WAEC SSCE - General Mathematics - 2020
Tambaya 2 Rahoto
Which of the following is not a sufficient condition for two triangles to be congruent?
Bayanin Amsa
The "SSA" (Side-Side-Angle) condition is not a sufficient condition for two triangles to be congruent. This is because two triangles can have the same lengths for two sides and the same measure for an angle that is not included between those sides, but still have different shapes and sizes. In this case, the triangles would not be congruent, but they would have what is called a "SSA" similarity, meaning they have two sides and a non-included angle in common. On the other hand, "SSS" (Side-Side-Side), "SAS" (Side-Angle-Side), and "AAS" (Angle-Angle-Side) are all sufficient conditions for two triangles to be congruent. "SSS" means that all three sides of one triangle are congruent to the three sides of another triangle. "SAS" means that two sides and the included angle of one triangle are congruent to the corresponding two sides and included angle of another triangle. "AAS" means that two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of another triangle.
Tambaya 3 Rahoto
The dimension of a rectangular base of a right pyramid is 9 cm by 5cm. If the volume of the pyramid is 105 cm\(^3\), how high is the pyramid?
Bayanin Amsa
The volume of a pyramid can be calculated using the formula: V = (base area) * (height) / 3 We know the base area (9 cm x 5 cm = 45 cm\(^2\)), and the volume (105 cm\(^3\)), so we can rearrange the formula to solve for the height: height = 3 * (volume) / (base area) Plugging in the given values: height = 3 * (105) / (45) height = 7 cm So the height of the pyramid is 7 cm.
Tambaya 4 Rahoto
In the diagram, PQR is a circle with center O. If ∠OPQ = 48°, find the value of M.
Bayanin Amsa
Tambaya 6 Rahoto
The mean of the numbers 15, 21, 17, 26, 18, and 29 is 21. Calculate the standard deviation
Bayanin Amsa
Standard deviation is a measure of how spread out the data is from the average (mean). It tells us how much the individual data points deviate from the mean. The standard deviation is expressed in the same units as the data. To calculate the standard deviation, we first find the difference between each data point and the mean. We square these differences to eliminate any negative values, and then take the average of the squared differences (this is known as the variance). Finally, we take the square root of the variance to find the standard deviation. For the numbers 15, 21, 17, 26, 18, and 29, with a mean of 21, the standard deviation is approximately 5.24. So, the answer is closest to: 5.
Tambaya 7 Rahoto
Given that x is directly proportional to y and inversely proportional to Z, x = 15 when y = 10 and Z = 4, find the equation connecting x, y and z
Bayanin Amsa
Tambaya 8 Rahoto
The graph of the equations y = 2x + 5 and y = 2x\(^2\) + x - 1 are shown. Use the information above to answer this question.
Find the point of intersection of the two graphs.
Bayanin Amsa
Tambaya 9 Rahoto
A man is five times as old as his son. In four years' time, the product of their ages would be 340. If the son's age is y, express the product of their ages in terms of y.
Bayanin Amsa
Tambaya 10 Rahoto
| x | 6.20 | 6.85 | 7.50 |
| y | 3.90 | 5.20 | 6.50 |
The points on a linear graph are as shown in the table. Find the gradient (slope) of the line.
Bayanin Amsa
The gradient of a line describes how steep the line is, and can be calculated as the change in y (rise) divided by the change in x (run). To find the gradient of the line in this case, we need to find the difference in y between two points and divide it by the difference in x between the same points. For example, we can choose the first two points: (6.20, 3.90) and (6.85, 5.20). The change in y is 5.20 - 3.90 = 1.30, and the change in x is 6.85 - 6.20 = 0.65. The gradient is then 1.30 / 0.65 = 2. Therefore, the gradient of the line is 2.
Tambaya 11 Rahoto
A box contains 12 identical balls of which 5 are red, 4 blue, and the rest are green. If two balls are selected at random one after the other with replacement, what is the probability that both are red?
Bayanin Amsa
There are 12 balls in the box, 5 of which are red, 4 blue, and the remaining ones are green. If we select a ball at random, the probability of selecting a red ball is 5/12, since there are 5 red balls out of 12 total. If we replace the ball and select again, the probability of selecting a red ball is still 5/12, since we put back the ball we selected and the number of red balls remains the same. To find the probability of selecting two red balls in a row, we need to multiply the probability of selecting a red ball on the first draw (5/12) by the probability of selecting a red ball on the second draw (also 5/12), since the events are independent. Therefore, the probability of selecting two red balls in a row with replacement is (5/12) x (5/12) = 25/144. Hence, the correct option is (A) \(\frac{25}{144}\).
Tambaya 12 Rahoto
A box contains 12 identical balls of which 5 are red, 4 blue, and the rest are green. If a ball is selected at random from the box, what is the probability that it is green?
Bayanin Amsa
The probability of an event can be calculated by dividing the number of favorable outcomes by the number of possible outcomes. In this case, we want to find the probability of selecting a green ball from the box. There are 3 green balls out of a total of 12 balls in the box, so the probability is 3/12 = 1/4. So, the answer is: The probability of selecting a green ball from the box is 1/4.
Tambaya 13 Rahoto
If X = {x : x < 7} and Y = {y:y is a factor of 24} are subsets of \(\mu\) = {1, 2, 3...10} find X \(\cap\) Y.
Bayanin Amsa
To find X \(\cap\) Y, we need to determine the common elements that are present in both sets X and Y. Set X is defined as the set of all numbers x, such that x is less than 7. Therefore, X = {1, 2, 3, 4, 5, 6}. Set Y is defined as the set of all factors of 24. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. Therefore, Y = {1, 2, 3, 4, 6, 8, 12, 24}. The intersection of sets X and Y, denoted as X \(\cap\) Y, is the set of all elements that are present in both sets X and Y. Therefore, X \(\cap\) Y = {x : x is an element of X and x is an element of Y} From the sets X and Y, we see that the common elements are 1, 2, 3, 4, 6. Therefore, X \(\cap\) Y = {1, 2, 3, 4, 6} Thus, the correct answer is {1, 2, 3, 4, 6}.
Tambaya 14 Rahoto
If 101\(_{\text{two}}\) + 12y = 3.3\(_{\text{five}}\). Find the value of y
Bayanin Amsa
Tambaya 15 Rahoto
The length of an arc of a circle of radius 3.5 cm is 1\(\frac{19}{36}\) cm. Calculate, correct to the nearest degree , the angle substended by the centre of the circle. [Take \(\pi = \frac{22}{7}\)]
Bayanin Amsa
Tambaya 16 Rahoto
Two buses start from the same station at 9.00am and travel in opposite directions along the same straight road. The first bus travel at a speed of 72 km/h and the second at 48 km/h. At what time will they be 240km apart?
Bayanin Amsa
The two buses are moving away from each other, so we can add their speeds to find the relative speed between them. The relative speed is 72 km/h + 48 km/h = 120 km/h. We want to know when the buses will be 240 km apart, which is twice the distance they cover in the first hour, since they are moving away from each other at a relative speed of 120 km/h. To cover a distance of 240 km at a relative speed of 120 km/h, it will take them (240 km) / (120 km/h) = 2 hours. Since they started at 9:00 am, they will be 240 km apart at 9:00 am + 2 hours = 11:00 am. Therefore, the correct option is (C) 11:00 am.
Tambaya 17 Rahoto
The pie chart shows the population of men, women, and children in a city. If the population of the city is 1,800,000, how many men are in the city?
Bayanin Amsa
Tambaya 18 Rahoto
Given that \(\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}}\)
= x + y\(\sqrt{15}\), find the value of (x + y)
Bayanin Amsa
Tambaya 19 Rahoto
Evaluate and correct to two decimal places, 75.0785 - 34.624 + 9.83
Tambaya 20 Rahoto
In the diagram, \(\overline{PU}\)/\(\overline{SR}\), \(\overline{PS}\)/\(\overline{TR}\), \(\overline{QS}\)/\(\overline{UR}\), \(\overline{UR}\) = 15cm, \(\overline{SR}\) = 8 cm, \(\overline{PS}\) = 10 cm and area of ?SUR = 24 cm\(^2\). Calculate the area of PTRS.
Bayanin Amsa
Tambaya 21 Rahoto
An amount of N550,000.00 was realized when a principal, x was saved at 2% simple interest for 5 years. Find the value of x
Bayanin Amsa
The formula to calculate simple interest is: Simple Interest = Principal x Rate x Time where Rate is the interest rate per year and Time is the number of years. In this case, we are given the amount and the interest rate, and we need to find the principal. We know that: Amount = Principal + Simple Interest Substituting the values given: N550,000 = x + (x x 0.02 x 5) Simplifying: N550,000 = x + 0.1x N550,000 = 1.1x Dividing both sides by 1.1: x = N500,000 Therefore, the value of the principal is N500,000. Hence, the correct answer is N500,000.
Tambaya 23 Rahoto
Fati buys milk at ₦x per tin sells each at a profit of ₦y. If she sells 10 tins of milk, how much does she receives from the sales?
Bayanin Amsa
To find out how much money Fati receives from the sales of 10 tins of milk, we need to know two things: the cost of each tin of milk and the profit she makes from selling each tin. Let's call the cost of each tin of milk "x" and the profit she makes from selling each tin "y". So, if she buys each tin of milk for ₦x and sells it at a profit of ₦y, then the total amount of money she receives for each tin of milk is ₦x + ₦y. Now, if she sells 10 tins of milk, then the total amount of money she receives from the sales of these 10 tins is 10 * (₦x + ₦y) = ₦10(x + y). So, the answer is ₦10(x + y).
Tambaya 24 Rahoto
Find the quadratic equation whose roots are \(\frac{1}{2}\) and -\(\frac{1}{3}\)
Tambaya 26 Rahoto
A woman received a discount of 20% on a piece of cloth she purchased from a shop. If she paid $525.00, what was the original price?
Bayanin Amsa
Tambaya 27 Rahoto
Simplify; \(\frac{a}{b} - \frac{a}{a} - \frac{c}{b}\)
Tambaya 28 Rahoto
If x = 3 and y = -1, evaluate 2(x\(^2\) - y\(^2\))
Bayanin Amsa
The expression 2(x\(^2\) - y\(^2\)) means 2 times the difference of x squared and y squared. First, we need to calculate x\(^2\) and y\(^2\) x\(^2\) = 3\(^2\) = 9 y\(^2\) = (-1)\(^2\) = 1 Next, we subtract y\(^2\) from x\(^2\) to get the difference x\(^2\) - y\(^2\) = 9 - 1 = 8 Finally, we multiply the difference by 2 2(x\(^2\) - y\(^2\)) = 2 * 8 = 16 So, the answer is 16.
Tambaya 29 Rahoto
If tan y is positive and sin y is negative, in which quadrant would y lie?
Bayanin Amsa
The value of y can only lie in the third quadrant. In the third quadrant, the sine values are negative and the tangent values are positive. This means that if sine is negative and tangent is positive, the angle must be in the third quadrant. It's important to remember that the sine and tangent functions have specific signs in each quadrant, which can be used to determine the location of an angle in the coordinate plane.
Tambaya 30 Rahoto
The graph of the equations y = 2x + 5 and y = 2x\(^2\) + x - 1 are shown. Use the information above to answer this question.
If x = -2.5, what is the value of u on the curve?
Bayanin Amsa
Tambaya 31 Rahoto
The expression \(\frac{5x + 3}{6x (x + 1)}\) will be undefined when x equals
Bayanin Amsa
The expression \(\frac{5x + 3}{6x (x + 1)}\) will be undefined when the denominator of the fraction equals zero. In this case, the denominator is \(6x(x+1)\), which will equal zero if either \(x=0\) or \(x=-1\). Therefore, the expression will be undefined when \(x\) is equal to either 0 or -1. To understand why this is the case, we can think of the denominator as representing the area of a rectangle with a width of \(6x\) and a height of \(x+1\). If either the width or the height of the rectangle is equal to zero, then the rectangle has zero area, and the fraction becomes undefined.
Tambaya 33 Rahoto
In the diagram, \(\overline{MN}\)//\(\overline{PQ}\), < MNP = 2x, and < NPQ = (3x - 50)°. Find the value of < NPQ
Tambaya 34 Rahoto
In the diagram, XYZ is an equilateral triangle of side 6cm and Y is the midpoint of \(\overline{XY}\). Find tan (< XZT)
Bayanin Amsa
Tambaya 35 Rahoto
In the diagram, PQ // SR. Find the value of x
Bayanin Amsa
Given the diagram, if PQ is parallel to SR, that means their corresponding angles are equal, and in this case, the angle marked "x" is equal in both triangles PQR and SRT. Since the two triangles are similar, we can use the fact that corresponding sides are proportional to find the value of x. Let's say the length of PQ is a and the length of SR is b. Since the two triangles are similar, we have the following proportion: a/PQ = b/SR Since we know the value of a and b, we can use this proportion to find the value of x. For example, if a = 8 and b = 10, we have: 8/PQ = 10/SR PQ = 10 * (8/10) PQ = 8 So, the length of PQ is 8 and the length of SR is 10, and since the angles marked x are equal in both triangles, we can conclude that x is equal to 46 degrees. Therefore, the value of x is 46.
Tambaya 36 Rahoto
If (x + 2) is a factor of x\(^2\) +px - 10, find the value of P.
Bayanin Amsa
To determine the value of P, we need to use the factor theorem, which states that if (x + a) is a factor of a polynomial, then the polynomial evaluated at -a will equal zero. In this case, we know that (x + 2) is a factor of x\(^2\) + px - 10, so we can set x = -2 and solve for P as follows: (-2)\(^2\) + P(-2) - 10 = 0 4 - 2P - 10 = 0 -2P - 6 = 0 -2P = 6 P = -3 Therefore, the value of P is -3.
Tambaya 37 Rahoto
The interquartile range of distribution is 7. If the 25th percentile is 16, find the upper quartile.
Bayanin Amsa
The interquartile range (IQR) is the difference between the upper quartile and the lower quartile of a distribution. We are given that the IQR is 7, which means the upper quartile (Q3) is 7 units away from the lower quartile (Q1). We are also given that the 25th percentile (Q1) is 16. The 25th percentile means that 25% of the data is below 16, which is the same as saying that the first quartile (Q1) is 16. To find the upper quartile (Q3), we need to add the IQR to Q1. So, Q3 = Q1 + IQR = 16 + 7 = 23. Therefore, the correct option is (C) 23.
Tambaya 38 Rahoto
Solve 3x - 2y = 10 and x + 3y = 7 simultaneously
Bayanin Amsa
To solve the two equations 3x - 2y = 10 and x + 3y = 7 simultaneously, we need to find the values of x and y that satisfy both equations. To do this, we'll use a method called substitution. First, we'll solve one of the equations for one of the variables. Let's solve the second equation, x + 3y = 7, for x: x = 7 - 3y Next, we'll substitute this expression for x into the first equation: 3x - 2y = 10 3(7 - 3y) - 2y = 10 21 - 9y - 2y = 10 21 - 11y = 10 11y = 11 y = 1 Now that we have y = 1, we can substitute this value back into the expression we found for x: x = 7 - 3y x = 7 - 3(1) x = 4 So the solution to the two equations is x = 4 and y = 1.
Tambaya 39 Rahoto
Find the equation of the line parallel to 2y = 3(x - 2) and passes through the point (2, 3)
Bayanin Amsa
To find the equation of a line that is parallel to another line and passes through a specific point, we can use the slope-point form of a line. The slope-point form of a line is given by: y - y1 = m(x - x1) where m is the slope of the line and (x1, y1) is a point on the line. The first step is to find the slope of the line 2y = 3(x - 2), which can be found by rearranging the equation into slope-intercept form (y = mx + b). To do this, we first isolate y on one side of the equation: 2y = 3x - 6 y = (3/2)x - 3 The slope of the line is (3/2). Next, we use the point (2, 3) and the slope of the line to write the equation in slope-point form: y - 3 = (3/2)(x - 2) This is the equation of the line that is parallel to 2y = 3(x - 2) and passes through the point (2, 3). So, the correct answer is y = (3/2)x - 3.
Tambaya 44 Rahoto
The diameter of a sphere is 12 cm. Calculate, correct of the nearest cm\(^3\), the volume of the sphere, [Take \(\pi = \frac{22}{7}\)]
Tambaya 45 Rahoto
In the diagram, O is the center of the circle. \(\overline{PQ}\) and \(\overline{RS}\) are tangents to the circle. Find the value of (M + N).
Bayanin Amsa
Tambaya 46 Rahoto
The first term of a geometric progression (G.P) is 3 and the 5th term is 48. Find the common ratio.
Tambaya 47 Rahoto
Find the sum of the interior angle of a pentagon.
Bayanin Amsa
The sum of the interior angles of a polygon can be found using the formula: (n-2) x 180 degrees, where "n" represents the number of sides of the polygon. In this case, the polygon is a pentagon, which has five sides. Substituting "n" with 5, we get: (5-2) x 180 degrees = 3 x 180 degrees = 540 degrees. Therefore, the sum of the interior angles of a pentagon is 540 degrees. Hence, the correct option is 540\(^o\).
Tambaya 48 Rahoto
A fence 2.4 m tall, is 10m away from a tree of height 16m. Calculate the angle of elevation of the top of the tree from the top of the fence.
Bayanin Amsa
To solve this problem, we can use trigonometry and the concept of similar triangles. Let's draw a diagram to visualize the situation. We have a fence that is 2.4 meters tall and a tree that is 16 meters tall. The tree is 10 meters away from the fence. We want to find the angle of elevation of the top of the tree from the top of the fence, which is the angle between the horizontal line passing through the top of the fence and the line connecting the top of the fence and the top of the tree. To start, we can create a right triangle using the height of the fence and the distance between the fence and the tree as the base and the hypotenuse, respectively. Let's call this triangle ABC, where A is the top of the fence, B is the bottom of the fence, and C is the point on the ground directly below the top of the tree. The angle between AB and AC is 90 degrees. Next, we can create another right triangle using the height of the tree and the same distance as the base and the hypotenuse, respectively. Let's call this triangle ACD, where D is the top of the tree. The angle between AD and AC is also 90 degrees. Since triangles ABC and ACD share the angle at A, they are similar triangles. This means that their corresponding angles are equal, and their corresponding sides are proportional. In particular, we can use the ratio of their heights to find the tangent of the angle we're looking for. Let's call the angle of elevation we're looking for x. Then, we have: tan(x) = CD/AB = (16-2.4)/10 = 1.36 To solve for x, we take the inverse tangent of both sides: x = tan^-1(1.36) ≈ 53.67° Therefore, the angle of elevation of the top of the tree from the top of the fence is approximately 53.67 degrees.
Tambaya 49 Rahoto
Simplify; [(\(\frac{16}{9}\))\(^{\frac{-3}{2}}\) x 16\(^{\frac{-3}{2}}\)]\(^{\frac{1}{3}}\)
Bayanin Amsa
Tambaya 50 Rahoto
The total surface area of a cone of slant height 1cm and base radius rcm is 224\(\pi\) cm\(^2\). If r : 1 = 2.5, find:
(a) correct to one decimal place, the value of r
(b) correct to the nearest whole number, the volume of the cone [Take \(\pi\) = \(\frac{22}{7}\)]
Setting up. The total surface area of a cone is \(\pi r(r + l) = 224\pi\), so
\[r(r + l) = 224.\]
The given ratio of slant height to base radius is \(l : r = 2.5\), i.e. \(l = 2.5r\) (the slant height must exceed the radius). Substituting,
\[r(r + 2.5r) = 224 \Rightarrow 3.5r^2 = 224 \Rightarrow r^2 = 64.\]
(a) \(r = 8.0\text{ cm}\), and \(l = 2.5\times8 = 20\text{ cm}\).
(b) Height \(h = \sqrt{l^2 - r^2} = \sqrt{20^2 - 8^2} = \sqrt{336} = 18.33\text{ cm}.\)
\[V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\times\tfrac{22}{7}\times64\times18.33 = 1229\text{ cm}^3\text{ (to the nearest whole number)}.\]
Bayanin Amsa
Setting up. The total surface area of a cone is \(\pi r(r + l) = 224\pi\), so
\[r(r + l) = 224.\]
The given ratio of slant height to base radius is \(l : r = 2.5\), i.e. \(l = 2.5r\) (the slant height must exceed the radius). Substituting,
\[r(r + 2.5r) = 224 \Rightarrow 3.5r^2 = 224 \Rightarrow r^2 = 64.\]
(a) \(r = 8.0\text{ cm}\), and \(l = 2.5\times8 = 20\text{ cm}\).
(b) Height \(h = \sqrt{l^2 - r^2} = \sqrt{20^2 - 8^2} = \sqrt{336} = 18.33\text{ cm}.\)
\[V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\times\tfrac{22}{7}\times64\times18.33 = 1229\text{ cm}^3\text{ (to the nearest whole number)}.\]
Tambaya 51 Rahoto
(a) Given that P = (\(\frac{rk}{Q} - ms\))\(^{\frac{2}{3}}\)
(i) Make Q the subject of the relation;
(ii) find, correct to two decimal places, the value of Q when P = 3, m = 15, s = 0.2, k = 4 and r = 10.
(b) Given that \(\frac{x + 2y}{5}\) = x - 2y, find x : y
(a)(i) Make Q the subject of \(P = \left(\dfrac{rk}{Q} - ms\right)^{\frac{2}{3}}\).
Raise both sides to the power \(\tfrac{3}{2}\):
\[P^{\frac{3}{2}} = \frac{rk}{Q} - ms.\]
\[\frac{rk}{Q} = P^{\frac{3}{2}} + ms \Rightarrow Q = \frac{rk}{P^{\frac{3}{2}} + ms}.\]
(a)(ii) With \(P = 3, m = 15, s = 0.2, k = 4, r = 10\):
- \(ms = 15 \times 0.2 = 3\)
- \(rk = 10 \times 4 = 40\)
- \(P^{\frac{3}{2}} = 3^{1.5} = \sqrt{27} = 5.196\)
\[Q = \frac{40}{5.196 + 3} = \frac{40}{8.196} = 4.88\ (\text{2 d.p.}).\]
(b) Given \(\dfrac{x + 2y}{5} = x - 2y\). Cross-multiplying:
\[x + 2y = 5(x - 2y) = 5x - 10y.\]
\[12y = 4x \Rightarrow \frac{x}{y} = \frac{12}{4} = 3.\]
Therefore \(x : y = 3 : 1\).
Bayanin Amsa
(a)(i) Make Q the subject of \(P = \left(\dfrac{rk}{Q} - ms\right)^{\frac{2}{3}}\).
Raise both sides to the power \(\tfrac{3}{2}\):
\[P^{\frac{3}{2}} = \frac{rk}{Q} - ms.\]
\[\frac{rk}{Q} = P^{\frac{3}{2}} + ms \Rightarrow Q = \frac{rk}{P^{\frac{3}{2}} + ms}.\]
(a)(ii) With \(P = 3, m = 15, s = 0.2, k = 4, r = 10\):
- \(ms = 15 \times 0.2 = 3\)
- \(rk = 10 \times 4 = 40\)
- \(P^{\frac{3}{2}} = 3^{1.5} = \sqrt{27} = 5.196\)
\[Q = \frac{40}{5.196 + 3} = \frac{40}{8.196} = 4.88\ (\text{2 d.p.}).\]
(b) Given \(\dfrac{x + 2y}{5} = x - 2y\). Cross-multiplying:
\[x + 2y = 5(x - 2y) = 5x - 10y.\]
\[12y = 4x \Rightarrow \frac{x}{y} = \frac{12}{4} = 3.\]
Therefore \(x : y = 3 : 1\).
Tambaya 52 Rahoto
The table shows the distribution of marks obtained by students in an examination.
| Marks (%) | 0 - 9 | 10 - 19 | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 | 80 - 89 | 90 - 99 |
| Frequency | 7 | 11 | 17 | 20 | 29 | 34 | 30 | 25 | 21 | 6 |
(a) Construct a cumulative frequency table for the distribution.
(b) Draw the cumulative frequency curve for the distribution.
(c) Using the curve, find correct to one decimal place, the:
(i) median mark;
(ii) lowest mark for the distinction if 5% of the students passed with distinction
(a) Cumulative Frequency Table
| Marks (%) | Upper Class Boundary | Frequency | Cumulative Frequency |
|---|---|---|---|
| 0 – 9 | 9.5 | 7 | 7 |
| 10 – 19 | 19.5 | 11 | 18 |
| 20 – 29 | 29.5 | 17 | 35 |
| 30 – 39 | 39.5 | 20 | 55 |
| 40 – 49 | 49.5 | 29 | 84 |
| 50 – 59 | 59.5 | 34 | 118 |
| 60 – 69 | 69.5 | 30 | 148 |
| 70 – 79 | 79.5 | 25 | 173 |
| 80 – 89 | 89.5 | 21 | 194 |
| 90 – 99 | 99.5 | 6 | 200 |
(b) Cumulative Frequency Curve
Using the upper class boundaries, plot the points:
(9.5, 7), (19.5, 18), (29.5, 35), (39.5, 55), (49.5, 84),
(59.5, 118), (69.5, 148), (79.5, 173), (89.5, 194) and (99.5, 200).
Join the points with a smooth cumulative frequency curve (ogive).
(c)
(i) Median mark
Total frequency, N = 200.
Median position = N⁄2 = 200⁄2 = 100th value.
The 100th value lies in the class 50
Bayanin Amsa
(a) Cumulative Frequency Table
| Marks (%) | Upper Class Boundary | Frequency | Cumulative Frequency |
|---|---|---|---|
| 0 – 9 | 9.5 | 7 | 7 |
| 10 – 19 | 19.5 | 11 | 18 |
| 20 – 29 | 29.5 | 17 | 35 |
| 30 – 39 | 39.5 | 20 | 55 |
| 40 – 49 | 49.5 | 29 | 84 |
| 50 – 59 | 59.5 | 34 | 118 |
| 60 – 69 | 69.5 | 30 | 148 |
| 70 – 79 | 79.5 | 25 | 173 |
| 80 – 89 | 89.5 | 21 | 194 |
| 90 – 99 | 99.5 | 6 | 200 |
(b) Cumulative Frequency Curve
Using the upper class boundaries, plot the points:
(9.5, 7), (19.5, 18), (29.5, 35), (39.5, 55), (49.5, 84),
(59.5, 118), (69.5, 148), (79.5, 173), (89.5, 194) and (99.5, 200).
Join the points with a smooth cumulative frequency curve (ogive).
(c)
(i) Median mark
Total frequency, N = 200.
Median position = N⁄2 = 200⁄2 = 100th value.
The 100th value lies in the class 50
Tambaya 53 Rahoto
(a) If A = {multiples of 2}, B = {multiples of 3} and C = {factors of 6} are subsets of \(\mu\) = {x: \(1 \leq x \leq 10\)} find A′ \(\cap\) B′ \(\cap\) C′
(b) Tickets for a movie premiere cost $18.50 each while the bulk purchase price for 5 tickets is $80.00. If 4 gentlemen decide to get a fifth person to join them so that they can share the bulk purchase price equally, how much would each person save?
(a) The universal set is \(\mu = \{1,2,3,4,5,6,7,8,9,10\}\).
- \(A = \{\text{multiples of }2\} = \{2,4,6,8,10\}\)
- \(B = \{\text{multiples of }3\} = \{3,6,9\}\)
- \(C = \{\text{factors of }6\} = \{1,2,3,6\}\)
By De Morgan's law, \(A' \cap B' \cap C' = (A \cup B \cup C)'\).
\[A \cup B \cup C = \{1,2,3,4,6,8,9,10\}.\]
Therefore the elements of \(\mu\) not in this union are:
\[A' \cap B' \cap C' = \{5, 7\}.\]
(b) Buying singly, one ticket costs \(\$18.50\). The bulk price for 5 tickets is \(\$80.00\), shared equally among the 5 people:
\[\text{Cost per person} = \frac{80.00}{5} = \$16.00.\]
Each person's saving compared with buying a single ticket:
\[18.50 - 16.00 = \$2.50.\]
Each person would save \(\mathbf{\$2.50}\).
Bayanin Amsa
(a) The universal set is \(\mu = \{1,2,3,4,5,6,7,8,9,10\}\).
- \(A = \{\text{multiples of }2\} = \{2,4,6,8,10\}\)
- \(B = \{\text{multiples of }3\} = \{3,6,9\}\)
- \(C = \{\text{factors of }6\} = \{1,2,3,6\}\)
By De Morgan's law, \(A' \cap B' \cap C' = (A \cup B \cup C)'\).
\[A \cup B \cup C = \{1,2,3,4,6,8,9,10\}.\]
Therefore the elements of \(\mu\) not in this union are:
\[A' \cap B' \cap C' = \{5, 7\}.\]
(b) Buying singly, one ticket costs \(\$18.50\). The bulk price for 5 tickets is \(\$80.00\), shared equally among the 5 people:
\[\text{Cost per person} = \frac{80.00}{5} = \$16.00.\]
Each person's saving compared with buying a single ticket:
\[18.50 - 16.00 = \$2.50.\]
Each person would save \(\mathbf{\$2.50}\).
Tambaya 54 Rahoto
(a) In the diagram, AB is a tangent to the circle with centre O, and COB is a straight line. If CD//AB and < ABE = 40°, find: < ODE.
(b) ABCD is a parallelogram in which |\(\overline{CD}\)| = 7 cm, I\(\overline{AD}\)I = 5 cm and < ADC= 125°.
(i) Illustrate the information in a diagram.
(ii) Find, correct to one decimal place, the area of the parallelogram.
(c) If x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\)). Evaluate (2x\(^2\) - 2x).
(a)
<ABE = <OBC (tangent and radius form a right angle)
<ABE = 40° (given)
Therefore, <OBC = 40°.
Since CD//AB, then <CDE = <ABE = 40° (alternate angles)
<ODE = <OBC + <CDE = 40° + 40° = 80°.
(b)
(i)
A ----------- B
\ /
\ /
\ /
\ /
\ /
\ /
D-------C
(125°)
(ii)
The area of the parallelogram is given by A = base x height. Since AD || BC, then the height of the parallelogram is given by the perpendicular distance between AD and BC, which is 7 cm. To find the length of the base, we use the cosine rule:
|\(\overline{AD}\)|² = |\(\overline{AB}\)|² + |\(\overline{BD}\)|² - 2|\(\overline{AB}\)||\(\overline{BD}\)| cos(ADC)
5² = x² + (x+7)² - 2x(x+7)cos(125°)
25 = 2x² + 14x + 49 + 2x(x+7)(-0.5736)
25 = 2x² + 14x + 49 - 1.1472x² - 10.0317x - 20.7032
0 = 0.8528x² + 4.0317x + 4.7032
Using the quadratic formula: x = (-b ± ?(b² - 4ac))/2a, we have:
x = (-4.0317 ± ?(4.0317² - 4(0.8528)(4.7032)))/(2(0.8528))
x = (-4.0317 ± 4.9988)/1.7056
x = 0.6559 or x = -3.6582
Since x represents a length, we discard the negative solution. Therefore, x = 0.6559 cm.
So, the area of the parallelogram is A = base x height = 0.6559 cm x 7 cm = 4.5913 cm² (to one decimal place).
(c)
x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\))
2x² - 2x = 2\(\left(\frac{1}{2}\)(1 - \(\sqrt{2}\))²\) - 2(1 - \(\sqrt{2}\))
= (1 - 2\(\sqrt{2}\) + 2) - 2 + 2\(\sqrt{2}\)
= 1 + 2\(\sqrt{2}\).
Therefore, 2x² - 2x = 1 + 2\(\sqrt{2}\).
Bayanin Amsa
(a)
<ABE = <OBC (tangent and radius form a right angle)
<ABE = 40° (given)
Therefore, <OBC = 40°.
Since CD//AB, then <CDE = <ABE = 40° (alternate angles)
<ODE = <OBC + <CDE = 40° + 40° = 80°.
(b)
(i)
A ----------- B
\ /
\ /
\ /
\ /
\ /
\ /
D-------C
(125°)
(ii)
The area of the parallelogram is given by A = base x height. Since AD || BC, then the height of the parallelogram is given by the perpendicular distance between AD and BC, which is 7 cm. To find the length of the base, we use the cosine rule:
|\(\overline{AD}\)|² = |\(\overline{AB}\)|² + |\(\overline{BD}\)|² - 2|\(\overline{AB}\)||\(\overline{BD}\)| cos(ADC)
5² = x² + (x+7)² - 2x(x+7)cos(125°)
25 = 2x² + 14x + 49 + 2x(x+7)(-0.5736)
25 = 2x² + 14x + 49 - 1.1472x² - 10.0317x - 20.7032
0 = 0.8528x² + 4.0317x + 4.7032
Using the quadratic formula: x = (-b ± ?(b² - 4ac))/2a, we have:
x = (-4.0317 ± ?(4.0317² - 4(0.8528)(4.7032)))/(2(0.8528))
x = (-4.0317 ± 4.9988)/1.7056
x = 0.6559 or x = -3.6582
Since x represents a length, we discard the negative solution. Therefore, x = 0.6559 cm.
So, the area of the parallelogram is A = base x height = 0.6559 cm x 7 cm = 4.5913 cm² (to one decimal place).
(c)
x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\))
2x² - 2x = 2\(\left(\frac{1}{2}\)(1 - \(\sqrt{2}\))²\) - 2(1 - \(\sqrt{2}\))
= (1 - 2\(\sqrt{2}\) + 2) - 2 + 2\(\sqrt{2}\)
= 1 + 2\(\sqrt{2}\).
Therefore, 2x² - 2x = 1 + 2\(\sqrt{2}\).
Tambaya 55 Rahoto
A die was rolled a number of times. The outcomes are as shown in the table
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
| Outcomes | 32 | m | 25 | 40 | 28 | 45 |
If the probability of obtaining 2 is 0.15, find the:
(a) value of m;
(b) number of times the die was rolled;
(c) probability of obtaining an even number.
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Outcomes | 32 | m | 25 | 40 | 28 | 45 |
(a) Value of m. The total number of rolls is \(N=32+m+25+40+28+45=170+m\). Since \(P(2)=0.15\):
\[ \frac{m}{170+m}=0.15 \;\Rightarrow\; m=0.15(170+m)=25.5+0.15m \] \[ 0.85m=25.5 \;\Rightarrow\; m=30 \](b) Number of times the die was rolled.
\[ N=170+30=200 \](c) Probability of an even number. Even outcomes are 2, 4 and 6:
\[ 30+40+45=115 \] \[ P(\text{even})=\frac{115}{200}=\frac{23}{40}=0.575 \]Bayanin Amsa
| Number | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Outcomes | 32 | m | 25 | 40 | 28 | 45 |
(a) Value of m. The total number of rolls is \(N=32+m+25+40+28+45=170+m\). Since \(P(2)=0.15\):
\[ \frac{m}{170+m}=0.15 \;\Rightarrow\; m=0.15(170+m)=25.5+0.15m \] \[ 0.85m=25.5 \;\Rightarrow\; m=30 \](b) Number of times the die was rolled.
\[ N=170+30=200 \](c) Probability of an even number. Even outcomes are 2, 4 and 6:
\[ 30+40+45=115 \] \[ P(\text{even})=\frac{115}{200}=\frac{23}{40}=0.575 \]Tambaya 56 Rahoto
(a) The diagram shows a wooden structure in the form of a cone, mounted on a hemispherical base. The vertical height of the cone is 48 m and the base radius is 14. Calculate, correct to three significant figures, the surface area of the structure, [Take \(\pi = \frac{22}{7}\)]
(b) Five years ago, Musah was twice as old as Sesay. If the sum of their ages is 100, find Sesay's present age.
(a) Surface Area of the Wooden Structure
The structure consists of a cone mounted on a hemispherical base. We need to find the total surface area of this structure.
Given:
- Radius of both the cone and the hemisphere, r=14 m
- Height of the cone, h=48 m
- π=722
Step 1: Calculate the Slant Height of the Cone
The slant height
l of the cone can be found using the Pythagorean theorem:
l=h2+r2=482+142
l=2304+196=2500=50 m
Step 2: Calculate the Surface Area of the Cone
The surface area of the cone (excluding the base) is:
Acone=πrl=722×14×50 Acone=22×50=1100 m2
Step 3: Calculate the Surface Area of the Hemisphere
The surface area of a hemisphere (excluding the base) is:
Ahemisphere=2πr2=2×722×142
Ahemisphere=2×722×196
Ahemisphere=2×22×28=1232 m2
Step 4: Calculate the Total Surface Area
The total surface area of the structure is the sum of the surface area of the cone and the surface area of the hemisphere:
Atotal=Acone+Ahemisphere=1100+1232=2332 m2
Thus, the surface area of the structure, correct to three significant figures, is: 2330 m2
(b) Finding Sesay's Present Age
Given:
- Five years ago, Musah was twice as old as Sesay.
- The sum of their current ages is 100.
Let Musah's present age be M and Sesay's present age be S.
Five years ago:
- Musah's age was M−5
- Sesay's age was S−5
According to the problem, five years ago, Musah was twice as old as Sesay:
M−5=2(S−5)
M−5=2S−10
M=2S−5
We are also given that the sum of their current ages is 100: M+S=100
Substitute:
M=2S−5 into
M+S=100:
(2S−5)+S=100
3S−5=100
3S=105
S=35
So, Sesay's present age is:
S=35
To verify, calculate Musah's present age:
M=2S−5=2(35)−5=70−5=65
Check the sum: M+S=65+35=100
Thus, Sesay's present age is 35 years.
Bayanin Amsa
(a) Surface Area of the Wooden Structure
The structure consists of a cone mounted on a hemispherical base. We need to find the total surface area of this structure.
Given:
- Radius of both the cone and the hemisphere, r=14 m
- Height of the cone, h=48 m
- π=722
Step 1: Calculate the Slant Height of the Cone
The slant height
l of the cone can be found using the Pythagorean theorem:
l=h2+r2=482+142
l=2304+196=2500=50 m
Step 2: Calculate the Surface Area of the Cone
The surface area of the cone (excluding the base) is:
Acone=πrl=722×14×50 Acone=22×50=1100 m2
Step 3: Calculate the Surface Area of the Hemisphere
The surface area of a hemisphere (excluding the base) is:
Ahemisphere=2πr2=2×722×142
Ahemisphere=2×722×196
Ahemisphere=2×22×28=1232 m2
Step 4: Calculate the Total Surface Area
The total surface area of the structure is the sum of the surface area of the cone and the surface area of the hemisphere:
Atotal=Acone+Ahemisphere=1100+1232=2332 m2
Thus, the surface area of the structure, correct to three significant figures, is: 2330 m2
(b) Finding Sesay's Present Age
Given:
- Five years ago, Musah was twice as old as Sesay.
- The sum of their current ages is 100.
Let Musah's present age be M and Sesay's present age be S.
Five years ago:
- Musah's age was M−5
- Sesay's age was S−5
According to the problem, five years ago, Musah was twice as old as Sesay:
M−5=2(S−5)
M−5=2S−10
M=2S−5
We are also given that the sum of their current ages is 100: M+S=100
Substitute:
M=2S−5 into
M+S=100:
(2S−5)+S=100
3S−5=100
3S=105
S=35
So, Sesay's present age is:
S=35
To verify, calculate Musah's present age:
M=2S−5=2(35)−5=70−5=65
Check the sum: M+S=65+35=100
Thus, Sesay's present age is 35 years.
Tambaya 57 Rahoto
(a) Copy and complete the table of values for the relation \(y = 3 \sin 2x\).
| x | \(o^o\) | \(15^o\) | \(30^o\) | \(45^o\) | \(60^o\) | \(75^o\) | \(90^o\) | \(105^o\) | \(120^o\) | \(135^o\) | \(^o\) |
| y | 0.0 | 1.5 | -2.6 |
(b) Using a scale of 2 cm to 15° on the x-axis and 2cm to I unit on the y-axis, draw the graph of \(y = 3 \sin 2x\) for \(0° \geq x \geq 150°\).
(c) Use the graph to find the truth set of;
(i) \(3 \sin 2x + 2 = 0\);
(ii ) \(\frac{3}{2} \sin 2x = 0.25\).
(a) For each value of \(x\), calculate \(y=3\sin 2x\), correct to 1 decimal place where necessary.
| \(x\) (degrees) | 0 | 15 | 30 | 45 | 60 | 75 | 90 | 105 | 120 | 135 | 150 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y=3\sin2x\) | 0.0 | 1.5 | 2.6 | 3.0 | 2.6 | 1.5 | 0.0 | -1.5 | -2.6 | -3.0 | -2.6 |
For example, \(3\sin 2(30^\circ)=3\sin60^\circ=2.598\ldots\approx2.6\).
(b) The plotted graph of \(y=3\sin2x\) is shown below. The curve is drawn through the tabulated points, using the stated scales.
(c)(i) \(3\sin2x+2=0\) gives \(y=-2\). Reading the intersection of the curve with \(y=-2\) gives
\[x\approx111^\circ.\]
Hence the truth set is \(\{111^\circ\}\), approximately.
(c)(ii) \(\frac32\sin2x=0.25\). Since \(y=3\sin2x\), this is equivalent to \(y=0.5\). Reading the intersections of the curve with \(y=0.5\) gives
\[x\approx5^\circ\quad\text{or}\quad x\approx85^\circ.\]
Therefore, the truth set is \(\{5^\circ,\ 85^\circ\}\), approximately.
Bayanin Amsa
(a) For each value of \(x\), calculate \(y=3\sin 2x\), correct to 1 decimal place where necessary.
| \(x\) (degrees) | 0 | 15 | 30 | 45 | 60 | 75 | 90 | 105 | 120 | 135 | 150 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| \(y=3\sin2x\) | 0.0 | 1.5 | 2.6 | 3.0 | 2.6 | 1.5 | 0.0 | -1.5 | -2.6 | -3.0 | -2.6 |
For example, \(3\sin 2(30^\circ)=3\sin60^\circ=2.598\ldots\approx2.6\).
(b) The plotted graph of \(y=3\sin2x\) is shown below. The curve is drawn through the tabulated points, using the stated scales.
(c)(i) \(3\sin2x+2=0\) gives \(y=-2\). Reading the intersection of the curve with \(y=-2\) gives
\[x\approx111^\circ.\]
Hence the truth set is \(\{111^\circ\}\), approximately.
(c)(ii) \(\frac32\sin2x=0.25\). Since \(y=3\sin2x\), this is equivalent to \(y=0.5\). Reading the intersections of the curve with \(y=0.5\) gives
\[x\approx5^\circ\quad\text{or}\quad x\approx85^\circ.\]
Therefore, the truth set is \(\{5^\circ,\ 85^\circ\}\), approximately.
Tambaya 58 Rahoto
(a) In the diagram, MNPQ is a circle with centre O, |MN| = |NP| and < OMN = 50°. Find:
(I) < MNP
(ii) < POQ
(b) Find the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12)
(a) Circle \(MNPQ\) with centre \(O\), \(|MN|=|NP|\) and \(\angle OMN=50^{\circ}\).
From the diagram \(M\) and \(Q\) are the ends of a diameter through \(O\), and the points lie in the order \(M, N, P, Q\) round the circle.
(i) Find \(\angle MNP\).
\(OM\) and \(ON\) are radii, so triangle \(OMN\) is isosceles with \(OM=ON\). Hence the base angles are equal:
\[\angle ONM=\angle OMN=50^{\circ}\]Angle at the centre of triangle \(OMN\):
\[\angle MON=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]Equal chords subtend equal angles at the centre, and \(|MN|=|NP|\), so:
\[\angle NOP=\angle MON=80^{\circ}\]The central angle standing on the minor arc \(MNP\) is therefore
\[\angle MON+\angle NOP=80^{\circ}+80^{\circ}=160^{\circ}.\]The reflex central angle on the major arc \(MP\) (the arc not containing \(N\)) is
\[360^{\circ}-160^{\circ}=200^{\circ}.\]The inscribed angle \(\angle MNP\) stands on this major arc \(MP\), so it is half of it:
\[\angle MNP=\frac{1}{2}\times 200^{\circ}=100^{\circ}\](ii) Find \(\angle POQ\).
\(M\), \(O\), \(Q\) are collinear (diameter), so the central angles along the arc \(M\to N\to P\to Q\) add up to a straight angle:
\[\angle MON+\angle NOP+\angle POQ=180^{\circ}\]\[80^{\circ}+80^{\circ}+\angle POQ=180^{\circ}\]\[\angle POQ=20^{\circ}\](b) Line with the same gradient as \(8y+4x=24\) through \((-8,\,12)\).
Make \(y\) the subject to read off the gradient:
\[8y=24-4x\]\[y=3-\tfrac{1}{2}x\]The gradient is \(m=-\dfrac{1}{2}\). Using \(y-y_1=m(x-x_1)\) with \((x_1,y_1)=(-8,12)\):
\[y-12=-\tfrac{1}{2}(x+8)\]\[y-12=-\tfrac{1}{2}x-4\]\[y=-\tfrac{1}{2}x+8\]The required line is \(y=-\dfrac{1}{2}x+8\), i.e. \(x+2y=16\).
Bayanin Amsa
(a) Circle \(MNPQ\) with centre \(O\), \(|MN|=|NP|\) and \(\angle OMN=50^{\circ}\).
From the diagram \(M\) and \(Q\) are the ends of a diameter through \(O\), and the points lie in the order \(M, N, P, Q\) round the circle.
(i) Find \(\angle MNP\).
\(OM\) and \(ON\) are radii, so triangle \(OMN\) is isosceles with \(OM=ON\). Hence the base angles are equal:
\[\angle ONM=\angle OMN=50^{\circ}\]Angle at the centre of triangle \(OMN\):
\[\angle MON=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]Equal chords subtend equal angles at the centre, and \(|MN|=|NP|\), so:
\[\angle NOP=\angle MON=80^{\circ}\]The central angle standing on the minor arc \(MNP\) is therefore
\[\angle MON+\angle NOP=80^{\circ}+80^{\circ}=160^{\circ}.\]The reflex central angle on the major arc \(MP\) (the arc not containing \(N\)) is
\[360^{\circ}-160^{\circ}=200^{\circ}.\]The inscribed angle \(\angle MNP\) stands on this major arc \(MP\), so it is half of it:
\[\angle MNP=\frac{1}{2}\times 200^{\circ}=100^{\circ}\](ii) Find \(\angle POQ\).
\(M\), \(O\), \(Q\) are collinear (diameter), so the central angles along the arc \(M\to N\to P\to Q\) add up to a straight angle:
\[\angle MON+\angle NOP+\angle POQ=180^{\circ}\]\[80^{\circ}+80^{\circ}+\angle POQ=180^{\circ}\]\[\angle POQ=20^{\circ}\](b) Line with the same gradient as \(8y+4x=24\) through \((-8,\,12)\).
Make \(y\) the subject to read off the gradient:
\[8y=24-4x\]\[y=3-\tfrac{1}{2}x\]The gradient is \(m=-\dfrac{1}{2}\). Using \(y-y_1=m(x-x_1)\) with \((x_1,y_1)=(-8,12)\):
\[y-12=-\tfrac{1}{2}(x+8)\]\[y-12=-\tfrac{1}{2}x-4\]\[y=-\tfrac{1}{2}x+8\]The required line is \(y=-\dfrac{1}{2}x+8\), i.e. \(x+2y=16\).
Tambaya 59 Rahoto
(a} In the diagram, O is the centre of the circle ABCDE, = I\(\overline{BC}\)I = |\(\overline{CD}\)| and < BCD = 108°. Find < CDE.
(b) Given that tan x = \(\sqrt{3}\), 0\(^o\) \(\geq\) x \(\geq\) 90\(^o\), evaluate
\(\frac{(cos x)^2 - sin x}{(sin x)^2 + cos x}\)
(a) To find < CDE, we need to first find < BOC.
Since O is the center of the circle, < BOC is twice the angle < BAC. Therefore,
< BAC = 1/2 * < BOC
= 1/2 * (360° - < BCD - < ABC)
= 1/2 * (360° - 108° - 36°)
= 1/2 * 216°
= 108°
Similarly, we can find that < BDC = 1/2 * (360° - < BCD - < CDE)
= 1/2 * (360° - 108° - < CDE)
= 126° - 1/2 * < CDE
Since < BDC and < BCD are equal, we have:
126° - 1/2 * < CDE = 108°
1/2 * < CDE = 18°
< CDE = 36°
Therefore, < CDE is 36°.
(b) We are given that tan x = √3, and we need to find the value of:
(cos x)^2 - sin x
---------------------
(sin x)^2 + cos x
Using the identity (cos x)^2 + (sin x)^2 = 1, we can write:
(cos x)^2 = 1 - (sin x)^2
Substituting this into the expression, we get:
[(1 - (sin x)^2) - sin x] / [(sin x)^2 + cos x]
Simplifying, we get:
(1 - 2(sin x)^2 - sin x) / [(sin x)^2 + cos x]
Substituting tan x = √3, we have:
sin x / cos x = √3
sin x = √3 cos x
Substituting this into the expression, we get:
(1 - 2(3cos^2 x) - √3 cos x) / (3cos^2 x + cos x)
Multiplying both the numerator and denominator by -1, we get:
(2(3cos^2 x) + √3 cos x - 1) / (3cos^2 x + cos x)
Simplifying, we get:
(6cos^2 x + √3 cos x - 1) / (3cos^2 x + cos x)
Substituting √3 for sin x / cos x, we have:
(6/4 + √3/4 - 1) / (3/4 + 1/4)
= (9/4 + √3/4) / 1
= 9 + √3
Therefore, the value of the expression is
Bayanin Amsa
(a) To find < CDE, we need to first find < BOC.
Since O is the center of the circle, < BOC is twice the angle < BAC. Therefore,
< BAC = 1/2 * < BOC
= 1/2 * (360° - < BCD - < ABC)
= 1/2 * (360° - 108° - 36°)
= 1/2 * 216°
= 108°
Similarly, we can find that < BDC = 1/2 * (360° - < BCD - < CDE)
= 1/2 * (360° - 108° - < CDE)
= 126° - 1/2 * < CDE
Since < BDC and < BCD are equal, we have:
126° - 1/2 * < CDE = 108°
1/2 * < CDE = 18°
< CDE = 36°
Therefore, < CDE is 36°.
(b) We are given that tan x = √3, and we need to find the value of:
(cos x)^2 - sin x
---------------------
(sin x)^2 + cos x
Using the identity (cos x)^2 + (sin x)^2 = 1, we can write:
(cos x)^2 = 1 - (sin x)^2
Substituting this into the expression, we get:
[(1 - (sin x)^2) - sin x] / [(sin x)^2 + cos x]
Simplifying, we get:
(1 - 2(sin x)^2 - sin x) / [(sin x)^2 + cos x]
Substituting tan x = √3, we have:
sin x / cos x = √3
sin x = √3 cos x
Substituting this into the expression, we get:
(1 - 2(3cos^2 x) - √3 cos x) / (3cos^2 x + cos x)
Multiplying both the numerator and denominator by -1, we get:
(2(3cos^2 x) + √3 cos x - 1) / (3cos^2 x + cos x)
Simplifying, we get:
(6cos^2 x + √3 cos x - 1) / (3cos^2 x + cos x)
Substituting √3 for sin x / cos x, we have:
(6/4 + √3/4 - 1) / (3/4 + 1/4)
= (9/4 + √3/4) / 1
= 9 + √3
Therefore, the value of the expression is
Tambaya 60 Rahoto
(a) Two cyclists X and Y leave town Q at the same time. Cyclist X travels at the rate of 5 km/h on a bearing of 049° and cyclist Y travels at the rate of 9 km/h on a bearing of 319°.
(a) Illustrate the information on a diagram.
(b) After travelling for two hours, calculate. correct to the nearest whole number, the:
(i) distance between cyclist X and Y;
(ii) bearing of cyclist X from Y.
(c) Find the average speed at which cyclist X will get to Y in 4 hours.
(a) Diagram. From Q draw QX on bearing 049° and QY on bearing 319°. The angle between the two paths is \(\angle XQY = 360^\circ - (319^\circ - 49^\circ) = 360^\circ - 270^\circ = 90^\circ\), a right angle at Q.
(b) After two hours: \(QX = 5\times2 = 10\) km and \(QY = 9\times2 = 18\) km.
(i) Distance XY (right angle at Q, so use Pythagoras):
\[XY = \sqrt{10^2 + 18^2} = \sqrt{424} = 20.6 \approx 21\text{ km}.\]
(ii) Bearing of X from Y. Taking Q as origin, \(X = (10\sin49^\circ, 10\cos49^\circ) = (7.55, 6.56)\) and \(Y = (18\sin319^\circ, 18\cos319^\circ) = (-11.81, 13.59)\). Then \(\vec{YX} = (19.36, -7.02)\), which points into the south-east region:
\[\text{bearing} = 180^\circ - \tan^{-1}\!\frac{19.36}{7.02} = 180^\circ - 70^\circ = 110^\circ.\]
(c) Average speed to cover XY in 4 hours:
\[\text{speed} = \frac{20.6}{4} = 5.1 \approx 5\text{ km/h}.\]
Bayanin Amsa
(a) Diagram. From Q draw QX on bearing 049° and QY on bearing 319°. The angle between the two paths is \(\angle XQY = 360^\circ - (319^\circ - 49^\circ) = 360^\circ - 270^\circ = 90^\circ\), a right angle at Q.
(b) After two hours: \(QX = 5\times2 = 10\) km and \(QY = 9\times2 = 18\) km.
(i) Distance XY (right angle at Q, so use Pythagoras):
\[XY = \sqrt{10^2 + 18^2} = \sqrt{424} = 20.6 \approx 21\text{ km}.\]
(ii) Bearing of X from Y. Taking Q as origin, \(X = (10\sin49^\circ, 10\cos49^\circ) = (7.55, 6.56)\) and \(Y = (18\sin319^\circ, 18\cos319^\circ) = (-11.81, 13.59)\). Then \(\vec{YX} = (19.36, -7.02)\), which points into the south-east region:
\[\text{bearing} = 180^\circ - \tan^{-1}\!\frac{19.36}{7.02} = 180^\circ - 70^\circ = 110^\circ.\]
(c) Average speed to cover XY in 4 hours:
\[\text{speed} = \frac{20.6}{4} = 5.1 \approx 5\text{ km/h}.\]
Tambaya 61 Rahoto
(a) Ms. Maureen spent \(\frac{1}{4}\) of her monthly income at a shopping mall, \(\frac{1}{3}\) at an open market and \(\frac{2}{5}\) of the remaining amount at a Mechanic workshop. If she had N222,000.00 left, find:
(i) her monthly income.
(ii) the amount spent at the open market.
(b) The third term of an Arithmetic Progression (A. P.) is 4m - 2n. If the ninth term of the progression is 2m - 8n. find the common difference in terms of m and n.
a) (i) To find Ms. Maureen's monthly income:
To find Ms. Maureen's monthly income, we can set up an equation using the information given. Let's call her monthly income "M".
Ms. Maureen spent 1/4 of her monthly income at the shopping mall, so: M * 1/4 = M/4
She spent 1/3 of her monthly income at the open market, so: M * 1/3 = M/3
The remaining amount after spending at the shopping mall and open market is M - M/4 - M/3 = M * (3/4) - M/3.
She then spent 2/5 of this remaining amount at the mechanic workshop, so: (M * (3/4) - M/3) * 2/5 = 2/5 * M * (3/4) - 2/5 * M/3.
We know that she had N222,000 left, so we can set this equal to the amount spent at the mechanic workshop: N222,000 = 2/5 * M * (3/4) - 2/5 * M/3.
Solving for M, we can find her monthly income:
M = N222,000 * 5/2 * 4/3 / (3/4 - 2/5) = N222,000 * 5/2 * 4/3 / (5/4 - 6/5) = N222,000 * 20/3 / (25/20 - 24/20) = N222,000 * 20/3 / (5/20) = N222,000 * 20/3 * 20/5 = N222,000 * 4 = N888,000.
So Ms. Maureen's monthly income is N888,000.
(ii) To find the amount spent at the open market:
To find the amount spent at the open market, we use the equation M * 1/3 = M/3 and substitute M = N888,000:
N888,000 * 1/3 = N888,000/3
N888,000/3 = N296,000
So the amount spent at the open market is N296,000.
b) The third term of an arithmetic progression (A.P.) is 4m - 2n, and the ninth term is 2m - 8n. To find the common difference:
To find the common difference, we can subtract the third term from the ninth term and divide by 6, since there are 6 terms between the third and ninth terms:
(2m - 8n) - (4m - 2n) = -2m + 6n
The common difference is -2m + 6n / 6 = -m + n.
So the common difference in terms of m and n is -m + n.
Bayanin Amsa
a) (i) To find Ms. Maureen's monthly income:
To find Ms. Maureen's monthly income, we can set up an equation using the information given. Let's call her monthly income "M".
Ms. Maureen spent 1/4 of her monthly income at the shopping mall, so: M * 1/4 = M/4
She spent 1/3 of her monthly income at the open market, so: M * 1/3 = M/3
The remaining amount after spending at the shopping mall and open market is M - M/4 - M/3 = M * (3/4) - M/3.
She then spent 2/5 of this remaining amount at the mechanic workshop, so: (M * (3/4) - M/3) * 2/5 = 2/5 * M * (3/4) - 2/5 * M/3.
We know that she had N222,000 left, so we can set this equal to the amount spent at the mechanic workshop: N222,000 = 2/5 * M * (3/4) - 2/5 * M/3.
Solving for M, we can find her monthly income:
M = N222,000 * 5/2 * 4/3 / (3/4 - 2/5) = N222,000 * 5/2 * 4/3 / (5/4 - 6/5) = N222,000 * 20/3 / (25/20 - 24/20) = N222,000 * 20/3 / (5/20) = N222,000 * 20/3 * 20/5 = N222,000 * 4 = N888,000.
So Ms. Maureen's monthly income is N888,000.
(ii) To find the amount spent at the open market:
To find the amount spent at the open market, we use the equation M * 1/3 = M/3 and substitute M = N888,000:
N888,000 * 1/3 = N888,000/3
N888,000/3 = N296,000
So the amount spent at the open market is N296,000.
b) The third term of an arithmetic progression (A.P.) is 4m - 2n, and the ninth term is 2m - 8n. To find the common difference:
To find the common difference, we can subtract the third term from the ninth term and divide by 6, since there are 6 terms between the third and ninth terms:
(2m - 8n) - (4m - 2n) = -2m + 6n
The common difference is -2m + 6n / 6 = -m + n.
So the common difference in terms of m and n is -m + n.
