{1}{(x + 1)^2}\] Therefore, the correct answer is $\frac{1}{(x + 1)^2}$.

View Answer

Question 11 Report

Find the values of x at the point of intersection of the curve $y = x^{2} + 2x - 3$ and the lines $y + x = 1$.

(1, -2)

(0, 4)

(2, -3)

(1, -4) Correct answer

Answer Details

To find the point of intersection of the curve and the line, we need to solve the system of equations formed by equating the two equations: \begin{align*} y &= x^2 + 2x - 3 \\ y &= -x + 1 \end{align*} Setting the right-hand sides equal to each other, we get: \begin{align*} x^2 + 2x - 3 &= -x + 1 \\ x^2 + 3x - 4 &= 0 \\ (x + 4)(x - 1) &= 0 \end{align*} Thus, the values of x at the points of intersection are x = -4 and x = 1. To find the corresponding y-values, we substitute these values of x back into either equation. Using the equation y = x^2 + 2x - 3, we get: \begin{align*} y &= (-4)^2 + 2(-4) - 3 \\ &= 7 \end{align*} and \begin{align*} y &= (1)^2 + 2(1) - 3 \\ &= 0 \end{align*} Therefore, the points of intersection are (-4, 7) and (1, 0). So, the correct answer is (1, -4).

Read lesson note on Integration (WAEC)

Read lesson note on Polynomial Functions (WAEC)

Integration

Polynomial Functions

View Answer

Question 12 Report

A stone is dropped from a height of 45m. Find the time it takes to hit the ground. $[g = 10 ms^{-2}]$

3.0 seconds Correct answer

4.5 seconds

5.0 seconds

9.0 seconds

Answer Details

To solve this problem, we can use the formula: \[ s = ut + \frac{1}{2}at^2 \] where s is the distance, u is the initial velocity, t is the time, and a is the acceleration due to gravity. In this case, the initial velocity is zero because the stone is dropped from rest. We also know that the distance is 45m and the acceleration due to gravity is 10 $ms^{-2}$. Thus, we have: \begin{align*} s &= ut + \frac{1}{2}at^2 \\ 45 &= 0t + \frac{1}{2}(10)t^2 \\ 45 &= 5t^2 \\ t^2 &= 9 \\ t &= 3 \end{align*} Therefore, the time it takes for the stone to hit the ground is 3 seconds. Hence, the answer is (a) 3.0 seconds.

Read lesson note on Probability (WAEC)

Probability

View Answer

Question 13 Report

A straight line makes intercepts of -3 and 2 on the x- and y- axes respectively. Find the equation of the line.

2x + 3y + 6 = 0

3x - 2y - 6 = 0

-3x + 2y - 6 = 0

-2x + 3y - 6 =0 Correct answer

Answer Details

Read lesson note on Co-ordinate Geometry (WAEC)

Co-ordinate Geometry

View Answer

Question 14 Report

From the diagram above, which of the following represents the vector V in component form?

$-5\sqrt{3} i + 5j$

$3i + 3\sqrt{3}j$ Correct answer

$i + 2\sqrt{3}j$

$-2\sqrt{3}i - j$

Answer Details

Read lesson note on Vectors (WAEC)

Vectors

View Answer

Question 15 Report

A binary operation, $\Delta$, is defined on the set of real numbers by $a \Delta b = a + b + 4$. Find the identity element.

$\frac{1}{4}$

-4 Correct answer

Answer Details

An identity element in a binary operation is an element such that when it operates with any other element of the set, it does not change the other element. In this case, we need to find an element, say "x," such that for any real number "a," $$a \Delta x = a$$ Substituting the given definition of the operation, we get: $$a + x + 4 = a$$ Solving for x, we get: $$x = -4$$ Thus, -4 is the identity element for the given binary operation.

Read lesson note on Binary Operations (WAEC)

Binary Operations

View Answer

Question 16 Report

Two forces 10N and 6N act in the directions 060° and 330° respectively. Find the x- component of their resultant.

$5\sqrt{3} - 3$ Correct answer

$3 - 5\sqrt{3}$

$5 - 3\sqrt{3}$

$3\sqrt{3} - 5$

Answer Details

Read lesson note on Vectors (WAEC)

Vectors

View Answer

Question 17 Report

If the midpoint of the line joining (1 - k, -4) and (2, k + 1) is (-k, k), find the value of k.

-4

-3

-2

-1 Correct answer

Answer Details

We can start by using the midpoint formula which states that the midpoint of a line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$. So, the midpoint of the line joining (1-k, -4) and (2, k+1) is: \begin{align*} &\left(\frac{(1-k)+2}{2},\frac{(-4)+(k+1)}{2}\right) \\ &\Rightarrow \left(\frac{3-k}{2},\frac{k-3}{2}\right) \end{align*} We are given that the midpoint is (-k, k), so we can equate the x and y coordinates: \begin{align*} \frac{3-k}{2} &=-k \\ \frac{k-3}{2} &=k \end{align*} Solving for k, we get: \begin{align*} k &= 2 \\ \end{align*} Therefore, the value of k is 2, which is the answer option labeled as "-2".

Read lesson note on Surds (WAEC)

Read lesson note on Vectors (WAEC)

Surds

Vectors

View Answer

Question 18 Report

Given that $^{n}P_{r} = 90$ and $^{n}C_{r} = 15$, find the value of r.

3 Correct answer

Answer Details

We know that: $$^{n}P_{r} = \frac{n!}{(n-r)!} = 90$$ and $$^{n}C_{r} = \binom{n}{r} = \frac{n!}{r!(n-r)!} = 15$$ To find the value of r, we can use the formula: $$^{n}C_{r} = \frac{^{n}P_{r}}{r!}$$ Substituting the given values, we get: $$15 = \frac{90}{r!}$$ Simplifying the equation, we get: $$r! = 6$$ The only integer value of r that satisfies this equation is 3, since 3! = 6. Therefore, the answer is r = 3.

Read lesson note on Permutation And Combinations (WAEC)

Permutation And Combinations

View Answer

Question 19 Report

Given that $\sin x = \frac{-\sqrt{3}}{2}$ and $\cos x > 0$, find x.

300° Correct answer

240°

120°

60°

Answer Details

Read lesson note on Trigonometry (WAEC)

Trigonometry

View Answer

Question 20 Report

Given that $\sqrt{6}, 3\sqrt{2}, 3\sqrt{6}, 9\sqrt{2},...$ are the first four terms of an exponential sequence (G.P), find in its simplest form the 8th term.

$27\sqrt{2}$

$27\sqrt{6}$

$81\sqrt{2}$ Correct answer

$81\sqrt{6}$

Answer Details

Read lesson note on Surds (WAEC)

Surds

View Answer

Question 21 Report

If $y = x^{3} - x^{2} - x + 6$, find the values of x at the turning point.

$\frac{1}{2}, 3$

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

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

$1, \frac{1}{3}$

Answer Details

Read lesson note on Polynomial Functions (WAEC)

Polynomial Functions

View Answer

Question 22 Report

If r denotes the correlation coefficient between two variables, which of the following is always true?

$0 < r \leq 1$

$-1 \leq r < 1$

$-1 < r \leq 0$

$-1 \leq r \leq 1$ Correct answer

Answer Details

The correct answer is: $-1 \leq r \leq 1$. The correlation coefficient (r) measures the degree of linear association between two variables. The value of r ranges from -1 to 1, where -1 indicates a perfect negative correlation, 0 indicates no correlation, and 1 indicates a perfect positive correlation. Since r can take on any value between -1 and 1 (including -1 and 1), the correct statement is $-1 \leq r \leq 1$, which means that the correlation coefficient is always between -1 and 1, inclusive. Therefore, options (A), (B), and (C) are incorrect.

Read lesson note on Statistics (WAEC)

Statistics

View Answer

Question 23 Report

The marks obtained by 10 students in a test are as follows: 3, 7, 6, 2, 8, 5, 9, 1, 4 and 10. Find the variance.

8.25 Correct answer

8.50

9.00

9.17

Answer Details

Read lesson note on Statistics (WAEC)

Statistics

View Answer

Question 24 Report

A polynomial is defined by $f(x + 1) = x^{3} + px^{2} - 4x + 2$, find f(2).

-8

-2

2 Correct answer

Answer Details

Read lesson note on Polynomial Functions (WAEC)

Polynomial Functions

View Answer

Question 25 Report

Solve the inequality $x^{2} - 2x \geq 3$

$-1 \leq x \leq 3$

$x \geq 3$ and $x \leq -1$ Correct answer

$x \geq 3$ or $x < -1$

$-1 \leq x < 3$

Answer Details

Read lesson note on Polynomial Functions (WAEC)

Polynomial Functions

View Answer

Question 26 Report

Find the unit vector in the direction of the vector $-12i + 5j$.

$\frac{-12i}{13} - \frac{5j}{13}$

$\frac{-1i}{13} + \frac{5j}{13}$

$\frac{-12i}{13} + \frac{5j}{13}$ Correct answer

$\frac{-5i}{13} + \frac{12j}{13}$

Answer Details

To find the unit vector in the direction of the vector $-12i + 5j$, we need to divide the vector by its magnitude. The magnitude of a vector with components $a$ and $b$ is given by the formula $\sqrt{a^2+b^2}$. So, the magnitude of the vector $-12i + 5j$ is $\sqrt{(-12)^2+5^2} = 13$. Now, to get the unit vector, we divide each component of the vector by its magnitude: \[\frac{-12}{13}i + \frac{5}{13}j\] This is the unit vector in the direction of the vector $-12i + 5j$. Therefore, the correct option is $\frac{-12i}{13} + \frac{5j}{13}$.

Read lesson note on Vectors (WAEC)

Vectors

View Answer

Question 27 Report

Simplify: $\frac{\cos 2\theta - 1}{\sin 2\theta}$

$-\tan \theta$ Correct answer

$-\cos \theta$

$\tan \theta$

$\cos \theta$

Answer Details

Read lesson note on Trigonometry (WAEC)

Trigonometry

View Answer

Question 28 Report

In computing the mean of 8 numbers, a boy mistakenly used 17 instead of 25 as one of the numbers and obtained 20 as the mean. Find the correct mean

21 Correct answer

Answer Details

Let's call the sum of the 8 correct numbers "S" and the incorrect number that was added "x". The mean of the 8 numbers is: S/8 But the boy used 17 instead of 25, so the sum he used was: S + 17 - 25 = S - 8 And he got a mean of 20, so we can set up the equation: (S - 8)/8 = 20 Solving for S, we get: S = 168 So the correct mean is: S/8 = 168/8 = 21 Therefore, the correct mean is 21, which is.

Read lesson note on Statistics (WAEC)

Statistics

View Answer

Question 29 Report

Which of the following is nor a measure of central tendency?

Mean

Variance Correct answer

Median

Mode

Answer Details

Variance is not a measure of central tendency. Measures of central tendency are used to describe the typical or central value of a set of data, while variance is a measure of how spread out the data is from the mean. Variance is a measure of variability, not centrality. Therefore, the answer is "Variance."

Read lesson note on Statistics (WAEC)

Statistics

View Answer

Question 30 Report

A box contains 4 red and 3 blue identical balls. If two are picked at random, one after the other without replacement, find the probability that one is red and the other is blue.

$\frac{4}{7}$ Correct answer

$\frac{2}{7}$

$\frac{1}{7}$

$\frac{1}{12}$

Answer Details

Read lesson note on Probability (WAEC)

Probability

View Answer

Question 31 Report

The diagram above is a velocity- time graph of a moving object. Calculate the distance travelled when the acceleration is zero.

300 m

90 m

150 m Correct answer

210 m

Answer Details

Read lesson note on Surds (WAEC)

Surds

View Answer

Question 32 Report

Given that $P = \begin{pmatrix} 2 & 1 \\ 5 & -3 \end{pmatrix}$ and $Q = \begin{pmatrix} 4 & -8 \\ 1 & -2 \end{pmatrix}$, Find (2P - Q).

$\begin{pmatrix} -6 & 17 \\ 3 & 1 \end{pmatrix}$

$\begin{pmatrix} -2 & 9 \\ 4 & 1 \end{pmatrix}$

$\begin{pmatrix} 0 & -6 \\ 9 & -8 \end{pmatrix}$

$\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}$ Correct answer

Answer Details

To find the value of (2P - Q), we first need to compute 2P and Q, and then subtract Q from 2P. To compute 2P, we multiply each element of matrix P by 2: 2P = $\begin{pmatrix} 4 & 2 \\ 10 & -6 \end{pmatrix}$ To subtract Q from 2P, we subtract each corresponding element of matrix Q from matrix 2P: 2P - Q = $\begin{pmatrix} 4-4 & 2+8 \\ 10-1 & -6+2 \end{pmatrix}$ = $\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}$ Therefore, the answer is $\begin{pmatrix} 0 & 10 \\ 9 & -4 \end{pmatrix}$.

Read lesson note on Matrices And Linear Transformation (WAEC)

Matrices And Linear Transformation

View Answer

Question 33 Report

Find the constant term in the binomial expansion of $(2x - \frac{3}{x})^{8}$.

90720 Correct answer

1296

1120

672

Answer Details

Read lesson note on Binomial Theorem (WAEC)

Binomial Theorem

View Answer

Question 34 Report

Evaluate $\int_{-2}^{3} (3x^{2} - 2x - 12) \mathrm {d} x$

-30 Correct answer

-18

-6

Answer Details

Read lesson note on Integration (WAEC)

Integration

View Answer

Question 35 Report

QRS is a triangle such that $\overrightarrow{QR} = (3i + 2j)$ and $\overrightarrow{SR} = (-5i + 3j)$, find $\overrightarrow{SQ}$.

8i + j Correct answer

2i - j

-2i - 3j

-8i - j

Answer Details

To find $\overrightarrow{SQ}$, we can use the fact that $\overrightarrow{SR} = \overrightarrow{SQ} + \overrightarrow{QR}$. Rearranging this equation to solve for $\overrightarrow{SQ}$ gives us: $$\overrightarrow{SQ} = \overrightarrow{SR} - \overrightarrow{QR} = (-5i + 3j) - (3i + 2j) = -8i + j$$ Therefore, the value of $\overrightarrow{SQ}$ is -8i + j. So, the correct option is (A) 8i + j.

Read lesson note on Vectors (WAEC)

Vectors

View Answer

Question 36 Report

The angle of a sector of a circle is 0.9 radians. If the radius of the circle is 4cm, find the length of the arc of the sector.

3.6 cm Correct answer

7.6 cm

8.0 cm

11.6 cm

Answer Details

The formula for finding the length of an arc of a sector is given by L = rθ, where L is the length of the arc, r is the radius of the circle, and θ is the angle in radians. Using this formula and the given values, we have: L = 4 x 0.9 L = 3.6 cm Therefore, the length of the arc of the sector is 3.6 cm. Answer: 3.6 cm.

Read lesson note on Trigonometry (WAEC)

Trigonometry

View Answer

Question 37 Report

Which of the following sets is equivalent to $(P \cup Q) \cap (P \cup Q')$?

P Correct answer

$P \cap Q$

$P \cup Q$

$\emptyset$

Answer Details

Read lesson note on Sets (WAEC)

Sets

View Answer

Question 38 Report

Simplify $\frac{x^{3n + 1}}{x^{2n + \frac{5}{2}}(x^{2n - 3})^{\frac{1}{2}}}$

$0$

$-\frac{1}{2}$

$1$ Correct answer

$10$

Answer Details

Read lesson note on Rational Functions (WAEC)

Rational Functions

View Answer

Question 39 Report

Find the acute angle between the lines 2x + y = 4 and -3x + y + 7 = 0.

40°

44°

45° Correct answer

54°

Answer Details

To find the acute angle between two lines, we need to find the angle between their respective direction vectors. The direction vector of the line 2x + y = 4 is v = i + 2j and the direction vector of the line -3x + y + 7 = 0 is w = -3i + j. The acute angle θ between two vectors a and b is given by the formula cos(θ) = (a.b) / (|a|.|b|), where a.b is the dot product of the vectors and |a| and |b| are their respective magnitudes. Using this formula, we can find cos(θ) for the given direction vectors as follows: cos(θ) = (v.w) / (|v|.|w|) = (-5) / (√5.√10) = -1/√2 Since we are looking for the acute angle, we need to take the inverse cosine of -1/√2 in the range [0, π/2]: θ = cos^-1(-1/√2) ≈ 45° Therefore, the acute angle between the given lines is approximately 45°, which is.

Read lesson note on Trigonometry (WAEC)

Trigonometry

View Answer

Question 40 Report

Find the number of different arrangements of the word IKOTITINA.

30240 Correct answer

60840

120960

362880

Answer Details

Read lesson note on Permutation And Combinations (WAEC)

Permutation And Combinations

View Answer






Press & Hold to Drag Around
Click Here to Close

WAEC SSCE - Further Mathematics - 2012 (Objective)