WAEC SSCE - General Mathematics - 1997
Question 2 Report
A bag contains red, black and green identical balls. A ball is picked and replaced. The table shows the result of 100 trials. Find the experimental probability of picking a green ball.
Question 3 Report
The eleventh term of an AP, is 25 and its first term is -3. Find its common difference.
Answer Details
Question 5 Report
If P = {3, 5, 6} and Q = {4, 5, 6} then P∩Q equals
Answer Details
P∩Q represents the intersection of sets P and Q, which means finding the common elements between them. The common elements between P and Q are 5 and 6, so P∩Q equals {5, 6}. Therefore, the correct answer is (D) {5, 6}.
Question 6 Report
If x varies over the set of real numbers, which of the following is illustrated in the diagram above?
Question 7 Report
Answer Details
To make q the subject of the formula, we need to isolate q on one side of the equation. Starting with: t = √pq/r - r2 1. Add r^2 to both sides of the equation: t + r^2 = √pq/r 2. Square both sides of the equation to eliminate the radical: (t + r^2)^2 = pq/r 3. Multiply both sides of the equation by r: r(t + r^2)^2 = pq 4. Divide both sides of the equation by p: q = r(t + r^2)^2 / p Therefore, the answer is: q = rt2 / (p - r3)
Question 9 Report
Find the value of t for which \(\frac{64}{27} = (\frac{3}{4})^{t - 1}\)
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Question 10 Report
The diagram above is a triangle XYZ whose area is 23.5cm2. XY =1Ocm and YZ = 8cm, calculate ? correct to the nearest degree.
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Question 11 Report
The cross-section of a prism is a right-angled triangle 3cm by 4cm by 5cm. The height of the prism is 8cm. Calculate its volume
Answer Details
To find the volume of the prism, we need to multiply the area of the base by the height. Since the cross-section of the prism is a right-angled triangle, the area of the triangle is given by: Area = 1/2 x base x height = 1/2 x 3cm x 4cm = 6cm^2 The height of the prism is given as 8cm. Therefore, the volume of the prism is given by: Volume = Area of base x height = 6cm^2 x 8cm = 48cm^3 Hence, the volume of the prism is 48cm^3. Therefore, the correct option is (a) 48cm^3.
Question 12 Report
Calculate the total surface area of a solid cone of slant height 15cm and base radius 8cm in terms of π
Answer Details
The total surface area of a cone is the sum of the curved surface area and the area of the base. The curved surface area of a cone can be found using the formula: πrl where r is the radius of the base and l is the slant height of the cone. The area of the base can be found using the formula: πr² where r is the radius of the base. Given the slant height of the cone as 15cm and base radius as 8cm, we can find the height of the cone using the Pythagorean theorem as follows: height² + 8² = 15² height² + 64 = 225 height² = 225 - 64 height² = 161 height = √161 Therefore, the height of the cone is √161 cm. Now we can find the curved surface area of the cone: πrl = π(8)(√161) = 8π√161 And we can find the area of the base: πr² = π(8)² = 64π The total surface area of the cone is the sum of the curved surface area and the area of the base: total surface area = curved surface area + area of base total surface area = 8π√161 + 64π total surface area = 8π(√161 + 8) Therefore, the total surface area of the cone is 8π(√161 + 8) cm², which is approximately equal to 184π cm². So the answer is (c) 184πcm².
Question 13 Report
Which of the following is not necessarily sufficient for the construction of a triangle?
Answer Details
To construct a triangle, we need to satisfy the triangle inequality theorem which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side. Based on this theorem, the option that is not necessarily sufficient for the construction of a triangle is option (C) - two sides and a right angle. This is because having two sides and a right angle does not guarantee that the remaining side satisfies the triangle inequality theorem. For example, if the two sides are very short, the remaining side (hypotenuse) will also be short, and the sum of the lengths of the two shorter sides will be less than the length of the remaining side, violating the triangle inequality theorem.
Question 14 Report
Solve the inequality 3x - 8 ≥ 5x
Answer Details
To solve the inequality 3x - 8 ≥ 5x, we need to isolate the variable x on one side of the inequality sign. First, we can simplify the inequality by subtracting 3x from both sides: 3x - 8 - 3x ≥ 5x - 3x Simplifying further, we get: -8 ≥ 2x To isolate x, we divide both sides by 2: -4 ≥ x This is the same as x ≤ -4. Therefore, the correct option is: - x ≤ -4
Question 15 Report
A bag contains 3 white, 6 red, 5 blue identical balls. A ball is picked at random from bag. What is the probability that is either white or blue?
Answer Details
There are a total of 3+6+5=14 balls in the bag. The probability of selecting a white ball is 3/14 since there are 3 white balls in the bag. Similarly, the probability of selecting a blue ball is 5/14 since there are 5 blue balls in the bag. To find the probability of selecting either a white or a blue ball, we add the probabilities of selecting a white ball and a blue ball. However, we need to subtract the probability of selecting a ball that is both white and blue since we cannot count it twice. Since there are no balls that are both white and blue, we don't need to subtract anything. Therefore, the probability of selecting either a white or a blue ball is: 3/14 + 5/14 = 8/14 = 4/7 So, the answer is 4/7.
Question 16 Report
In the triangle XYZ , XM is the altitude from X to YZ.XY = 13cm, XZ = 15cm and YM = 5cm. Find the length of YZ
Answer Details
Question 17 Report
A chord is 5cm from the centre of a circle of diameter 26cm. Find the length of the chord
Answer Details
To find the length of the chord, we can use the Pythagorean theorem.
First, we can find the radius of the circle, which is half of the diameter:
radius = 26cm / 2 = 13cm
The chord divides the circle into two parts, each with its own radius, as shown below:
O / \ / \ / \ -------- R R
We can draw a line from the center of the circle to the midpoint of the chord, which will bisect the chord and form a right angle with the chord:
O / \ / \ / . \ -------- R R
The line from the center of the circle to the midpoint of the chord is also the perpendicular bisector of the chord, and so it divides the chord into two equal segments.
Let the length of each of these segments be x. Then, we can use the Pythagorean theorem to find x:
x^2 + 5^2 = 13^2 x^2 = 169 - 25 x^2 = 144 x = 12
Therefore, the length of the chord is twice the length of one of the segments:
length of chord = 2x = 2(12) = 24 cm
So the correct option is (c) 24cm.
Question 18 Report
If two triangles are similar, which of the following is true? Their
Answer Details
If two triangles are similar, their corresponding angles are equal. This is a fundamental property of similar triangles. When two triangles have the same shape but different sizes, they are said to be similar. The corresponding sides of similar triangles are proportional, which means that if one side of a triangle is multiplied by a certain number, then all the corresponding sides of the other triangle must be multiplied by the same number. However, the sides are not necessarily equal. Similarly, the altitudes and areas of the triangles are also proportional, but not necessarily equal. Therefore, the correct answer is that the corresponding angles of similar triangles are equal.
Question 19 Report
Find the radius of a circle in which an arc of length 44cm subtends angle 200° at the centre of the circle. [Take π = 22/7]
Answer Details
In a circle, there are 360 degrees at the center. If an arc subtends an angle of 200 degrees at the center of the circle, then the length of that arc can be found as follows: Length of arc = (angle/360) x 2πr where angle is in degrees, r is the radius of the circle and π = 22/7 Substituting the given values we get: 44 = (200/360) x 2 x (22/7) x r 44 = (5/9) x (44/7) x r r = (44 x 7 x 9) / (5 x 44) r = 12.6 cm Therefore, the radius of the circle is 12.6 cm. Answer is correct.
Question 21 Report
What is the probability of having an odd number in a single throw of a fair die with the faces numbered 1,2,3,4,5,6?
Answer Details
There are six possible outcomes when a fair die is rolled, which are the numbers 1, 2, 3, 4, 5, and 6. Since three of these numbers are odd (1, 3, and 5), and the die is fair (meaning that each outcome is equally likely), the probability of rolling an odd number is 3/6, which simplifies to 1/2. Therefore, the answer is 1/2.
Question 23 Report
The interior angles of a pentagon are 126o, 114o, y, 92o and 83o. Find the value of y.
Answer Details
The sum of the interior angles of a pentagon is given by the formula (n-2) × 180o where n is the number of sides of the polygon. In this case, n=5 since it's a pentagon. So, sum of interior angles = (5-2) × 180o = 540o We know that the other four angles measure 126o, 114o, 92o, and 83o. Therefore, 126o + 114o + y + 92o + 83o = 540o Simplifying the above equation, we get: y = 125o Therefore, the value of y is 125o.
Question 25 Report
Divide 3.6721 by 4
Answer Details
To divide 3.6721 by 4, we can use long division. The first step is to place the dividend, 3.6721, under the division symbol, and the divisor, 4, outside the division symbol. We then ask, how many times does 4 go into 3? Since 4 is greater than 3, we cannot divide 3 by 4, so we bring down the next digit, 6, and ask how many times 4 goes into 36. The answer is 9, so we write 9 above the 6, and subtract 36 - 36 to get 0. We then bring down the next digit, 7, and ask how many times 4 goes into 0. Since 4 is greater than 0, we cannot divide 0 by 4, so we add a decimal point and a 0, and ask how many times 4 goes into 7. The answer is 1 with a remainder of 3, so we write 1 above the 7 and bring down the next digit, 2. We add a decimal point to the quotient and ask how many times 4 goes into 32. The answer is 8, so we write 8 above the 2, and subtract 32 - 32 to get 0. We then bring down the next digit, 1, and ask how many times 4 goes into 1. Since 4 is greater than 1, we cannot divide 1 by 4, so we add a decimal point and a 0, and ask how many times 4 goes into 10. The answer is 2 with a remainder of 2, so we write 2 above the 1, and bring down the final digit, 0. We add a decimal point to the quotient and ask how many times 4 goes into 20. The answer is 5, so we write 5 above the 0, and subtract 20 - 20 to get 0. Therefore, the final answer is 0.9180. So, 3.6721 divided by 4 is 0.9180.
Question 26 Report
P varies inversely as the square of W. When W = 4, P = 9. Find the value of P when W = 9
Answer Details
The problem states that P varies inversely as the square of W, which means that as W increases, P decreases, and vice versa. We can write this relationship as P = k/W^2, where k is the constant of proportionality. To solve for k, we can use the given information that when W = 4, P = 9. Plugging these values into the equation, we get: 9 = k/4^2 9 = k/16 k = 9 x 16 k = 144 Now that we know k, we can use the equation to find P when W = 9: P = 144/9^2 P = 144/81 P = 16/9 Therefore, the value of P when W = 9 is 16/9.
Question 27 Report
Find the common ratio in the GP: log 3, log 9, log 81 ..........
Answer Details
In a geometric progression (GP), the ratio between any two successive terms is constant. Let's find the ratio between the second and first terms of the given GP: log 9 / log 3 = log (3^2) / log 3 = 2 log 3 / log 3 = 2 Similarly, the ratio between the third and second terms is: log 81 / log 9 = log (9^2) / log 9 = 2 log 9 / log 9 = 2 Since the ratio between any two successive terms is the same, the common ratio of the given GP is 2. Therefore, the answer is 2.
Question 28 Report
A and B are two sets. The number of elements in A∪B is 49, the number in A is 22 and number in B is 34. How many elements are in A ∩ E?
Answer Details
To find the number of elements in the intersection of A and B, we can use the formula: |A ∪ B| = |A| + |B| - |A ∩ B| We are given |A ∪ B| = 49, |A| = 22, and |B| = 34. Substituting these values into the formula gives: 49 = 22 + 34 - |A ∩ B| Simplifying, we get: |A ∩ B| = 22 + 34 - 49 = 7 Therefore, there are 7 elements in the intersection of A and B. The correct answer is 7.
Question 29 Report
The table shows the scores of a group of students in a test. If the average score is 3.5, find the value of x
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Question 30 Report
Two sides of a triangle are perpendicular. If the two sides are 8cm and 6cm, calculate correct to the nearest degree, the smallest angle of the triangle.
Answer Details
If two sides of a triangle are perpendicular, then those sides are the legs of a right-angled triangle. We can use Pythagoras' theorem to find the hypotenuse of this right-angled triangle which will be the longest side of the original triangle. Using Pythagoras' theorem: a2 + b2 = c2 where a and b are the lengths of the legs and c is the length of the hypotenuse. In this case, a = 6cm and b = 8cm. So, 62 + 82 = c2 36 + 64 = c2 100 = c2 c = 10cm So, the longest side of the triangle is 10cm. Now, we can use trigonometry to find the smallest angle of the triangle. sin A = opposite / hypotenuse where A is the smallest angle of the triangle. We know that the opposite side to A is 6cm (the shorter leg). So, sin A = 6 / 10 sin A = 0.6 Using a calculator or a trigonometric table, we can find the angle whose sine is 0.6. A = 36.87o (rounded to the nearest degree) Therefore, the smallest angle of the triangle is approximately 37o. So, the answer is (c) 37o.
Question 31 Report
PQRS and GHRS are parallelograms on the same base SR and between the same parallel straight line PH and SR. Which of the following is true?
Answer Details
Question 32 Report
Express (0.0425 / 2.5) as a fraction
Answer Details
To express (0.0425 / 2.5) as a fraction, we need to simplify the expression as much as possible. First, we can convert the decimal 0.0425 to a fraction by putting the decimal numerator over 1 and multiplying both the numerator and denominator by 10000 to get 425/10000. Then we can simplify the fraction by dividing both the numerator and denominator by their greatest common factor, which is 25, to get 17/400. Finally, we can simplify further by dividing both the numerator and denominator by 4, giving us a final answer of 17/1000. Therefore, the correct option is:
Question 33 Report
The length of the two parallel sides of trapezium are 6cm and 10cm and the perpendicular distance between them is 5cm. Find the area of the trapezium
Question 35 Report
A student found the approximate value of 0.02548 correct to two places of decimal instead of two significant figures. Find the percentage error.
Answer Details
Question 36 Report
lf sin θ= \(\frac{-1}{2}\), find all the values of θ between 0° and 360°.
Question 37 Report
O is the centre of the circle PQRS. PR and QS intersect at T POR is a diameter, ?PQT = 42o and ?QTR = 64o; Find ?QRT
Question 38 Report
Given that \(\frac{1}{2} \log_{10} P = 1\), find the value of P.
Answer Details
The equation \(\frac{1}{2} \log_{10} P = 1\) can be simplified by multiplying both sides by 2: \(\log_{10} P = 2\). We know that \(\log_{10} P = x\) is equivalent to \(10^x = P\). So, in this case, \(10^2 = P\), which means that P is equal to 100. Therefore, the correct answer is 102.
Question 39 Report
Two points P and Q are on longitude 67°W. Their latitudes differ by 90°. Calculate their distance apart in terms of π. (Take radius of the earth = 6400km).
Answer Details
Question 41 Report
Solve me equation: 2/3 (x + 5) = 1/4(5x - 3)
Answer Details
To solve the given equation: 2/3(x + 5) = 1/4(5x - 3) We can simplify both sides of the equation by getting rid of the fractions. To do this, we can multiply both sides by the least common multiple of the denominators, which is 12: 12 * 2/3(x + 5) = 12 * 1/4(5x - 3) 8(x + 5) = 3(5x - 3) Now, we can distribute the terms on both sides: 8x + 40 = 15x - 9 We can then isolate the variable on one side by subtracting 8x from both sides: 40 = 7x - 9 Finally, we can isolate the variable x by adding 9 to both sides and then dividing by 7: 49/7 = x Simplifying, we get: x = 7 Therefore, the solution to the equation is x = 7.
Question 42 Report
The angle of elevation of a point T on a tower from a point U on the horizontal ground is 30o. If TU = 54m, how high is T above the horizontal ground?
Answer Details
Question 43 Report
Which of the following statements is/are true when two straight lines intersect?
I. Adjacent angles are equal II. Vertically opposite angles are equal III. Adjacent angles are supplementary
Answer Details
When two straight lines intersect, they form four angles at the point of intersection. - Vertically opposite angles are opposite to each other and are equal in measure. This is true and corresponds to statement II. - Adjacent angles are two angles that share a common vertex and a common side. They are not necessarily equal, but their sum is equal to 180 degrees. This is true and corresponds to statement III. - Statement I is not true because adjacent angles need not be equal in measure. For example, two adjacent angles in a scalene triangle are not equal in measure. Therefore, the correct answer is II and III only.
Question 44 Report
TR is the tangent to the circle PQR with centre O. Find the size of ?PRT
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Question 45 Report
The graph is the cumulative frequency curve for the weight distribution of 100 workers in a factory. Which of the points P,Q,R,S and T indicates the median weight?
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Question 46 Report
(a) Use logarithm tables to evaluate \(\frac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\).
(b) The first 5 students to arrive in a school on a Monday morning were 2 boys and 3 girls. Of these, two were chosen at random for an assignment. Find the probability that :
(i) both were boys ; (ii) the two were of different sexes.
(a) Evaluate \(\dfrac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\) with logarithms.
First \(\sqrt{0.00695}\): \(\log 0.00695 = \bar{3}.8420\), half of it \(= \bar{2}.9210\), so \(\sqrt{0.00695} = 0.08337\).
Numerator: \(\log 15.05 = 1.1775\); \(1.1775 + \bar{2}.9210 = 0.0985\), giving \(15.05 \times 0.08337 = 1.2547\).
Divide by \(695\): \(\log 695 = 2.8420\); \(0.0985 - 2.8420 = \bar{3}.2565\).
\[\text{answer} = 1.806 \times 10^{-3} \approx 0.00181\]
(b) Two boys and three girls; two are chosen at random from the five. Total ways \(= \binom{5}{2} = 10\).
(i) Both boys: \(\dfrac{\binom{2}{2}}{\binom{5}{2}} = \dfrac{1}{10}\).
(ii) Different sexes (one boy, one girl): \(\dfrac{2 \times 3}{10} = \dfrac{6}{10} = \dfrac{3}{5}\).
Answer Details
(a) Evaluate \(\dfrac{15.05 \times \sqrt{0.00695}}{6.95 \times 10^{2}}\) with logarithms.
First \(\sqrt{0.00695}\): \(\log 0.00695 = \bar{3}.8420\), half of it \(= \bar{2}.9210\), so \(\sqrt{0.00695} = 0.08337\).
Numerator: \(\log 15.05 = 1.1775\); \(1.1775 + \bar{2}.9210 = 0.0985\), giving \(15.05 \times 0.08337 = 1.2547\).
Divide by \(695\): \(\log 695 = 2.8420\); \(0.0985 - 2.8420 = \bar{3}.2565\).
\[\text{answer} = 1.806 \times 10^{-3} \approx 0.00181\]
(b) Two boys and three girls; two are chosen at random from the five. Total ways \(= \binom{5}{2} = 10\).
(i) Both boys: \(\dfrac{\binom{2}{2}}{\binom{5}{2}} = \dfrac{1}{10}\).
(ii) Different sexes (one boy, one girl): \(\dfrac{2 \times 3}{10} = \dfrac{6}{10} = \dfrac{3}{5}\).
Question 47 Report
(a)
PQRST is a circle with centre C. PCS is a straight line, RS // QT, |QR| = |RS| and < QTS = 56°. Find (i) SQT (ii) PQT.
(b) 
In the diagram, points B and C are on a horizontal plane and |BC| = 30cm. A and D are points vertically above B and C respectively. |DC| = 40 cm and |AB| = 26 cm. Calculate the angles of depression of : (i) B from D ; (ii) A from D ; correct to the nearest degree.
Answer Details
None
Question 48 Report
(a) Solve the simultaneous equation:
\[ \log_{10} x + \log_{10} y = 4 \]
\[ \log_{10} x + 2\log_{10} y = 3 \]
(b) The time, \(t\), taken to buy fuel at a petrol station varies directly as the number of vehicles \(V\) on queue and jointly varies inversely as the number of pumps \(P\) available in the station. In a station with 5 pumps, it took 10 minutes to fuel 20 vehicles. Find:
(i) the relationship between \(t\), \(P\) and \(V\); (ii) the time it will take to fuel 50 vehicles in the station with 2 pumps; (iii) the number of pumps required to fuel 40 vehicles in 20 minutes.
(a) Let \(\log_{10}x = u\) and \(\log_{10}y = v\).
\[u + v = 4 \quad\text{...(1)}\qquad u + 2v = 3 \quad\text{...(2)}\]
Subtract (1) from (2): \(v = -1\). Then \(u = 4 - (-1) = 5\).
\[\log_{10}x = 5 \Rightarrow x = 10^5 = 100000, \qquad \log_{10}y = -1 \Rightarrow y = 10^{-1} = 0.1\]
(b) \(t\) varies directly as \(V\) and inversely as \(P\): \(t = \dfrac{kV}{P}\).
With \(P = 5,\, V = 20,\, t = 10\): \(10 = \dfrac{k(20)}{5} = 4k \Rightarrow k = 2.5\).
(i) \(t = \dfrac{2.5V}{P}\) (equivalently \(t = \dfrac{5V}{2P}\)).
(ii) \(V = 50,\, P = 2\): \(t = \dfrac{2.5(50)}{2} = 62.5\) minutes.
(iii) \(V = 40,\, t = 20\): \(20 = \dfrac{2.5(40)}{P} \Rightarrow 20 = \dfrac{100}{P} \Rightarrow P = 5\) pumps.
Answer Details
(a) Let \(\log_{10}x = u\) and \(\log_{10}y = v\).
\[u + v = 4 \quad\text{...(1)}\qquad u + 2v = 3 \quad\text{...(2)}\]
Subtract (1) from (2): \(v = -1\). Then \(u = 4 - (-1) = 5\).
\[\log_{10}x = 5 \Rightarrow x = 10^5 = 100000, \qquad \log_{10}y = -1 \Rightarrow y = 10^{-1} = 0.1\]
(b) \(t\) varies directly as \(V\) and inversely as \(P\): \(t = \dfrac{kV}{P}\).
With \(P = 5,\, V = 20,\, t = 10\): \(10 = \dfrac{k(20)}{5} = 4k \Rightarrow k = 2.5\).
(i) \(t = \dfrac{2.5V}{P}\) (equivalently \(t = \dfrac{5V}{2P}\)).
(ii) \(V = 50,\, P = 2\): \(t = \dfrac{2.5(50)}{2} = 62.5\) minutes.
(iii) \(V = 40,\, t = 20\): \(20 = \dfrac{2.5(40)}{P} \Rightarrow 20 = \dfrac{100}{P} \Rightarrow P = 5\) pumps.
Question 49 Report
(a) Given that \(\sin x = \frac{5}{13}, 0° \leq x \leq 90°\), find \(\frac{\cos x - 2 \sin x }{2\tan x}\).
(b) 
The diagram represents the vertical cross-section of a mountain with height NQ standing on a horizontal ground PRN. If the angles of elevation of the top of the mountain from P and R are 30° and 70° respectively and PR = 500m, calculate, correct to 3 significant figures :
(i) |QP| ; (ii) the height of the mountain.
(a) Given \(\sin x=\dfrac{5}{13}\) with \(0^\circ\le x\le 90^\circ\), evaluate \(\dfrac{\cos x-2\sin x}{2\tan x}\).
Using a \(5\)-\(12\)-\(13\) right triangle (adjacent \(=\sqrt{13^2-5^2}=\sqrt{144}=12\)):
\[\cos x=\frac{12}{13},\qquad \tan x=\frac{5}{12}.\]Substitute:
\[\frac{\cos x-2\sin x}{2\tan x}=\frac{\dfrac{12}{13}-2\cdot\dfrac{5}{13}}{2\cdot\dfrac{5}{12}}=\frac{\dfrac{12-10}{13}}{\dfrac{10}{12}}=\frac{\dfrac{2}{13}}{\dfrac{5}{6}}.\]\[=\frac{2}{13}\times\frac{6}{5}=\mathbf{\frac{12}{65}}.\](b) \(NQ\) is the vertical height of the mountain on horizontal ground \(PRN\) (with \(R\) between \(P\) and \(N\)). The angle of elevation of the top \(Q\) is \(30^\circ\) from \(P\) and \(70^\circ\) from \(R\), and \(PR=500\text{ m}\).
(i) \(|QP|\). In triangle \(PQR\), \(\angle QPR=30^\circ\). Since \(\angle QRN=70^\circ\) is exterior to the triangle at \(R\), the interior angle is
\[\angle QRP=180^\circ-70^\circ=110^\circ,\]\[\angle PQR=180^\circ-30^\circ-110^\circ=40^\circ.\]By the sine rule:
\[\frac{|QP|}{\sin\angle QRP}=\frac{|PR|}{\sin\angle PQR}\]\[|QP|=\frac{500\times\sin 110^\circ}{\sin 40^\circ}=\frac{500\times0.9397}{0.6428}\approx\mathbf{731\text{ m}}.\](ii) Height of the mountain \(NQ\). In right-angled triangle \(QNP\) (right angle at \(N\)), with \(\angle QPN=30^\circ\):
\[NQ=|QP|\sin 30^\circ=730.95\times0.5\approx\mathbf{365\text{ m}}.\](Check via triangle \(QRN\): \(QR=\dfrac{500\sin30^\circ}{\sin40^\circ}=388.9\text{ m}\), and \(NQ=QR\sin70^\circ=388.9\times0.9397\approx365\text{ m}\).)
Answer Details
(a) Given \(\sin x=\dfrac{5}{13}\) with \(0^\circ\le x\le 90^\circ\), evaluate \(\dfrac{\cos x-2\sin x}{2\tan x}\).
Using a \(5\)-\(12\)-\(13\) right triangle (adjacent \(=\sqrt{13^2-5^2}=\sqrt{144}=12\)):
\[\cos x=\frac{12}{13},\qquad \tan x=\frac{5}{12}.\]Substitute:
\[\frac{\cos x-2\sin x}{2\tan x}=\frac{\dfrac{12}{13}-2\cdot\dfrac{5}{13}}{2\cdot\dfrac{5}{12}}=\frac{\dfrac{12-10}{13}}{\dfrac{10}{12}}=\frac{\dfrac{2}{13}}{\dfrac{5}{6}}.\]\[=\frac{2}{13}\times\frac{6}{5}=\mathbf{\frac{12}{65}}.\](b) \(NQ\) is the vertical height of the mountain on horizontal ground \(PRN\) (with \(R\) between \(P\) and \(N\)). The angle of elevation of the top \(Q\) is \(30^\circ\) from \(P\) and \(70^\circ\) from \(R\), and \(PR=500\text{ m}\).
(i) \(|QP|\). In triangle \(PQR\), \(\angle QPR=30^\circ\). Since \(\angle QRN=70^\circ\) is exterior to the triangle at \(R\), the interior angle is
\[\angle QRP=180^\circ-70^\circ=110^\circ,\]\[\angle PQR=180^\circ-30^\circ-110^\circ=40^\circ.\]By the sine rule:
\[\frac{|QP|}{\sin\angle QRP}=\frac{|PR|}{\sin\angle PQR}\]\[|QP|=\frac{500\times\sin 110^\circ}{\sin 40^\circ}=\frac{500\times0.9397}{0.6428}\approx\mathbf{731\text{ m}}.\](ii) Height of the mountain \(NQ\). In right-angled triangle \(QNP\) (right angle at \(N\)), with \(\angle QPN=30^\circ\):
\[NQ=|QP|\sin 30^\circ=730.95\times0.5\approx\mathbf{365\text{ m}}.\](Check via triangle \(QRN\): \(QR=\dfrac{500\sin30^\circ}{\sin40^\circ}=388.9\text{ m}\), and \(NQ=QR\sin70^\circ=388.9\times0.9397\approx365\text{ m}\).)
Question 50 Report
(a) The 6th term of an A.P is 35 and the 13th term is 77. Find the 20th term.
(b) 
The Venn diagram represents three subsets P, Q and R of the universal set U. Copy the Venn diagram. Shade and indicate the regions represented by (i) \(P \cap Q' \cap R\) ; (ii) \(P' \cap Q \cap R'\).
(a) The 20th term. For an A.P., \(T_n=a+(n-1)d\).
\[T_6=a+5d=35,\qquad T_{13}=a+12d=77.\]
Subtracting the first from the second: \(7d=42\Rightarrow d=6\). Then \(a=35-5(6)=5\).
\[T_{20}=a+19d=5+19(6)=5+114=119.\]
(b) Shading the Venn diagram. The three circles \(P\) (top left), \(Q\) (top right) and \(R\) (bottom) divide \(U\) into eight regions.
(i) \(P\cap Q'\cap R\): shade the region lying inside both \(P\) and \(R\) but outside \(Q\). This is the overlap (lens) of the \(P\) and \(R\) circles, with the central piece that also lies in \(Q\) left unshaded.
(ii) \(P'\cap Q\cap R'\): shade the region lying inside \(Q\) only, that is, the part of the \(Q\) circle that is outside both \(P\) and \(R\).
Answer Details
(a) The 20th term. For an A.P., \(T_n=a+(n-1)d\).
\[T_6=a+5d=35,\qquad T_{13}=a+12d=77.\]
Subtracting the first from the second: \(7d=42\Rightarrow d=6\). Then \(a=35-5(6)=5\).
\[T_{20}=a+19d=5+19(6)=5+114=119.\]
(b) Shading the Venn diagram. The three circles \(P\) (top left), \(Q\) (top right) and \(R\) (bottom) divide \(U\) into eight regions.
(i) \(P\cap Q'\cap R\): shade the region lying inside both \(P\) and \(R\) but outside \(Q\). This is the overlap (lens) of the \(P\) and \(R\) circles, with the central piece that also lies in \(Q\) left unshaded.
(ii) \(P'\cap Q\cap R'\): shade the region lying inside \(Q\) only, that is, the part of the \(Q\) circle that is outside both \(P\) and \(R\).
Question 51 Report
(a) Using a ruler and a pair of compasses only, construst \(\Delta\) ABC in which |AB| = 7cm, |BC| = 5cm and < ABC = 75°. Measure |AC|.
(b) In (a) above, locate by construction, a point D such that CD is parallel to AB and D is equidistant from points A and C. Measure < BAD.
(a) Construction: Draw \(AB = 7\text{ cm}\). At \(B\), construct an angle of \(75^\circ\) by constructing \(60^\circ\) and \(90^\circ\), then bisecting the angle between them. On the \(75^\circ\) ray, mark \(C\) such that \(BC = 5\text{ cm}\). Join \(AC\).
\(|AC| = 7.5\text{ cm}\).
(b) Construction: Through \(C\), construct a line parallel to \(AB\). Construct the perpendicular bisector of \(AC\). Let it meet the parallel line at \(D\). Join \(AD\).
\(\angle BAD = 80^\circ\).
Answer Details
(a) Construction: Draw \(AB = 7\text{ cm}\). At \(B\), construct an angle of \(75^\circ\) by constructing \(60^\circ\) and \(90^\circ\), then bisecting the angle between them. On the \(75^\circ\) ray, mark \(C\) such that \(BC = 5\text{ cm}\). Join \(AC\).
\(|AC| = 7.5\text{ cm}\).
(b) Construction: Through \(C\), construct a line parallel to \(AB\). Construct the perpendicular bisector of \(AC\). Let it meet the parallel line at \(D\). Join \(AD\).
\(\angle BAD = 80^\circ\).
Question 52 Report
(a) Copy and complete the binary multiplication table:
| x | 10 | 11 | 100 | 101 |
| 10 | 100 | 1000 | ||
| 11 | 110 | 1100 | ||
| 100 | 10000 | 10100 |
(b) Convert \(11.011_{two}\) to a number in base ten.
(c) Simplify \(\frac{9.6 \times 10^{18}}{0.24 \times 10^{5}}\) and express your answer in the form \(P \times 10^{m}\) where 1 < P < 10 and m is an integer.
(a) Binary multiplication table. Multiply in base ten, then convert back to base two (e.g. \(11\times 11 = 3\times 3 = 9 = 1001_{two}\), \(11\times 101 = 3\times 5 = 15 = 1111_{two}\)).
| \(\times\) | 10 | 11 | 100 | 101 |
|---|---|---|---|---|
| 10 | 100 | 110 | 1000 | 1010 |
| 11 | 110 | 1001 | 1100 | 1111 |
| 100 | 1000 | 1100 | 10000 | 10100 |
(b) Convert \(11.011_{two}\) to base ten.
\[1\times 2 + 1\times 1 + 0\times\tfrac{1}{2} + 1\times\tfrac{1}{4} + 1\times\tfrac{1}{8} = 2 + 1 + 0.25 + 0.125 = \mathbf{3.375}\](c) Simplify.
\[\frac{9.6\times 10^{18}}{0.24\times 10^{5}} = \frac{9.6}{0.24}\times 10^{18-5} = 40\times 10^{13} = 4.0\times 10^{14}\]So \(P = 4.0\) and \(m = 14\).
Answer Details
(a) Binary multiplication table. Multiply in base ten, then convert back to base two (e.g. \(11\times 11 = 3\times 3 = 9 = 1001_{two}\), \(11\times 101 = 3\times 5 = 15 = 1111_{two}\)).
| \(\times\) | 10 | 11 | 100 | 101 |
|---|---|---|---|---|
| 10 | 100 | 110 | 1000 | 1010 |
| 11 | 110 | 1001 | 1100 | 1111 |
| 100 | 1000 | 1100 | 10000 | 10100 |
(b) Convert \(11.011_{two}\) to base ten.
\[1\times 2 + 1\times 1 + 0\times\tfrac{1}{2} + 1\times\tfrac{1}{4} + 1\times\tfrac{1}{8} = 2 + 1 + 0.25 + 0.125 = \mathbf{3.375}\](c) Simplify.
\[\frac{9.6\times 10^{18}}{0.24\times 10^{5}} = \frac{9.6}{0.24}\times 10^{18-5} = 40\times 10^{13} = 4.0\times 10^{14}\]So \(P = 4.0\) and \(m = 14\).
Question 53 Report
Above is the graph of the quadratic function \(y = ax^{2} + bx + c\) where a, b and c are constants. Using the graph, find :
(a)(i) the scales on both axes ; (ii) the equation of the line of symmetry of the curve ; (iii) the roots of the quadratic equation \(ax^{2} + bx + c = 0\)
(b) Use the coordinates of D, E and G to find the values of the constants a, b and c hence write down the quadratic function illustrated in the graph.
(c) Find the greatest value of y within the range \(-3 \leq x \leq 5\).
(a)(i) The scales are:
- -axis: 2 cm represents 1 unit.
- -axis: 2 cm represents 5 units.
(ii) The axis of symmetry is midway between the roots:
\[x=\frac{0.25+2.25}{2}=1.25.\]
Hence, the equation of the line of symmetry is
\[\boxed{x=1.25}.\]
(iii) The roots are the \(x\)-coordinates of the points where the curve cuts the \(x\)-axis:
\[\boxed{x=0.25\text{ and }x=2.25}.\]
(b) From the graph,
\[D(0,1),\qquad E(1,-2),\qquad G(3,4).\]
Since \(y=ax^2+bx+c\):
Using \(D(0,1)\),
\[1=a(0)^2+b(0)+c\]
\[c=1.\]
Using \(E(1,-2)\),
\[-2=a+b+1\]
\[a+b=-3 \qquad \cdots (1)\]
Using \(G(3,4)\),
\[4=9a+3b+1\]
\[9a+3b=3\]
\[3a+b=1 \qquad \cdots (2)\]
Subtracting (1) from (2),
\[2a=4\]
\[a=2.\]
Then,
\[a+b=-3\]
\[2+b=-3\]
\[b=-5.\]
Therefore,
\[\boxed{a=2,\quad b=-5,\quad c=1}\]
and the quadratic function is
\[\boxed{y=2x^2-5x+1}.\]
(c) The parabola opens upwards, so its greatest value on \(-3\leq x\leq5\) occurs at an endpoint. From the graph, the greatest value is approximately
\[\boxed{33.5}.\]
Answer Details
(a)(i) The scales are:
- -axis: 2 cm represents 1 unit.
- -axis: 2 cm represents 5 units.
(ii) The axis of symmetry is midway between the roots:
\[x=\frac{0.25+2.25}{2}=1.25.\]
Hence, the equation of the line of symmetry is
\[\boxed{x=1.25}.\]
(iii) The roots are the \(x\)-coordinates of the points where the curve cuts the \(x\)-axis:
\[\boxed{x=0.25\text{ and }x=2.25}.\]
(b) From the graph,
\[D(0,1),\qquad E(1,-2),\qquad G(3,4).\]
Since \(y=ax^2+bx+c\):
Using \(D(0,1)\),
\[1=a(0)^2+b(0)+c\]
\[c=1.\]
Using \(E(1,-2)\),
\[-2=a+b+1\]
\[a+b=-3 \qquad \cdots (1)\]
Using \(G(3,4)\),
\[4=9a+3b+1\]
\[9a+3b=3\]
\[3a+b=1 \qquad \cdots (2)\]
Subtracting (1) from (2),
\[2a=4\]
\[a=2.\]
Then,
\[a+b=-3\]
\[2+b=-3\]
\[b=-5.\]
Therefore,
\[\boxed{a=2,\quad b=-5,\quad c=1}\]
and the quadratic function is
\[\boxed{y=2x^2-5x+1}.\]
(c) The parabola opens upwards, so its greatest value on \(-3\leq x\leq5\) occurs at an endpoint. From the graph, the greatest value is approximately
\[\boxed{33.5}.\]
Question 54 Report
The table below shows the mark distribution of candidates in an aptitude test for selection into the public service.
| Marks (in %) | Freq |
| 44 - 46 | 2 |
| 47 - 49 | 5 |
| 50 - 52 | 11 |
| 53 - 55 | 20 |
| 56 - 61 | 42 |
| 62 - 64 | 46 |
| 65 - 67 | 36 |
| 68 - 70 | 9 |
| 71 - 73 | 3 |
(a) Make a cumulative frequency for the distribution
(b) Draw the cumulative frequency curve.
(c) From your graph, estimate the median mark.
(d) The cut-off mark was 63%. What percentage of the candidates was selected?
(a) Cumulative frequency table
Total frequency, \(N=174\).
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 44–46 | 43.5–46.5 | 2 | 2 |
| 47–49 | 46.5–49.5 | 5 | 7 |
| 50–52 | 49.5–52.5 | 11 | 18 |
| 53–55 | 52.5–55.5 | 20 | 38 |
| 56–61 | 55.5–61.5 | 42 | 80 |
| 62–64 | 61.5–64.5 | 46 | 126 |
| 65–67 | 64.5–67.5 | 36 | 162 |
| 68–70 | 67.5–70.5 | 9 | 171 |
| 71–73 | 70.5–73.5 | 3 | 174 |
(b) Cumulative frequency curve (ogive)
Plot the upper class boundaries against the cumulative frequencies, beginning with \((43.5,0)\), and join the points with a smooth curve.
(c) Median mark
The median corresponds to \(\frac{N}{2}=\frac{174}{2}=87\) on the cumulative-frequency axis. From the ogive, this gives a mark of approximately \(\boxed{62\%}\).
(d) Percentage selected
At a mark of \(63\%\), the cumulative frequency read from the curve is approximately \(103\). Hence, the number selected is
\[174-103=71.\]
\[\text{Percentage selected}=\frac{71}{174}\times100\%=40.8\%\approx\boxed{41\%}.\]
Answer Details
(a) Cumulative frequency table
Total frequency, \(N=174\).
| Marks (%) | Class boundaries | Frequency | Cumulative frequency |
|---|---|---|---|
| 44–46 | 43.5–46.5 | 2 | 2 |
| 47–49 | 46.5–49.5 | 5 | 7 |
| 50–52 | 49.5–52.5 | 11 | 18 |
| 53–55 | 52.5–55.5 | 20 | 38 |
| 56–61 | 55.5–61.5 | 42 | 80 |
| 62–64 | 61.5–64.5 | 46 | 126 |
| 65–67 | 64.5–67.5 | 36 | 162 |
| 68–70 | 67.5–70.5 | 9 | 171 |
| 71–73 | 70.5–73.5 | 3 | 174 |
(b) Cumulative frequency curve (ogive)
Plot the upper class boundaries against the cumulative frequencies, beginning with \((43.5,0)\), and join the points with a smooth curve.
(c) Median mark
The median corresponds to \(\frac{N}{2}=\frac{174}{2}=87\) on the cumulative-frequency axis. From the ogive, this gives a mark of approximately \(\boxed{62\%}\).
(d) Percentage selected
At a mark of \(63\%\), the cumulative frequency read from the curve is approximately \(103\). Hence, the number selected is
\[174-103=71.\]
\[\text{Percentage selected}=\frac{71}{174}\times100\%=40.8\%\approx\boxed{41\%}.\]
Question 55 Report
The table below shows how a company's sales manager spent his 1995 annual salary.
| Food | 30% |
| Rent | 18% |
| Car Maintenance | 25% |
| Savings | 12% |
| Taxes | 5% |
| Others | 10% |
(a) Represent this information on a pie chart.
(b) Find his savings at the end of the year if his annual salary was N60,000.00.
(a) Pie chart of the manager's annual expenditure
The angle for each sector is calculated using:
\[\text{Sector angle}=\frac{\text{Percentage}}{100}\times360^\circ\]
| Item | Percentage | Angle of sector |
|---|---|---|
| Food | 30% | \(108^\circ\) |
| Rent | 18% | \(64.8^\circ\) |
| Car Maintenance | 25% | \(90^\circ\) |
| Savings | 12% | \(43.2^\circ\) |
| Taxes | 5% | \(18^\circ\) |
| Others | 10% | \(36^\circ\) |
| Total | 100% | \(360^\circ\) |
(b)
\[\text{Savings}=12\%\text{ of }\text{₦}60,000\]
\[=\frac{12}{100}\times 60,000=\text{₦}7,200.00\]
His savings at the end of the year was \(\text{₦}7,200.00\).
Answer Details
(a) Pie chart of the manager's annual expenditure
The angle for each sector is calculated using:
\[\text{Sector angle}=\frac{\text{Percentage}}{100}\times360^\circ\]
| Item | Percentage | Angle of sector |
|---|---|---|
| Food | 30% | \(108^\circ\) |
| Rent | 18% | \(64.8^\circ\) |
| Car Maintenance | 25% | \(90^\circ\) |
| Savings | 12% | \(43.2^\circ\) |
| Taxes | 5% | \(18^\circ\) |
| Others | 10% | \(36^\circ\) |
| Total | 100% | \(360^\circ\) |
(b)
\[\text{Savings}=12\%\text{ of }\text{₦}60,000\]
\[=\frac{12}{100}\times 60,000=\text{₦}7,200.00\]
His savings at the end of the year was \(\text{₦}7,200.00\).
Question 56 Report
(a) Given that \(\frac{5y - x}{8y + 3x} = \frac{1}{5}\), find the value of \(\frac{x}{y}\) to two decimal places.
(b) If 3 is a root of the quadratic equation \(x^{2} + bx - 15 = 0\), determine the value of b. Find the other root.
(a) \(\dfrac{5y - x}{8y + 3x} = \dfrac15\). Cross-multiplying:
\[5(5y - x) = 8y + 3x \;\Rightarrow\; 25y - 5x = 8y + 3x\]
\[17y = 8x \;\Rightarrow\; \frac{x}{y} = \frac{17}{8} = 2.13 \text{ (2 d.p.)}\]
(b) Since \(3\) is a root of \(x^2 + bx - 15 = 0\):
\[3^2 + 3b - 15 = 0 \;\Rightarrow\; 9 + 3b - 15 = 0 \;\Rightarrow\; 3b = 6 \;\Rightarrow\; b = 2\]
The equation becomes \(x^2 + 2x - 15 = 0 = (x + 5)(x - 3)\). Hence the other root is \(x = -5\).
(Check by product of roots: \(3 \times (-5) = -15\), matching the constant term.)
Answer Details
(a) \(\dfrac{5y - x}{8y + 3x} = \dfrac15\). Cross-multiplying:
\[5(5y - x) = 8y + 3x \;\Rightarrow\; 25y - 5x = 8y + 3x\]
\[17y = 8x \;\Rightarrow\; \frac{x}{y} = \frac{17}{8} = 2.13 \text{ (2 d.p.)}\]
(b) Since \(3\) is a root of \(x^2 + bx - 15 = 0\):
\[3^2 + 3b - 15 = 0 \;\Rightarrow\; 9 + 3b - 15 = 0 \;\Rightarrow\; 3b = 6 \;\Rightarrow\; b = 2\]
The equation becomes \(x^2 + 2x - 15 = 0 = (x + 5)(x - 3)\). Hence the other root is \(x = -5\).
(Check by product of roots: \(3 \times (-5) = -15\), matching the constant term.)
Question 57 Report
The solid is a cylinder surmounted by a hemispherical bowl. Calculate its
(a) total surface area ;
(b) volume (Take \(\pi = \frac{22}{7}\))
From the diagram the solid is a closed cylinder capped on top by a hemisphere. The hemisphere has height \(7\text{ cm}\); since a hemisphere's height equals its radius, the radius is \(r = 7\text{ cm}\), which is also the radius of the cylinder. The cylinder height is \(h = 10\text{ cm}\). Take \(\pi = \dfrac{22}{7}\).
(a) Total surface area.
The exposed surfaces are: the curved surface of the cylinder, the flat circular base of the cylinder, and the curved surface of the hemisphere. (The flat top of the cylinder is exactly covered by the flat face of the hemisphere, so it is not counted.)
Curved surface of cylinder:
\[2\pi r h = 2 \times \frac{22}{7} \times 7 \times 10 = 440 \text{ cm}^2\]Base circle of cylinder:
\[\pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2\]Curved surface of hemisphere:
\[2\pi r^2 = 2 \times 154 = 308 \text{ cm}^2\]Total surface area:
\[440 + 154 + 308 = 902 \text{ cm}^2\]Total surface area \(= 902\text{ cm}^2\).
(b) Volume.
Volume of cylinder:
\[\pi r^2 h = \frac{22}{7} \times 49 \times 10 = 22 \times 7 \times 10 = 1540 \text{ cm}^3\]Volume of hemisphere:
\[\frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2}{3} \times \frac{22}{7} \times 343 = \frac{2}{3} \times 22 \times 49 = \frac{2156}{3} \approx 718.67 \text{ cm}^3\]Total volume:
\[1540 + 718.67 = 2258.67 \text{ cm}^3\]Total volume \(\approx 2258.67\text{ cm}^3\) (i.e. \(2258\tfrac{2}{3}\text{ cm}^3\)).
Answer Details
From the diagram the solid is a closed cylinder capped on top by a hemisphere. The hemisphere has height \(7\text{ cm}\); since a hemisphere's height equals its radius, the radius is \(r = 7\text{ cm}\), which is also the radius of the cylinder. The cylinder height is \(h = 10\text{ cm}\). Take \(\pi = \dfrac{22}{7}\).
(a) Total surface area.
The exposed surfaces are: the curved surface of the cylinder, the flat circular base of the cylinder, and the curved surface of the hemisphere. (The flat top of the cylinder is exactly covered by the flat face of the hemisphere, so it is not counted.)
Curved surface of cylinder:
\[2\pi r h = 2 \times \frac{22}{7} \times 7 \times 10 = 440 \text{ cm}^2\]Base circle of cylinder:
\[\pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154 \text{ cm}^2\]Curved surface of hemisphere:
\[2\pi r^2 = 2 \times 154 = 308 \text{ cm}^2\]Total surface area:
\[440 + 154 + 308 = 902 \text{ cm}^2\]Total surface area \(= 902\text{ cm}^2\).
(b) Volume.
Volume of cylinder:
\[\pi r^2 h = \frac{22}{7} \times 49 \times 10 = 22 \times 7 \times 10 = 1540 \text{ cm}^3\]Volume of hemisphere:
\[\frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 7^3 = \frac{2}{3} \times \frac{22}{7} \times 343 = \frac{2}{3} \times 22 \times 49 = \frac{2156}{3} \approx 718.67 \text{ cm}^3\]Total volume:
\[1540 + 718.67 = 2258.67 \text{ cm}^3\]Total volume \(\approx 2258.67\text{ cm}^3\) (i.e. \(2258\tfrac{2}{3}\text{ cm}^3\)).
