WAEC SSCE - General Mathematics - 2004
Tambaya 1 Rahoto
Solve for t in the equation \(\frac{3}{4}t+\frac{1}{3}(21-t)\) = 11,
Bayanin Amsa
First, we simplify the left-hand side of the equation: \begin{align*} \frac{3}{4}t + \frac{1}{3}(21-t) &= 11 \\ \frac{3}{4}t + 7 - \frac{1}{3}t &= 11 \\ \frac{5}{12}t &= 4 \\ t &= \frac{4 \times 12}{5} \\ t &= \frac{48}{5} \end{align*} Therefore, the value of t is \(9\frac{3}{5}\). So, the correct answer is (d).
Tambaya 2 Rahoto
The locus of a point which moves in a plane such that it is equidistant from two fixed points X and Y is
Bayanin Amsa
Tambaya 3 Rahoto
In the diagram, KS is a tangent to the circle centre O at R and ?ROQ = 80o. Find ?QRS.
Bayanin Amsa
Tambaya 4 Rahoto
The table above gives the marks scored by a group of students in a test Use the table to answer the Question below
What is the probability of selecting a student from the group that scored 2 or 3?
Bayanin Amsa
Tambaya 5 Rahoto
Solve the equation 10-3x-x2 = 0
Bayanin Amsa
To solve the equation 10-3x-x2 = 0, we can first rearrange it to get x2+3x-10=0. We can then factorize this quadratic equation as (x+5)(x-2)=0. This means that either x+5=0 or x-2=0, which gives us the solutions x=-5 and x=2. Therefore, the correct answer is x=2 or -5, which is the first option.
Tambaya 6 Rahoto
If \(y = 23_{five} + 101_{three}\), find y, leaving your answer in base two
Bayanin Amsa
Tambaya 7 Rahoto
There are m boys and 12 girls in a class. What is the probability of selecting at random a girl from the class?
Bayanin Amsa
The probability of selecting a girl at random from the class is the ratio of the number of girls in the class to the total number of students in the class. Given that there are m boys and 12 girls in the class, the total number of students in the class is m + 12. Therefore, the probability of selecting a girl at random is: \begin{align*} \frac{12}{m+12} \end{align*} Therefore, the correct option is (c) \(\frac{12}{m+12}\).
Tambaya 8 Rahoto
In the diagram, \(R\hat{P}Q = Q\hat{R}Y, \hspace{1mm} P\hat{Q}R = R\hat{Y}Q, \\ \hspace{1mm}|QP| = 3cm \hspace{1mm}|QY| = 4cm \hspace{1mm}and \hspace{1mm}|RY | = 5cm. \hspace{1mm} Find \hspace{1mm}|QR|\)
Bayanin Amsa
Tambaya 9 Rahoto
If \(\frac{x}{a+1}+\frac{y}{b}\) 1. Make y the subject of the relation
Tambaya 12 Rahoto
If \(\left(\frac{1}{4}\right)^{(2-y)} = 1\), find y.
Bayanin Amsa
We know that any number raised to the power of zero is equal to 1. Therefore, we can rewrite the equation \(\left(\frac{1}{4}\right)^{(2-y)} = 1\) as \(\left(\frac{1}{4}\right)^{(2-y)} = \left(\frac{1}{4}\right)^0\). Using the rule of exponents that states when we have the same base raised to different powers, we can multiply the bases and subtract the exponents. So, we get \(\left(\frac{1}{4}\right)^{(2-y)} = \left(\frac{1}{4}\right)^0 \Rightarrow \frac{1}{4^{(2-y)}} = \frac{1}{4^0}\). Since \(4^0 = 1\), we can simplify the right-hand side to 1. Therefore, we have \(\frac{1}{4^{(2-y)}} = 1\). Multiplying both sides by \(4^{(2-y)}\) gives us \(1 = 4^{(2-y)}\). We can rewrite the left-hand side as \(4^0\) because any number raised to the power of zero is equal to 1. So, we have \(4^0 = 4^{(2-y)}\). Using the rule of exponents again, we can set the exponents equal to each other, which gives us \(0 = 2 - y\). Solving for y, we get \(y = 2\). Therefore, the value of y that satisfies the equation \(\left(\frac{1}{4}\right)^{(2-y)} = 1\) is 2.
Tambaya 13 Rahoto
A bag contains 3 red and 2 white identical balls. lf 2 balls are picked at random from the bag, one after the other and with replacement, find the probability that they are of different colours
Bayanin Amsa
There are two possible ways to get two balls of different colors: first picking a red ball, then picking a white ball, or first picking a white ball, then picking a red ball. Since we replace the first ball before picking the second one, the probability of picking a red ball on the first draw is $\frac{3}{5}$, and the probability of picking a white ball on the second draw is also $\frac{2}{5}$, hence the probability of getting a red ball followed by a white ball is $\frac{3}{5}\cdot \frac{2}{5} = \frac{6}{25}$. The probability of picking a white ball first and a red ball second is also $\frac{6}{25}$. Therefore, the probability of getting two balls of different colors is the sum of the probabilities of the two cases, which is $\frac{6}{25}+\frac{6}{25} = \frac{12}{25}$. So, the correct option is: - $\frac{12}{25}$
Tambaya 14 Rahoto
A boy walks 800m in 20 minutes. Calculate his average speed in km per hour
Bayanin Amsa
To calculate the average speed of the boy, we need to convert the distance and time to the same units. The boy walked 800m in 20 minutes. To convert minutes to hours, we divide by 60: 20 minutes ÷ 60 = 0.333 hours Now we can calculate the average speed: Speed = Distance ÷ Time Speed = 800m ÷ 0.333 hours Speed ≈ 2400m/h To convert meters per hour to kilometers per hour, we divide by 1000: Speed ≈ 2.4 km/h Therefore, the boy's average speed is 2.4 km/h. So the answer is (a) 2.4.
Tambaya 15 Rahoto
A point X is on the bearing 342o from a point Y. What is the bearing of Y from X?
Bayanin Amsa
If point X is on a bearing of 342o from point Y, then point Y is on a bearing of 162o from point X. This is because the bearing from X to Y is the opposite direction from the bearing from Y to X. Therefore, we subtract the original bearing of 342o from 180o to get 162o.
Tambaya 17 Rahoto
Which of the following quadratic equations has \(-\frac{1}{2}\) and \(\frac{3}{4}\) as its roots?
Bayanin Amsa
The quadratic equation with roots \(-\frac{1}{2}\) and \(\frac{3}{4}\) can be written in factored form as: $$a(x+\frac{1}{2})(x-\frac{3}{4}) = 0$$ where a is a constant. Expanding this expression, we get: $$a(x+\frac{1}{2})(x-\frac{3}{4}) = ax^2+\frac{1}{8}a = 0$$ Simplifying the equation, we get: $$8ax^2 + a = 0$$ Now we can compare the coefficients of this equation with those in the given options to find the answer. Comparing the coefficients of the given options with our equation, we can see that only the equation: 8x2 - 2x - 3 = 0 has the same coefficients, and therefore, the same roots, as the equation we derived. Therefore, the answer is: 8x2 - 2x - 3 = 0.
Tambaya 19 Rahoto
Given that ξ = {1, 2, 3, . . . . . . ,10}, P= (x : x is prime) and Q = {y : y is odd}, find Pl∩Q
Bayanin Amsa
Tambaya 20 Rahoto
Find the mean deviation of 6, 7, 8, 9, 10
Bayanin Amsa
To find the mean deviation, we first need to find the mean (average) of the given numbers. The mean is calculated by adding up all the numbers and then dividing by the total number of numbers: \[\text{Mean} = \frac{6 + 7 + 8 + 9 + 10}{5} = 8\] Next, we find the deviation of each number from the mean. To do this, we subtract the mean from each number: \[\begin{aligned} \text{Deviation of 6} &= 6 - 8 = -2 \\ \text{Deviation of 7} &= 7 - 8 = -1 \\ \text{Deviation of 8} &= 8 - 8 = 0 \\ \text{Deviation of 9} &= 9 - 8 = 1 \\ \text{Deviation of 10} &= 10 - 8 = 2 \end{aligned}\] Note that any negative deviations should be treated as positive values, so we need to ignore the negative signs when we calculate the mean deviation. To find the mean deviation, we add up all the deviations (ignoring any negative signs) and then divide by the total number of numbers: \[\begin{aligned} \text{Mean Deviation} &= \frac{|-2| + |-1| + |0| + |1| + |2|}{5} \\ &= \frac{2 + 1 + 0 + 1 + 2}{5} \\ &= \frac{6}{5} \\ &= 1.2 \end{aligned}\] Therefore, the mean deviation of 6, 7, 8, 9, 10 is 1.2.
Tambaya 21 Rahoto
A school girl spends \(\frac{1}{4}\) of her pocket money on books and \(\frac{1}{3}\) on dress. What fraction remains?
Bayanin Amsa
The girl spends \(\frac{1}{4}\) of her pocket money on books and \(\frac{1}{3}\) on dress, which means she has spent a total of \(\frac{1}{4}+\frac{1}{3}\) of her money. To find the fraction that remains, we need to subtract this amount from 1 (since 1 represents the whole amount of money). \(\frac{1}{4}+\frac{1}{3} = \frac{3}{12}+\frac{4}{12} = \frac{7}{12}\) So, the fraction that remains is: \(1-\frac{7}{12} = \frac{12}{12}-\frac{7}{12} = \frac{5}{12}\) Therefore, the answer is \(\frac{5}{12}\).
Tambaya 22 Rahoto
Given that p = 2, q = -5 and r = - 4, evaluate 3p2 - q2 - r3
Bayanin Amsa
Substitute the given values of p, q and r into the expression 3p2 - q2 - r3 to get: 3(2)2 - (-5)2 - (-4)3 = 12 - 25 + 64 = 51 Therefore, the answer is 51.
Tambaya 23 Rahoto
What is the volume of a solid cylinder of diameter 7cm and height 7cm? (Take \(\pi = \frac{22}{7}\))
Bayanin Amsa
The volume of a cylinder can be calculated by multiplying the area of its base by its height. The base of this cylinder is a circle with diameter 7cm. The radius of the circle is half of the diameter, which is 3.5cm. Using the formula for the area of a circle, we can find that the area of the base is: \[\text{Area of base} = \pi r^2 = \frac{22}{7} \times 3.5^2 = 38.5\text{ cm}^2\] The height of the cylinder is also given as 7cm. Using the formula for the volume of a cylinder, we can find the volume of the solid: \[\text{Volume of cylinder} = \text{Area of base} \times \text{height} = 38.5\text{ cm}^2 \times 7\text{ cm} = 269.5\text{ cm}^3\] Therefore, the volume of the solid cylinder is 269.5cm3.
Tambaya 24 Rahoto
A Cooperative Society, charges an interest of 51/2% per annum on any amount borrowed by its members. If a member borrows N125,000, how much does he pay back after one year?
Bayanin Amsa
If the Cooperative Society charges an interest of 51/2% per annum, then the total amount to be paid back by the member at the end of the year will be the principal amount plus 51/2% of the principal amount. 51/2% can be written as a fraction of 5/2 in its simplest form. Therefore, the interest on N125,000 will be (5/2) * N125,000/100 = N3125. The total amount to be paid back by the member after one year will be the sum of the principal and the interest, which is N125,000 + N3125 = N128,125. Therefore, the member will pay back N128,125 after one year. The option that matches this result is N131,875, which is not correct.
Tambaya 25 Rahoto
Given that \(P\propto \frac{1}{\sqrt{r}}\) and p = 3 when r = 16, find the value of r when p =-5
Bayanin Amsa
Tambaya 26 Rahoto
A point on the ground is 5m away from the foot of a vertical wall 7 m high, Calculate, correct to the nearest degree, the angle of depression of the point from the top of the wall
Tambaya 27 Rahoto
Each interior angle of a regular polygon is 108°. How many sides has it?
Bayanin Amsa
To find the number of sides of a regular polygon with interior angle 108°, we need to use the formula for the sum of interior angles of a polygon, which is (n-2) x 180°, where n is the number of sides. Since the polygon is regular, all its interior angles are equal, so we can use the fact that the sum of the interior angles of a polygon is also equal to the number of sides times the interior angle. Therefore, we have: (n-2) x 180° = n x 108° Simplifying this equation, we get: 180n - 360 = 108n 72n = 360 n = 5 Therefore, the regular polygon has 5 sides, and the answer is option A.
Tambaya 28 Rahoto
Given that \(\sqrt{128}+\sqrt{18}-\sqrt{K} = 7\sqrt{2}\), find K,
Bayanin Amsa
We have the expression: \(\sqrt{128}+\sqrt{18}-\sqrt{K} = 7\sqrt{2}\). We can simplify the two radicals to get: \(\sqrt{128}=8\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\). Substituting these values into the expression, we get: \(8\sqrt{2}+3\sqrt{2}-\sqrt{K} = 7\sqrt{2}\) Combining like terms on the left side, we have: \(11\sqrt{2}-\sqrt{K} = 7\sqrt{2}\) Isolating the square root on the left side, we get: \(-\sqrt{K} = -4\sqrt{2}\) Squaring both sides, we have: \(K = (-4\sqrt{2})^2\) Simplifying the expression, we get: \(K = 32\) Therefore, the value of K is 32.
Tambaya 29 Rahoto
In the diagram, POQ is the diameter of the circle centre O. Calculate ∠QRS
Bayanin Amsa
Since POQ is the diameter of the circle, we know that angle POR = 90 degrees. Therefore, angle QOR is also 90 degrees because it is vertically opposite to angle POR. Since QR is a chord of the circle, angle QRS is half of the angle subtended by the same chord at the circumference of the circle, which is angle QOR. Thus, angle QRS = 1/2 * angle QOR = 1/2 * 90 degrees = 45 degrees. Therefore, the answer is 45 degrees, which is 35o.
Tambaya 30 Rahoto
Simplify \(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2}\)and correct your answer to the nearest whole number
Bayanin Amsa
To simplify the expression \(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2}\), we need to perform the arithmetic operations in the following order: division, addition, and subtraction. First, we need to simplify the expression inside the parentheses: \begin{align*} 2\frac{1}{2}+3 &= \frac{5}{2} + 3 \\ &= \frac{5}{2} + \frac{6}{2} \\ &= \frac{11}{2} \end{align*} Next, we need to divide $\frac{11}{2}$ by $16\frac{1}{2}$: \begin{align*} \frac{11}{2} \div 16\frac{1}{2} &= \frac{\frac{11}{2}}{\frac{33}{2}} \\ &= \frac{11}{33} \\ &= \frac{1}{3} \end{align*} Substituting $\frac{1}{3}$ into the original expression, we have: \begin{align*} 7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2} &= 7\frac{1}{2}-\frac{1}{3} \\ &= \frac{22}{3}-\frac{1}{3} \\ &= \frac{21}{3} \\ &= 7 \end{align*} Therefore, the answer is 7, corrected to the nearest whole number. Answer is correct.
Tambaya 31 Rahoto
Find the next two terms of the sequence
1, 5, 14, 30, 55, ...
Tambaya 33 Rahoto
The angle of elevation of the top of a tower from a point on the ground which is 36m away from the foot of the tower is 30o. Calculate the height of the tower.
Bayanin Amsa
Let h be the height of the tower. We can form a right triangle with the tower, the ground point and the point directly below the top of the tower. Let's call the latter point P. Then, we have: - Angle of elevation of the top of the tower from point on ground = 30o - Length of base of the right triangle (i.e., distance from the point on the ground to the foot of the tower) = 36m Therefore, we can use tangent function to find h: tan(30o) = opposite / adjacent = h / 36 Solving for h, we have: h = 36 * tan(30o) ≈ 20.78m Therefore, the height of the tower is approximately 20.78m. So, the correct option is (b) 20.78m.
Tambaya 34 Rahoto
In the diagram, PQRW is a circle. Line P, V and QR are produced to meet at M, where ?WMR = 30o and |WM| = |MR| Find the value of x
Bayanin Amsa
Tambaya 35 Rahoto
Given that x + y = 7 and 3x-y = 5, evaluate \(\frac{y}{2}-3\).
Bayanin Amsa
To solve this problem, we first need to find the values of x and y, and then substitute them into the expression \(\frac{y}{2}-3\). From the given equations, we can use the elimination method to solve for x and y. Multiplying the second equation by 2 and adding it to the first equation, we get: 2(3x-y) + (x+y) = 11x = 17 x = 17/11 Substituting x into the first equation, we get: y = 7 - x = 7 - 17/11 = 60/11 Now we can substitute these values into the expression \(\frac{y}{2}-3\): \(\frac{y}{2}-3 = \frac{60/11}{2} - 3 = \frac{30}{11} - \frac{33}{11} = -\frac{3}{11}\) Therefore, the answer is -1.
Tambaya 36 Rahoto
Simplify \(\frac{2}{a+b}-\frac{1}{a-b}\)
Bayanin Amsa
To simplify the expression, we need to find a common denominator for the two fractions. We can use the difference of two squares identity \((a+b)(a-b) = a^2-b^2\) to find a common denominator. \begin{aligned} \frac{2}{a+b}-\frac{1}{a-b} &= \frac{2(a-b)}{(a+b)(a-b)} - \frac{1(a+b)}{(a-b)(a+b)}\\ &= \frac{2(a-b)-1(a+b)}{a^2-b^2}\\ &= \frac{2a-2b-a-b}{a^2-b^2}\\ &= \frac{a-3b}{a^2-b^2} \end{aligned} Therefore, the simplified expression is \(\frac{a-3b}{a^2-b^2}\), which corresponds to.
Tambaya 37 Rahoto
Find the area of a rectangle of length 4cm and whose diagonal is 6cm, (Leave your answer in surd form)
Tambaya 38 Rahoto
Given that sin (5x - 28)o = cos (3x - 50)o,0 < x < 90o, find the value of x
Tambaya 39 Rahoto
Find the average of the first four prime numbers greater than10
Bayanin Amsa
The first four prime numbers greater than 10 are 11, 13, 17, and 19. To find their average, we add them up and divide by the total number of primes, which is 4. \begin{align*} \text{Average} &= \frac{11 + 13 + 17 + 19}{4} \\ &= \frac{60}{4} \\ &= 15 \end{align*} Therefore, the average of the first four prime numbers greater than 10 is 15. Answer: 15
Tambaya 40 Rahoto
The table above gives the marks scored by a group of students in a test Use the table to answer the Question below
What is the median mark?
Bayanin Amsa
Tambaya 41 Rahoto
Factorise 27p2x2 - 48y2.
Tambaya 42 Rahoto
Calculate the total surface area of a cupboard which measures 12cm by 10cm by 8cm
Tambaya 43 Rahoto
Use the graph to answer the Question below
What are the roots of the equation x2 + 3x - 4 = 0?
Tambaya 44 Rahoto
Find, correct to two decimal places, the mean of 9, 13, 16, 17, 19, 23, 24.
Bayanin Amsa
To find the mean of a set of numbers, we need to add up all the numbers and divide by the total number of numbers. Adding the given numbers, we get: 9 + 13 + 16 + 17 + 19 + 23 + 24 = 121 There are 7 numbers in the set, so to find the mean, we divide the sum by 7: Mean = 121/7 = 17.29 (correct to two decimal places) Therefore, the answer is (b) 17.29.
Tambaya 45 Rahoto
Which of the following is/are true? In a plane, the locus of points I. Equidistant from a straight line is a circle radius d where d is the distance between the point and the straight line.
II Equidistant from two given points P and Q is a circle of radius |PQ|. III Equidistant from two points is the perpendicular bisector of the line joining the two points.
Bayanin Amsa
Option III is true, while options I and II are not true. Explanation: I. The locus of points equidistant from a straight line is actually a pair of parallel lines, not a circle. To see this, consider any point on one side of the line, and draw the perpendicular from that point to the line. The set of all points equidistant from the line will be the line that passes through all of those perpendiculars. II. The locus of points equidistant from two given points is actually the perpendicular bisector of the line segment joining the two points, not a circle. To see this, note that any point on the perpendicular bisector is equidistant from the two points by definition. Conversely, any point equidistant from the two points must lie on the perpendicular bisector. III. The locus of points equidistant from two points is indeed the perpendicular bisector of the line segment joining the two points. To see this, consider any point on the perpendicular bisector. By definition, it is equidistant from the two points. Conversely, any point equidistant from the two points must lie on the perpendicular bisector.
Tambaya 46 Rahoto
The sides of two cubes are in the ratio 2:5. What is the ratio of their volumes?
Bayanin Amsa
The ratio of the sides of the two cubes is 2:5. Let us assume that the length of the sides of the first cube is 2x, then the length of the sides of the second cube will be 5x, since the ratio of their sides is 2:5. The volume of the first cube will be (2x)^3 = 8x^3, and the volume of the second cube will be (5x)^3 = 125x^3. The ratio of the volumes of the two cubes will be: Volume of the first cube : Volume of the second cube = 8x^3 : 125x^3 = 8 : 125 Therefore, the ratio of their volumes is 8:125. Hence, the correct answer is "8:125".
Tambaya 48 Rahoto
Find the range of values of x for which\(\frac{x+2}{4}-\frac{x+1}{3}>\frac{1}{2}\)
Bayanin Amsa
We can start solving the inequality by first finding a common denominator for the fractions: \begin{align*} \frac{x+2}{4}-\frac{x+1}{3}&>\frac{1}{2}\\ \frac{3(x+2)}{12}-\frac{4(x+1)}{12}&>\frac{6}{12}\\ \frac{3x+6-4x-4}{12}&>\frac{1}{2}\\ -\frac{x-2}{12}&>\frac{1}{2}\\ \end{align*} Multiplying both sides by $-1$ changes the direction of the inequality: \begin{align*} \frac{x-2}{12}&<-\frac{1}{2}\\ \end{align*} Multiplying both sides by $12$ gives: \begin{align*} x-2&<-6\\ x&< -4\\ \end{align*} Therefore, the answer is: - x < -4
Tambaya 50 Rahoto
In the diagram, ?XYZ is similar to ?PRQ, |XY| = 5cm, |XZ| =3.5cm and |PR| = 8cm. Find |PQ|
Bayanin Amsa
We can use the property of similar triangles that corresponding sides are proportional. First, let's find the ratio of corresponding sides between the two triangles. Ratio of corresponding sides: |XY|/|PR| = 5/8 |XZ|/|PQ| = 3.5/|PQ| Since the two triangles are similar, the ratio of corresponding sides must be equal. Therefore: 5/8 = 3.5/|PQ| Cross-multiplying: 5 x |PQ| = 8 x 3.5 |PQ| = 28/5 Simplifying: |PQ| = 5.6 cm Therefore, |PQ| is 5.6 cm.
Tambaya 51 Rahoto
(a) Simplify : \((\frac{x^{2}}{2} - x + \frac{1}{2})(\frac{1}{x - 1})\)
(b) A point P is 40km from Q on a bearing 061°. Calculate, correct to one decimal place, the distance of P to (i) north of Q ; (ii) east of Q.
(c) A man left N5,720 to be shared among his son and three daughters. Each daughter's share was \(\frac{3}{4}\) of the son's share. How much did the son receive?
None
Bayanin Amsa
None
Tambaya 52 Rahoto
(a) The sides of an isosceles triangle triangle are in the ratio \(7 : 5 : 7\). Calculate, correct to the nearest degree, the angle included between the equal sides.
(b) The sum of the interior angles of a regular polygon is 1440°. Calculate : (i) the number of sides ; (ii) the size of one exterior angle of the polygon.
Tambaya 53 Rahoto
(a) The angles of depression of the top and bottom of a building are 51° and 62° respectively from the top of a tower 72m high. The base of the building is on the same horizontal level as the foot of the tower. Calculate the height of the building correct to 2 significant figures.
(b) 
In the diagram, PR is a chord of the circle centre O and radius 30cm, < POR = 120°. Calculate correct to three significant figures : (i) the length of chord PR ; (ii) the length of arc PQR ; (iii) the perimeter of the shaded portion. (Take \(\pi = 3.142\)).
Bayanin Amsa
None
Tambaya 54 Rahoto
The table shows the marks scored by a group of students in a class test.
| Marks | 0 | 1 | 2 | 3 | 4 | 5 |
| Frequency | 1 | 4 | 9 | 8 | 5 | 3 |
(a)(i) Calculate the mean mark ; (ii) Find the median.
(b) If the information were to be represented in a pie chart, what would be the sectorial angle for the mark 2?
Bayanin Amsa
None
Tambaya 55 Rahoto
(a) Simplify : \(\sqrt{1001_{two}}\), leaving your answer in base two.
(b) 
In the diagram, O is the centre of the circle radius x. /PQ/ = z, /OK/ = y and < OKP = 90°. Find the value of z in terms of x and y.
(c) 
In the diagram, P, Q, R and S are points of the circle centre O. \(\stackrel\frown{POQ} = 160°\), \(\stackrel\frown{QSR} = 45°\) and \(\stackrel\frown{PQS} = 40°\). Calculate, (i) < QPS ; (ii) < RQS.
None
Bayanin Amsa
None
Tambaya 56 Rahoto
(a) Copy and complete the following table of values for the relation \(y = 2x^{2} - 7x - 3\).
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 19 | -3 | -9 |
(b) Using 2 cm to 1 unit on the x- axis and 2 cm to 5 units on the y- axis, draw the graph of \(y = 2x^{2} - 7x - 3\) for \(-2 \leq x \leq 5\).
(c) From your graph, find the : (i) minimum value of y ;
(ii) gradient of the curve at x = 1.
(d) By drawing a suitable straight line, find the values of x for which \(2x^{2} - 7x - 5 = x + 4\).
Tambaya 57 Rahoto
(a) In the simultaneous equations : \(px + qy = 5 ; qx + py = -10\); p and q are constants. If x = 1 and y = -2 is a solution of the equations, find p and q.
(b) Solve : \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\).
Tambaya 58 Rahoto
(a) 
In the diagram, XY is a chord of a circle of radius 5cm. The chord subtends an angle 96° at the centre. Calculate, correct to three significant figures, the area of the minor segment cut-off. (Take \(\pi = \frac{22}{7}\)).
(b) 
The figure shows a circle inscribed in a square. If a portion of the circle is shaded with some portions of the square, calculate the total area of the shaded portions. [Take \(\pi = \frac{22}{7}\)].
Bayanin Amsa
None
Tambaya 59 Rahoto
K(lat. 60°N, long. 50°W) is a point on the eart's surface. L is another point due East of K and the third point N is due North of K. The distance KL is 3520km and KN is 10951km.
(a) Calculate: (i) The longitude of L ; (ii) The latitude of N. (Take \(\pi = \frac{22}{7}\) and the radius of the earth = 6400km).
(b) A man was allowed 20% of his income as tax free. He then paid 25 kobo in the naira on the remainder. If he paid N1,200.00 as tax, calculate his total income.
None
Bayanin Amsa
None
Tambaya 60 Rahoto
The frequency distribution shows tha marks of 100 students in a Mathematics test.
| Marks | 1-10 | 11-20 | 21-30 | 31-40 | 41-50 | 51-60 | 61-70 | 71-80 | 81-90 | 91-100 |
| No. of Students |
2 | 4 | 9 | 13 | 18 | 32 | 13 | 5 | 3 | 1 |
(a) Draw cumulative frequency curve for the distribution .
(b) Use your curve to estimate : (i) the median ; (ii) the lower quartile ; (iii) the 60th percentile.
None
Bayanin Amsa
None
Tambaya 61 Rahoto
Using a ruler and a pair of compasses only,
(a) Construct : (i) \(\Delta PQR\) such that /PQ/ = 8cm, /PR/ = 7cm and < QPR = 105°. (ii) locus \(L_{1}\) of points equidistant from P and Q. (iii) locus \(l_{2}\) of points equidistant Q and R.
(b)(i) Label the point T where \(l_{1}\) and \(l_{2}\) intersect ; (ii) With centre T and radius /TQ/, construct a circle \(l_{3}\). (iii) Complete quadrilateral PQSR such that /RS/ = /QS/ and /RQ/ = /TS/.
Bayanin Amsa
None
Tambaya 62 Rahoto
(a) Solve \(\frac{1}{81^{(x - 2)}} = 27^{(1 - x)}\)
(b) Simplify \(\frac{5}{\sqrt{7} - \sqrt{3}} + \frac{1}{\sqrt{7} + \sqrt{3}}\), leaving your answer in surd form.
Bayanin Amsa
None
Tambaya 63 Rahoto
(a) AB is a chord of a circle centre O. If |AB| = 24.2 cm and the perimeter of \(\Delta\) AOB is 52.2 cm, calculate < AOB, correct to the nearest degree.
(b) A rectangular tank 60cm by 80cm by 100cm is half filled with water. How many litres of water is it holding?
Bayanin Amsa
None
