WAEC SSCE - General Mathematics - 2010
Ajụjụ 1 Ripọtì
The sum of the exterior of an n-sided convex polygon is half the sum of its interior angle. find n
Akọwa Nkọwa
An exterior angle of a polygon is the angle between any side of the polygon and its adjacent extended side. The sum of the exterior angles of any polygon, regardless of the number of sides, is always equal to 360 degrees. The sum of the interior angles of a polygon can be found using the formula (n-2) x 180, where n is the number of sides of the polygon. So if we have an n-sided polygon, the sum of its interior angles is (n-2) x 180. According to the problem, the sum of the exterior angles is half the sum of the interior angles. Mathematically, we can express this as: Sum of exterior angles = 1/2 x sum of interior angles Substituting the formulas we have for the sum of exterior and interior angles, we get: 360 = 1/2 x (n-2) x 180 Simplifying, we get: 2 x 360 = (n-2) x 180 720 = (n-2) x 180 Dividing both sides by 180, we get: 4 = n-2 n = 6 Therefore, the number of sides of the polygon is 6.
Ajụjụ 2 Ripọtì
A casting is made up of Copper and Zinc. If 65% of the casting is Zinc and there are 147g of Copper. What is the mass of the casting?
Akọwa Nkọwa
If 65% of the casting is Zinc, then the percentage of Copper in the casting is (100% - 65%) = 35%. Let's assume that the mass of the casting is x grams. Then, the mass of Zinc in the casting is 0.65x grams and the mass of Copper in the casting is 0.35x grams. We know that the mass of Copper in the casting is 147g, so we can write the following equation: 0.35x = 147 Solving for x, we get: x = 147 / 0.35 = 420 Therefore, the mass of the casting is 420g. Hence, the answer is 420g.
Ajụjụ 3 Ripọtì
If the sum of the roots of the equation (x - p)(2x - 1) - 0 is 1, find the value of x
Akọwa Nkọwa
Ajụjụ 5 Ripọtì
An empty rectangular tank is 250cm long and 120cm wide. If 180 litres of water is poured into the tank. Calculate the height of the water
Akọwa Nkọwa
We can begin by calculating the volume of water that has been poured into the tank. We know that the volume of a rectangular tank is given by the formula V = lwh, where l is the length, w is the width, and h is the height. In this case, we are given that l = 250cm, w = 120cm, and V = 180 litres. However, we need to convert the volume to cubic centimeters, since the dimensions of the tank are given in centimeters. To do this, we can multiply 180 litres by 1000 cubic centimeters per liter, which gives us: V = 180 x 1000 = 180000 cubic centimeters Now we can use the formula for the volume of a rectangular tank to solve for the height of the water. We rearrange the formula to get: h = V / lw Substituting the given values, we get: h = 180000 / (250 x 120) = 6 cm Therefore, the height of the water in the tank is 6cm. The answer is 6.0cm.
Ajụjụ 6 Ripọtì
The shaded portion in the diagram is the solution of
Akọwa Nkọwa
The shaded portion in the diagram represents the region where the values of x and y satisfy the inequality x + y < 3. This inequality represents a boundary line passing through (3, 0) and (0, 3) and the shaded region lies below this line. Therefore, the answer is x + y < 3.
Ajụjụ 7 Ripọtì
Simplify 0.000215 x 0.000028 and express your answer in standard form
Akọwa Nkọwa
To simplify 0.000215 x 0.000028, we multiply the two numbers to obtain: 0.000215 x 0.000028 = 0.00000000602 To express this in standard form, we move the decimal point 9 places to the right, since the number is less than 1. This gives: 0.00000000602 = 6.02 x 10-9 Therefore, the answer is 6.02 x 10-9.
Ajụjụ 8 Ripọtì
The sum of 2 consecutive whole numbers is \(\frac{5}{6}\) of their product, find the numbers
Akọwa Nkọwa
Let the two consecutive whole numbers be x and x+1. Then, according to the problem statement: x + (x+1) = \(\frac{5}{6}\)(x(x+1)) 2x+1 = \(\frac{5}{6}\)(x² + x) Multiplying both sides by 6 to remove the fraction: 12x + 6 = 5x² + 5x 5x² - 7x - 6 = 0 Solving the quadratic equation above by factoring or using the quadratic formula, we get: x = -1 or x = 1.2 Since we're looking for consecutive whole numbers, x must be 1. Therefore, the two consecutive whole numbers are 1 and 2. So the answer is (2, 3).
Ajụjụ 9 Ripọtì
The subtraction below is in base seven. Find the missing number.
5 1 6 2seven
-2 6 4 4seven
--------
2 * 1 5
--------
Akọwa Nkọwa
Ajụjụ 10 Ripọtì
If x km/h = y m/s, then y =
Akọwa Nkọwa
To convert from km/h to m/s, we need to multiply by a conversion factor. We know that 1 km = 1000 m and 1 hour = 3600 seconds. So, 1 km/h = \(\frac{1000\text{m}}{3600\text{s}}\) = \(\frac{5}{18}\) m/s. Therefore, to convert x km/h to m/s, we multiply by \(\frac{5}{18}\) to get y. So, y = \(\frac{5}{18}\)x. Therefore, option D, \(\frac{5}{18}\)x, is the correct answer.
Ajụjụ 11 Ripọtì
If sin 3y = cos 2y and 0o \(\leq\) 90o, find the value of y
Akọwa Nkọwa
Using the identity sin(90° - θ) = cos θ, we can write cos 2y as sin (90° - 2y), and so sin 3y = sin (90° - 2y). Since the sine function is periodic with a period of 360°, we can write: 3y = 90° - 2y + 360°k or 3y = 270° - 2y + 360°k for some integer k. Solving for y in each equation gives: 5y = 90° + 360°k or 5y = 270° + 360°k Dividing both sides by 5 gives: y = 18° + 72°k or y = 54° + 72°k Since the problem specifies that 0° ≤ y ≤ 90°, the only solution that works is y = 18°. Therefore, the answer is (a) 18°.
Ajụjụ 12 Ripọtì
The bar chart shows the marks distribution in am English test. If 50% is the pass mark, how many students passed the test?
Akọwa Nkọwa
To determine how many students passed the test, we need to calculate the total number of students who scored 50 or more. From the bar chart, we can see that the number of students who scored 50 or more in the test is the sum of the heights of the bars for the grades C, B, and A. The height of the bar for grade C is 20, the height of the bar for grade B is 30, and the height of the bar for grade A is 35. Therefore, the total number of students who scored 50 or more is 20 + 30 + 35 = 85. Hence, the answer is 85, and 85 students passed the test.
Ajụjụ 13 Ripọtì
If x2 + kx + \(\frac{16}{9}\) is a perfect square, find the value of k
Akọwa Nkọwa
We can approach this problem using a well-known algebraic identity: (a + b)^2 = a^2 + 2ab + b^2 If we let a = x and b = \(\frac{4}{3}\), we can write: x^2 + kx + \(\frac{16}{9}\) = (x + \(\frac{4}{3}\))^2 Expanding the right side of the equation, we get: x^2 + 2x(\(\frac{4}{3}\)) + \(\frac{16}{9}\) = x^2 + kx + \(\frac{16}{9}\) Simplifying, we get: 2x(\(\frac{4}{3}\)) = kx k = 2(\(\frac{4}{3}\)) = \(\frac{8}{3}\) Therefore, the value of k is \(\frac{8}{3}\).
Ajụjụ 14 Ripọtì
How many times, correct to the nearest whole number, will a man run round circular track of diameter 100m to cover a distance of 1000m?
Akọwa Nkọwa
The distance covered in one complete lap around a circular track of diameter 100m is equal to the circumference of the circle. The circumference of a circle is given by the formula C = 2πr, where r is the radius of the circle. The diameter of the circle is 100m, so the radius is half of that, which is 50m. Thus, the circumference of the circle is C = 2π(50) = 100π m. To cover a distance of 1000m, the man needs to run around the track 1000/100π = 10/π times. To the nearest whole number, this is equal to 3. So the man will need to run around the track 3 times to cover a distance of 1000m. Therefore, the answer is 3.
Ajụjụ 16 Ripọtì
Given that = {x: -2 < x \(\leq\) 9}, where x is an integer what is n(T)?
Akọwa Nkọwa
Ajụjụ 17 Ripọtì
\(\begin{array}{c|c} \text{No. of pets} & 0 & 1 & 2 & 3 & 4 \\ \hline \text{No. of students} & 8 & 4 & 5 & 10 & 3\end{array}\) The table shows the number of pets kept by 30 students in a class. If a student is picked at random ftom the class. What is the probability that he/she kept more than one pet?
Akọwa Nkọwa
Ajụjụ 18 Ripọtì
The diagram is a net right rectangular pyramid. Calculate the total surface area
Ajụjụ 19 Ripọtì
Simplify \(\frac{2}{2 + x} + \frac{2}{2 - x}\)
Akọwa Nkọwa
To simplify the expression, we need to find a common denominator first. The denominators are \((2+x)\) and \((2-x)\), whose product is \((2+x)(2-x)=4-x^2\). So, we can write: $$ \frac{2}{2 + x} + \frac{2}{2 - x} = \frac{2(2-x)}{(2+x)(2-x)} + \frac{2(2+x)}{(2+x)(2-x)} = \frac{4 - 2x + 4 + 2x}{4 - x^2} = \frac{8}{4 - x^2} $$ Therefore, the simplified expression is \(\frac{8}{4 - x^2}\), which is option (B).
Ajụjụ 21 Ripọtì
The sum of 6 and one-third of x is one more than twice x, find x
Akọwa Nkọwa
The given information can be written in the form of an equation as follows: 6 + (1/3)x = 2x + 1 To solve for x, we need to isolate x on one side of the equation. First, we can subtract 6 from both sides: (1/3)x = 2x - 5 Next, we can subtract 2x from both sides: (1/3)x - 2x = -5 Simplifying the left side: (1/3 - 6/3)x = -5 -5/3x = -5 Multiplying both sides by -3/5: x = 3 Therefore, the solution is x = 3.
Ajụjụ 22 Ripọtì
Given that P = {x : 1 \(\leq x \leq 6\)}, and Q = {x : 2 \(\leq x \leq 10\)} where x is an integer. Find n(P \(\cap\) Q)
Akọwa Nkọwa
The intersection of two sets, denoted by A ∩ B, is the set of elements that are in both set A and set B. In this case, P ∩ Q represents the set of integers that are between 1 and 6, and also between 2 and 10. The smallest integer in the intersection is 2, and the largest is 6. Therefore, the set P ∩ Q is {2, 3, 4, 5, 6}. The number of elements in this set is 5. Hence, the answer is: n(P ∩ Q) = 5.
Ajụjụ 23 Ripọtì
What is the value of x when y = 5?
y = \(\frac{1}{2}\) x + 1
Akọwa Nkọwa
We are given a linear equation y = (1/2)x + 1 and asked to find the value of x when y = 5. Substituting y = 5 into the equation, we get: 5 = (1/2)x + 1 Subtracting 1 from both sides, we get: 4 = (1/2)x Multiplying both sides by 2, we get: 8 = x Therefore, when y = 5, x = 8. Hence, the answer is (a) 8.
Ajụjụ 24 Ripọtì
A car uses one litre of petrol for every 14km. If one of petrol cost N63.00, how far can the car go with N900.00 worth of petrol?
Akọwa Nkọwa
The car uses 1 litre of petrol for every 14km, so for N63.00 (the cost of one litre of petrol), the car can travel 14km. We can use this information to find how far the car can go with N900.00 worth of petrol. To do this, we need to find how many litres of petrol N900.00 can buy. We can set up a proportion using the cost of petrol and the amount of petrol used per kilometre: 1 litre of petrol costs N63.00, so x litres cost N900.00 1 litre of petrol takes the car 14km, so x litres of petrol will take the car 14x km. Putting these together, we get: 1/63 = x/900 x = (1/63) * 900 x = 14.29 So, N900.00 can buy 14.29 litres of petrol. To find how far the car can go, we multiply the amount of petrol by the distance travelled per litre: 14.29 litres of petrol * 14km per litre = 200.06 km Therefore, the car can go approximately 200km with N900.00 worth of petrol. The closest option is 200km.
Ajụjụ 25 Ripọtì
In the diagram, GI is a tangent to the circle at H. If EF//GI, calculate the size of < EHF
Akọwa Nkọwa
Ajụjụ 26 Ripọtì
Simplify 1\(\frac{1}{2} + 2\frac{1}{3} \times \frac{3}{4} - \frac{1}{2}\)
Akọwa Nkọwa
Ajụjụ 29 Ripọtì
If y = \(\frac{y(2\sqrt{x^2 + m})}{3N}\), make x the subject of the formular
Ajụjụ 30 Ripọtì
The diagram shows a rectangular cardboard from which a semi-circle is cut off. Calculate the area of the remaining part
Akọwa Nkọwa
Ajụjụ 31 Ripọtì
The mean of the numbers 2, 5, 2x and 7 is less than or equal to 5. Find the range of the values of x
Akọwa Nkọwa
To find the range of values of x, we first need to find the mean of the numbers 2, 5, 2x, and 7, and then solve for x. The mean of the four numbers is given by: \[\frac{2 + 5 + 2x + 7}{4} = \frac{14 + 2x}{4} = \frac{7 + x}{2}\] Since the mean is less than or equal to 5, we can write: \[\frac{7 + x}{2} \leq 5\] Multiplying both sides by 2, we get: \[7 + x \leq 10\] Subtracting 7 from both sides, we get: \[x \leq 3\] Therefore, the range of values of x that satisfy the condition is: \[x \leq 3\] So the answer is: x ≤ 3.
Ajụjụ 32 Ripọtì
The mean age of R men in a club is 50 years, Two men aged 55 and 63, left the club and the mean age reduced by 1 year. Find the value of R
Akọwa Nkọwa
Let the total age of all R men in the club be T. Then, we know that the mean age of R men is 50 years. Therefore, we have: T/R = 50 When two men aged 55 and 63 leave the club, the total age of the remaining men becomes T - 55 - 63, and the total number of remaining men becomes R - 2. We also know that the new mean age is reduced by 1 year. Therefore, we have: (T - 55 - 63)/(R - 2) = 49 Expanding this equation and substituting T/R = 50, we get: (50R - 118)/(R - 2) = 49 Multiplying both sides by (R - 2), we get: 50R - 118 = 49(R - 2) 50R - 118 = 49R - 98 R = 20 Therefore, there were originally 20 men in the club.
Ajụjụ 33 Ripọtì
Bola sold an article for N6,900.00 and made a profit of 15%. If he sold it for N6,600.00 he would make a
Akọwa Nkọwa
If Bola sold the article for N6,900.00 and made a profit of 15%, then the cost price of the article would be: Cost price = (100 / (100 + 15)) x N6,900.00 = N6,000.00 If Bola sold the same article for N6,600.00, he would be selling it at a loss. To find out the percentage of loss, we need to calculate the selling price that would result in a profit of 0%. Let's call the new selling price "S". We know that the cost price is N6,000.00 and the desired profit is 0%, so: S - N6,000.00 = 0 S = N6,000.00 This means that if Bola sells the article for N6,000.00, he will break even and make no profit or loss. But Bola is actually selling the article for N6,600.00, which is N600.00 more than the break-even price. To find the percentage of loss, we need to calculate what percentage of the cost price N600.00 represents: Percentage loss = (600 / 6000) x 100% = 10% Therefore, if Bola sells the article for N6,600.00, he would make a loss of 10%. So the answer is "loss of 10%".
Ajụjụ 34 Ripọtì
In the diagram, the tangent MN makes an angle of 55o with the chord PS. IF O is the centre of the circle, find < RPS
Akọwa Nkọwa
Since MN is tangent to the circle, it must be perpendicular to the radius OP drawn to the point of contact. Therefore, angle MOP is 90 degrees. Let x be the measure of angle RPS. Then, angle RPO is also x degrees because RP is a radius and angles at the center are twice those at the circumference. Since MN is tangent to the circle, angle MPN is also 90 degrees. Since angle MPN and angle NPS form a linear pair (as they add up to 180 degrees) and angle MPN is 90 degrees, we have that angle NPS is 90 - x degrees. Finally, since angle MNQ and angle NPS are alternate angles, they are equal. Thus, angle MNQ is also 90 - x degrees. Now, we have a triangle MPN with angles 90, 90 - x, and 55 degrees. The angles of a triangle add up to 180 degrees, so we can write: 90 + (90 - x) + 55 = 180 Simplifying this equation gives: x = 35 Therefore, < RPS = 35 degrees.
Ajụjụ 37 Ripọtì
In the diagram, < WOX = 60o, < YOE = 50o and < OXY = 30o. What is the bearing of x from y?
Akọwa Nkọwa
Ajụjụ 38 Ripọtì
In the diagram, O is the centre of the circle, < SQR = 60o, < SPR = y and < SOR = 3x. Find the value of (x + y)
Ajụjụ 40 Ripọtì
What is the length of a rectangular garden whose perimeter is 32cm and area 39cm2?
Akọwa Nkọwa
Let the length and width of the rectangular garden be L and W, respectively. The perimeter is the sum of all sides of the rectangle, so we have: P = 2L + 2W = 32cm The area is given by: A = LW = 39cm2 We can use the first equation to solve for one of the variables in terms of the other. For example, solving for L we get: L = 16 - W Substituting this expression for L in the equation for the area we obtain: (16 - W)W = 39 Expanding and rearranging, we get a quadratic equation: W2 - 16W + 39 = 0 Solving for W using the quadratic formula, we get: W = 3 or W = 13 Since the length cannot be smaller than the width, we discard the solution W = 3. Therefore, W = 13cm, and L = 16 - W = 3cm. Therefore, the length of the rectangular garden is 13cm. So the answer is (c) 13cm.
Ajụjụ 41 Ripọtì
A rectangle has length xcm and width (x - 1)cm. If the perimeter is 16cm. Find the value of x
Akọwa Nkọwa
Let's start by finding an equation that relates the length and width of the rectangle to its perimeter. We know that the perimeter of a rectangle is the sum of the lengths of all its sides, which in this case is given as 16cm. The perimeter, P = sum of all sides of rectangle = 2(length + width) Therefore, we can write the equation as: 2(x + (x - 1)) = 16 Simplifying the equation, we have: 2x + 2(x - 1) = 16 4x - 2 = 16 4x = 18 x = 4.5cm Therefore, the value of x is 4\(\frac{1}{2}\)cm, which is.
Ajụjụ 42 Ripọtì
In the diagram, triangles HKL and HIJ are similar. Which of the following ratios is equal to \(\frac{LH}{JH}\)
Akọwa Nkọwa
Since triangles HKL and HIJ are similar, their corresponding sides are proportional. We know that LH and JH are corresponding sides. Therefore, the ratio \(\frac{LH}{JH}\) can be expressed as the ratio of the corresponding sides of the two triangles that form the fraction. In other words, we need to find the ratio of the side lengths of the triangles that share the same vertex as L and J, respectively. That would be the ratio of KL to JI. Thus, the answer is (a) \(\frac{KL}{JI}\).
Ajụjụ 43 Ripọtì
In an athletic composition, the probability that an athlete wins a 100m race is \(\frac{1}{8}\) and the probability that he wins in high jump is \(\frac{1}{4}\). What is the probability that he wins only one of the events?
Akọwa Nkọwa
Ajụjụ 44 Ripọtì
Given that tan x = 1, where 0o \(\geq\) x 90o, evaluate \(\frac{1 - \sin^2 x}{\cos x}\)
Akọwa Nkọwa
Since tan x = 1, we can determine that x = 45 degrees or \(\frac{\pi}{4}\) radians. Using the identity \(\tan^2x + 1 = \sec^2x\), we can determine that \(\sec x = \sqrt{2}\). Next, we can use the identity \(\sin^2x + \cos^2x = 1\) to determine that \(\cos x = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\). Now we can substitute the values of \(\sec x\) and \(\cos x\) into the expression for \(\frac{1 - \sin^2 x}{\cos x}\) to get: \[\frac{1 - \sin^2 x}{\cos x} = \frac{1 - \frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{\sqrt{2}} = \boxed{\frac{\sqrt{2}}{2}}\]
Ajụjụ 45 Ripọtì
In the diagram, QR//ST, /PQ/ = /PR/ and < PST = 75o. Find the value of y
Ajụjụ 46 Ripọtì
Correct 0.002473 to 3 significant figure
Akọwa Nkọwa
To round off 0.002473 to 3 significant figures, we count the number of digits from the first non-zero digit to the third significant figure. In this case, the first non-zero digit is 2, and the third significant figure is 4. So, we count 1, 2, 3 digits from the first non-zero digit, which means the third digit is the last significant figure. The fourth digit, which is 7 in this case, is greater than or equal to 5, so we round up the last significant figure by adding 1 to it. Therefore, the final answer is 0.00247, which is the third option.
Ajụjụ 47 Ripọtì
\(\begin{array}{c|c} x & 0 & 2 & 4 & 6\\ \hline y & 1 & 2 & 3 & 4\end{array}\).
The table is for the relation y = mx + c where m and c are constants. What is the equation of the line described in the tablet?
Akọwa Nkọwa
We can use the values in the table to find the values of m and c. Since y = mx + c, we can write: When x = 0, y = c, so c = 1 When x = 2, y = 2m + 1 When x = 4, y = 4m + 1 When x = 6, y = 6m + 1 To find the value of m, we can use the values for x = 2 and x = 4: 2m + 1 = 2 4m + 1 = 3 Solving these equations simultaneously gives m = 1/2. Substituting this value of m and c = 1 into y = mx + c gives: y = (1/2)x + 1 So the equation of the line described in the table is y = (1/2)x + 1. Therefore, the answer is (d) y = \(\frac{1}{2}x + 1\).
Ajụjụ 48 Ripọtì
The bar chart shows the marks distribution in am English test. What percentage of the students had marks ranging from 35 to 50?
Akọwa Nkọwa
Ajụjụ 49 Ripọtì
Given that \(\frac{5^{n +3}}{25^{2n -2}}\) = 5o, find n
Ajụjụ 50 Ripọtì
(a) 
In the diagram, < PTQ = < PSR = 90°, /PQ/ = 10 cm, /PS/ = 14.4 cm and /TQ/ = 6 cm. Calculate the area of the quadrilateral QRST.
(b) Two opposite sides of a square are each decreased by 10% while the other two are each increased by 15% to form a rectangle. Find the ratio of the area of the rectangle to that of the square.
None
Akọwa Nkọwa
None
Ajụjụ 51 Ripọtì
A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x : 3 < x < 9} are subsets of the universal set U = {2, 3, 4, 5, 6, 7, 8, 9}. Find
(a) \(A \cap (B' \cap C')\) ;
(b) \((A \cup B) \cap (B \cup C)\).
Ajụjụ 52 Ripọtì
Two fair die are thrown. M is the event described by "The sum of the scores is 10" and N is the event described by "The difference between the scores is 3".
(a) Write out the elements of M and N.
(b) Find the probability of M or N.
(c) Are M and N mutually exclusive? Give reasons.
Ajụjụ 53 Ripọtì
In the diagram, /AB/ = 8 km, /BC/ = 13 km, the bearing of A from B is 310° and the bearing of B from C is 230°. Calculate, correct to 3 significant figures,
(a) the distance AC ;
(b) the bearing of C from A ;
(c) how far east of B, C is.
Ajụjụ 55 Ripọtì
Using ruler and a pair of compasses only,
(a) construct (i) a quadrilateral PQRS with /PS/ = 6 cm, < RSP = 9 cm, /QR/ = 8.4 cm and /PQ/ = 5.4 cm; (ii) the bisectors of < RSP and < SPQ to meet at X ; (iii) the perpendicular XT to meet PS at T.
(b) Measure /XT/.
Akọwa Nkọwa
None
Ajụjụ 56 Ripọtì
(a) The third term of a Geometric Progression (G.P) is 24 and its seventh term is \(4\frac{20}{27}\). Find its first term.
(b) Given that y varies directly as x and inversely as the square of z. If y = 4, when x = 3 and z = 1, find y when x = 3 and z = 2.
Akọwa Nkọwa
None
Ajụjụ 57 Ripọtì
(a) 
Curved Surface Area = \(\pi rl\)
\(115.5 = \frac{22}{7} \times r \times 10.5\)
\(115.5 = 33r\)
\(r = \frac{115.5}{33} = 3.5 cm\)
(b) 
\(\therefore h^{2} + (3.50)^{2} = (10.5)^{2}\)
\(h^{2} = 10.5^{2} - 3.5^{2}\)
\(h^{2} = 98 \implies h = \sqrt{98}\)
\(h = 9.8994 cm \approxeq 9.90 cm\)
(c) Volume of a cone = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 9.90\)
= \(\frac{23.1 \times 11}{2}\)
= \(127.05 cm^{3} \approxeq 127 cm^{3}\)
None
Akọwa Nkọwa
None
Ajụjụ 58 Ripọtì
(a) The scale of a map is 1 : 20,000. Calculate the area, in square centimetres, on the map of a forest reserve which covers 85\(km^{2}\).
(b) A rectangular playing field is 18m wide. It is surrounded by a path 6m wide such that its area is equal to the perimeter of the path. Calculate the length of the field.
(c) 
The diagram shows a circle centre O. If < POQ = x°, the diameter of the circle is 7 cm and the area of the shaded portion is 27.5\(cm^{2}\). Find, correct to the nearest degree, the value of x. [Take \(\pi = \frac{22}{7}\)].
Akọwa Nkọwa
None
Ajụjụ 59 Ripọtì
(a) Madam Kwakyewaa imported a quantity of frozen fish costing GH¢ 400.00. The goods attracted an import duty of 15% of its cost. She also paid a sales tax of 10% of the total cost of the goods including the import duty and then sold the goods for GH¢ 660.00. Calculate the percentage profit.
(b) In a school, there are 1000 boys and a number of girls. The 48% of the total number of students that were successful in an examination was made up of 50%of the boys and 40% of the girls. Find the number of girls in the school.
None
Akọwa Nkọwa
None
Ajụjụ 60 Ripọtì
(a) The angle of depression of a boat from the mid-point of a vertical cliff is 35°. If the boat is 120m from the foot of the cliff, calculate the height of the cliff.
(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.
Akọwa Nkọwa
None
Ajụjụ 61 Ripọtì
The frequency distribution of the weight of 100 participants in a high jump competition is as shown below :
| Weight (kg) | 20 - 29 | 30 - 39 | 40 - 49 | 50 - 59 | 60 - 69 | 70 - 79 |
| Number of Participants |
10 | 18 | 22 | 25 | 16 | 9 |
(a) Construct the cumulative frequency table.
(b) Draw the cumulative frequency curve.
(c) From the curve, estimate the : (i) median ; (ii) semi- interquartile range ; (iii) probability that a participant chosen at random weighs at least 60 kg.
Akọwa Nkọwa
None
Ajụjụ 62 Ripọtì
(a) Copy and complete the table of values for the relation \(y = -x^{2} + x + 2; -3 \leq x \leq 3\).
| x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| y | -4 | 2 | -4 |
(b) Using scales of 2 cm to 1 unit on the x- axis and 2 cm to 2 units on the y- axis, draw a graph of the relation \(y = -x^{2} + x + 2\).
(c) From the graph, find the : (i) minimum value of y ; (ii) roots of equation \(x^{2} - x - 2 = 0\) ; (iii) gradient of the curve at x = -0.5.
