WAEC SSCE - General Mathematics - 1999
Ajụjụ 2 Ripọtì
In the diagram, O is the centre of the circle ?PQR = 75o, ?OPS = yo and \(\bar{OR}\) is parallel to \(\bar{PS}\). Find y
Akọwa Nkọwa
Ajụjụ 3 Ripọtì
The pie chart illustrate the amount of private time a student spends in a week studying various subjects. use it to answer the question below
Find the value of K
Akọwa Nkọwa
Ajụjụ 4 Ripọtì
The sides of a right angle triangle in ascending order of magnitude are 8cm, (x-2)cm and x cm. Find x
Akọwa Nkọwa
In a right angle triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. Using this theorem, we can write the following equation: x² = 8² + (x-2)² Expanding the brackets, we get: x² = 64 + x² - 4x + 4 Simplifying and rearranging, we get: 4x = 68 x = 17 Therefore, the value of x is 17, which is the second option.
Ajụjụ 5 Ripọtì
If \(y = \sqrt{ax-b}\) express x in terms of y, a and b
Akọwa Nkọwa
Given, $y = \sqrt{ax-b}$. Squaring both sides, we get: $$y^2 = ax - b$$ Adding b on both sides, we get: $$ax = y^2 + b$$ Dividing both sides by a, we get: $$x = \frac{y^2 + b}{a}$$ Hence, the correct option is $\boxed{\textbf{(D)}\ x = \frac{y^2 + b}{a}}$.
Ajụjụ 6 Ripọtì
Which of the following pairs of inequalities is represented on the number line?
Akọwa Nkọwa
The pie chart illustrate the amount of private time a student spends in a week studying various subjects. use it to answer the question belowAjụjụ 8 Ripọtì
A group of 11 people can speak either English or French or Both. Seven can speak English and six can speak French. What is the probability that a person chosen at random can speak both English and French?
Akọwa Nkọwa
There are 11 people in total, and we know that some of them can speak English only, some of them can speak French only, and some of them can speak both English and French. Let's denote the number of people who can speak both languages as "x". From the information given, we know that: - 7 people can speak English, including those who can speak both languages (which means that some of these 7 people can also speak French) - 6 people can speak French, including those who can speak both languages (which means that some of these 6 people can also speak English) To find the number of people who can speak both languages, we can use the formula: Total = English only + French only + Both - Neither We know that there are no people who can speak neither language, so we can simplify the formula to: Total = English only + French only + Both Substituting the given values, we get: 11 = 7 - x + 6 - x + x 11 = 13 - x x = 2 So, there are 2 people who can speak both English and French. The probability of choosing a person who can speak both languages out of the 11 people is: Probability = Number of people who can speak both / Total number of people Probability = 2 / 11 Therefore, the answer is \(\frac{2}{11}\).
Ajụjụ 10 Ripọtì
From the top of a cliff 20m high, a boat can be sighted at sea 75m from the foot of the cliff. Calculate the angle of depression of the boat from the top of the cliff
Akọwa Nkọwa
We can solve this problem using trigonometry. The angle of depression is the angle between the horizontal line and the line of sight from the top of the cliff to the boat. To find this angle, we need to first find the length of the line of sight. Let's call the angle of depression θ, the height of the cliff h, and the distance from the foot of the cliff to the boat d. From the problem statement, we know that h = 20m and d = 75m. We can use the tangent function to find the length of the line of sight: tan(θ) = h / d tan(θ) = 20 / 75 tan(θ) = 0.2667 To find θ, we need to take the inverse tangent (or arctangent) of both sides: θ = tan-1(0.2667) Using a calculator, we get: θ ≈ 14.9° Therefore, the angle of depression of the boat from the top of the cliff is approximately 14.9°. Answer: 14.9°
Ajụjụ 11 Ripọtì
If \(tan x = \frac{1}{\sqrt{3}}\), find cos x - sin x such that \(0^o \leq x \leq 90^o\)
Akọwa Nkọwa
Given that \(\tan x = \frac{1}{\sqrt{3}}\), we can draw a right-angled triangle where the opposite side is equal to 1 and the adjacent side is equal to \(\sqrt{3}\), and the hypotenuse is equal to 2. Therefore, sin x = \(\frac{1}{2}\) and cos x = \(\frac{\sqrt{3}}{2}\). Now, cos x - sin x = \(\frac{\sqrt{3}}{2}\) - \(\frac{1}{2}\) = \(\frac{\sqrt{3}-1}{2}\). Hence, the correct option is \(\frac{\sqrt{3}-1}{2}\).
Ajụjụ 12 Ripọtì
A pole of length L leans against a vertical wall so that it makes an angle of 60o with the horizontal ground. If the top of the pole is 8m above the ground, calculate L.
Akọwa Nkọwa
Ajụjụ 13 Ripọtì
In the diagram, POR is a circle with center O. ∠QPR = 50°, ∠PQO = 30° and ∠ORP = m. Find m.
Ajụjụ 15 Ripọtì
The side of a rhombus is 10cm long, correct to the nearest whole number. Between what limits should the perimeter P lie?
Akọwa Nkọwa
Ajụjụ 16 Ripọtì
If \(K\sqrt{28}+\sqrt{63}-\sqrt{7}=0\), find K.
Akọwa Nkọwa
We can simplify the given equation as follows: $$K\sqrt{28}+\sqrt{63}-\sqrt{7}=0$$ $$K\sqrt{4\cdot7}+\sqrt{9\cdot7}-\sqrt{7}=0$$ $$2K\sqrt{7}+3\sqrt{7}-\sqrt{7}=0$$ $$2K\sqrt{7}+2\sqrt{7}=0$$ $$2\sqrt{7}(K+1)=0$$ Since $\sqrt{7}$ is not zero, we can divide both sides by $2\sqrt{7}$ to get: $$K+1=0$$ Therefore, $K=-1$. So the answer is (B) -1.
Ajụjụ 18 Ripọtì
From a set \(A = [3, \sqrt{2}, 2\sqrt{3}, \sqrt{9}, \sqrt{7}]\), a number is selected at random. Find the probability that is a rational number
Akọwa Nkọwa
Ajụjụ 19 Ripọtì
Simplify \(\frac{4}{x+1}-\frac{3}{x-1}\)
Akọwa Nkọwa
To simplify \(\frac{4}{x+1}-\frac{3}{x-1}\), we first need to find a common denominator for the two fractions. The common denominator is \((x+1)(x-1)\), so we can rewrite the expression as follows: $$\frac{4(x-1)}{(x+1)(x-1)} - \frac{3(x+1)}{(x+1)(x-1)}$$ Next, we can combine the two fractions by subtracting the second fraction from the first, giving us: $$\frac{4(x-1)-3(x+1)}{(x+1)(x-1)}$$ Simplifying the numerator, we get: $$\frac{4x-4-3x-3}{(x+1)(x-1)} = \frac{x-7}{x^2-1}$$ Therefore, the answer is $\frac{x-7}{x^2-1}$
Ajụjụ 20 Ripọtì
If \(x^2 +15x + 50 = ax^2 + bx + c = 0\). Which of the following statement is not true?
Akọwa Nkọwa
Given the quadratic equation: \(x^2 +15x + 50 = ax^2 + bx + c = 0\), where a, b, and c are constants. The question is asking which of the following statements is not true. We can first use the quadratic formula to find the values of x in terms of a, b, and c: $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Comparing this with the given equation, we have: $$a = 1, \quad b = 15, \quad c = 50$$ Using the quadratic formula, we have: $$x = \dfrac{-15 \pm \sqrt{15^2 - 4(1)(50)}}{2(1)} = -5, -10$$ So, statement (a) x = -5 is true. To find out which statement is not true, we can check each option. Statement (b) x = 10: We can see that this is not a solution to the quadratic equation. Statement (c) x + 10 = 0: This can be rewritten as x = -10, which is a solution to the quadratic equation. Statement (d) bc = 750: Multiplying the coefficients of x gives: b*c = a*(-50) = -50a, so this statement is true. Therefore, the statement that is not true is (b) x = 10.
Ajụjụ 21 Ripọtì
Simplify \(\frac{2-18m^2}{1+3m}\)
Akọwa Nkọwa
To simplify the given expression, we need to factorize the numerator and denominator, then cancel out any common factors. The numerator can be factorized as the difference of two squares: 2 - 18m^2 = 2(1 - 9m^2) = 2(1 + 3m)(1 - 3m) The denominator is already factorized: 1 + 3m Now we can cancel out the common factor of (1 + 3m) from the numerator and denominator: \[\frac{2-18m^2}{1+3m} = \frac{2(1+3m)(1-3m)}{1+3m} = 2(1-3m)\] Therefore, the simplified form of the expression is 2(1-3m).
Ajụjụ 23 Ripọtì
Water flows from a tap into cylindrical container at the rate 5πcm\(^3\) per second. If the radius of the container is 3cm, calculate the level of water in the container at the end of 9 seconds.
Akọwa Nkọwa
To solve the problem, we can use the formula for the volume of a cylinder: V = πr\(^2\)h where V is the volume of the water in the cylinder, r is the radius of the cylinder, and h is the height of the water in the cylinder. We know that water is flowing into the cylinder at a rate of 5π cm\(^3\) per second, which means that the volume of the water in the cylinder is increasing at a rate of 5π cm\(^3\) per second. We want to find the height of the water in the cylinder at the end of 9 seconds. Let's use h to represent the height of the water in the cylinder after 9 seconds. The increase in volume of the water in the cylinder after 9 seconds is: ΔV = 5π x 9 ΔV = 45π So, the volume of the water in the cylinder after 9 seconds is: V = πr\(^2\)h + 45π Since the radius of the cylinder is 3cm, we can substitute this value into the formula: V = π x 3\(^2\) x h + 45π V = 9πh + 45π We know that the volume of the water in the cylinder after 9 seconds is equal to the height of the water multiplied by the area of the base of the cylinder (which is πr\(^2\)): V = πr\(^2\)h 9πh + 45π = π x 3\(^2\) x h 9πh + 45π = 9πh 45π = 0 This equation has no solution, which means that something went wrong in our calculations. We made a mistake in assuming that the volume of the water in the cylinder was increasing at a constant rate of 5π cm\(^3\) per second. In reality, the rate of increase of the volume of water is not constant, because the height of the water in the cylinder is also increasing. To calculate the correct height of the water in the cylinder after 9 seconds, we need to integrate the rate of increase of the volume of water over time. The rate of increase of the volume of water at any time t is given by: dV/dt = πr\(^2\) dh/dt We know that the radius of the cylinder is 3cm and the rate of flow of water is 5π cm\(^3\) per second. So, we can write: dV/dt = π x 3\(^2\) x dh/dt 5π = 9π dh/dt dh/dt = 5/9 cm/s This means that the height of the water in the cylinder is increasing at a rate of 5/9 cm/s. We want to find the height of the water in the cylinder after 9 seconds. Let's use h to represent the height of the water in the cylinder after 9 seconds. The increase in height of the water in the cylinder after 9 seconds is: Δh = (5/9) x 9 Δh = 5 So, the height of the water in the cylinder after 9 seconds is: h = 5 + 0 h = 5cm Therefore, the correct answer is 5cm.
Ajụjụ 25 Ripọtì
Two numbers 24\(_{x}\) and 31\(_y\) are equal in value when converted to base ten. Find the equation connecting x and y
Akọwa Nkọwa
Ajụjụ 26 Ripọtì
The diagram shows the position of three ships A, B and C at sea. B is due north of C such that |AB| = |BC| and the bearing of B from A = 040°. What is the bearing of A from C
Akọwa Nkọwa
Ajụjụ 28 Ripọtì
The height of a pyramid on square base is 15cm. If the volume is 80cm\(^3\), find the length of the side of the base
Akọwa Nkọwa
The volume of a pyramid is given by the formula: V = 1/3 * Base Area * Height The base of a square pyramid is a square, so the Base Area is equal to the side length squared (B = s²). In this problem, the height is given as 15cm, and the volume is given as 80cm³. We are asked to find the length of a side of the base. Let's call that length "s". So we can set up an equation using the formula for the volume of a pyramid: 80 = 1/3 * s² * 15 To solve for s, we can start by simplifying the equation: 240 = s² * 15 Dividing both sides by 15: 16 = s² Taking the square root of both sides (and remembering to consider both the positive and negative square roots): s = ±4 Since we are dealing with the length of a side of a pyramid, the answer cannot be negative. Therefore, the length of a side of the base is 4cm (option C). Answer: 4.0cm
Ajụjụ 29 Ripọtì
Solve the inequality 2x + 3 < 5x
Akọwa Nkọwa
To solve the inequality 2x + 3 < 5x, we want to isolate x on one side of the inequality sign. We can do this by subtracting 2x from both sides of the inequality: 2x + 3 < 5x -2x -2x 3 < 3x Now we have 3 < 3x. To isolate x, we can divide both sides of the inequality by 3: 3 < 3x /3 /3 1 < x So the solution to the inequality 2x + 3 < 5x is x > 1. This means that any value of x that is greater than 1 will make the inequality true. For example, if x = 2, then 2(2) + 3 = 7, which is indeed less than 5(2) = 10. Therefore, the correct option is "x > 1".
Ajụjụ 30 Ripọtì
The interior angle of a regular polygon is twice the exterior angle. How many sides has the polygon?
Akọwa Nkọwa
Ajụjụ 31 Ripọtì
From the diagram above. ABC is a triangle inscribed in a circle center O. ?ACB = 40o and |AB| = x cm. calculate the radius of the circle.
Akọwa Nkọwa
Ajụjụ 33 Ripọtì
Which of the following figures have one line of symmetry only? I. Isosceles triangle II. Rhombus III. Kite
Akọwa Nkọwa
Ajụjụ 35 Ripọtì
Given that \(81\times 2^{2n-2} = K, find \sqrt{K}\)
Akọwa Nkọwa
To find K given that 81×22n−2=K, we need to simplify the
expression step-by-step.
First,
rewrite 81 in terms of powers of 3: 81=34
So,
the given equation becomes: K=34×22n−2
Next,
we need to find K: K=34×22n−2
Using the property of square roots a×b=a×b: K=34×22n−2
Since 34=32=9 and 22n−2=2(2n−2)/2=2n−1: K=9×2n−1
Thus, the correct answer is: =
9×2n−1
Ajụjụ 36 Ripọtì
In the diagram above, RST is a tangent to circle VSU center O ∠SVU = 50° and UV is a diameter. Calculate ∠RSV.
Akọwa Nkọwa
Since UV is a diameter of circle VSU, we know that ∠VUS = 90°. Also, since RST is tangent to circle VSU at S, then ∠VST = 90°. Therefore, ∠VUS + ∠VST = 90° + 90° = 180°. Since VSU is a straight line, then ∠SUV = 180° - ∠SVU - ∠VUS = 180° - 50° - 90° = 40°. Since RST is tangent to circle VSU at S, then ∠RST = ∠SVU = 50° (tangent and radius form a right angle). Finally, we can calculate ∠RSV using the fact that the angles of a triangle sum up to 180°: ∠RSV = 180° - ∠VUS - ∠SUV - ∠RST = 180° - 90° - 40° - 50° = 0° Therefore, the answer is (d) 40°. Note that this is a trick question, as ∠RSV is not defined in this case since R, S, and V are collinear.
Ajụjụ 39 Ripọtì
A car travel at x km per hour for 1 hour and at y km per hour for 2 hours. Find its average speed
Akọwa Nkọwa
The average speed of a car is given by the total distance traveled divided by the total time taken. Let's assume that the car traveled a distance of D km in the given time. Then, we can write: Total time taken = 1 hour + 2 hours = 3 hours Distance traveled in the first hour = x km Distance traveled in the next 2 hours = y × 2 = 2y km Total distance traveled = D = x + 2y km Therefore, the average speed of the car is: Average speed = Total distance / Total time = D / 3 = (x + 2y) / 3 km/h Hence, the answer is option (C): \(\frac{x + 2y}{3}kmh^{-1}\).
Ajụjụ 40 Ripọtì
Simplify 3.72 x 0.025 and express your answer in the standard form
Akọwa Nkọwa
To multiply 3.72 by 0.025, we can first ignore the decimal points and treat the numbers as if they were whole numbers, i.e., 372 and 25. We multiply these numbers to get 9300. Next, we need to determine the power of 10 that represents the decimal point in the product. To do this, we count the total number of decimal places in the two original numbers (three in 3.72 and two in 0.025), which gives us five decimal places. Therefore, our answer needs to be in scientific notation with a power of 10 that is negative five. Finally, we can insert the decimal point into the product and express the answer in scientific notation as follows: 3.72 x 0.025 = 0.093 = 9.3 x 10^(-2) Therefore, the answer is \(9.3\times 10^{-2}\).
Ajụjụ 42 Ripọtì
Amina had m mangoes. She ate 3 and shared the remainder equally with her brother Uche. Each had at least 10. Which of the following inequalities represents the statements above.
Akọwa Nkọwa
Amina started with m mangoes, ate 3 and shared the remainder equally with her brother Uche, which means she gave \(\frac{m-3}{2}\) mangoes to Uche. The statement "each had at least 10" means that the number of mangoes each of them had is greater than or equal to 10. Therefore, the correct inequality is: \[\frac{m-3}{2}\ge10\] This means that the number of mangoes Amina started with minus 3, divided by 2, must be greater than or equal to 10, which implies that each of them had at least 10 mangoes.
Ajụjụ 43 Ripọtì
The height and base of a triangle are in ratio 1:3 respectively. If the area of the triangle is 216 cm\(^2\), find the length of the base.
Akọwa Nkọwa
We know that the area of a triangle is given by the formula: Area = (1/2) x base x height Let the height of the triangle be h and the base be b. We are given that h:b = 1:3, which means that h = (1/3)b. Substituting this value of h in the formula for the area, we get: 216 = (1/2) x b x (1/3)b Multiplying both sides by 6 to get rid of the fraction, we get: 1296 = b^2 Taking the square root of both sides, we get: b = ±36 Since the base of a triangle cannot be negative, we take the positive value of b, which is b = 36. Therefore, the length of the base of the triangle is 36 cm, and the correct option is "36cm".
Ajụjụ 44 Ripọtì
(a) If \(\varepsilon\) is the set \({1, 2, 3,..., 19, 20}\) and A, B and C are subsets of \(\varepsilon\) such that A = { multiples of five}, B = {multiples of four} and C = {multiples of three}, list the elements of (i) A ; (ii) B ; (iii) C ;
(b) Find : (i) \(A \cap B\) ; (ii) \(A \cap C\) ; (iii) \(B \cup C\).
(c) Using your results in (b), show that \((A \cap B) \cup (A \cap C) = A \cap (B \cup C)\).
Ajụjụ 45 Ripọtì
(a) Simplify \((\frac{4}{25})^{-\frac{1}{2}} \times 2^{4} \div (\frac{15}{2})^{-2}\)
(b) Evaluate \(\log_{5} (\frac{3}{5}) + 3 \log_{5} (\frac{5}{2}) - \log_{5} (\frac{81}{8})\).
Akọwa Nkọwa
None
Ajụjụ 46 Ripọtì
In the diagram, a ladder TF, 10 metres long is placed against a wall at an angle of 70° to the horizontal.
(a) How high up the wall, correct to the nearest metre, does the ladder reach?
(b) If the foot (F) of the ladder is pulled from the wall to F\(^{1}\) by 1 metre, (i) how far, correct to 2 significant figures, does the top T slide down the wall to T\(^{1}\).
(ii) Calculate, correct to the nearest degree, \(QF^{1}T^{1}\).
Akọwa Nkọwa
None
Ajụjụ 47 Ripọtì
The table gives the frequency distribution of marks obtained by a group of students in a test.
| Marks | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 5 | x - 1 | x | 9 | 4 | 1 |
If the mean is 5,
(a) Calculate the value of x;
(b) Find the : (i) mode ; (ii) median of the distribution.
(c) If one of the students is selected at random, find the probability that he scored at least 7 marks.
None
Akọwa Nkọwa
None
Ajụjụ 48 Ripọtì
(a) Copy and complete the following table of values for \(y = 9 \cos x + 5 \sin x\) to one decimal place.
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° | 210° |
| y | 10.3 | -0.2 | -5.3 | -10.3 |
(b) Using a scale of 2cm to 30° on the x- axis and 2 cm to 1 unit on the y- axis, draw the graph of \(y = 9 \cos x + 5 \sin x\) for \(0° \leq x \leq 210°\).
(c) Use your graph to solve the equation: (i) \(9\cos x + 5\sin x = 0\); (ii) \(9\cos x+ 5\sin x = 3.5\), correct to the nearest degree.
(d) Find the maximum value of y correct to one decimal place.
Ajụjụ 49 Ripọtì
ABC is a triangle, right-angled at C. P is the mid-point of AC, < PBC = 37° and |BC| = 5 cm. Calculate :
(a) |AC|, correct to 3 significant figures ;
(b) < PBA.
Akọwa Nkọwa
None
Ajụjụ 50 Ripọtì
(a) Solve the simultaneous equations 3y - 2x = 21 ; 4y + 5x = 5.
(b) Six identical cards numbered 1 - 6 are placed face down. A card is to be picked at random. A person wins $60.00 if he picks the card numbered 6. If he picks any of the other cards, he loses $10.00 times the number on the card. Calculate the probability of (i) losing ; (ii) losing $20.00 after two picks.
Akọwa Nkọwa
None
Ajụjụ 51 Ripọtì
In the diagram, ABCD is a trapezium in which \(AD \parallel BC\) and \(< ABC\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate :
(a) |AB| ;
(b) the area of the triangle BCD ;
(c) |CD| ;
(d) perimeter of the trapezium.
Akọwa Nkọwa
None
Ajụjụ 52 Ripọtì
(a) Given that \(\sin(A + B) = \sin A \cos B + \cos A \sin B\). Without using mathematical tables or calculator, evaluate \(\sin 105°\), leaving your answer in the surd form.
(You may use 105° = 60° + 45°)
(b) The houses on one side of a particular street are assigned odd numbers, starting from 11. If the sum of the numbers is 551, how many houses are there?
(c) The 1st and 3rd terms of a Geometric Progression (G.P) are \(2\) and \(\frac{2}{9}\) respectively. Find :
(i) the common difference ; (ii) the 5th term.
Akọwa Nkọwa
None
Ajụjụ 53 Ripọtì
The cost of maintaining a school is partly constant and partly varies as the number of pupils. With 50 pupils, the cost is $15,705.00 and with 40 pupils, it is $13,305.00.
(a) Find the cost when there are 44 pupils.
(b) If the fee per pupil is $360.00, what is the least number of pupils for which the school can run without a loss?
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
(a) A shop owner marked a shirt at a price to enable him to make a gain of 20%. During a special sales period, the shirt was sold at 10% reduction to a customer at N864.00. What was the original cost to the shop owner?
(b) A rectangular lawn of length (x + 5) metres is (x - 2) metres wide. If the diagonal is (x + 6) metres, find ;
(i) the value of x ; (ii) the area of lawn.
Akọwa Nkọwa
None
Ajụjụ 55 Ripọtì
(a) A cylindrical pipe is 28 metres long. Its internal radius is 3.5 cm and external radius 5 cm. Calaulate : (i) the volume, in cm\(^{3}\), of metal used in making the pipe ; (ii) the volume of water in litres that the pipe can hold when full, correct to 1 decimal place. [Take \(\pi = \frac{22}{7}\)]
(b) 
In the diagram, MP is a tangent to the circle LMN at M. If the chord LN is parallel to MP, show that the triangle LMN is isosceles.
Ajụjụ 56 Ripọtì
On a graph sheet, using a scale of 2cm to 2 units on both axes,
(a) Draw the straight line joining points P(-5, 3) and Q(2, 3);
(b) construct the locus L of points equidistant from P and Q;
(c) by construction, locate points R and S on L, such that PRQS forms a rhombus of sides 5cm;
(d) find : (i) coordinates of R and S; (ii) area of the rhombus in cm\(^{2}\).
None
Akọwa Nkọwa
None
