WAEC SSCE - General Mathematics - 2004
Ajụjụ 2 Ripọtì
Given that x + y = 7 and 3x-y = 5, evaluate \(\frac{y}{2}-3\).
Akọwa Nkọwa
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.
Ajụjụ 3 Ripọtì
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.
Akọwa Nkọwa
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.
Ajụjụ 4 Ripọtì
Find the range of values of x for which\(\frac{x+2}{4}-\frac{x+1}{3}>\frac{1}{2}\)
Akọwa Nkọwa
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
Ajụjụ 5 Ripọtì
If \(\frac{x}{a+1}+\frac{y}{b}\) 1. Make y the subject of the relation
Ajụjụ 6 Ripọtì
Calculate the total surface area of a cupboard which measures 12cm by 10cm by 8cm
Ajụjụ 8 Ripọtì
Factorise 27p2x2 - 48y2.
Ajụjụ 9 Ripọtì
Find the sum of the roots of the equation 2x2 + 3x - 9 = 0
Ajụjụ 10 Ripọtì
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?
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
Use the graph to answer the Question below
What are the roots of the equation x2 + 3x - 4 = 0?
Akọwa Nkọwa
Ajụjụ 12 Ripọtì
A boy walks 800m in 20 minutes. Calculate his average speed in km per hour
Akọwa Nkọwa
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.
Ajụjụ 13 Ripọtì
If \(\left(\frac{1}{4}\right)^{(2-y)} = 1\), find y.
Akọwa Nkọwa
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.
Ajụjụ 14 Ripọtì
Solve the equation 10-3x-x2 = 0
Akọwa Nkọwa
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.
Ajụjụ 15 Ripọtì
Find the average of the first four prime numbers greater than10
Akọwa Nkọwa
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
Ajụjụ 17 Ripọtì
The locus of a point which moves in a plane such that it is equidistant from two fixed points X and Y is
Akọwa Nkọwa
Ajụjụ 18 Ripọtì
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?
Akọwa Nkọwa
Ajụjụ 19 Ripọtì
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
Akọwa Nkọwa
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.
Ajụjụ 20 Ripọtì
What is the volume of a solid cylinder of diameter 7cm and height 7cm? (Take \(\pi = \frac{22}{7}\))
Akọwa Nkọwa
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.
Ajụjụ 22 Ripọtì
In the diagram, POQ is the diameter of the circle centre O. Calculate ∠QRS
Akọwa Nkọwa
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.
Ajụjụ 23 Ripọtì
In the diagram, KS is a tangent to the circle centre O at R and ?ROQ = 80o. Find ?QRS.
Akọwa Nkọwa
Ajụjụ 24 Ripọtì
Find the area of a rectangle of length 4cm and whose diagonal is 6cm, (Leave your answer in surd form)
Ajụjụ 25 Ripọtì
Solve for t in the equation \(\frac{3}{4}t+\frac{1}{3}(21-t)\) = 11,
Akọwa Nkọwa
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).
Ajụjụ 26 Ripọtì
A point X is on the bearing 342o from a point Y. What is the bearing of Y from X?
Akọwa Nkọwa
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.
Ajụjụ 27 Ripọtì
Which of the following quadratic equations has \(-\frac{1}{2}\) and \(\frac{3}{4}\) as its roots?
Akọwa Nkọwa
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.
Ajụjụ 28 Ripọtì
Given that p = 2, q = -5 and r = - 4, evaluate 3p2 - q2 - r3
Akọwa Nkọwa
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.
Ajụjụ 29 Ripọtì
Find, correct to two decimal places, the mean of 9, 13, 16, 17, 19, 23, 24.
Akọwa Nkọwa
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.
Ajụjụ 30 Ripọtì
Each interior angle of a regular polygon is 108°. How many sides has it?
Akọwa Nkọwa
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.
Ajụjụ 31 Ripọtì
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
Akọwa Nkọwa
Ajụjụ 32 Ripọtì
Given that sin (5x - 28)o = cos (3x - 50)o,0 < x < 90o, find the value of x
Ajụjụ 33 Ripọtì
Given that ξ = {1, 2, 3, . . . . . . ,10}, P= (x : x is prime) and Q = {y : y is odd}, find Pl∩Q
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
Given that \(P\propto \frac{1}{\sqrt{r}}\) and p = 3 when r = 16, find the value of r when p =-5
Ajụjụ 35 Ripọtì
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?
Akọwa Nkọwa
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.
Ajụjụ 37 Ripọtì
If \(y = 23_{five} + 101_{three}\), find y, leaving your answer in base two
Ajụjụ 38 Ripọtì
In the diagram, \(R\hat{P}Q = Q\hat{R}Y, \hspace{1mm} P\hat{Q}R = R\hat{Y}Q, \\ \hspace{1mm}|QP| = 3cm \hspace{1mm}|QY| = 4cm \hspace{1mm}and \hspace{1mm}|RY | = 5cm. \hspace{1mm} Find \hspace{1mm}|QR|\)
Ajụjụ 40 Ripọtì
Simplify \(\frac{2}{a+b}-\frac{1}{a-b}\)
Akọwa Nkọwa
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.
Ajụjụ 41 Ripọtì
There are m boys and 12 girls in a class. What is the probability of selecting at random a girl from the class?
Akọwa Nkọwa
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) \(\frac{12}{m+12}\).
Ajụjụ 42 Ripọtì
In the diagram, ?XYZ is similar to ?PRQ, |XY| = 5cm, |XZ| =3.5cm and |PR| = 8cm. Find |PQ|
Akọwa Nkọwa
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.
Ajụjụ 43 Ripọtì
Find the mean deviation of 6, 7, 8, 9, 10
Akọwa Nkọwa
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.
Ajụjụ 44 Ripọtì
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
Akọwa Nkọwa
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: - $\frac{12}{25}$
Ajụjụ 46 Ripọtì
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.
Akọwa Nkọwa
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.
Ajụjụ 47 Ripọtì
The sides of two cubes are in the ratio 2:5. What is the ratio of their volumes?
Akọwa Nkọwa
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".
Ajụjụ 48 Ripọtì
A school girl spends \(\frac{1}{4}\) of her pocket money on books and \(\frac{1}{3}\) on dress. What fraction remains?
Akọwa Nkọwa
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 \(\frac{5}{12}\).
Ajụjụ 49 Ripọtì
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
Ajụjụ 50 Ripọtì
Given that \(\sqrt{128}+\sqrt{18}-\sqrt{K} = 7\sqrt{2}\), find K,
Akọwa Nkọwa
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.
Ajụjụ 51 Ripọtì
(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 : \(\frac{4r - 3}{6r + 1} = \frac{2r - 1}{3r + 4}\).
Ajụjụ 52 Ripọtì
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/.
Akọwa Nkọwa
None
Ajụjụ 53 Ripọtì
(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 \(\pi = \frac{22}{7}\)].
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
(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?
Akọwa Nkọwa
None
Ajụjụ 55 Ripọtì
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?
Akọwa Nkọwa
None
Ajụjụ 56 Ripọtì
(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.
Akọwa Nkọwa
None
Ajụjụ 57 Ripọtì
(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.
Ajụjụ 58 Ripọtì
(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
Akọwa Nkọwa
None
Ajụjụ 59 Ripọtì
(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 \(2x^{2} - 7x - 5 = x + 4\).
Ajụjụ 60 Ripọtì
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.
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Akọwa Nkọwa
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Ajụjụ 61 Ripọtì
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.
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Akọwa Nkọwa
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Ajụjụ 62 Ripọtì
(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?
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Akọwa Nkọwa
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Ajụjụ 63 Ripọtì
(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 \(\pi = 3.142\)).
Akọwa Nkọwa
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