WAEC SSCE - General Mathematics - 1988

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Since BOC is a diameter, then angle BAC is 90 degrees (as it is an inscribed angle that subtends a diameter). Also, angle ADC is 37 degrees (given). Therefore, angle ACB can be found using the fact that the angles in a triangle sum to 180 degrees. Thus: angle ACB = 180 - angle BAC - angle ADC = 180 - 90 - 37 = 53 degrees Therefore, the answer is 53 degrees.

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Let's use the fact that cosine is an even function, meaning that cos(-x) = cos(x). Therefore, we can find the angle with cosine of -1/2 by finding the angle with cosine of 1/2 and then taking its negative. We know that cos(60°) = 1/2. The cosine function is positive in the first and fourth quadrants, so we need to look for angles in those quadrants where cos(x) = 1/2. In the first quadrant, the reference angle for which cosine is 1/2 is 60°. In the fourth quadrant, the reference angle is 360° - 60° = 300°. Taking the negative of these reference angles, we get that the angles with cosine of -1/2 are: - 60° - 180° = -120° - 300° - 180° = 120° Therefore, the answer is 120°.

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Abidjan is 4^o west of Accra, and they are on the same circle of latitude. Since the circumference of a circle of radius r is given by 2πr, the distance between two points on the same circle of latitude is proportional to the angle between them. Since Abidjan is 4^o west of Accra, the distance between them is 4/360 times the circumference of the circle of latitude. Using the formula for the circumference of a circle, we have: C = 2πr = 2 x 22/7 x 6370 = 40,014 km (approx.) Therefore, the distance between Accra and Abidjan is: 4/360 x 40,014 km = 445 km (approx.) Hence, the answer is 445km, correct to the nearest kilometer.

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The sales girl gave a change of N1.15 instead of N1.25, which means that she made an error of N0.10. To find the percentage error made by the sales girl, we can use the formula: Percentage error = (Error / Actual value) x 100% In this case, the error is N0.10 and the actual value is N1.25. Substituting these values into the formula, we get: Percentage error = (0.10 / 1.25) x 100% Percentage error = 0.08 x 100% Percentage error = 8% Therefore, the percentage error made by the sales girl is 8%.

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A fair die has six faces numbered from 1 to 6. Each face has an equal chance of showing up when the die is rolled. To find the probability of rolling an odd number, we need to count the number of odd faces on the die, which are 1, 3, and 5. Since there are three odd faces out of the six possible faces, the probability of rolling an odd number is: Probability of rolling an odd number = Number of odd faces / Total number of faces Probability of rolling an odd number = 3 / 6 Probability of rolling an odd number = 1/2 Therefore, the probability of having an odd number in a single toss of a fair die is 1/2.

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To simplify this expression, we can use the rules of exponents. - 36\(^{\frac{1}{2}}\) can be written as the square root of 36, which equals 6. - 64\(^{-\frac{1}{3}}\) can be written as the cube root of 64 raised to the power of -1, which equals 1/4. - 5\(^0\) equals 1. So, the expression becomes: 6 x 1/4 x 1 = 6/4 = 3/2 = 1.5. Therefore, the answer is 1¹/₂.

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The volume of the tank in cubic meters is 3008 m^3, as we found in question 22. To convert this volume to liters, we need to multiply by 1000, since there are 1000 liters in one cubic meter: V_liters = V_meters^3 * 1000 V_liters = 3008 * 1000 V_liters = 3,008,000 liters Therefore, we need to multiply our answer from question 22 by 1000 to determine the capacity of the tank in liters. The closest answer choice is 1000, which is the correct answer.

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The set Q contains all perfect squares less than 30. These are 1, 4, 9, 16, and 25. The set P contains all odd numbers from 1 to 10. These are 1, 3, 5, 7, and 9. The symbol ∩ means "intersection," which represents the common elements in two or more sets. To find Q ∩ P, we need to identify the elements that are present in both sets Q and P. Looking at the sets Q and P, we can see that the only element they have in common is 9. Therefore, Q ∩ P = {9}. Therefore, the answer is (1, 9).

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The correct method for constructing an angle of 60^o at Q is option II only, which involves using a compass to draw an arc from point Q, cutting the line segment PQ at point R. Then, a new arc is drawn from point R, cutting the previous arc at point S. The line segment QS passing through Q and S forms a 60^o angle with the line segment PQ. Option I only involves bisecting an angle of 120^o formed by the line segments PQ and PR, which would result in an angle of 60^o at Q only if the original angle was 120^o. However, the original angle is not given in the question. Option III only involves using a ruler to measure a distance of 3 cm from Q on the line segment PQ, which would not necessarily result in an angle of 60^o at Q.

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We can use the Pythagorean theorem to find the distance between points P and Q. The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In this case, we can consider P, Q, and R as the three vertices of a right-angled triangle, with PQ as the hypotenuse. Since P is 24 meters north of R and Q is 7 meters east of R, we can draw a right-angled triangle with vertical side 24 meters and horizontal side 7 meters. Then, using the Pythagorean theorem, we have: |PQ| = sqrt(24^2 + 7^2) |PQ| = sqrt(576 + 49) |PQ| = sqrt(625) |PQ| = 25 meters Therefore, the distance between points P and Q is 25 meters. Hence the answer is 25.

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In an arithmetic progression (A.P.), each term is obtained by adding a fixed constant (d) to the previous term. The common difference (d) between any two consecutive terms of an A.P. is constant. In this A.P., the first term is -3 and the common difference is 2, since each term is obtained by adding 2 to the previous term (-3 + 2 = -1, -1 + 2 = 1, and so on). To find the 8th term, we can use the formula: nth term = a + (n-1)d where nth term is the term we want to find (in this case, the 8th term), a is the first term, n is the position of the term we want to find, and d is the common difference. Substituting the given values, we get: 8th term = -3 + (8-1)*2 8th term = -3 + 14 8th term = 11 Therefore, the 8th term of the A.P. -3, -1, 1...... is 11.

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The expression (27^(1/3))^2 means we take the cube root of 27 first, which is 3, and then we square the result. (27^(1/3))^2 = 3^2 = 9 Therefore, the answer is 9.

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To find the value of tanθ, we can use the relationship: tanθ = sinθ / cosθ We are given sinθ = 3/5. To find cosθ, we can use the identity: sin^2θ + cos^2θ = 1 Substituting sinθ = 3/5, we get: (3/5)^2 + cos^2θ = 1 9/25 + cos^2θ = 1 cos^2θ = 16/25 cosθ = ±4/5 Since 0 < θ < 90°, we know that cosθ is positive. Therefore, cosθ = 4/5. Substituting sinθ = 3/5 and cosθ = 4/5 into the equation for tanθ, we get: tanθ = sinθ / cosθ tanθ = (3/5) / (4/5) tanθ = 3/4 Therefore, the value of tanθ for sinθ = 3/5 is 3/4.

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We can use the identity for the difference of squares: a^2 - b^2 = (a + b)(a - b) In this case, we have: (101.5)^2 - (100.5)^2 = [(101.5 + 100.5)(101.5 - 100.5)] = (202)(1) = 202 Therefore, the answer is option (D) 202.

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If the roots of a quadratic equation are given, we can express the quadratic equation in factored form. For the given problem, the roots are x = -2 or x = 7. This means that the factors of the quadratic equation are (x + 2) and (x - 7). Multiplying these factors, we get: (x + 2)(x - 7) = x² - 5x - 14 = 0 Therefore, the quadratic equation whose roots are x = -2 or x = 7 is x² - 5x - 14 = 0.

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To find the difference in longitude between P (50°W) and Q (150°W), we subtract the longitude of P from the longitude of Q: Longitude of Q - Longitude of P = 150°W - 50°W = 100° Therefore, the answer is 100°.

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To find the product of x and y, we can use the formula for the nth term of a geometric progression: an = a1 * r^(n-1) where an is the nth term, a1 is the first term, r is the common ratio, and n is the number of terms. Since x, 3/2, 6/7, y are in geometric progression, we can write: 3/2 = x * r 6/7 = 3/2 * r y = 6/7 * r Solving for r in the second equation: 6/7 = 3/2 * r r = (6/7) / (3/2) r = 4/7 Substituting r in the first and third equations: 3/2 = x * (4/7) x = (3/2) / (4/7) x = 21/8 y = 6/7 * (4/7) y = 24/49 Therefore, the product xy is: xy = (21/8) * (24/49) xy = 9/7 Therefore, the answer is: 9/7.

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To multiply two numbers in standard form, we multiply the decimal parts of the numbers and add the exponents of 10. So, to find the product of 0.06 and 0.09, we multiply 6 and 9 to get 54, and count the total number of decimal places in the factors, which is 4. Therefore, we express the product in standard form as 5.4 x 10^-3. Answer: 5.4 x 10^-3

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There are a total of 40 children in the class. The probability of choosing a child with a surname beginning with O is 16/40, and the probability of choosing a child with a surname beginning with A is 9/40. However, we have to be careful not to double-count the surnames that begin with both O and A. Therefore, we need to subtract the probability of choosing a child with a surname beginning with both O and A, which is the intersection of the two events. The intersection is the number of surnames that begin with both O and A, which is zero in this case because it is not given that any surname starts with both O and A. So the probability of choosing a child with a surname beginning with either O or A is: P(O or A) = P(O) + P(A) - P(O and A) P(O or A) = 16/40 + 9/40 - 0 P(O or A) = 25/40 Simplifying, we get: P(O or A) = 5/8 Therefore, the probability of choosing a child with a surname beginning with either O or A is 5/8. Answer: (A) 5/8.

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Radius \(R=6400\) km, \(\pi=3.142\), latitude \(46^{\circ}\N\), \(\cos46^{\circ}=0.6947\).

(i) Length of the parallel of latitude \(46^{\circ}\)N

\[L=2\pi R\cos46^{\circ}=2(3.142)(6400)(0.6947)=27{,}940\text{ km}\approx\mathbf{27{,}900\text{ km}}\ (\text{nearest }100).\]

(ii) Distance K to Q (same latitude). \(K\) at \(130^{\circ}\W\), \(Q\) at \(103^{\circ}\W\): difference \(=27^{\circ}\).

\[d_{KQ}=\frac{27}{360}\times27{,}940=2095\text{ km}\approx\mathbf{2100\text{ km}}\ (\text{nearest }100).\]

(iii) Distance Q to P along the parallel. \(Q\) at \(103^{\circ}\W\), \(P\) at \(23^{\circ}\E\): difference \(=103^{\circ}+23^{\circ}=126^{\circ}\).

\[d_{QP}=\frac{126}{360}\times27{,}940=9779\text{ km}\approx\mathbf{9780\text{ km}}\ (\text{nearest }10).\]

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(i) Bar chart of company production

The bar chart below represents the number of items produced in each year. The horizontal axis shows the years and the vertical axis shows the number of items produced.

(ii) Average production

\[\text{Average production}=\frac{4100+2500+1500+1800+9200}{5}=\frac{19100}{5}=3820\]

Therefore, the average production for the five-year period is 3,820 items.

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Construction

1. Draw the line segment \(AB=5\text{ cm}\).
2. At \(A\), construct an angle of \(120^\circ\). With centre \(A\) and radius \(7.5\text{ cm}\), cut this ray at \(C\). Join \(BC\).
3. With centres \(A\) and \(B\), and with the same compass radius greater than \(2.5\text{ cm}\), draw arcs above and below \(AB\) to intersect. Join the arc-intersection points. This is \(L_1\), the perpendicular bisector of \(AB\).
4. Construct the internal angle bisector of \(\angle BAC\). This is \(L_2\), the locus of points equidistant from the lines \(AB\) and \(AC\).
5. Label the intersection of \(L_1\) and \(L_2\) as \(P\).

Measurement:

Measuring the constructed line gives

\[|CP|\approx 6.6\text{ cm}.\]

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(a) \(A=\{m,a,p,e\}\), \(B=\{a,e,i,o,u\}\), \(C=\{l,m,n,o,p,q,r,s,t,u\}\).

• (i) \(A\cup B=\{a,e,i,m,o,p,u\}\).
• (ii) \(A\cup C=\{a,e,l,m,n,o,p,q,r,s,t,u\}\).
• (iii) \(B\cap C=\{o,u\}\), so \(A\cup(B\cap C)=\{a,e,m,o,p,u\}\).

(b) Total \(=400\); Biology \(n(B)=300\); Chemistry \(n(C)=190\); neither \(=70\).

(ii) At least one \(=400-70=\mathbf{330}\) students.

(i) By inclusion-exclusion, both \(=n(B)+n(C)-n(B\cup C)=300+190-330=\mathbf{160}\) students.

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(a) Using the scale 2 cm to 1 unit on both axes, plot the given points and draw a smooth curve through them. The completed graph is shown below.

\(x\)	-2	-1.5	-1	-0.5	0	0.5	1	1.5	2	2.5	3
\(y=2+x-x^2\)	-4	-1.75	0	1.25	2	2.25	2	1.25	0	-1.75	-4

(b) The greatest value of \(y\) is \(2.25\). It occurs when \(x=0.5\).

(c) For \(y=1-x\), use, for example, the points \((-2,3)\), \((0,1)\), \((1,0)\), \((2,-1)\) and \((3,-2)\). Join them with a straight line on the same axes, as shown.

(d) The required equation may be written as

\[1+2x-x^2=0\iff 2+x-x^2=1-x.\]

Hence its solutions are the \(x\)-coordinates of the intersections of \(y=2+x-x^2\) and \(y=1-x\). Reading from the graph gives

\[\boxed{x\approx -0.4\quad\text{or}\quad x\approx 2.4}.\]

(The exact values are \(x=1\pm\sqrt2\).)

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Rewrite the boundary equations as

\[ -x+5y=10\quad\Rightarrow\quad y=2+\frac{x}{5}, \]

\[ 3x-4y=8\quad\Rightarrow\quad y=-2+\frac{3x}{4}, \]

together with the vertical boundary \(x=-1\).

Use the following values to plot the two straight lines.

For \(-x+5y=10\)	\(x=-4\)	\(x=0\)	\(x=4\)
\(y=2+\frac{x}{5}\)	\(1.2\)	\(2\)	\(2.8\)

For \(3x-4y=8\)	\(x=-4\)	\(x=0\)	\(x=4\)
\(y=-2+\frac{3x}{4}\)	\(-5\)	\(-2\)	\(1\)

Plot the graph below. Draw the first two boundary lines solid, since their inequalities include equality. Draw \(x=-1\) as a broken line, since \(x>-1\) does not include the boundary.

Using \((0,0)\) as a test point:

\[0\leq10,\qquad 0\leq8,\qquad 0>-1.\]

Hence select the side containing \((0,0)\) for each inequality. Therefore the required region is

\[\boxed{\;x>-1,\qquad -2+\frac{3x}{4}\leq y\leq2+\frac{x}{5}\;}\]

The two sloping lines meet at

\[2+\frac{x}{5}=-2+\frac{3x}{4}\Rightarrow x=\frac{80}{11},\quad y=\frac{38}{11}.\]

Thus shade the triangular region between the two solid lines, to the right of the broken line \(x=-1\). The broken left-hand edge is not part of the region.

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(a)(i) Cumulative frequency table

Marks (%)	Frequency	Cumulative frequency
0–9	10	10
10–19	40	50
20–29	80	130
30–39	140	270
40–49	170	440
50–59	130	570
60–69	100	670
70–79	70	740
80–89	40	780
90–99	20	800

(a)(ii) Ogive

Plot the upper class boundaries against the cumulative frequencies and join successive points with a smooth increasing curve.

(a)(iii) 50th percentile

The 50th percentile corresponds to cumulative frequency

\[\frac{50}{100}\times800=400.\]

From the ogive, the mark corresponding to cumulative frequency 400 is approximately \(47\) marks.

Equivalently, by interpolation in the \(40\text{–}49\) class,

\[P_{50}=39.5+\frac{400-270}{170}\times10=47.15\approx47\text{ marks}.\]

(b)

Those withdrawn scored less than \(25\%\). From the ogive, at \(24.5\) marks,

\[50+\frac{24.5-19.5}{10}\times80=90.\]

Thus, approximately \(90\) candidates are withdrawn.

Those awarded scholarships scored more than \(75\%\). The cumulative frequency at \(75.5\) marks is

\[670+\frac{75.5-69.5}{10}\times70=712.\]

Hence, the number awarded scholarships is

\[800-712=88.\]

Therefore, the number retained but not awarded a scholarship is

\[800-(90+88)=\boxed{622\text{ candidates (approximately)}}.\]

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When a sector is rolled into a cone, the arc length of the sector becomes the circumference of the cone's base, and the sector radius becomes the slant height.

(a) Arc length \(=\dfrac{270}{360}\times2\pi\times7=\dfrac{3}{4}\times2\times\dfrac{22}{7}\times7=\dfrac{3}{4}\times44=33\text{ cm}.\)

Base circumference \(=2\pi r=33\):

\[2\times\frac{22}{7}\times r=33\Rightarrow\frac{44}{7}r=33\Rightarrow r=\frac{33\times7}{44}=5.25\text{ cm}.\]

Base radius \(=5.25\) cm.

(b) Base area \(=\pi r^{2}=\dfrac{22}{7}\times(5.25)^{2}=\dfrac{22}{7}\times27.5625=86.625\text{ cm}^{2}\).

\(\approx 87\text{ cm}^{2}\) (nearest square centimetre).

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From the diagram, S lies on \(PR\) with \(\angle PQR = 90^\circ\) (right angle at Q) and \(\angle PSQ = 90^\circ\), so \(QS\) is the altitude from Q onto the hypotenuse \(PR\). The given lengths are \(|PS| = 9\text{ cm}\), \(|SR| = 16\text{ cm}\) and \(|SQ| = x\).

(a) Value of x by a trigonometric ratio

Let \(\angle QPS = \alpha\). In right triangle \(PSQ\):

In right triangle \(QSR\), the angle \(\angle SQR\) equals \(\alpha\) (both equal \(90^\circ - \angle R\)), so

Equating the two expressions:

(b)(i) Size of \(\angle QRS\) to the nearest degree

In right triangle \(QSR\):

(b)(ii) \(|PQ|\)

In right triangle \(PSQ\), by Pythagoras:

Therefore \(x = 12\text{ cm}\), \(\angle QRS \approx 37^\circ\) and \(|PQ| = 15\text{ cm}\).

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(a) \(\sin\alpha=0.3907\Rightarrow\alpha=23^{\circ}\).

• (i) \(\tan\alpha=\tan23^{\circ}\approx0.4245.\)
• (ii) \(\cos\alpha=\cos23^{\circ}\approx0.9205\) (also \(\cos\alpha=\sqrt{1-0.3907^{2}}=\sqrt{0.8474}=0.9205\)).

(b) The ladder reaches a height up the wall of \[h=4.5\sin50^{\circ}=4.5\times0.7660=3.447\text{ m}.\] The bottom of the window is \(4\text{ m}\) above the ground, so the gap between the top of the ladder and the window bottom is \[4-3.447=0.553\text{ m}\approx55\text{ cm}.\]

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Total number of lesson periods:

\[10+7+3+4+3+9=36\]

Hence, the angle for each lesson period is:

\[\frac{360^\circ}{36}=10^\circ\]

Subject	Number of periods	Angle of sector
English	10	\(10\times10^\circ=100^\circ\)
Mathematics	7	\(7\times10^\circ=70^\circ\)
Biology	3	\(3\times10^\circ=30^\circ\)
Statistics	4	\(4\times10^\circ=40^\circ\)
Ibo	3	\(3\times10^\circ=30^\circ\)
Others	9	\(9\times10^\circ=90^\circ\)
Total	36	\(360^\circ\)

The required pie chart is:

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(a) The angle sum of a triangle is two right angles.

Let \(\triangle ABC\) have interior angles \(a,b,c\) at \(A,B,C\). Through \(A\) draw a line \(XY\) parallel to \(BC\).

• \(\angle XAB=\angle ABC=b\) (alternate angles, \(XY\parallel BC\)).
• \(\angle YAC=\angle ACB=c\) (alternate angles, \(XY\parallel BC\)).

Angles on the straight line \(XY\) at \(A\): \(\angle XAB+\angle BAC+\angle YAC=180^{\circ}\), i.e. \(b+a+c=180^{\circ}\). Hence \(a+b+c=180^{\circ}=\) two right angles. \(\blacksquare\)

(b) In \(\triangle LMN\): \(\angle LMN=46^{\circ}\), \(\angle MLN=80^{\circ}\), so \(\angle LNM=180^{\circ}-46^{\circ}-80^{\circ}=54^{\circ}\).

\(NM\) is produced to \(P\), so \(\angle LNP=180^{\circ}-\angle LNM=126^{\circ}\) (angles on a straight line). Its bisector \(NQ\) gives \(\angle LNQ=\tfrac12(126^{\circ})=63^{\circ}\).

Since \(M,L,Q\) are collinear (\(Q\) on \(ML\) produced), \(\angle NLQ=180^{\circ}-\angle NLM=180^{\circ}-80^{\circ}=100^{\circ}\).

In \(\triangle LNQ\): \(\angle LQN=180^{\circ}-\angle LNQ-\angle NLQ=180^{\circ}-63^{\circ}-100^{\circ}=\mathbf{17^{\circ}}.\)