WAEC SSCE - General Mathematics - 2008

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We can use the formula for simple interest, which is: I = P * r * t Where I is the interest earned, P is the principal (initial amount invested), r is the annual interest rate, and t is the time in years. In this problem, we know that the interest earned is 0.36P, the principal is P, and the time is 4 years. Substituting these values into the formula, we get: 0.36P = P * r * 4 Simplifying the equation, we can divide both sides by P * 4: 0.36 = r * 4 r = 0.36 / 4 r = 0.09 or 9% Therefore, the answer is option (C) 9.

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The sum of the angles of a quadrilateral is 360^o. Therefore, (x + 10)^o + 2y^o + 90^o + (100 - y)^o = 360^o. Simplifying the equation, we have: x + 10 + 2y + 90 + 100 - y = 360 x + 200 + y = 360 y = 160 - x Therefore, y is equal to 160 subtracted by x. Hence, the correct option is y = 160 - x.

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To simplify the expression, we can start by getting a common denominator for the two fractions in the numerator. The common denominator is xy. So, we can write: \begin{align*} \frac{\frac{1}{x} + \frac{1}{y}}{x + y} &= \frac{\frac{y}{xy} + \frac{x}{xy}}{x + y} \\ &= \frac{\frac{x + y}{xy}}{x + y} \\ &= \frac{1}{xy} \end{align*} Therefore, the simplified expression is \(\frac{1}{xy}\).

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To find the length of a side of a cube, we need to take the cube root of its volume. In this case, the volume is given as 343cm³. Taking the cube root of 343, we get: ∛343 = 7 Therefore, the length of the side of the cube is 7cm. So the correct answer is 7cm.

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We know that cos (90^o - A) = sin A. So, cos (x + 25^o) = cos (90^o - 45^o) = cos 45^o. Since cos A = cos B implies A = ±B + 360^on, we have: x + 25^o = ±45^o + 360^on x = -25^o ± 45^o + 360^on x = 20^o + 360^on or x = 70^o + 360^on Since we are given the range of x, we take x = 20^o. Therefore, the value of x is 20^o.

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To simplify the given expression, we need to factor the numerator and the denominator and then simplify the fraction by canceling out any common factors. The numerator can be factored as follows: \begin{align*} 2x^2 - 5x - 12 &= 2x^2 - 8x + 3x - 12 \\ &= 2x(x - 4) + 3(x - 4) \\ &= (2x + 3)(x - 4) \end{align*} The denominator can be factored using the difference of squares formula as: \begin{align*} 4x^2 - 9 &= (2x)^2 - 3^2 \\ &= (2x - 3)(2x + 3) \end{align*} Now we can simplify the fraction as follows: \begin{align*} \frac{2x^2 - 5x - 12}{4x^2 - 9} &= \frac{(2x + 3)(x - 4)}{(2x - 3)(2x + 3)} \\ &= \frac{x - 4}{2x - 3} \end{align*} Therefore, the simplified form of the expression is \(\frac{x - 4}{2x - 3}\). So, option (B) is the correct answer.

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Substitute p = 2 in the equation p - 2g + 1 = g + 3p to get: 2 - 2g + 1 = g + 3(2) -2g = 3 - 2 -2g = 1 g = -1/2 Therefore, the answer is -1.

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Let's start by translating the sentence into an equation: 4y = 9 + y + 3x We can simplify this equation by subtracting y from both sides: 3y = 9 + 3x Now, let's isolate x by subtracting 9 from both sides and then dividing by 3: x = (3y - 9)/3 = y - 3 So we see that x is 3 less than y. Therefore, the answer is (A) 3.

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Given that p = \(\frac{1}{2}\) and \(\frac{1}{p - 1} = \frac{2}{p + x}\), we want to find the value of x. We can start by substituting the value of p in the second equation: \begin{aligned} \frac{1}{p - 1} &= \frac{2}{p + x} \\ \\ \frac{1}{\frac{1}{2} - 1} &= \frac{2}{\frac{1}{2} + x} && \text{(Substitute } p = \frac{1}{2} \text{)}\\ \\ \frac{1}{-\frac{1}{2}} &= \frac{2}{\frac{1}{2} + x} && \text{(Simplify)}\\ \\ -2 &= \frac{2}{\frac{1}{2} + x} && \text{(Simplify)}\\ \\ -2\left(\frac{1}{2} + x\right) &= 2 && \text{(Cross-multiply)}\\ \\ -1 - 2x &= 1 && \text{(Simplify)}\\ \\ -2x &= 2 \\ \\ x &= -1 \end{aligned} Therefore, the value of x is -1.5, which corresponds to option B: -1\(\frac{1}{2}\).

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Given that p = \(\frac{3}{5} \sqrt{\frac{q}{r}}\), we need to express q in terms of p and r. Let's first isolate the term containing q on one side of the equation: \begin{align*} p &= \frac{3}{5} \sqrt{\frac{q}{r}} \\ \frac{5}{3}p &= \sqrt{\frac{q}{r}} \\ \left(\frac{5}{3}p\right)^2 &= \frac{q}{r} \\ \frac{25}{9} p^2r &= q \\ \end{align*} Therefore, q is equal to \(\frac{25}{9} p^2r\). So the correct option is: \(\frac{25}{9} p^2r\)

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The formula for the length of an arc is L = rθ where L is the length of the arc, r is the radius of the circle, and θ is the angle subtended by the arc at the centre of the circle in radians. To find the angle in degrees, we need to convert radians to degrees by multiplying by 180/π. In this problem, the radius of the circle is 14cm and the length of the arc is 18.33cm. Plugging these values into the formula, we get: 18.33 = 14θ Solving for θ, we get: θ = 18.33/14 θ = 1.3093 radians To convert to degrees, we multiply by 180/π: θ = 1.3093 × 180/π θ ≈ 75 degrees Therefore, the angle which the arc subtends at the centre of the circle is approximately 75 degrees. Answer: 75^o

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To find the total number of staff in the office, we need to add the number of staff who own a Mercedes and the number of staff who own a Toyota, but we must subtract the number of staff who own both cars to avoid counting them twice. Using the given information, we know that there are 20 staff who own a Mercedes, 15 staff who own a Toyota, and 5 staff who own both. Therefore, the number of staff who own either a Mercedes or a Toyota is: 20 + 15 - 5 = 30 Therefore, there are 30 staff in the office who own either a Mercedes or a Toyota.

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The average speed of the train is given as 400km/hr. This means that in one hour, the train travels 400km. We can convert this to km/minute by dividing by 60 (the number of minutes in an hour). Therefore, the speed of the train in km/minute is: 400km/hr ÷ 60 = 6.67km/min (rounded to two decimal places) We are given that the train travels 60km in M minutes. We can use the formula: speed = distance/time to find the value of M. Rearranging the formula gives: time = distance/speed Substituting the given values, we get: M = 60km ÷ 6.67km/min M = 9 minutes (rounded to the nearest minute) Therefore, the value of M is 9.

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We can start by simplifying the numerator first: \begin{align*} \log \sqrt{8} &= \log \left(\sqrt{4}\cdot\sqrt{2}\right) \\ &= \log 4 + \log \sqrt{2} \\ &= 2 + \frac{1}{2} \log 2 \end{align*} Similarly, we can simplify the denominator: \begin{align*} \log 4 - \log 2 &= \log\frac{4}{2} \\ &= \log 2 \end{align*} Now we can substitute these simplified expressions back into the original expression: \begin{align*} \frac{\log \sqrt{8}}{\log 4 - \log 2} &= \frac{2 + \frac{1}{2} \log 2}{\log 2} \\ &= \frac{2}{\log 2} + \frac{1}{2} \\ &= \boxed{\frac{3}{2}} \end{align*} Therefore, the correct option is (c) \(\frac{3}{2}\).

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We can start solving the problem by setting up an equation based on the information given. The problem states that "x% of 240 equals 12". In other words, we can translate this into an equation: x/100 * 240 = 12 To solve for x, we can isolate it by dividing both sides of the equation by 240/100, which simplifies to 2.4: x/100 * 240 = 12 x = 12 / 2.4 x = 5 Therefore, the value of x that satisfies the given equation is 5.

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To factorize 5y^2 + 2ay - 3a^2, we need to find two numbers that multiply to give -15a^2 and add up to 2a. These numbers are 5a and -3a. We can then rewrite the expression as: 5y^2 + 5ay - 3ay - 3a^2 Factorising by grouping: (5y^2 + 5ay) - (3ay + 3a^2) Factorising out the common factor in each group: 5y(y + a) - 3a(y + a) Factorising out the common factor (y + a): (5y - 3a)(y + a) Therefore, the answer is (5y - 3a)(y + a).

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To find the mean deviation of a set of numbers, we first find the mean (average) of the numbers. The mean of the numbers 10, 12, 14, 15, 17, and 19 is: (mean) = (10 + 12 + 14 + 15 + 17 + 19) / 6 = 87 / 6 = 14.5 Next, we find the deviation of each number from the mean by subtracting the mean from each number: 10 - 14.5 = -4.5 12 - 14.5 = -2.5 14 - 14.5 = -0.5 15 - 14.5 = 0.5 17 - 14.5 = 2.5 19 - 14.5 = 4.5 We take the absolute value of each deviation to ensure that they are all positive: |-4.5| = 4.5 |-2.5| = 2.5 |-0.5| = 0.5 |0.5| = 0.5 |2.5| = 2.5 |4.5| = 4.5 Then we find the mean of the absolute deviations: (mean deviation) = (4.5 + 2.5 + 0.5 + 0.5 + 2.5 + 4.5) / 6 = 2.5 Therefore, the mean deviation of the numbers is 2.5. Therefore, the correct option is (a) 2.5.

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The given inequality is: 1 - 2x < -\(\frac{1}{3}\) To solve this inequality, we need to isolate the variable, which in this case is x. First, we can simplify the inequality by adding \(\frac{1}{3}\) to both sides: 1 - 2x + \(\frac{1}{3}\) < 0 Next, we can combine like terms: \(\frac{4}{3}\) - 2x < 0 Now, we can isolate x by subtracting \(\frac{4}{3}\) from both sides: -2x < -\(\frac{4}{3}\) Finally, we can solve for x by dividing both sides by -2, remembering to flip the inequality because we are dividing by a negative number: x > \(\frac{2}{3}\) Therefore, the solution to the inequality is x > \(\frac{2}{3}\).

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If \(\tan x = 1\), then we can find the values of sin x and cos x using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\). Since \(\tan x = \frac{\sin x}{\cos x} = 1\), we can say that \(\sin x = \cos x\). So we have, \[\sin^2 x + \cos^2 x = 2\cos^2 x = 1 \implies \cos x = \pm\frac{1}{\sqrt{2}}\] Using \(\sin x = \cos x\), we can say that \(\sin x = \pm\frac{1}{\sqrt{2}}\) Now, \[\sin x + \cos x = \pm\frac{1}{\sqrt{2}} \pm \frac{1}{\sqrt{2}} = \pm\sqrt{2}\] Since \(\tan x = 1\), x must be in the first quadrant, where both \(\sin x\) and \(\cos x\) are positive. Therefore, we have \[\sin x + \cos x = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}\] Hence, the answer is \(\sqrt{2}\). Therefore, the option that represents this answer is: $\sqrt{2}$.

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To find the area of the sector PQR, we need to know the angle at the center of the circle that the sector subtends. Since the radius of the circle is 4 cm and PQR = 30^o, the circumference of the circle is 2πr = 8π cm. The fraction of the circle that the sector PQR subtends is 30/360, or 1/12. Therefore, the area of the sector PQR is 1/12 of the total area of the circle: Area of sector PQR = (1/12) × πr² = (1/12) × π × (4 cm)² ≈ 3.347 cm² Rounded to 3 significant figures, the area of sector PQR is approximately 3.35 cm², which corresponds to the first option: 4.19 cm².

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If the roots of the quadratic equation are c and -c, then the equation can be expressed in the factored form as: (x - c)(x + c) = 0 Expanding the above expression: x^2 - c^2 = 0 Therefore, the quadratic equation whose roots are c and -c is x^2 - c^2 = 0. Hence, the correct option is: x² - c² = 0.

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(a) \(\dfrac25(x - 2) - \dfrac16(x + 5) \le 0\). Multiply through by the LCM \(30\):

\[12(x - 2) - 5(x + 5) \le 0\]

\[12x - 24 - 5x - 25 \le 0 \;\Rightarrow\; 7x - 49 \le 0 \;\Rightarrow\; 7x \le 49 \;\Rightarrow\; x \le 7\]

(b)(i) Factorise numerator and denominator of \(P = \dfrac{x^2 - y^2}{x^2 + xy}\):

\[P = \frac{(x - y)(x + y)}{x(x + y)} = \frac{x - y}{x}\]

(ii) With \(x = -4,\, y = -6\):

\[P = \frac{-4 - (-6)}{-4} = \frac{2}{-4} = -\frac12\]

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(a) Express both equations to base 2.

\(2^{x+y}=16=2^{4}\Rightarrow x+y=4\)

Since \(4=2^{2}\) and \(\frac{1}{32}=2^{-5}\): \(4^{x-y}=2^{2(x-y)}=2^{-5}\Rightarrow 2(x-y)=-5\Rightarrow x-y=-2.5\)

Solve the pair:

Adding: \(2x=1.5\Rightarrow x=0.75\). Then \(y=4-0.75=3.25\).

\(x=0.75,\ y=3.25\)

(b) \(P\propto\dfrac{Q^{2}}{R}\Rightarrow P=\dfrac{kQ^{2}}{R}\).

Using \(P=36,\ Q=3,\ R=4\): \(36=\dfrac{k(3)^{2}}{4}=\dfrac{9k}{4}\Rightarrow k=16\).

When \(P=200,\ R=2\): \(200=\dfrac{16Q^{2}}{2}=8Q^{2}\Rightarrow Q^{2}=25\Rightarrow Q=5\).

\(Q=5\)

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(a) For \(y=3\sin x+2\cos x\), the completed table is:

\(x\)	\(0^\circ\)	\(60^\circ\)	\(120^\circ\)	\(180^\circ\)	\(240^\circ\)	\(300^\circ\)	\(360^\circ\)
\(y\)	2.00	3.60	1.60	-2.00	-3.60	-1.60	2.00

For example, at \(x=60^\circ\), \(y=3(0.8660)+2(0.5)=3.598\approx3.60\).

(b) The graph, drawn using the stated scales, is shown below.

(c) Draw the horizontal line \(y=1.5\). Its intersections with the curve give

\[x\approx122^\circ\quad\text{or}\quad x\approx351^\circ.\]

(d) Draw the horizontal line \(y=-1\). The curve is below this line between its two points of intersection. Hence,

\[\boxed{162^\circ<x<310^\circ}.\]

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(a) Write \(3.6 = \dfrac{36}{10} = \dfrac{2^2 \times 3^2}{10}\):

\[\log 3.6 = 2\log 2 + 2\log 3 - \log 10\]

\[= 2(0.3010) + 2(0.4771) - 1 = 0.6020 + 0.9542 - 1 = 0.5562\]

(b) All numbers are in base two. Convert to base ten: \(101_2 = 5\), \(11_2 = 3\), \(10010_2 = 18\). The equation \(\dfrac{y}{y + 101} = \dfrac{11}{10010}\) becomes

\[\frac{y}{y + 5} = \frac{3}{18} = \frac16\]

\[6y = y + 5 \;\Rightarrow\; 5y = 5 \;\Rightarrow\; y = 1_{10}\]

Converting back to base two, \(y = 1_2\).

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(a) Construction

1. Draw the line segment \(PQ=7\text{ cm}\).
2. At \(P\), construct an angle of \(60^\circ\). On this ray, mark \(S\) such that \(PS=6.5\text{ cm}\).
3. Join \(Q\) to \(S\). With centre \(Q\) and radius \(QS\), draw an arc. At \(Q\), construct \(\angle PQR=135^\circ\); let this ray meet the arc at \(R\). Join \(R\) to \(S\). Thus \(QR=QS\).
4. Construct the perpendicular bisector of \(PQ\) and label it \(l_1\). This is the locus of points equidistant from \(P\) and \(Q\).
5. Construct the perpendicular bisector of \(PS\) and label it \(l_2\). This is the locus of points equidistant from \(P\) and \(S\).

(b)

1. The point of intersection of \(l_1\) and \(l_2\) is \(T\).
2. With centre \(T\) and radius \(TP\), draw the circle \(l_3\), as shown above. The circle passes through \(P\), \(Q\) and \(S\), since \(TP=TQ=TS\).

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(a) \(3,\ x,\ y,\ 18\) are consecutive terms of an A.P. The first term is \(3\) and the fourth term is \(18\):

So \(x=3+5=8\) and \(y=8+5=13\).

\(x=8,\ y=13\)

(b)(i) For a G.P. with first term \(a\) and ratio \(r\), terms are \(a,\ ar,\ ar^{2},\ ar^{3},\ldots\)

"Sum of second and third terms \(=6\times\) fourth term":

Divide through by \(ar\ (a,r\neq0)\): \(1+r=6r^{2}\Rightarrow 6r^{2}-r-1=0\).

\(r=\tfrac{1}{2}\) or \(r=-\tfrac{1}{3}\).

(ii) The ratio is positive, so \(r=\tfrac{1}{2}\). The second term is \(ar=8\):

First six terms: \(16,\ 8,\ 4,\ 2,\ 1,\ \tfrac{1}{2}\).

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(a) Finding x and y. From the diagram \(AB\parallel CD\), \(BC\parallel FE\), \(\widehat{CDE}=75^\circ\) and \(\widehat{DEF}=26^\circ\).

Draw a line through \(D\) parallel to \(BC\) and \(FE\). Because \(DE\) is a transversal of the parallel lines \(FE\) and this new line, the angle \(DE\) makes with the parallel family at \(D\) equals the angle it makes at \(E\):

Since \(\widehat{CDE}=75^\circ\) is the angle between \(DC\) and \(DE\), the angle between \(DC\) and the \(BC\)-family is

Now \(x=\widehat{ABC}\) is the angle at \(B\) between \(BA\) (in the \(CD\)-family, since \(AB\parallel CD\)) and \(BC\). This is exactly that same angle between the two families:

By alternate angles (\(AB\parallel CD\), transversal \(BC\)), \(\widehat{BCD}=\widehat{ABC}=49^\circ\). The marked angle \(y\) is the reflex angle at \(C\) (the loop over the peak):

(b) Circle \(ABCD\), centre \(O\), radius \(7\text{ cm}\). Reflex \(\widehat{AOC}=190^\circ\), so the ordinary \(\widehat{AOC}=360^\circ-190^\circ=170^\circ\).

(i) \(\widehat{ABC}\) stands on the arc \(ADC\) (the arc corresponding to the reflex \(190^\circ\)). Angle at circumference \(=\tfrac12\) angle at centre:

(ii) \(\widehat{ADC}\) stands on the arc \(ABC\) (central angle \(170^\circ\)):

(Check: \(ABCD\) is cyclic, \(95^\circ+85^\circ=180^\circ\).)

(c)(i) Arc \(ABC\). Central angle \(=170^\circ\), \(r=7\), \(\pi=3.142\):

(c)(ii) Chord \(AD\). In \(\triangle OAD\), \(OA=OD=7\) so it is isosceles with base angles \(\widehat{DAO}=\widehat{ODA}=35^\circ\). Hence \(\widehat{AOD}=180^\circ-2(35^\circ)=110^\circ\).

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(a) The exterior angles of any polygon add up to \(360^\circ\). The pentagon has five exterior angles: one is \(60^\circ\), two are \((90-m)^\circ\) each, and the remaining two are \((30+2m)^\circ\) each. Therefore:

(b) From the diagram PQR is a straight line, \(\overline{QR}=\sqrt{3}\text{ cm}\), \(\overline{SQ}=2\text{ cm}\), and the right-angle mark at R shows \(SR\perp QR\). So triangle SQR is right-angled at R, with SQ the hypotenuse.

In right-angled triangle SQR, using \(\angle SQR\):

Since PQR is a straight line, \(\angle PQS\) and \(\angle SQR\) are angles on a straight line at Q:

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None

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(a) \(4x^{2}=11x+21\Rightarrow 4x^{2}-11x-21=0\).

Using the formula with \(a=4,\ b=-11,\ c=-21\):

\(\sqrt{457}=21.378\). So \(x=\dfrac{11+21.378}{8}=4.047\) or \(x=\dfrac{11-21.378}{8}=-1.297\).

\(x\approx 4.05\) or \(x\approx -1.30\)

(b) Principal \(=\pounds1500\); after one year the amount is \(\pounds1560\).

(i) First-year interest \(=1560-1500=\pounds60\).

(ii) Interest for the second year is charged on \(\pounds1560\):

(c) Car costs \(N300{,}000\), depreciating \(25\%\) then \(20\%\).

After year 1: \(300000\times(1-0.25)=300000\times0.75=N225{,}000\).

After year 2: \(225000\times(1-0.20)=225000\times0.80=N180{,}000\).

Value after 2 years \(=N180{,}000.00\)

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(a)(i) Largest angle. The largest angle faces the longest side \(AB=17\) m, so it is angle \(C\), between sides \(CA=10\) and \(CB=12\). By the cosine rule:

\(C=\cos^{-1}(-0.1875)\approx 100.8^{\circ}\).

Largest angle \(\approx 100.8^{\circ}\).

(ii) Area.

(b) The tower \(RS\) is vertical with \(RS=38\) m; from \(T\) on the ground the elevation of the top \(R\) is \(63^{\circ}\). \(TS\) is the horizontal distance sought.

Distance \(TS\approx 19\) m (nearest metre).

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Total monthly income

\[4500+3000+2500+2000+5000+3000=N20\,000\]

(a) Pie chart

The angle for each sector is:

\[\text{Sector angle}=\frac{\text{amount spent}}{20\,000}\times360^\circ\]

Item	Amount (₦)	Sector angle
Food	4,500	\(\frac{4500}{20000}\times360^\circ=81^\circ\)
House Rent	3,000	\(54^\circ\)
Provisions	2,500	\(45^\circ\)
Electricity	2,000	\(36^\circ\)
Transportation	5,000	\(90^\circ\)
Others	3,000	\(54^\circ\)
Total	20,000	\(360^\circ\)

The required pie chart is:

(b)

\[\text{Percentage spent on transportation}=\frac{5000}{20000}\times100\%=25\%\]

Therefore, 25% of his income is spent on transportation.

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(a) Frequency distribution

Age (years)	Tally	Class mark, \(x\)	Frequency, \(f\)	\(fx\)
21 - 25	||||| ||	23	7	161
26 - 30	||||| ||||| |	28	11	308
31 - 35	||||| ||||	33	9	297
36 - 40	||||| ||||	38	9	342
41 - 45	||||| |	43	6	258
46 - 50	||||| |||	48	8	384
Total			50	1750

(b) Histogram

The continuous class boundaries are \(20.5\), \(25.5\), \(30.5\), \(35.5\), \(40.5\), \(45.5\) and \(50.5\). Since all class widths are 5 years, the bar heights are the frequencies.

(c) Modal age

The modal class is \(26\text{ - }30\), for which \(L=25.5\), \(f_1=11\), \(f_0=7\), \(f_2=9\), and class width \(h=5\).

\[\text{Mode}=L+\left(\frac{f_1-f_0}{(f_1-f_0)+(f_1-f_2)}\right)h\]

\[=25.5+\left(\frac{11-7}{(11-7)+(11-9)}\right)\times5=25.5+\frac{4}{6}\times5=28.8\]

Hence, the estimated mode is \(28.8\) years.

(d) Mean age

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{1750}{50}=35\]

Therefore, the mean age is 35 years.