WAEC SSCE - General Mathematics - 2007
Question 1 Report
Solve the simultaneous equations x + y = 2 and 3x - 2y = 1
Answer Details
To solve the simultaneous equations x + y = 2 and 3x - 2y = 1, we can use the method of substitution or elimination. Method of Substitution: We can solve for one variable in terms of the other from the first equation and substitute it into the second equation, then solve for the other variable. Let's solve for y in terms of x from the first equation: x + y = 2 y = 2 - x Now substitute y = 2 - x into the second equation: 3x - 2y = 1 3x - 2(2 - x) = 1 3x - 4 + 2x = 1 5x = 5 x = 1 Now substitute x = 1 into either equation to find y: x + y = 2 1 + y = 2 y = 1 Therefore, the solution is x = 1 and y = 1. So, the correct answer is: x = 1 and y = 1.
Question 2 Report
Find the LCM of \(2^{3}\times 3\times 5^{2}, 2\times 3^{2}\times 5 \hspace{1mm}and \hspace{1mm}2^{2}\times 3^{2}\times 5\)
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Question 3 Report
Find the smallest value of k such that 2\(^2\) x 3\(^3\) x 5 x k is a perfect square.
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Question 4 Report
If p \(\alpha \frac{I}{Q}\) which of the following is true?
Answer Details
Given that p \(\alpha \frac{I}{Q}\), where \(\alpha\) means 'is proportional to'. To determine the relationship between q and p, we need to manipulate the equation so that q is isolated on one side. We can write p = k\(\frac{I}{Q}\), where k is the constant of proportionality. Multiplying both sides by Q, we get pQ = kI. Dividing both sides by p, we get Q = \(\frac{k}{p}\)I. Since k is a constant of proportionality, we can write it as k = cp for some other constant c. Therefore, Q = \(\frac{c}{p}\)I. This means that q is inversely proportional to p. So, the correct option is q \(\alpha \frac{1}{p}\).
Question 5 Report
Expand the expression(3a - xy)(3a + xy)
Answer Details
To expand the expression (3a - xy)(3a + xy), we can use the formula for multiplying two binomials: (a + b)(a - b) = a2 - b2 If we let a = 3a and b = xy, then we have: (3a - xy)(3a + xy) = (3a)2 - (xy)2 Expanding the terms on the right side, we get: (3a)2 - (xy)2 = 9a2 - x2y2 Therefore, the answer is 9a2 - x2y2.
Question 6 Report
Simplify(0.3 x 105) \(\div\) (0.4 \(\times\) 107), leaving you answer in the standard form.
Answer Details
To simplify this expression, we need to perform division and then express the answer in standard form. (0.3 x 105) \(\div\) (0.4 x 107) = (0.3/0.4) x 105−7 = 0.75 x 10-2 = 7.5 x 10-3 Therefore, the answer is 7.5 x 10-3.
Question 7 Report
The table shows the ages(in years) of twenty children chosen at random from a community. What is the mean age? \(\begin{array}{c|c} Age(years) & 1 & 2 & 3 & 4 & 5 \\ \hline {\text {Number of children}} & 2 & 3 & 5 & 6 & 4 \end{array}\)
Answer Details
To find the mean age of the twenty children, we need to calculate the sum of the ages and divide by the total number of children. Sum of the ages = (1 x 2) + (2 x 3) + (3 x 5) + (4 x 6) + (5 x 4) = 2 + 6 + 15 + 24 + 20 = 67 Total number of children = 2 + 3 + 5 + 6 + 4 = 20 Mean age = sum of the ages/total number of children = 67/20 = 3.35 years Therefore, the mean age of the twenty children is 3.35 years. The correct option is (b) 3.35 years.
Question 8 Report
The table above gives the distribution of the marks of a number of students in a test.
\(\begin{array}{c|c} Mark &1 & 2 & 3 & 4 & 5 & 6 \\ \hline Frequency & 5 & 3 & 6 & 4 & 2 & 5\end{array}\), find the mode of the distribution.
Answer Details
Question 9 Report
For what range of values of x is 4x - 3(2x - 1) > 1?
Answer Details
To solve the inequality 4x - 3(2x - 1) > 1, we can simplify it as follows: 4x - 6x + 3 > 1 -2x + 3 > 1 -2x > -2 x < 1 Therefore, the range of values of x that satisfy the inequality is x < 1. So, the correct option is "x < 1".
Question 10 Report
Simplify \(3\sqrt{45}-12\sqrt{5}+16\sqrt{20}\)leaving your answer in surd form
Answer Details
First, we need to simplify each term under the square root sign. We can write: \begin{align*} 3\sqrt{45} &= 3\sqrt{9\times 5} = 3\times 3\sqrt{5} = 9\sqrt{5}\\ 12\sqrt{5} &= 2\times 2\times 3\sqrt{5} = 4\sqrt{5}\times 3\\ 16\sqrt{20} &= 16\sqrt{4\times 5} = 16\times 2\sqrt{5} = 32\sqrt{5} \end{align*} Now we can substitute these simplified expressions back into the original expression and simplify: \begin{align*} 3\sqrt{45}-12\sqrt{5}+16\sqrt{20} &= 9\sqrt{5}-4\sqrt{5}\times 3+32\sqrt{5}\\ &= (9-12+32)\sqrt{5}\\ &= 29\sqrt{5} \end{align*} Therefore, the answer is \(29\sqrt{5}\).
Question 12 Report
Which of the following statement is true for the ste P = {-3.2 \(\leq\) x < 5} where x is an integer?
Answer Details
The given set P = {-3.2 ≤ x < 5} contains all the integers that are greater than or equal to -3.2 but less than 5. However, since x has to be an integer, the least possible value of x is -3 (since -3 is the greatest integer less than or equal to -3.2) and the greatest possible value of x is 4 (since 4 is the greatest integer less than 5). Therefore, the statement "least value of x is -3" is true for the set P, and the statement "greatest value of x is 5" is false.
Question 13 Report
Simplify\(\frac{3x^{3}}{(3x)^{3}}\)
Answer Details
To simplify the expression, we need to first expand the denominator of the fraction, which is \((3x)^3\). This can be written as \(3^3 x^3\) or \(27x^3\). Substituting this in the original fraction gives: $$\frac{3x^3}{(3x)^3} = \frac{3x^3}{27x^3} = \frac{1}{9}$$ Therefore, the simplified form of the expression is \(\frac{1}{9}\). Hence, the correct option is \(\frac{1}{9}\).
Question 14 Report
Divide the sum of 8, 6, 7, 2, 0, 4, 7, 2, 3, by their mean
Answer Details
To divide the sum of the given numbers by their mean, we need to first find their sum and mean. Sum = 8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3 = 39 Mean = (8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3) / 9 = 4.33 (rounded to two decimal places) Now, to divide the sum by the mean, we simply divide the sum by the mean: 39 / 4.33 = 9 (rounded to the nearest whole number) Therefore, the answer is 9.
Question 15 Report
The area of a square field is 110.25m2. Find the cost of fencing it round at N75.00 per meter square
Answer Details
Question 17 Report
Which of the following is true for the set \(P = \{-3.2\leq x< 5\}\) where x is an integer
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Question 18 Report
The table shows the ages(in years) of twenty children chosen at random from a community. What is the median of the distribution? \(\begin{array}{c|c} Age(years) & 1 & 2 & 3 & 4 & 5 \\ \hline {\text {Number of children}} & 2 & 3 & 5 & 6 & 4 \end{array}\)
Question 19 Report
I am x years old and my brother is 3 years older. How old was my brother last year?
Answer Details
If you are currently x years old, and your brother is 3 years older than you, then your brother's current age is x + 3. To find out how old your brother was last year, you need to subtract 1 from his current age: (x + 3) - 1 = x + 2 Therefore, your brother was x + 2 years old last year. So, the correct option is (b) (x + 2) years.
Question 20 Report
Simplify \((0.3\times 10^{5})\div (0.4\times 10^{7})\)leaving your answer in standard form
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Question 21 Report
Evaluate(111two)2 and leave your answer in base 2.
Answer Details
To evaluate (111two)2, we need to convert the binary number 111two to its decimal equivalent, then square the result, and finally convert the answer back to binary. 111two = 1 × 22 + 1 × 21 + 1 × 20 = 7ten (111two)2 = (7ten)2 = 49ten To convert 49ten back to binary, we repeatedly divide by 2 and write down the remainders in reverse order: 49 ÷ 2 = 24 remainder 1 24 ÷ 2 = 12 remainder 0 12 ÷ 2 = 6 remainder 0 6 ÷ 2 = 3 remainder 0 3 ÷ 2 = 1 remainder 1 1 ÷ 2 = 0 remainder 1 Reading the remainders in reverse order gives us: 49ten = 110001two Therefore, the answer is (B) 110001two.
Question 22 Report
Solve the simultaneous equation: x+y=2 and 3x-2y=1
Answer Details
To solve this system of simultaneous equations, we need to find the values of x and y that satisfy both equations at the same time. One way to do this is to use the method of substitution. From the first equation, we have x + y = 2, which we can rearrange as x = 2 - y. We can then substitute this expression for x into the second equation, giving 3(2-y) - 2y = 1. Simplifying the left-hand side, we get 6 - 5y = 1, and solving for y, we find y = 1. We can then substitute this value for y into either equation to find x. Using the first equation, we get x + 1 = 2, so x = 1. Therefore, the solution to the simultaneous equations is x = 1 and y = 1.
Question 23 Report
The bearing of a point P from another point Q is 310o. If |PQ| = 200m, how far west of Q is P?
Answer Details
Question 24 Report
make w the subject of the relation \(\frac{a + bc}{wd + f}\) = g
Answer Details
To make w the subject of the relation \(\frac{a + bc}{wd + f} = g\), we can follow these steps: 1. Multiply both sides by \((wd + f)\) 2. Divide both sides by g 3. Divide both sides by (a + bc) This gives: \begin{align*} \frac{a + bc}{wd + f} &= g \\ (a + bc) &= g(wd + f) \\ a + bc &= gwd + gf \\ gwd &= a + bc - gf \\ w &= \frac{a + bc - gf}{gd} \end{align*} Therefore, the solution is: \begin{equation*} w = \frac{a + bc - gf}{gd} \end{equation*} Hence, the option that matches this answer is: - \(\frac{a + bc - fg}{dg}\)
Question 25 Report
Evaluate \((111_{two})^2\) and leave your answer in base 2
Answer Details
To solve this question, we need to convert the number 111 from binary to decimal, then square the result and convert the answer back to binary. Starting with the binary number 111, we can convert it to decimal using the place value system. The rightmost digit represents 1, the second digit from the right represents 2, and the leftmost digit represents 4. Adding these values together, we get: 1 + 2 + 4 = 7 So 111 in binary is equal to 7 in decimal. To square 7, we simply multiply it by itself: 7 x 7 = 49 So the decimal equivalent of (111two)2 is 49. To convert this back to binary, we use the same place value system but in reverse. Starting with the largest power of 2 that is less than or equal to 49, we subtract that value and place a 1 in the corresponding digit. We then repeat this process with the remainder until we reach 0. 49 is greater than or equal to 32, so we subtract 32 and place a 1 in the 6th digit from the right. The remainder is 17. 17 is greater than or equal to 16, so we subtract 16 and place a 1 in the 5th digit from the right. The remainder is 1. 1 is less than 2, so we place a 1 in the 1st digit from the right. The remainder is 0, so we have our final answer: 49 in binary is equal to 110001two Therefore, the correct answer is (b) 110001two.
Question 26 Report
A sector of a circle of radius 14cm containing an angle 60o is folded to form a cone. Calculate the radius of the base of the cone
Answer Details
Question 28 Report
Simplify 3\(\sqrt{45} - 12\sqrt{5} + 16\sqrt{20}\), leaving your answer in surd form.
Answer Details
First, we need to simplify each of the surds in the expression. \[\begin{aligned} 3\sqrt{45} &= 3\sqrt{9\times 5} = 3\times 3\sqrt{5} = 9\sqrt{5} \\ 16\sqrt{20} &= 16\sqrt{4\times 5} = 16\times 2\sqrt{5} = 32\sqrt{5} \end{aligned}\] Now we can substitute these simplified surds back into the original expression and simplify it further: \[\begin{aligned} 3\sqrt{45} - 12\sqrt{5} + 16\sqrt{20} &= 9\sqrt{5} - 12\sqrt{5} + 32\sqrt{5} \\ &= (9-12+32)\sqrt{5} \\ &= 29\sqrt{5} \end{aligned}\] Therefore, the simplified expression is 29\(\sqrt{5}\). The correct option is (a).
Question 31 Report
N140,000 is shared between ABU, Kayode and Uche. Abu has twice as much as Kayode, and Kayode has twice as much as Uche. What is Kayode's share?
Answer Details
Let's start by using variables to represent the amount of money each person gets. Let Uche's share be x. Then, we know that Kayode's share is twice that amount, so his share is 2x. And we also know that Abu's share is twice Kayode's share, so his share is 2(2x) = 4x. The sum of their shares is given as N140,000, so we can set up an equation: x + 2x + 4x = 140,000 Simplifying the left side of the equation, we get: 7x = 140,000 Dividing both sides by 7, we get: x = 20,000 So Uche's share is N20,000. We can then find Kayode's share, which is twice that amount: 2x = 2(20,000) = 40,000 Therefore, Kayode's share is N40,000. So the answer is, N40,000.
Question 32 Report
Find the product of 0.0409 and 0.0021 leaving your answer in the standard form
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Question 33 Report
At what rate per cent per annum will N520.00 yield a simple interest of N39.00 in three years?
Answer Details
We can use the formula for simple interest: Simple Interest = (P * r * t) / 100 where P is the principal, r is the rate of interest per annum, and t is the time period in years. We are given that P = N520.00 and the simple interest, SI = N39.00. We are also given that the time period is 3 years. Substituting these values into the formula, we get: N39.00 = (N520.00 * r * 3) / 100 Simplifying this equation, we get: r = (N39.00 * 100) / (N520.00 * 3) = 2.5% Therefore, the rate of interest per annum is 2.5%. So, the correct option is: - 2\(\frac{1}{2}\)%
Question 34 Report
The venn diagram shows the choice of food of a number of visitors to a canteen. How many people took at least two kinds of food? If there were 35 visitors in all
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Question 35 Report
In the diagram, O is the centre of the circle and PQ is a diameter. Triangle RSO is an equilateral triangle of side 4cm. Find the area of the shaded region
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Question 36 Report
The venn diagram shows the choice of food of a number of visitors to a canteen. If there were 35 visitors in all, find the value of x
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Question 37 Report
Simplify \(\frac{3x^3}{(3x)^3}\)
Answer Details
We can simplify this expression by using the rule that states \(\frac{a^m}{a^n}=a^{m-n}\) for any non-zero real number a and integers m and n. Applying this rule, we get: \[\frac{3x^3}{(3x)^3} = \frac{3x^3}{27x^3} = \frac{1}{9}\] Therefore, the answer is \(\frac{1}{9}\).
Question 38 Report
A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the centre of the circle.
Answer Details
When a chord is drawn in a circle, it divides the circle into two equal parts. The line connecting the midpoint of the chord and the center of the circle is perpendicular to the chord. Therefore, to find the distance of the chord from the center of the circle, we need to draw a perpendicular bisector to the chord and measure the distance from the center of the circle to the perpendicular bisector. Let O be the center of the circle, AB be the chord of length 6cm and M be the midpoint of AB. Then OM is the perpendicular bisector of AB, and AM = MB = 3cm. Using the Pythagorean theorem in triangle OAM, we have: $$OA^2 = OM^2 + AM^2 = 5^2 - 3^2 = 16$$ Taking the square root of both sides, we have OA = 4cm. Therefore, the distance of the chord from the center of the circle is 4cm. So the correct option is (d) 4.0cm.
Question 39 Report
Find the value of x in the diagram
Question 40 Report
In the diagram |LN| = 4cm, LNM = 90o and tan y = \(\frac{2}{3}\). What is the area of the \(\bigtriangleup\)LMN?
Answer Details
Question 41 Report
For what value of x is the expression \(\frac{2x - 1}{x + 3}\) not defined?
Answer Details
The expression \(\frac{2x-1}{x+3}\) is undefined if the denominator is equal to zero, since division by zero is not allowed in mathematics. Therefore, we need to solve the equation \(x+3=0\) to find the value of x for which the expression is undefined. Solving for x, we have: \begin{align*} x+3 &= 0 \\ x &= -3 \end{align*} Therefore, the expression is not defined for x = -3. Thus, the correct answer is (-3).
Question 42 Report
Each of the interior angle of a regular polygon is 162o. How many sides has the polygon?
Answer Details
The sum of the interior angles of a polygon can be found by using the formula (n-2) x 180, where n is the number of sides of the polygon. For a regular polygon, all interior angles are equal. Let's call the measure of each interior angle x. Since the polygon is regular, we can use the fact that the sum of the measures of the interior angles of a polygon with n sides is (n-2) x 180, and that each interior angle of this polygon has measure x. Thus, we can set up the equation: n x x = (n-2) x 180 Simplifying this equation, we get: nx = 180n - 360 nx - 180n = -360 n(x - 180) = -360 n = -360 / (x - 180) We know that each interior angle of the polygon has a measure of 162 degrees, so x = 162. Substituting this value into the equation we just derived, we get: n = -360 / (162 - 180) n = -360 / (-18) n = 20 Therefore, the polygon has 20 sides. Answer: 20.
Question 43 Report
Q is 32 km away from P on a bearing 042o and R is 25km from P on a bearing of 132o. Calculate the bearing of R from Q.
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Question 45 Report
The area of a square field is 110.25m2. Find the cost of fencing it round at N75.00 per metre
Answer Details
The area of the square field is given as 110.25m2. We know that the area of a square is given by the formula A = s2, where A is the area and s is the side length. Therefore, we can find the side length of the square as follows: s2 = A s2 = 110.25 s = √110.25 = 10.5m The perimeter of the square is given by the formula P = 4s, where P is the perimeter and s is the side length. Therefore, we can find the perimeter of the square as follows: P = 4s = 4(10.5m) = 42m To find the cost of fencing it round at N75.00 per metre, we need to multiply the perimeter by N75.00: Cost = P × N75.00 = 42m × N75.00 = N3,150.00 Therefore, the cost of fencing the square field at N75.00 per metre is N3,150.00. So the correct option is (B) N3,150.00.
Question 46 Report
A messenger was paid N2.50 an hour during the normal working hours and 4.00 n hour during overtime. If he received N31.00 for 10 hours work, how many hours are for overtime?
Answer Details
Let's call the number of normal working hours the messenger did as "x" and the number of overtime hours as "y". We can create two equations based on the given information: Equation 1: The messenger worked a total of 10 hours x + y = 10 Equation 2: The total amount paid is N31.00 2.5x + 4y = 31 We can use the first equation to solve for x in terms of y: x = 10 - y Now we can substitute this expression for x into the second equation: 2.5(10 - y) + 4y = 31 Simplifying and solving for y, we get: 25 - 2.5y + 4y = 31 1.5y = 6 y = 4 Therefore, the messenger worked 4 hours of overtime. Answer: (c) 4.
Question 48 Report
If x = 3, Y = 2 and z = 4, what is the value of 3x2 - 2y + z?
Answer Details
Substituting x = 3, y = 2, and z = 4 into the expression 3x2 - 2y + z gives: 3x2 - 2y + z = 3(3)2 - 2(2) + 4 = 27 - 4 + 4 = 27 Therefore, the value of 3x2 - 2y + z is 27.
Question 49 Report
If x =3, y = 2 and z = 4 what is the value of \(3x^{2}-2y+z\)
Answer Details
Substituting x = 3, y = 2 and z = 4 into the expression \(3x^{2}-2y+z\), we get: $$3(3)^2 - 2(2) + 4 = 27 - 4 + 4 = 27$$ Therefore, the value of \(3x^{2}-2y+z\) when x = 3, y = 2 and z = 4 is 27. The correct option is (b) 27.
Question 50 Report
A sector of a circle of radius 14cm containing an angle 60o is folded to form a cone. Calculate the radius of the base of the cone.
Answer Details
The sector of a circle can be folded to form a cone if and only if the arc length of the sector is equal to the circumference of the base of the cone. Given that the radius of the sector is 14 cm and the angle is 60o, the arc length of the sector is \(\frac{60}{360}\times2\pi(14)=\frac{14\pi}{3}\) cm. Let's denote the radius of the base of the cone by r. Then, the circumference of the base of the cone is \(2\pi r\) cm. Since the arc length of the sector is equal to the circumference of the base of the cone, we have: \[\frac{14\pi}{3}=2\pi r\] Dividing both sides by 2π gives: \[r=\frac{14}{3}\div2=\frac{7}{3}=2\frac{1}{3}\,\text{cm}\] Therefore, the radius of the base of the cone is \(2\frac{1}{3}\,\text{cm}\). Answer:.
Question 51 Report
For what value of x is the expression \(\frac{2x-1}{x+3}\)not defined?
Answer Details
The expression \(\frac{2x-1}{x+3}\) is not defined when the denominator, \(x+3\), is equal to zero. Therefore, we can solve the equation \(x+3=0\) to find the value of x that makes the expression undefined. \(x+3=0\) \(x=-3\) So, the value of x that makes the expression undefined is -3. Therefore, the correct answer is (-3).
Question 52 Report
If 8x-1 = \(\frac{1}{4}\), find x
Answer Details
We know that 8x-1 = \(\frac{1}{4}\). We can write 8x-1 as (23)x-1 = 23x-3. Similarly, we can write 1/4 as 2-2. Therefore, we can write the given equation as 23x-3 = 2-2. Since the bases are equal, we can equate the exponents to get: 3x - 3 = -2 Adding 3 to both sides, we get: 3x = 1 Dividing both sides by 3, we get: x = \(\frac{1}{3}\) Therefore, the value of x is \(\frac{1}{3}\).
Question 53 Report
In the diagram, |XR| = 4cm
|RZ| = 12cm, |SR| = n, |XZ| = m and SR||YZ. Find m in terms of n
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Question 54 Report
Convert 425 to base three numeral
Answer Details
To convert a number from base 5 to base 3, we need to first convert it to base 10, and then convert it to base 3. 425 can be written as: $42_{5} = 4\cdot5^{1} + 2\cdot5^{0} = 20 + 2 = 22_{10}$ Now we convert 2210 to base 3. We can do this by repeatedly dividing 22 by 3 and taking the remainders, until the quotient is zero. The remainders, read from bottom to top, give us the base 3 numeral. \begin{array}{c|c} \text{Dividend} & \text{Remainder}\\ \hline 22 & 1 \\ 7 & 2 \\ 2 & 2 \\ 0 & \\ \end{array} So, 2210 is equal to 2123. Therefore, the correct option is (3) 2113.
Question 55 Report
If the probability of an event occurring is x, what is the probability of the event not occurring?
Answer Details
The probability of an event occurring is the measure of the likelihood that the event will happen, given as a number between 0 and 1, inclusive. The probability of an event not occurring is the measure of the likelihood that the event will not happen. Since the sum of the probabilities of an event occurring and not occurring is 1, we can find the probability of an event not occurring by subtracting the probability of the event occurring from 1. Therefore, the probability of an event not occurring is equal to 1 minus the probability of the event occurring. Mathematically, if the probability of an event occurring is x, then the probability of the event not occurring is 1 - x. So, option A, 1 - x, is the correct answer.
Question 57 Report
In the diagram, O is the centre of the circle and < PQR = 106o, find the value of y
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Question 58 Report
The volume of a cylinder is 1200cm3 and the area of its base is 150cm2. Find the height of the cylinder.
Answer Details
We can use the formula for the volume of a cylinder to solve this problem, which is: V = πr²h where V is the volume of the cylinder, r is the radius of the base, and h is the height of the cylinder. We are given that the volume of the cylinder is 1200cm³, so we can substitute this value into the formula: 1200 = πr²h We are also given that the area of the base is 150cm², and the formula for the area of a circle is: A = πr² where A is the area of the circle and r is the radius. Since the base of the cylinder is a circle, we can use this formula to find the radius: 150 = πr² r² = 150/π r ≈ 6.12 cm (rounded to two decimal places) Now we can substitute this value for r in the first formula and solve for h: 1200 = π(6.12)²h h ≈ 8 cm (rounded to two decimal places) Therefore, the height of the cylinder is approximately 8cm. So the correct option is (b) 8.00cm.
Question 59 Report
I am x years old and my brother is 3 years older how old was my brother last year
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Question 60 Report
N140,000 is shared between Abu, Kayode and Uche. Abu has twice as much as Kayode and Kayode has twice as much as Uche. What is Kayode'sshare?
Answer Details
Let Uche's share be x. Since Kayode has twice as much as Uche, Kayode's share will be 2x. Also, since Abu has twice as much as Kayode, Abu's share will be 2(2x) = 4x. The sum of their shares is N140,000, so we have: x + 2x + 4x = N140,000 7x = N140,000 x = N20,000 Therefore, Kayode's share is 2x = 2(N20,000) = N40,000. Hence, the answer is N40,000.
Question 61 Report
A chord of length 6cm is drawn in a circle of radius 5cm. Find the distance of the chord from the center of the circle
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Question 62 Report
PQR is an equilateral triangle with sides 2\(\sqrt{3cm}\). calculate its height
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Question 63 Report
IN the diagram, |LN| = 4cm, LNM = 90o and tan y = 2/3. What is the area of the ?LMN?
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Question 64 Report
Find the volume of a solid cylinder with base radius 10cm and height 14cm.
Answer Details
The volume of a cylinder is given by the formula V = πr2h, where r is the radius of the base and h is the height of the cylinder. In this case, the base radius is 10cm and the height is 14cm. Substituting these values into the formula, we have: V = π(10cm)2(14cm) = π(100cm2)(14cm) = 1400πcm3 Using the value of π ≈ 3.14, we can approximate the answer to: V ≈ 1400(3.14) ≈ 4400cm3 Therefore, the volume of the cylinder is approximately 4400cm3. The correct option is (D) 4400cm3.
Question 65 Report
Two circles have radii 16cm and 23cm. What is the difference between their circumference? take \(\pi = \frac{22}{7}\)
Answer Details
The circumference of a circle is given by the formula 2πr, where r is the radius of the circle and π is a constant value approximately equal to 22/7 or 3.14. Using the given formula, the circumference of the circle with radius 16cm is 2 x (22/7) x 16cm = 100.57cm (to 2 decimal places). Similarly, the circumference of the circle with radius 23cm is 2 x (22/7) x 23cm = 144.57cm (to 2 decimal places). Therefore, the difference between the circumferences of the two circles is 144.57cm - 100.57cm = 44.00cm (to 2 decimal places). Hence, the correct option is (C) 44.00cm.
Question 66 Report
In a , < PQR = < PRQ = 45o. which of the following statements is/are correct? i. \(\bigtriangleup\)PQR is an equalateral triangle ii. \(\bigtriangleup\)PQR is an isosceles triangle iii. \(\bigtriangleup\)PQR is a right-angled triangle
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Question 67 Report
If \(8^{x+1}=\frac{1}{4}\), find x
Answer Details
We can solve the given equation by taking logarithms to the base 2 on both sides. This gives: \begin{align*} 8^{x+1} &= \frac{1}{4}\\ \Rightarrow \quad 2^{3(x+1)} &= 2^{-2}\\ \Rightarrow \quad 3(x+1) &= -2\\ \Rightarrow \quad 3x+3 &= -2\\ \Rightarrow \quad 3x &= -5\\ \Rightarrow \quad x &= -\frac{5}{3} \end{align*} Therefore, the value of $x$ is $-\frac{5}{3}$. So, the correct option is \(-\frac{5}{3}\).
Question 68 Report
(a) With the aid of four- figure logarithm tables, evaluate \((0.004592)^{\frac{1}{3}}\).
(b) If \(\log_{10} y + 3\log_{10} x = 2\), express y in terms of x.
(c) Solve the equations:
\[ \begin{aligned} 3x - 2y &= 21\\ 4x + 5y &= 5. \end{aligned} \](a) Using four-figure logarithm tables: \[\log(0.004592) = \bar{3}.6621 = -3 + 0.6621 = -2.3379.\] \[(0.004592)^{1/3}:\quad \frac{1}{3}(-2.3379) = -0.7793 = \bar{1}.2207.\] Antilog of \(0.2207\) is \(1.662\), so the value is \(1.662\times 10^{-1} \approx \mathbf{0.1662}.\)
(b) \(\log_{10} y + 3\log_{10} x = 2 \Rightarrow \log_{10}(yx^3) = 2 \Rightarrow yx^3 = 10^2 = 100.\) \[\mathbf{y = \frac{100}{x^3}}.\]
(c) \(3x - 2y = 21\ \ (1)\), \(4x + 5y = 5\ \ (2).\)
\((1)\times 5:\ 15x - 10y = 105.\quad (2)\times 2:\ 8x + 10y = 10.\) Adding: \(23x = 115 \Rightarrow x = 5.\)
From \((1)\): \(3(5) - 2y = 21 \Rightarrow -2y = 6 \Rightarrow y = -3.\)
Solution: \(\mathbf{x = 5,\ y = -3}.\)
Answer Details
(a) Using four-figure logarithm tables: \[\log(0.004592) = \bar{3}.6621 = -3 + 0.6621 = -2.3379.\] \[(0.004592)^{1/3}:\quad \frac{1}{3}(-2.3379) = -0.7793 = \bar{1}.2207.\] Antilog of \(0.2207\) is \(1.662\), so the value is \(1.662\times 10^{-1} \approx \mathbf{0.1662}.\)
(b) \(\log_{10} y + 3\log_{10} x = 2 \Rightarrow \log_{10}(yx^3) = 2 \Rightarrow yx^3 = 10^2 = 100.\) \[\mathbf{y = \frac{100}{x^3}}.\]
(c) \(3x - 2y = 21\ \ (1)\), \(4x + 5y = 5\ \ (2).\)
\((1)\times 5:\ 15x - 10y = 105.\quad (2)\times 2:\ 8x + 10y = 10.\) Adding: \(23x = 115 \Rightarrow x = 5.\)
From \((1)\): \(3(5) - 2y = 21 \Rightarrow -2y = 6 \Rightarrow y = -3.\)
Solution: \(\mathbf{x = 5,\ y = -3}.\)
Question 69 Report
(a) Simplify : \(\frac{x^{2} - y^{2}}{3x + 3y}\)
(b) 
In the diagram, PQRS is a rectangle. /PK/ = 15 cm, /SK/ = /KR/ and <PKS = 30°. Calculate, correct to three significant figures : (i) /PS/ ; (ii) /SK/ and (iii) the area of the shaded portion.
Answer Details
None
Question 70 Report
Y is 60 km away from X on a bearing of 135°. Z is 80 km away from X on a bearing of 225°. Find the :
(a) distance of Z from Y ;
(b) bearing of Z from Y.
\(Y\) is on bearing \(135^\circ\), \(Z\) on bearing \(225^\circ\) from \(X\). The angle between them at \(X\) is \[\angle YXZ = 225^\circ - 135^\circ = 90^\circ,\] so triangle \(YXZ\) is right-angled at \(X\).
(a) By Pythagoras: \[|YZ| = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = \mathbf{100\,\text{km}}.\]
(b) Using coordinates: \(Y = (60\sin135^\circ, 60\cos135^\circ) = (42.43, -42.43)\) and \(Z = (80\sin225^\circ, 80\cos225^\circ) = (-56.57, -56.57).\) \[\vec{YZ} = (-98.99,\ -14.14),\] which points South-West. The bearing satisfies \(\sin B = \dfrac{-98.99}{100},\ \cos B = \dfrac{-14.14}{100}\) (third quadrant), so \[B = 180^\circ + \tan^{-1}\!\frac{98.99}{14.14} = 180^\circ + 81.9^\circ \approx \mathbf{262^\circ}.\] Bearing of \(Z\) from \(Y\) is about \(262^\circ.\)
Answer Details
\(Y\) is on bearing \(135^\circ\), \(Z\) on bearing \(225^\circ\) from \(X\). The angle between them at \(X\) is \[\angle YXZ = 225^\circ - 135^\circ = 90^\circ,\] so triangle \(YXZ\) is right-angled at \(X\).
(a) By Pythagoras: \[|YZ| = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = \mathbf{100\,\text{km}}.\]
(b) Using coordinates: \(Y = (60\sin135^\circ, 60\cos135^\circ) = (42.43, -42.43)\) and \(Z = (80\sin225^\circ, 80\cos225^\circ) = (-56.57, -56.57).\) \[\vec{YZ} = (-98.99,\ -14.14),\] which points South-West. The bearing satisfies \(\sin B = \dfrac{-98.99}{100},\ \cos B = \dfrac{-14.14}{100}\) (third quadrant), so \[B = 180^\circ + \tan^{-1}\!\frac{98.99}{14.14} = 180^\circ + 81.9^\circ \approx \mathbf{262^\circ}.\] Bearing of \(Z\) from \(Y\) is about \(262^\circ.\)
Question 71 Report
(a) Simplify : \(\frac{x^{2} - 8x + 16}{x^{2} - 7x + 12}\).
(b) If \(\frac{1}{2}, \frac{1}{x}, \frac{1}{3}\) are successive terms of an arithmetic progression (A.P), show that \(\frac{2 - x}{x - 3} = \frac{2}{3}\).
(a) Factorise numerator and denominator: \[x^2 - 8x + 16 = (x-4)^2,\qquad x^2 - 7x + 12 = (x-3)(x-4).\] \[\frac{(x-4)^2}{(x-3)(x-4)} = \frac{x-4}{x-3}\quad (x\neq 4).\]
(b) If \(\tfrac{1}{2}, \tfrac{1}{x}, \tfrac{1}{3}\) are successive terms of an A.P., the common difference is constant: \[\frac{1}{x} - \frac{1}{2} = \frac{1}{3} - \frac{1}{x}.\] The left side is \(\dfrac{2 - x}{2x}\) and the right side is \(\dfrac{x - 3}{3x}\), so \[\frac{2 - x}{2x} = \frac{x - 3}{3x}.\] Multiplying both sides by \(3x\) and dividing by \((x-3)\): \[\frac{2 - x}{x - 3} = \frac{2x}{3x} = \frac{2}{3}.\] Hence \(\dfrac{2 - x}{x - 3} = \dfrac{2}{3}\), as required.
Answer Details
(a) Factorise numerator and denominator: \[x^2 - 8x + 16 = (x-4)^2,\qquad x^2 - 7x + 12 = (x-3)(x-4).\] \[\frac{(x-4)^2}{(x-3)(x-4)} = \frac{x-4}{x-3}\quad (x\neq 4).\]
(b) If \(\tfrac{1}{2}, \tfrac{1}{x}, \tfrac{1}{3}\) are successive terms of an A.P., the common difference is constant: \[\frac{1}{x} - \frac{1}{2} = \frac{1}{3} - \frac{1}{x}.\] The left side is \(\dfrac{2 - x}{2x}\) and the right side is \(\dfrac{x - 3}{3x}\), so \[\frac{2 - x}{2x} = \frac{x - 3}{3x}.\] Multiplying both sides by \(3x\) and dividing by \((x-3)\): \[\frac{2 - x}{x - 3} = \frac{2x}{3x} = \frac{2}{3}.\] Hence \(\dfrac{2 - x}{x - 3} = \dfrac{2}{3}\), as required.
Question 72 Report
(a) A cylinder with radius 3.5 cm has its two ends closed, if the total surface area is \(209 cm^{2}\), calculate the height of the cylinder. [Take \(\pi = \frac{22}{7}\)].
(b) 
In the diagram, O is the centre of the circle and ABC is a tangent at B. If \(\stackrel\frown{BDF} = 66°\) and \(\stackrel\frown{DBC} = 57°\), calculate, (i) \(\stackrel\frown{EBF}\) and (ii) \(\stackrel\frown{BGF}\).
(a) Height of the closed cylinder
For a cylinder closed at both ends, the total surface area is
\[TSA = 2\pi r(r+h)\]With \(r = 3.5\text{ cm}\), \(TSA = 209\text{ cm}^2\) and \(\pi = \tfrac{22}{7}\):
\[2\times\frac{22}{7}\times 3.5\,(3.5+h) = 209\]Now \(2\times\dfrac{22}{7}\times 3.5 = 22\), so
\[22(3.5+h) = 209 \;\Rightarrow\; 3.5+h = \frac{209}{22} = 9.5\]\[h = 9.5 - 3.5 = 6\text{ cm}\]The height of the cylinder is \(6\text{ cm}\).
(b) Circle with tangent ABC at B
From the diagram, \(EB\) is a diameter (it passes through the centre O). The given angles are the inscribed angle \(\angle BDF = 66^\circ\) at D and the tangent-chord angle \(\angle DBC = 57^\circ\) at B. G is the point where chord \(FD\) crosses the diameter \(EB\).
Finding the arcs
The inscribed angle \(\angle BDF\) stands on arc \(BF\) (not containing D), so
\[\text{arc }BF = 2\times 66^\circ = 132^\circ\]The tangent-chord angle \(\angle DBC\) equals half the intercepted arc \(BD\):
\[\text{arc }BD = 2\times 57^\circ = 114^\circ\]Since \(EB\) is a diameter, each semicircle is \(180^\circ\):
\[\text{arc }EF = 180^\circ - \text{arc }BF = 180^\circ - 132^\circ = 48^\circ\]\[\text{arc }ED = 180^\circ - \text{arc }BD = 180^\circ - 114^\circ = 66^\circ\](Check: \(114+66+48+132 = 360^\circ\). Correct.)
(i) \(\angle EBF\)
\(\angle EBF\) is the inscribed angle at B standing on arc \(EF\):
\[\angle EBF = \tfrac{1}{2}\,\text{arc }EF = \tfrac{1}{2}\times 48^\circ = 24^\circ\](ii) \(\angle BGF\)
G is the intersection of chords \(EB\) and \(FD\). The angle between two chords equals half the sum of the two arcs it intercepts (arc \(BF\) and the vertically opposite arc \(ED\)):
\[\angle BGF = \tfrac{1}{2}\big(\text{arc }BF + \text{arc }ED\big) = \tfrac{1}{2}(132^\circ + 66^\circ) = \tfrac{1}{2}\times 198^\circ = 99^\circ\]Therefore \(\angle EBF = 24^\circ\) and \(\angle BGF = 99^\circ\).
Answer Details
(a) Height of the closed cylinder
For a cylinder closed at both ends, the total surface area is
\[TSA = 2\pi r(r+h)\]With \(r = 3.5\text{ cm}\), \(TSA = 209\text{ cm}^2\) and \(\pi = \tfrac{22}{7}\):
\[2\times\frac{22}{7}\times 3.5\,(3.5+h) = 209\]Now \(2\times\dfrac{22}{7}\times 3.5 = 22\), so
\[22(3.5+h) = 209 \;\Rightarrow\; 3.5+h = \frac{209}{22} = 9.5\]\[h = 9.5 - 3.5 = 6\text{ cm}\]The height of the cylinder is \(6\text{ cm}\).
(b) Circle with tangent ABC at B
From the diagram, \(EB\) is a diameter (it passes through the centre O). The given angles are the inscribed angle \(\angle BDF = 66^\circ\) at D and the tangent-chord angle \(\angle DBC = 57^\circ\) at B. G is the point where chord \(FD\) crosses the diameter \(EB\).
Finding the arcs
The inscribed angle \(\angle BDF\) stands on arc \(BF\) (not containing D), so
\[\text{arc }BF = 2\times 66^\circ = 132^\circ\]The tangent-chord angle \(\angle DBC\) equals half the intercepted arc \(BD\):
\[\text{arc }BD = 2\times 57^\circ = 114^\circ\]Since \(EB\) is a diameter, each semicircle is \(180^\circ\):
\[\text{arc }EF = 180^\circ - \text{arc }BF = 180^\circ - 132^\circ = 48^\circ\]\[\text{arc }ED = 180^\circ - \text{arc }BD = 180^\circ - 114^\circ = 66^\circ\](Check: \(114+66+48+132 = 360^\circ\). Correct.)
(i) \(\angle EBF\)
\(\angle EBF\) is the inscribed angle at B standing on arc \(EF\):
\[\angle EBF = \tfrac{1}{2}\,\text{arc }EF = \tfrac{1}{2}\times 48^\circ = 24^\circ\](ii) \(\angle BGF\)
G is the intersection of chords \(EB\) and \(FD\). The angle between two chords equals half the sum of the two arcs it intercepts (arc \(BF\) and the vertically opposite arc \(ED\)):
\[\angle BGF = \tfrac{1}{2}\big(\text{arc }BF + \text{arc }ED\big) = \tfrac{1}{2}(132^\circ + 66^\circ) = \tfrac{1}{2}\times 198^\circ = 99^\circ\]Therefore \(\angle EBF = 24^\circ\) and \(\angle BGF = 99^\circ\).
Question 73 Report
(a) Evaluate, without using mathematical tables or calculator, \((3.69 \times 10^{5}) \div (1.64 \times 10^{-3})\), leaving your answer in standard form.
(b) A man invested N20,000 in bank A and N25,000 in bank B at the beginning of the year. Bank A pays simple interest at a rate of y% per annum and B pays 1.5y% per annum. If his total interest at the end of the year from the two banks was N4,600, find the value of y.
(a) \[\frac{3.69\times 10^{5}}{1.64\times 10^{-3}} = \frac{3.69}{1.64}\times 10^{5-(-3)} = 2.25\times 10^{8}.\] Answer in standard form: \(\mathbf{2.25\times 10^{8}}.\)
(b) Simple interest for one year. Bank A: \(\dfrac{20000\times y\times 1}{100} = 200y.\) Bank B: \(\dfrac{25000\times 1.5y\times 1}{100} = 375y.\)
Total interest: \[200y + 375y = 4600 \Rightarrow 575y = 4600 \Rightarrow y = \frac{4600}{575} = \mathbf{8}.\]
Answer Details
(a) \[\frac{3.69\times 10^{5}}{1.64\times 10^{-3}} = \frac{3.69}{1.64}\times 10^{5-(-3)} = 2.25\times 10^{8}.\] Answer in standard form: \(\mathbf{2.25\times 10^{8}}.\)
(b) Simple interest for one year. Bank A: \(\dfrac{20000\times y\times 1}{100} = 200y.\) Bank B: \(\dfrac{25000\times 1.5y\times 1}{100} = 375y.\)
Total interest: \[200y + 375y = 4600 \Rightarrow 575y = 4600 \Rightarrow y = \frac{4600}{575} = \mathbf{8}.\]
Question 74 Report
In a college, the number of absentees recorded over a period of 30 days was as shown in the frequency distribution table
| Number of absentees | 0-4 | 5-9 | 10-14 | 15-19 | 20-24 |
| Number of Days | 1 | 5 | 10 | 9 | 5 |
Calculate the : (a) Mean
(b) Standard deviation , correct to two decimal places.
Take the class mid-values \(x\) and build a working table (\(N=1+5+10+9+5=30\)):
| Class | Mid-value x | f | fx | fx² |
|---|---|---|---|---|
| 0 - 4 | 2 | 1 | 2 | 4 |
| 5 - 9 | 7 | 5 | 35 | 245 |
| 10 - 14 | 12 | 10 | 120 | 1440 |
| 15 - 19 | 17 | 9 | 153 | 2601 |
| 20 - 24 | 22 | 5 | 110 | 2420 |
| Total | 30 | 420 | 6710 |
(a) Mean.
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{420}{30}=14.\]
(b) Standard deviation.
\[s=\sqrt{\frac{\sum fx^{2}}{N}-\bar{x}^{2}}=\sqrt{\frac{6710}{30}-14^{2}}=\sqrt{223.67-196}=\sqrt{27.67}\approx 5.26.\]
The mean number of absentees is 14 and the standard deviation is 5.26 (to 2 d.p.).
Answer Details
Take the class mid-values \(x\) and build a working table (\(N=1+5+10+9+5=30\)):
| Class | Mid-value x | f | fx | fx² |
|---|---|---|---|---|
| 0 - 4 | 2 | 1 | 2 | 4 |
| 5 - 9 | 7 | 5 | 35 | 245 |
| 10 - 14 | 12 | 10 | 120 | 1440 |
| 15 - 19 | 17 | 9 | 153 | 2601 |
| 20 - 24 | 22 | 5 | 110 | 2420 |
| Total | 30 | 420 | 6710 |
(a) Mean.
\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{420}{30}=14.\]
(b) Standard deviation.
\[s=\sqrt{\frac{\sum fx^{2}}{N}-\bar{x}^{2}}=\sqrt{\frac{6710}{30}-14^{2}}=\sqrt{223.67-196}=\sqrt{27.67}\approx 5.26.\]
The mean number of absentees is 14 and the standard deviation is 5.26 (to 2 d.p.).
Question 75 Report
(a) 
In the diagram, AOB is a straight line. < AOC = 3(x + y)°, < COB = 45°, < AOD = (5x + y)° and < DOB = y°. Find the values of x and y.
(b) From two points on opposite sides of a pole 33m high, the angles of elevation of the top of the pole are 53° and 67°. If the two points and the base are on te same horizontal level, calculate, correct to three significant figures, the distance between the two points.
(a) From the diagram, AOB is a straight line, so the angles on the upper side of the line sum to \(180^\circ\), and the angles on the lower side also sum to \(180^\circ\).
Upper side (angles \(AOC\) and \(COB\)):
\[3(x+y) + 45 = 180\]\[3(x+y) = 135 \Rightarrow x + y = 45 \quad (1)\]Lower side (angles \(AOD\) and \(DOB\)):
\[(5x + y) + y = 180\]\[5x + 2y = 180 \quad (2)\]From (1), \(y = 45 - x\). Substitute into (2):
\[5x + 2(45 - x) = 180\]\[5x + 90 - 2x = 180\]\[3x = 90 \Rightarrow x = 30\]Then \(y = 45 - 30 = 15\).
\(x = 30,\quad y = 15\)
Check: \(3(x+y)=135^\circ\), \(135+45=180^\circ\) (upper). \(5x+y=165^\circ\), \(165+15=180^\circ\) (lower). Correct.
(b) Let the base of the pole be \(B\), and let the two points be \(P\) and \(Q\) on opposite sides. The pole \(BT = 33\ \text{m}\) is vertical.
For the point with elevation \(53^\circ\):
\[\tan 53^\circ = \frac{33}{PB} \Rightarrow PB = \frac{33}{\tan 53^\circ} = \frac{33}{1.3270} = 24.867\ \text{m}\]For the point with elevation \(67^\circ\):
\[\tan 67^\circ = \frac{33}{QB} \Rightarrow QB = \frac{33}{\tan 67^\circ} = \frac{33}{2.3559} = 14.008\ \text{m}\]Since the points are on opposite sides of the pole, the distance between them is:
\[PQ = PB + QB = 24.867 + 14.008 = 38.875\ \text{m}\]Distance \(\approx 38.9\ \text{m}\) (to 3 significant figures).
Answer Details
(a) From the diagram, AOB is a straight line, so the angles on the upper side of the line sum to \(180^\circ\), and the angles on the lower side also sum to \(180^\circ\).
Upper side (angles \(AOC\) and \(COB\)):
\[3(x+y) + 45 = 180\]\[3(x+y) = 135 \Rightarrow x + y = 45 \quad (1)\]Lower side (angles \(AOD\) and \(DOB\)):
\[(5x + y) + y = 180\]\[5x + 2y = 180 \quad (2)\]From (1), \(y = 45 - x\). Substitute into (2):
\[5x + 2(45 - x) = 180\]\[5x + 90 - 2x = 180\]\[3x = 90 \Rightarrow x = 30\]Then \(y = 45 - 30 = 15\).
\(x = 30,\quad y = 15\)
Check: \(3(x+y)=135^\circ\), \(135+45=180^\circ\) (upper). \(5x+y=165^\circ\), \(165+15=180^\circ\) (lower). Correct.
(b) Let the base of the pole be \(B\), and let the two points be \(P\) and \(Q\) on opposite sides. The pole \(BT = 33\ \text{m}\) is vertical.
For the point with elevation \(53^\circ\):
\[\tan 53^\circ = \frac{33}{PB} \Rightarrow PB = \frac{33}{\tan 53^\circ} = \frac{33}{1.3270} = 24.867\ \text{m}\]For the point with elevation \(67^\circ\):
\[\tan 67^\circ = \frac{33}{QB} \Rightarrow QB = \frac{33}{\tan 67^\circ} = \frac{33}{2.3559} = 14.008\ \text{m}\]Since the points are on opposite sides of the pole, the distance between them is:
\[PQ = PB + QB = 24.867 + 14.008 = 38.875\ \text{m}\]Distance \(\approx 38.9\ \text{m}\) (to 3 significant figures).
Question 76 Report
Out of the 24 apples in a box, 6 are bad. If three apples are taken from the box at random, with replacement, find the probability that :
(a) the first two are good and the third is bad ;
(b) all three are bad ;
(c) all the three are good.
Out of \(24\) apples, \(6\) are bad, so \(18\) are good. Because the drawing is with replacement, each draw is independent with \[P(\text{good}) = \frac{18}{24} = \frac{3}{4},\qquad P(\text{bad}) = \frac{6}{24} = \frac{1}{4}.\]
(a) First two good, third bad: \[\frac{3}{4}\times\frac{3}{4}\times\frac{1}{4} = \mathbf{\frac{9}{64}}.\]
(b) All three bad: \[\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4} = \mathbf{\frac{1}{64}}.\]
(c) All three good: \[\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4} = \mathbf{\frac{27}{64}}.\]
Answer Details
Out of \(24\) apples, \(6\) are bad, so \(18\) are good. Because the drawing is with replacement, each draw is independent with \[P(\text{good}) = \frac{18}{24} = \frac{3}{4},\qquad P(\text{bad}) = \frac{6}{24} = \frac{1}{4}.\]
(a) First two good, third bad: \[\frac{3}{4}\times\frac{3}{4}\times\frac{1}{4} = \mathbf{\frac{9}{64}}.\]
(b) All three bad: \[\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4} = \mathbf{\frac{1}{64}}.\]
(c) All three good: \[\frac{3}{4}\times\frac{3}{4}\times\frac{3}{4} = \mathbf{\frac{27}{64}}.\]
Question 77 Report
(a) Simplify : \(\frac{3\frac{1}{12} + \frac{7}{8}}{2\frac{1}{4} - \frac{1}{6}}\)
(b) If \(p = \frac{m}{2} - \frac{n^{2}}{5m}\) ;
(i) make n the subject of the relation ; (ii) find, correct to three significant figures, the value of n when p = 14 and m = -8.
(a) Numerator: \(3\tfrac{1}{12} + \tfrac{7}{8} = \tfrac{37}{12} + \tfrac{7}{8} = \tfrac{74}{24} + \tfrac{21}{24} = \tfrac{95}{24}.\)
Denominator: \(2\tfrac{1}{4} - \tfrac{1}{6} = \tfrac{9}{4} - \tfrac{1}{6} = \tfrac{27}{12} - \tfrac{2}{12} = \tfrac{25}{12}.\)
\[\frac{95/24}{25/12} = \frac{95}{24}\times\frac{12}{25} = \frac{95}{2\times 25} = \frac{95}{50} = \frac{19}{10} = \mathbf{1\tfrac{9}{10}}.\]
(b)(i) \(p = \dfrac{m}{2} - \dfrac{n^2}{5m}.\) Multiply through by \(5m\): \[5mp = \frac{5m^2}{2} - n^2 \Rightarrow n^2 = \frac{5m^2}{2} - 5mp \Rightarrow \mathbf{n = \sqrt{\frac{5m^2}{2} - 5mp}}.\]
(b)(ii) With \(p = 14,\ m = -8\): \[n^2 = \frac{5(-8)^2}{2} - 5(-8)(14) = \frac{320}{2} + 560 = 160 + 560 = 720.\] \[n = \sqrt{720} = \mathbf{26.8}\ (\text{to 3 s.f.}).\]
Answer Details
(a) Numerator: \(3\tfrac{1}{12} + \tfrac{7}{8} = \tfrac{37}{12} + \tfrac{7}{8} = \tfrac{74}{24} + \tfrac{21}{24} = \tfrac{95}{24}.\)
Denominator: \(2\tfrac{1}{4} - \tfrac{1}{6} = \tfrac{9}{4} - \tfrac{1}{6} = \tfrac{27}{12} - \tfrac{2}{12} = \tfrac{25}{12}.\)
\[\frac{95/24}{25/12} = \frac{95}{24}\times\frac{12}{25} = \frac{95}{2\times 25} = \frac{95}{50} = \frac{19}{10} = \mathbf{1\tfrac{9}{10}}.\]
(b)(i) \(p = \dfrac{m}{2} - \dfrac{n^2}{5m}.\) Multiply through by \(5m\): \[5mp = \frac{5m^2}{2} - n^2 \Rightarrow n^2 = \frac{5m^2}{2} - 5mp \Rightarrow \mathbf{n = \sqrt{\frac{5m^2}{2} - 5mp}}.\]
(b)(ii) With \(p = 14,\ m = -8\): \[n^2 = \frac{5(-8)^2}{2} - 5(-8)(14) = \frac{320}{2} + 560 = 160 + 560 = 720.\] \[n = \sqrt{720} = \mathbf{26.8}\ (\text{to 3 s.f.}).\]
Question 78 Report
(a) The 3rd and 8th terms of an arithmetic progression (A.P) are -9 and 26 respectively. Find the : (i) common difference ; (ii) first term.
(b) 
In the diagram \(\overline{PQ} || \overline{YZ}\), |XP| = 2cm, |PY| = 3 cm, |PQ| = 6 cm and the area of \(\Delta\) XPQ = 24\(cm^{2}\).Calculate the area of the trapezium PQZY.
(a) Arithmetic progression.
Let the first term be \(a\) and the common difference be \(d\). The \(n\)th term is \(T_n = a + (n-1)d\).
Given \(T_3 = -9\) and \(T_8 = 26\):
\[a + 2d = -9 \quad\text{...(1)}\]\[a + 7d = 26 \quad\text{...(2)}\](i) Common difference. Subtract (1) from (2):
\[5d = 35 \quad\Rightarrow\quad d = 7\](ii) First term. Substitute \(d = 7\) into (1):
\[a + 2(7) = -9 \quad\Rightarrow\quad a + 14 = -9 \quad\Rightarrow\quad a = -23\]Common difference \(= 7\); first term \(= -23\).
(b) Area of the trapezium PQZY.
From the diagram, \(PQ \parallel YZ\), with \(|XP| = 2\text{ cm}\), \(|PY| = 3\text{ cm}\), \(|PQ| = 6\text{ cm}\) and area of \(\triangle XPQ = 24\text{ cm}^2\). P lies on XY and Q lies on XZ.
Since \(PQ \parallel YZ\), triangle \(XPQ\) is similar to triangle \(XYZ\) (equal corresponding angles). The ratio of corresponding sides along XY is:
\[\frac{XP}{XY} = \frac{XP}{XP + PY} = \frac{2}{2 + 3} = \frac{2}{5}\]For similar triangles the ratio of areas is the square of the ratio of sides:
\[\frac{\text{Area }\triangle XPQ}{\text{Area }\triangle XYZ} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}\]Therefore:
\[\text{Area }\triangle XYZ = 24 \times \frac{25}{4} = 150 \text{ cm}^2\]The trapezium PQZY is what remains when \(\triangle XPQ\) is removed from \(\triangle XYZ\):
\[\text{Area of trapezium PQZY} = 150 - 24 = 126 \text{ cm}^2\]Area of trapezium PQZY \(= 126\text{ cm}^2\).
Answer Details
(a) Arithmetic progression.
Let the first term be \(a\) and the common difference be \(d\). The \(n\)th term is \(T_n = a + (n-1)d\).
Given \(T_3 = -9\) and \(T_8 = 26\):
\[a + 2d = -9 \quad\text{...(1)}\]\[a + 7d = 26 \quad\text{...(2)}\](i) Common difference. Subtract (1) from (2):
\[5d = 35 \quad\Rightarrow\quad d = 7\](ii) First term. Substitute \(d = 7\) into (1):
\[a + 2(7) = -9 \quad\Rightarrow\quad a + 14 = -9 \quad\Rightarrow\quad a = -23\]Common difference \(= 7\); first term \(= -23\).
(b) Area of the trapezium PQZY.
From the diagram, \(PQ \parallel YZ\), with \(|XP| = 2\text{ cm}\), \(|PY| = 3\text{ cm}\), \(|PQ| = 6\text{ cm}\) and area of \(\triangle XPQ = 24\text{ cm}^2\). P lies on XY and Q lies on XZ.
Since \(PQ \parallel YZ\), triangle \(XPQ\) is similar to triangle \(XYZ\) (equal corresponding angles). The ratio of corresponding sides along XY is:
\[\frac{XP}{XY} = \frac{XP}{XP + PY} = \frac{2}{2 + 3} = \frac{2}{5}\]For similar triangles the ratio of areas is the square of the ratio of sides:
\[\frac{\text{Area }\triangle XPQ}{\text{Area }\triangle XYZ} = \left(\frac{2}{5}\right)^2 = \frac{4}{25}\]Therefore:
\[\text{Area }\triangle XYZ = 24 \times \frac{25}{4} = 150 \text{ cm}^2\]The trapezium PQZY is what remains when \(\triangle XPQ\) is removed from \(\triangle XYZ\):
\[\text{Area of trapezium PQZY} = 150 - 24 = 126 \text{ cm}^2\]Area of trapezium PQZY \(= 126\text{ cm}^2\).
Question 79 Report
(a) Using a ruler and a pair of compasses only, construc :
(i) a triangle PQR such that /PQ/ = 10 cm, /QR/ = 7 cm and < PQR = 90° ; (ii) the locus \(l_{1}\) of points equidistant from Q and R ; (iii) the locus \(l_{2}\) of points equidistant from P and Q.
(b) Locate the point O equidistant from P, Q and R.
(c) With O as centre, draw the circumcircle of the triangle PQR.
(d) Measure the radius of the circumcircle.
Construction and measurements
- Draw the line segment \(PQ=10\text{ cm}\).
- At \(Q\), construct a perpendicular to \(PQ\). On this perpendicular, mark \(R\) such that \(QR=7\text{ cm}\). Join \(P\) to \(R\) to obtain \(\triangle PQR\), where \(\angle PQR=90^\circ\).
- Construct the perpendicular bisector of \(QR\). This is the locus \(l_1\) of points equidistant from \(Q\) and \(R\).
- Construct the perpendicular bisector of \(PQ\). This is the locus \(l_2\) of points equidistant from \(P\) and \(Q\).
- The two loci meet at \(O\). Hence \(O\) is equidistant from \(P\), \(Q\), and \(R\).
- With centre \(O\) and radius \(OP\), draw the circle through \(P\), \(Q\), and \(R\).
Since \(\triangle PQR\) is right-angled at \(Q\), \(PR\) is the diameter of the circumcircle.
\[PR=\sqrt{PQ^2+QR^2}=\sqrt{10^2+7^2}=\sqrt{149}\text{ cm}\]
Therefore, the radius is
\[OP=\frac{PR}{2}=\frac{\sqrt{149}}{2}=6.10\text{ cm}\approx \boxed{6.1\text{ cm}}.\]
Answer Details
Construction and measurements
- Draw the line segment \(PQ=10\text{ cm}\).
- At \(Q\), construct a perpendicular to \(PQ\). On this perpendicular, mark \(R\) such that \(QR=7\text{ cm}\). Join \(P\) to \(R\) to obtain \(\triangle PQR\), where \(\angle PQR=90^\circ\).
- Construct the perpendicular bisector of \(QR\). This is the locus \(l_1\) of points equidistant from \(Q\) and \(R\).
- Construct the perpendicular bisector of \(PQ\). This is the locus \(l_2\) of points equidistant from \(P\) and \(Q\).
- The two loci meet at \(O\). Hence \(O\) is equidistant from \(P\), \(Q\), and \(R\).
- With centre \(O\) and radius \(OP\), draw the circle through \(P\), \(Q\), and \(R\).
Since \(\triangle PQR\) is right-angled at \(Q\), \(PR\) is the diameter of the circumcircle.
\[PR=\sqrt{PQ^2+QR^2}=\sqrt{10^2+7^2}=\sqrt{149}\text{ cm}\]
Therefore, the radius is
\[OP=\frac{PR}{2}=\frac{\sqrt{149}}{2}=6.10\text{ cm}\approx \boxed{6.1\text{ cm}}.\]
Question 80 Report
The diagram shows the cross- section of a railway tunnel. If |AB| = 100m and the radius of the arc is 56m, calculate, correct to the nearest metre, the perimetre of the cross- section.
Perimeter of the tunnel cross-section.
The cross-section is made up of the straight base \(AB=100\text{ m}\) together with the arc that domes over the top. The arc has radius \(r=56\text{ m}\), and the diagram shows the arc is the major arc (it rises above and beyond a semicircle).
Step 1: Angle subtended by the chord at the centre.
Let \(O\) be the centre and let the perpendicular from \(O\) bisect the chord \(AB\). Half the chord is \(50\text{ m}\). If \(\theta\) is half the central angle of the minor arc:
\[\sin\theta=\frac{50}{56}=0.8929\Rightarrow\theta=63.26^{\circ}\]
So the minor-arc central angle is:
\[2\theta=126.52^{\circ}\]
Step 2: Reflex angle for the major (tunnel) arc.
\[\text{Reflex angle}=360^{\circ}-126.52^{\circ}=233.48^{\circ}\]
Step 3: Length of the major arc.
\[\text{Arc}=\frac{233.48}{360}\times 2\pi r=\frac{233.48}{360}\times 2\times\frac{22}{7}\times 56\]
\[=\frac{233.48}{360}\times 352=0.6486\times 352=228.3\text{ m}\]
Step 4: Perimeter.
\[\text{Perimeter}=AB+\text{arc}=100+228.3=328.3\text{ m}\]
Correct to the nearest metre, the perimeter of the cross-section is \(328\text{ m}\).
Answer Details
Perimeter of the tunnel cross-section.
The cross-section is made up of the straight base \(AB=100\text{ m}\) together with the arc that domes over the top. The arc has radius \(r=56\text{ m}\), and the diagram shows the arc is the major arc (it rises above and beyond a semicircle).
Step 1: Angle subtended by the chord at the centre.
Let \(O\) be the centre and let the perpendicular from \(O\) bisect the chord \(AB\). Half the chord is \(50\text{ m}\). If \(\theta\) is half the central angle of the minor arc:
\[\sin\theta=\frac{50}{56}=0.8929\Rightarrow\theta=63.26^{\circ}\]
So the minor-arc central angle is:
\[2\theta=126.52^{\circ}\]
Step 2: Reflex angle for the major (tunnel) arc.
\[\text{Reflex angle}=360^{\circ}-126.52^{\circ}=233.48^{\circ}\]
Step 3: Length of the major arc.
\[\text{Arc}=\frac{233.48}{360}\times 2\pi r=\frac{233.48}{360}\times 2\times\frac{22}{7}\times 56\]
\[=\frac{233.48}{360}\times 352=0.6486\times 352=228.3\text{ m}\]
Step 4: Perimeter.
\[\text{Perimeter}=AB+\text{arc}=100+228.3=328.3\text{ m}\]
Correct to the nearest metre, the perimeter of the cross-section is \(328\text{ m}\).
