WAEC SSCE - General Mathematics - 2006
Ajụjụ 1 Ripọtì
If x + y = 12 and 3x - y = 20, find the value of 2x - y
Akọwa Nkọwa
We are given two equations as follows: x + y = 12 .... (Equation 1) 3x - y = 20 .... (Equation 2) We need to find the value of 2x - y. To solve the problem, we can use the method of substitution or elimination. Here, we will use the method of substitution. From Equation 1, we can write: y = 12 - x Substituting this value of y in Equation 2, we get: 3x - (12 - x) = 20 Simplifying the equation, we get: 4x - 12 = 20 Adding 12 to both sides, we get: 4x = 32 Dividing both sides by 4, we get: x = 8 Now, substituting the value of x in Equation 1, we get: y = 12 - 8 y = 4 Therefore, we have found the values of x and y as 8 and 4, respectively. Finally, substituting these values in the expression 2x - y, we get: 2x - y = 2(8) - 4 = 12 Hence, the value of 2x - y is 12.
Ajụjụ 2 Ripọtì
What must be added to x2 - 3x to make it a perfect square?
Akọwa Nkọwa
To make a perfect square, we need to add the square of half of the coefficient of the x-term. In this case, the coefficient of x is -3, so we need to add \(\left(\frac{-3}{2}\right)^2 = \frac{9}{4}\) to make it a perfect square. Thus, the answer is \(\frac{9}{4}\).
Ajụjụ 4 Ripọtì
If p = {1, 3, 5, 7, 9} and Q = {2, 4, 6, 8, 10} are subsets of a universal set. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. What are the elements of P1 \(\cap\) Q1?
Akọwa Nkọwa
Ajụjụ 5 Ripọtì
The ratio of boys to girls in a class is 5:3. Find the probability of selecting at random, a girl from the class
Akọwa Nkọwa
The ratio of boys to girls in the class is 5:3. This means that for every 5 boys, there are 3 girls in the class. Let B be the number of boys in the class and G be the number of girls in the class. Then, we can write: B:G = 5:3 We can simplify this ratio by dividing both sides by the greatest common factor of 5 and 3, which is 1. B:G = 5/1 : 3/1 B:G = 5 : 3 This means that there are 5 parts for boys and 3 parts for girls in the class. The total number of parts is 5+3 = 8. To find the probability of selecting a girl at random, we need to find the number of parts that represent the girls and divide it by the total number of parts. The number of parts that represent the girls is 3, since there are 3 parts for girls out of a total of 8 parts. Therefore, the probability of selecting a girl at random is: P(selecting a girl) = number of parts that represent the girls / total number of parts P(selecting a girl) = 3/8 Therefore, the probability of selecting a girl at random from the class is 3/8.
Ajụjụ 6 Ripọtì
A rectangular carpet 2.5m long and 2.4m wide covers 5% of a rectangular floor. Calculate the area of the floor
Akọwa Nkọwa
Let's assume that the rectangular floor has a length of L meters and a width of W meters. We know that the rectangular carpet covers 5% of the rectangular floor, which means that the area of the carpet is 5% of the area of the floor. We can write this as an equation: 2.5 × 2.4 = 0.05 × L × W Multiplying the left-hand side, we get: 6 = 0.05LW Dividing both sides by 0.05, we get: 120 = LW Therefore, the area of the rectangular floor is LW, which is 120 square meters. Hence, the area of the floor is 120 m², which is the third option.
Ajụjụ 7 Ripọtì
In the diagram, the two circles have a common centre O. If the area of the larger circle is 100\(\pi\) and that of the smaller circle is 49\(\pi\), find x
Akọwa Nkọwa
Let the radius of the smaller circle be r, and the radius of the larger circle be R. The area of a circle is given by the formula A = \(\pi r^2\). Therefore, the radius of the smaller circle is \(\sqrt{\frac{49\pi}{\pi}} = 7\). Similarly, the radius of the larger circle is \(\sqrt{\frac{100\pi}{\pi}} = 10\). Since O is the centre of both circles, the distance between O and the point where the two circles intersect is R - r = 10 - 7 = 3. From the diagram, we can see that x is also equal to this distance, so x = 3. Therefore, the answer is 3.
Ajụjụ 8 Ripọtì
The values of three angles at a point are 3y - 45o, y + 25o and yo . Find the value of y.
Akọwa Nkọwa
When three angles at a point are added together, they sum up to 360 degrees. Therefore, we have: (3y - 45) + (y + 25) + y = 360 Simplifying the equation, we get: 5y - 20 = 360 Adding 20 to both sides, we get: 5y = 380 Dividing both sides by 5, we get: y = 76 Hence, the value of y is 76 degrees, and the correct option is (d) 76o.
Ajụjụ 9 Ripọtì
In the diagram, PQR is a straight line, QRST is a parallelogram, < TPQ = 72o and < RST = 126o. What type of triangle is \(\bigtriangleup\) PQT?
Ajụjụ 10 Ripọtì
The diagram shows an arc MN of a circle, centre O, with 10cm. If < MON = 72o. calculate the length of the arc, correct to three significant figures. [Take \(\pi = \frac{22}{7}\)]
Akọwa Nkọwa
Ajụjụ 11 Ripọtì
Simplify \(\frac{x - 4}{4} - \frac{x - 3}{6}\)
Akọwa Nkọwa
To simplify the given expression, we first need to find a common denominator for the two fractions. In this case, the least common multiple of 4 and 6 is 12. So, we can write: \[\frac{x - 4}{4} - \frac{x - 3}{6} = \frac{3(x - 4)}{12} - \frac{2(x - 3)}{12}\] Now, we can combine the two fractions by subtracting the numerators: \[\frac{3(x - 4)}{12} - \frac{2(x - 3)}{12} = \frac{3x - 12 - 2x + 6}{12}\] Simplifying the numerator, we get: \[\frac{3x - 2x - 12 + 6}{12} = \frac{x - 6}{12}\] Therefore, the simplified expression is \(\frac{x - 6}{12}\). So, the answer is option (B).
Ajụjụ 12 Ripọtì
If 2x + y = 10, and y \(\neq\) 0, which of the following is not a possible value of x?
Akọwa Nkọwa
The equation 2x + y = 10 can be rearranged as 2x = 10 - y. Dividing both sides by 2 gives x = 5 - (y/2). Therefore, the value of x will depend on the value of y. Now, if we substitute each of the given values of x into the equation 2x + y = 10, we can determine which ones are not possible. For x = 4, we get 2(4) + y = 10, which simplifies to y = 2. This is a possible value of y, so x = 4 is a possible value of x. For x = 5, we get 2(5) + y = 10, which simplifies to y = 0. But the question states that y ≠ 0, so x = 5 is not a possible value of x. For x = 8, we get 2(8) + y = 10, which simplifies to y = -6. This is a possible value of y, so x = 8 is a possible value of x. For x = 10, we get 2(10) + y = 10, which simplifies to y = -10. This is also a possible value of y, so x = 10 is a possible value of x. Therefore, the answer is x = 5, which is not a possible value of x.
Ajụjụ 13 Ripọtì
The temperature in a chemical plant was -5oC at 2.00 am. The temperature fell by 6oC and then rose again by 7oC. What was the final temperature?
Akọwa Nkọwa
The temperature in the chemical plant was -5oC at 2.00 am. The temperature fell by 6oC, so the temperature became (-5 - 6) = -11oC. The temperature then rose by 7oC, which means the temperature increased from -11oC to (-11 + 7) = -4oC. Therefore, the final temperature was -4oC. Hence, the correct option is (-4oC).
Ajụjụ 14 Ripọtì
\(\begin{array}{c|c} x & 1 & 4 & p \\ \hline y & 0.5 & 1 & 2.5\end{array}\). The table below satisfies the relation y - k\(\sqrt{x}\), where k is a positive constant. Find the value of P,
Akọwa Nkọwa
Ajụjụ 15 Ripọtì
Evaluate (0.13)\(^3\)correct to three significant figures
Akọwa Nkọwa
To evaluate \((0.13)^3\) correct to three significant figures, we need to first cube the number and then round the answer to three significant figures. \((0.13)^3 = 0.002197\) Rounding to three significant figures, we get 0.00220. Therefore, the correct answer is 0.00220.
Ajụjụ 17 Ripọtì
The wheel of a tractor has a diameter 1.4m. What distance does it cover in 100 complete revolutions? [Take \(\pi = \frac{22}{7}\)]
Akọwa Nkọwa
The circumference of the wheel is the distance it covers in one revolution. Circumference of the wheel = π × diameter Substituting the given values, Circumference of the wheel = π × 1.4m = 4.4m (rounded to one decimal place) Therefore, the distance covered in 100 revolutions is: Distance covered = Circumference of the wheel × Number of revolutions Substituting the given values, Distance covered = 4.4m × 100 = 440m Therefore, the wheel covers a distance of 440m in 100 complete revolutions. Hence, the correct option is (D) 440m.
Ajụjụ 19 Ripọtì
Given that (2x - 1)(x + 5) = 2x2 - mx - 5, what is the value of m
Ajụjụ 20 Ripọtì
A ladder 16m long leans against an electric pole. If the ladder makes an angle of 65o with the ground, how far up the electric pole does its top reach
Akọwa Nkọwa
We can use trigonometry to solve this problem. Let x be the distance from the bottom of the ladder to the electric pole. Then, we can create a right triangle with the ladder as the hypotenuse, x as one leg, and the distance we are trying to find as the other leg. Using trigonometric functions, we know that: sin(65o) = opposite/hypotenuse sin(65o) = (distance up the pole)/16 (distance up the pole) = 16*sin(65o) (distance up the pole) ≈ 14.5m Therefore, the answer is 14.5m.
Ajụjụ 21 Ripọtì
If loga270 - loga10 + loga \(\frac{1}{3}\) = 2, what is the value of a?
Akọwa Nkọwa
Using the logarithmic property that loga(b) - loga(c) = loga(b/c), we can simplify the given expression as: loga(270/10) + loga(1/3) = 2 Simplifying further, we get: loga(27) + loga(1/3) = 2 Using the logarithmic property that loga(b) + loga(c) = loga(bc), we can write: loga(27 x 1/3) = 2 loga(9) = 2 Now, using the definition of logarithm, we can write: a^2 = 9 Taking the square root on both sides, we get: a = 3 or -3 However, a cannot be negative, as the base of a logarithm must be positive. Therefore, the value of a is 3. Hence, the value of a is 3.
Ajụjụ 22 Ripọtì
If the simple interest on a sum of money invested at 3% per annum for 2\(\frac{1}{2}\) years is N123, find the principal.
Akọwa Nkọwa
The formula for simple interest is: I = PRT Where: I = Interest P = Principal R = Rate of Interest T = Time From the question: R = 3% per annum T = 2.5 years I = N123 We can substitute the values given into the formula and solve for P: 123 = P x 0.03 x 2.5 123 = 0.075P P = 123 ÷ 0.075 P = N1,640 Therefore, the principal is N1,640. Answer option (C).
Ajụjụ 24 Ripọtì
Simplify \(\frac{25 \frac{2}{3} \div 25 \frac{1}{6}}{( \frac{1}{5})^{-\frac{7}{6}} \times ( \frac{1}{5})^{\frac{1}{6}}}\)
Ajụjụ 25 Ripọtì
In the diagram, |QR| = 5cm, PQR = 60o and PSR = 45o. Find |PS|, leaving your answe in surd form.
Ajụjụ 26 Ripọtì
A fair coin is tossed three times. Find the probability of getting two heads and one tail.
Akọwa Nkọwa
When a fair coin is tossed, there are two possible outcomes, heads (H) or tails (T), each with a probability of 1/2. Since the coin is tossed three times, the total number of possible outcomes is 2 × 2 × 2 = 8. We need to find the probability of getting two heads and one tail, which can happen in three different ways: HHT, HTH, or THH. Each of these outcomes has a probability of (1/2) × (1/2) × (1/2) = 1/8. Therefore, the probability of getting two heads and one tail is the sum of the probabilities of the three possible outcomes: P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8. Hence, the answer is 3/8.
Ajụjụ 27 Ripọtì
In the diagram what is x + y in terms of z?
Akọwa Nkọwa
In a triangle, the sum of the angles is always 180 degrees. In the given diagram, we can see that the angles x, y, and z form a triangle. Therefore, we have: x + y + z = 180o Rearranging this equation, we get: x + y = 180o - z Hence, the value of x + y in terms of z is 180o - z. Therefore, the answer is (a) 180o - z.
Ajụjụ 29 Ripọtì
In the diagram |XY| = 12cm, |XZ| = 9cm, |ZN| = 3cm and ZY||NM, calculate |MY|
Ajụjụ 31 Ripọtì
A seller allows 20% discount for cash payment on the marked price of his goods. What is the ratio of the cash payment to the marked price
Akọwa Nkọwa
If the seller allows a 20% discount for cash payment, then the customer will only have to pay 80% of the marked price. Let MP be the marked price and CP be the cash payment. CP = 80% of MP CP/MP = 80/100 CP/MP = 4/5 Therefore, the ratio of the cash payment to the marked price is 4:5. So the correct answer is option (D) 4:5.
Ajụjụ 32 Ripọtì
A baker used 40% of a 50kg bag of flour. If \(\frac{1}{8}\) of the amount used was for the cake, how many kilogram of flour was used for the cake?
Akọwa Nkọwa
The baker used 40% of a 50kg bag of flour, which is equivalent to: $$ 40\% \times 50kg = \frac{40}{100} \times 50kg = 20kg $$ If \(\frac{1}{8}\) of the amount used was for the cake, then the amount of flour used for the cake is: $$ \frac{1}{8} \times 20kg = \frac{20}{8} kg = 2\frac{1}{2} kg $$ Therefore, the answer is \boxed{2\frac{1}{2}}.
Ajụjụ 33 Ripọtì
\(\begin{array}{c|c} x & 1 & 4 & p \\ \hline y & 0.5 & 1 & 2.5\end{array}\). The table above satisfies the relation y = k\(\sqrt{x}\), where k is a positive constant. Find the value of K.
Akọwa Nkọwa
Ajụjụ 34 Ripọtì
In the diagram, \(\frac{PQ}{RS}\), find xo + yo
Ajụjụ 35 Ripọtì
In the diagram, P, Q and R are three points in a plane such that the bearing of R from Q is 110o and the bearing of Q from P is 050o. Find angle PQR.
Akọwa Nkọwa
Ajụjụ 36 Ripọtì
In the diagram, PR is a diameter of the circle centre O. RS is a tangent at R and QPR = 58o. Find < QRS
Akọwa Nkọwa
Ajụjụ 37 Ripọtì
A machine valued at N20,000 depreciates by 10% every year. What will be the value of the machine at the end of two years?
Akọwa Nkọwa
At the end of the first year, the value of the machine will be 90% of N20,000 = N18,000. At the end of the second year, the value of the machine will be 90% of N18,000 = N16,200. Therefore, the value of the machine at the end of two years will be N16,200. So the correct option is: - N16,200
Ajụjụ 38 Ripọtì
Which of the following is not a measure of centre tendency?
Akọwa Nkọwa
Range is not a measure of central tendency. Measures of central tendency are statistical measures that represent the typical or central values of a data set. They include the mean, median, and mode. The mean is the arithmetic average of a data set, the median is the middle value when the data is arranged in order, and the mode is the most frequently occurring value in the data set. On the other hand, range is a measure of dispersion, which describes how spread out the data is. It is the difference between the highest and lowest values in the data set.
Ajụjụ 40 Ripọtì
a conical water-jug is 7cm in diameter and 6cm deep. find the volume of water it can hold. [Take \(\pi = \frac{22}{7}\)]
Akọwa Nkọwa
The volume of a cone can be calculated using the formula: V = (1/3)πr2h where V is the volume of the cone, r is the radius of the circular base, h is the height of the cone, and π is a mathematical constant approximately equal to 22/7. Given that the diameter of the circular base is 7cm, the radius r can be calculated as half of the diameter: r = 7/2 = 3.5cm. The height of the cone is given as 6cm. Substituting these values into the formula for the volume of a cone, we get: V = (1/3) x (22/7) x (3.5cm)2 x 6cm Simplifying this expression, we get: V = (1/3) x (22/7) x 12.25cm2 x 6cm V = (22/21) x 74.25cm3 V = 77cm3 Therefore, the volume of water the conical water-jug can hold is 77cm3. Answer option (C) is correct.
Ajụjụ 41 Ripọtì
What is the place value of 9 in the number 3.0492?
Akọwa Nkọwa
In the number 3.0492, 9 is in the thousandth place. The places in a decimal number are counted from left to right starting from the tenths place, which is immediately to the right of the decimal point. Each place value to the right of the tenths place is ten times smaller than the one to its left. So the first digit to the right of the decimal point is in the tenths place, the second digit is in the hundredths place, and the third digit is in the thousandths place, and so on. Therefore, the place value of 9 in the number 3.0492 is \(\frac{9}{1000}\).
Ajụjụ 42 Ripọtì
P is a point on the same plane with a fixed point A. If P moves such that it is always equidistant from A, the locus of P is
Akọwa Nkọwa
Ajụjụ 44 Ripọtì
If p = [\(\frac{Q(R - T)}{15}\)]\(^ \frac{1}{3}\), make T the subject of the relation
Ajụjụ 45 Ripọtì
Given that y = 1 - \(\frac{2x}{4x - 3}\), find the value of x for which y is undefined
Akọwa Nkọwa
To find the value of x for which y is undefined, we need to look for values of x that make the denominator of the expression for y equal to zero, since division by zero is undefined. The denominator of the expression for y is 4x - 3. Therefore, we need to find the value of x that makes 4x - 3 equal to zero. Solving the equation 4x - 3 = 0, we get: 4x = 3 x = 3/4 Therefore, the value of x for which y is undefined is x = 3/4. Explanation: To understand why we need to look for values of x that make the denominator of the expression for y equal to zero, we need to remember that division by zero is undefined. When we divide a number by zero, there is no answer because it is impossible to divide any number into equal parts of zero size. In the given expression for y, the denominator is 4x - 3. If we plug in x = 3/4, the denominator becomes: 4x - 3 = 4(3/4) - 3 = 3 - 3 = 0 This means that when x = 3/4, the denominator becomes zero, and therefore division by zero occurs, which makes the value of y undefined. Hence, the value of x for which y is undefined is x = 3/4.
Ajụjụ 46 Ripọtì
Which of these equations best describes the points of intersection of the curve and the line?
Akọwa Nkọwa
Ajụjụ 47 Ripọtì
In the diagram, ABCDEO is two- thirds of a circle centre O. The radius AO is 7cm and /AB/ = /BC/ = /CD/ = /DE/. Calculate, correct to the nearest whole number, the area of the shaded portion. [Take \(\pi = \frac{22}{7}\)].
Akọwa Nkọwa
None
Ajụjụ 48 Ripọtì
(a) Two lines AB and CD intersect at x such that \(\stackrel\frown{CAX}\) is equal to \(\stackrel\frown{BDX}\). If |AX| = 6 cm, |XB| = 4 cm and |CX| = 3 cm, find |XD|.
(b) 
The diagram shows the positions of three points X, Y and Z on a horizontal plane. The bearing of Y from X is 312° and that of Y from Z is 022°. If |XY| = 32 km and |ZY| = 50 km, calculate, correct to one decimal place : (i) |XZ| ; (ii) the bearing of Z from X.
Ajụjụ 49 Ripọtì
(a)(i) If \(4x < 2 + 3x\) and \(x - 8 < 3x\), what range of values of x satisfies both inequalities? ; (ii) Represent your result in (i) on the number line.
(b) A shop is sending out a bill for an amount less than £100. The accountant interchanges the two digits and so overcharges the customer by 45. Given that the sum of the two digits is 9, find how much the bill should be.
Akọwa Nkọwa
None
Ajụjụ 50 Ripọtì
Two fair dice are tossed together once.
(a) Draw a sample space for the possible outcomes ;
(b) Find the probability of getting a total : (i) of 7 or 8 ; (ii) less than 4.
Ajụjụ 51 Ripọtì
(a) A sector of a circle of radius 8cm subtends an angle of 90° at the centre of the circle. If the sector is folded without overlap to form the curved surface of a cone, find the :
(i) base radius ; (ii) height ; (iii) volume of the cone. [Take \(\pi = \frac{22}{7}\)].
(b) A map is drawn to a scale of 1 : 20,000. Use it to calculate the : (i) distance, in kilometres, represented by 4.5 cm on the map ;
Akọwa Nkọwa
None
Ajụjụ 52 Ripọtì
(a) Copy and complete the table of the relation \(y = 2\sin x - \cos 2x\).
| x | 0° | 30° | 60° | 90° | 120° | 150° | 180° |
| y | 0.5 | -1.0 |
Using a scale of 2 cm to 30° on the x- axis and 2 cm to 0.5 unit on the y- axis, draw the graph of \(y = 2\sin x - \cos 2x\), for \(0° \leq x \leq 180°\).
(b) Using the same axes, draw the graph of \(y = 1.25\).
(c) Use your graphs to find the : (i) values of x for which \(2\sin x - \cos 2x = 0\) ; (ii) the roots of the equation \(2\sin x - \cos 2x = 1.25\).
Akọwa Nkọwa
None
Ajụjụ 53 Ripọtì
(a) 
The diagram shows a pyramid standing on a cuboid. The dimensions of the cuboid are 4m by 3m by 2m and the slant edge of the pyramid is 5m. Calculet the volume of the shape.
(b) The 2nd, 3rd and 4th terms of an A.P are x - 2, 5 and x + 2 respectively. Calculate the value of x.
None
Akọwa Nkọwa
None
Ajụjụ 54 Ripọtì
The following table shows the distribution of test scores in a class.
| Scores | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 | 10 |
| No of pupils | 1 | 1 | 5 | 3 | \(k^{2} + 1\) | 6 | 2 | 3 | 4 |
(a) If the mean score of the class is 6, find the : (i) value of k (ii) median score.
(b) Draw a bar chart for the distribution.
(c) If a pupil is picked at random, what is the probability that he/ she will score less than 6?
Akọwa Nkọwa
None
Ajụjụ 55 Ripọtì
(a) Simplify : \(\frac{4\frac{2}{9} - 1\frac{13}{15}}{2\frac{1}{5} + \frac{4}{7} \times 2\frac{1}{3}}\)
(b) By rationalising the denominator, simplify : \(\frac{7\sqrt{5}}{\sqrt{7}}\), leaving your answer in surd form.
Akọwa Nkọwa
None
Ajụjụ 57 Ripọtì
(a) Using a ruler and a pair of compasses only, (i) construct \(\Delta\) XYZ such that |XY| = 8 cm and < YXZ = < ZYX = 45°. (ii) locate a point P inside the triangle equidistant from XY and XZ and also equidistance from YX and YZ. (iii) construct a circle touching the three sides of the triangle (iv) measure the radius of the circle.
(b) The length of the sides of a hexagon are x - 5, 2x, 2x, 2x + 7, 2x and 2x - 1. If the perimeter is 144 cm, find the value of x.
Ajụjụ 58 Ripọtì
(a) A dealer sold a car to a man and made a profit of 15%. The man then sold it to a woman for N120,175.00 at a loss of 5%. How much did the dealer buy the car?
(b) The diameter of the wheel of a car is 36cm. How many revolutions, correct to three significant figures, will it make to cover a distance of 1.05 km? [Take \(\pi = \frac{22}{7}\)].
Akọwa Nkọwa
None
Ajụjụ 59 Ripọtì
(a) In a class of 45 students, 32 offered Physics(P), 28 offered Government(G) and 12 did not offer any of the two subjects. (i) Draw the Venn diagram to represent the information ; (ii) How many students offered both subjects? (iii) What is \(n(P \cup G)\)?
(b) If \(p = \frac{2u}{1 - u}\) and \(q = \frac{1 + u}{1 - u}\) ; express \(\frac{p + q}{p - q}\) in terms of u.
Akọwa Nkọwa
None
