Complete Solutions for Mathematics Examination Paper 2
SECTION A [40 marks]
Question 1
Given that \( \mu = \{ x : 1 < x < 20,\; x \in \mathbb{Z} \} \),
\( P = \{ x : x \text{ is a multiple of } 3 \} \),
\( Q = \{ x : x \text{ is a prime number} \} \),
where \( P \) and \( Q \) are subsets of \( \mu \). Find:
- \( P' \cap Q' \)
- \( P' \cup Q \)
- \( (P \cup Q)' \)
Solution to Question 1
First, we list the elements of the universal set and the subsets P and Q.
Universal set \( \mu = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19\} \)
Set P (multiples of 3): \( P = \{3, 6, 9, 12, 15, 18\} \)
Set Q (prime numbers): \( Q = \{2, 3, 5, 7, 11, 13, 17, 19\} \)
-
Find \( P' \cap Q' \)
Using De Morgan's Law, we know that \( P' \cap Q' = (P \cup Q)' \).
First, find the union of P and Q: \[ P \cup Q = \{2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19\} \] Now, find the complement, which are the elements in \( \mu \) but not in \( P \cup Q \).\( P' \cap Q' = (P \cup Q)' = \{4, 8, 10, 14, 16\} \)
-
Find \( P' \cup Q \)
First, find \( P' \) (elements in \( \mu \) but not in P): \[ P' = \{2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19\} \] Now, find the union of \( P' \) and \( Q \): \[ P' \cup Q = \{2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19\} \cup \{2, 3, 5, 7, 11, 13, 17, 19\} \]\( P' \cup Q = \{2, 3, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19\} \)
-
Find \( (P \cup Q)' \)
This is identical to part (a).\( (P \cup Q)' = \{4, 8, 10, 14, 16\} \)
Question 2
The product of the ages of Adu and Tanko is 9 less than Akorfu's age. If Tanko is 4 years older than Adu and Akorfu's age is six times Tanko's age, find Akorfu's age.
Solution to Question 2
Let Adu's age be \( A \), Tanko's age be \( T \), and Akorfu's age be \( K \).
From the problem statement, we can form three equations:
- \( A \times T = K - 9 \)
- \( T = A + 4 \implies A = T - 4 \)
- \( K = 6T \)
Substitute equation (3) into equation (1):
\[ A \times T = (6T) - 9 \]Now substitute \( A = T - 4 \) into this new equation:
\[ (T - 4)T = 6T - 9 \] \[ T^2 - 4T = 6T - 9 \]Rearrange into a standard quadratic equation:
\[ T^2 - 10T + 9 = 0 \]Factor the quadratic equation:
\[ (T - 9)(T - 1) = 0 \]The possible values for Tanko's age are \( T = 9 \) or \( T = 1 \).
If \( T = 1 \), then Adu's age \( A = 1 - 4 = -3 \). Since age cannot be negative, this solution is not valid.
If \( T = 9 \), then Adu's age \( A = 9 - 4 = 5 \). This is a valid solution.
Using Tanko's age \( T = 9 \), we find Akorfu's age from equation (3):
\[ K = 6T = 6 \times 9 = 54 \]Akorfu's age is 54 years.
Question 3
A company installs solar panels and the monthly savings on electricity (\$) is modelled by: \( S = 200 + 50x - 2x^2 \), where \( x \) is the number of months after installation.
- At what time will the savings stop increasing?
- Find the maximum savings.
Solution to Question 3
The savings model \( S = -2x^2 + 50x + 200 \) is a quadratic function. Since the coefficient of \(x^2\) is negative (-2), the graph is a downward-opening parabola. The maximum value occurs at the vertex.
-
Time when savings stop increasing
This corresponds to the x-coordinate of the vertex. The formula for the x-coordinate of the vertex of a parabola \(ax^2+bx+c\) is \( x = -\frac{b}{2a} \).In this model, \( a = -2 \) and \( b = 50 \).
\[ x = -\frac{50}{2(-2)} = -\frac{50}{-4} = 12.5 \]The savings will stop increasing after 12.5 months.
-
Maximum savings
To find the maximum savings, substitute the value of \( x = 12.5 \) back into the savings equation: \[ S_{max} = 200 + 50(12.5) - 2(12.5)^2 \] \[ S_{max} = 200 + 625 - 2(156.25) \] \[ S_{max} = 825 - 312.5 = 512.5 \]The maximum possible savings are $512.50.
Question 5
The scores obtained by 9 applicants in ascending order:
\[
(3x + 2),\; 22,\; (4x - 2),\; 23,\; 25,\; (5x - 4),\; 29,\; 29,\; (x^2 - 7)
\]
- Given that the range is 9, find:
- Value of \( x \)
- Mean mark of the applicants
- If the four highest scores were selected, determine the pass mark.
Solution to Question 5
-
Given the range is 9:
-
Value of \( x \)
The range is the difference between the highest and lowest scores. \[ \text{Highest Score} - \text{Lowest Score} = \text{Range} \] \[ (x^2 - 7) - (3x + 2) = 9 \] \[ x^2 - 3x - 9 = 9 \] \[ x^2 - 3x - 18 = 0 \] We factor the quadratic equation: \[ (x - 6)(x + 3) = 0 \] The possible values for \(x\) are 6 and -3. We must check which value preserves the ascending order of the scores.- Test \( x = -3 \): The first score is \(3(-3) + 2 = -7\). The third score is \(4(-3) - 2 = -14\). Since \(-14 < -7\), this violates the ascending order. So, \(x \neq -3\).
- Test \( x = 6 \): The scores become:
\(3(6)+2 = 20\)
\(4(6)-2 = 22\)
\(5(6)-4 = 26\)
\(6^2-7 = 29\)
The list is: 20, 22, 22, 23, 25, 26, 29, 29, 29. This is a valid ascending order.
The value of \( x \) is 6.
-
Mean mark of the applicants
The final list of scores is: 20, 22, 22, 23, 25, 26, 29, 29, 29. \[ \text{Sum of scores} = 20+22+22+23+25+26+29+29+29 = 225 \] \[ \text{Number of applicants} = 9 \] \[ \text{Mean} = \frac{\text{Sum of scores}}{\text{Number of applicants}} = \frac{225}{9} = 25 \]The mean mark is 25.
-
Value of \( x \)
-
Determine the pass mark.
If the four highest scores were selected, we look at the ordered list: 20, 22, 22, 23, 25, 26, 29, 29, 29. The four highest scores are 26, 29, 29, and 29. The minimum score required to be in this group is 26.The pass mark is 26.
SECTION B [60 marks]
Question 6
Electricity charges: first 30 units at \$1/unit; next 30 units at \$7/unit; each additional unit at \$5.
- If Amaka used 420 units in January, calculate the amount paid.
- If Amaka paid \$2,740 in February, find the number of units consumed.
- Find, correct to two decimal places, the percentage change in units consumed between January and February.
Solution to Question 6
-
Calculate the amount for 420 units.
We break down the cost by tiers:- Cost of first 30 units = \( 30 \times \$1 = \$30 \)
- Cost of next 30 units = \( 30 \times \$7 = \$210 \)
- Units remaining = \( 420 - (30+30) = 360 \) units
- Cost of additional units = \( 360 \times \$5 = \$1800 \)
The amount paid in January was $2,040.
-
Find units consumed for $2,740.
First, subtract the cost of the initial tiers:- Cost of first 60 units = \( \$30 + \$210 = \$240 \).
- Amount paid for additional units = \( \$2,740 - \$240 = \$2,500 \).
- Number of additional units = \( \frac{\$2500}{\$5/\text{unit}} = 500 \) units.
Amaka consumed 560 units in February.
-
Percentage change in units consumed.
- January units (Old) = 420
- February units (New) = 560
- Change in units = \( 560 - 420 = 140 \) (This is an increase).
The percentage change was a 33.33% increase.
Question 7
Yaro drove from Gaja to Banga. After 2 hours, he had covered 80 km. At that speed he’d be 15 min late. By increasing speed by 10 km/h, he would arrive 36 min early. Find the distance from Gaja to Banga.
Solution to Question 7
Let \(D\) be the total distance (km), \(S\) be the initial speed (km/h), and \(T_{sch}\) be the scheduled time (hours).
Step 1: Find the initial speed.
Yaro covered 80 km in 2 hours, so his initial speed is: \[ S = \frac{\text{Distance}}{\text{Time}} = \frac{80 \text{ km}}{2 \text{ hr}} = 40 \text{ km/h} \] Generated codeStep 2: Set up equations based on the two scenarios.
Scenario 1: Continuing at 40 km/h.
The total time taken would be \( T_1 = \frac{D}{40} \). He would be 15 mins (0.25 hours) late. \[ \frac{D}{40} = T_{sch} + 0.25 \quad \text{(Equation 1)} \]Scenario 2: Increasing speed after the first 2 hours.
The new speed is \( S_{new} = 40 + 10 = 50 \) km/h. The time for the first 80 km is 2 hours. The distance remaining is \( D - 80 \) km. The time for the rest of the journey is \( \frac{D-80}{50} \) hours. The total time taken is \( T_2 = 2 + \frac{D-80}{50} \). He would be 36 mins (0.6 hours) early. \[ 2 + \frac{D-80}{50} = T_{sch} - 0.6 \quad \text{(Equation 2)} \]Step 3: Solve the system of equations.
From Equation 1, isolate \(T_{sch}\): \( T_{sch} = \frac{D}{40} - 0.25 \). Substitute this into Equation 2: \[ 2 + \frac{D-80}{50} = \left(\frac{D}{40} - 0.25\right) - 0.6 \] \[ 2 + \frac{D-80}{50} = \frac{D}{40} - 0.85 \] To eliminate fractions, multiply the entire equation by 200 (the LCM of 50 and 40): \[ 200(2) + 200\left(\frac{D-80}{50}\right) = 200\left(\frac{D}{40}\right) - 200(0.85) \] \[ 400 + 4(D - 80) = 5D - 170 \] \[ 400 + 4D - 320 = 5D - 170 \] \[ 80 + 4D = 5D - 170 \] \[ 5D - 4D = 80 + 170 \] \[ D = 250 \]The distance from Gaja to Banga is 250 km.
Question 9
- Mrs. Otoo spends \(\frac13\) of her salary on rent, \(\frac14\) on food, and \(\frac15\) on clothes, with \$195 left. Find her monthly salary.
- A sector of a circle has a radius of 6 cm and an angle of 105°. Calculate its:
- Perimeter
- Area
Solution to Question 9
-
Mrs. Otoo's monthly salary
Let the total monthly salary be \(S\).First, find the total fraction of the salary spent:
\[ \text{Fraction spent} = \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \] The least common multiple of 3, 4, and 5 is 60. \[ \text{Fraction spent} = \frac{20}{60} + \frac{15}{60} + \frac{12}{60} = \frac{47}{60} \] The fraction of the salary that is left is: \[ \text{Fraction left} = 1 - \frac{47}{60} = \frac{13}{60} \] We are told this remaining fraction equals $195. \[ \frac{13}{60} \times S = 195 \] To find the total salary \(S\), we solve for S: \[ S = \frac{195 \times 60}{13} = 15 \times 60 = 900 \]Mrs. Otoo's monthly salary is $900.
-
Sector of a circle (Given: \(r=6\) cm, \(\theta=105^\circ\), \(\pi = \frac{22}{7}\))
-
Perimeter of the sector
The perimeter of a sector is the sum of the arc length and two radii (\(2r\)).Arc length = \( \left(\frac{\theta}{360}\right) \times 2\pi r \)
\[ \text{Arc length} = \left(\frac{105}{360}\right) \times 2 \times \frac{22}{7} \times 6 = \frac{7}{24} \times \frac{2 \times 22 \times 6}{7} = \frac{264}{24} = 11 \text{ cm} \] Perimeter = Arc length + \(2r\) = \(11 + 2(6) = 11 + 12 = 23\) cm.The perimeter of the sector is 23 cm.
-
Area of the sector
Area = \( \left(\frac{\theta}{360}\right) \times \pi r^2 \) \[ \text{Area} = \left(\frac{105}{360}\right) \times \frac{22}{7} \times 6^2 = \frac{7}{24} \times \frac{22}{7} \times 36 = \frac{22 \times 36}{24} = 33 \text{ cm}^2 \]The area of the sector is 33 cm².
-
Perimeter of the sector
Question 10
The cost \( C \) of feeding students is partly constant and partly varies as number of students \( n \).
For \( n=8 \), \( C=\$70 \); for \( n=10 \), \( C=\$90 \).
- Find expression for \( C \) in terms of \( n \)
- Cost of feeding 12 students
Solution to Question 10
The relationship is a partial variation, which can be expressed as a linear equation: \[ C = k + an \] where \(k\) is the constant cost and \(a\) is the variation constant (cost per student).
We are given two data points to form a system of simultaneous equations:
- When \(n=8\), \(C=70\): \( 70 = k + 8a \) (Equation 1)
- When \(n=10\), \(C=90\): \( 90 = k + 10a \) (Equation 2)
-
Find the expression for \(C\) in terms of \(n\).
Subtract Equation 1 from Equation 2 to eliminate \(k\): \[ (90 - 70) = (k + 10a) - (k + 8a) \] \[ 20 = 2a \implies a = 10 \] Substitute \(a=10\) back into Equation 1 to find \(k\): \[ 70 = k + 8(10) \] \[ 70 = k + 80 \implies k = -10 \] Now, substitute the values of \(a\) and \(k\) back into the general formula:The expression is \( C = 10n - 10 \).
-
Cost of feeding 12 students.
Use the expression found in part (a) with \(n=12\): \[ C = 10(12) - 10 \] \[ C = 120 - 10 = 110 \]The cost of feeding 12 students is $110.
Question 11
- Given: \[ P = \begin{pmatrix} 2 & -9 \\ 4 & 1 \end{pmatrix}, \quad Q = \begin{pmatrix} 1 & -1 \\ 3 & -2 \end{pmatrix} \] Find \( PQ + 2Q \)
- A bag contains 8 red balls and some white balls. If the probability of drawing a white ball is half of the probability of drawing a red ball, find the number of white balls.
Solution to Question 11
-
Find \( PQ + 2Q \).
Step 1: Calculate the matrix product \(PQ\). \[ PQ = \begin{pmatrix} 2 & -9 \\ 4 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 3 & -2 \end{pmatrix} = \begin{pmatrix} (2)(1)+(-9)(3) & (2)(-1)+(-9)(-2) \\ (4)(1)+(1)(3) & (4)(-1)+(1)(-2) \end{pmatrix} \] \[ PQ = \begin{pmatrix} 2-27 & -2+18 \\ 4+3 & -4-2 \end{pmatrix} = \begin{pmatrix} -25 & 16 \\ 7 & -6 \end{pmatrix} \] Step 2: Calculate the scalar product \(2Q\). \[ 2Q = 2 \begin{pmatrix} 1 & -1 \\ 3 & -2 \end{pmatrix} = \begin{pmatrix} 2(1) & 2(-1) \\ 2(3) & 2(-2) \end{pmatrix} = \begin{pmatrix} 2 & -2 \\ 6 & -4 \end{pmatrix} \] Step 3: Add the two resulting matrices. \[ PQ + 2Q = \begin{pmatrix} -25 & 16 \\ 7 & -6 \end{pmatrix} + \begin{pmatrix} 2 & -2 \\ 6 & -4 \end{pmatrix} = \begin{pmatrix} -25+2 & 16+(-2) \\ 7+6 & -6+(-4) \end{pmatrix} \]\( PQ + 2Q = \begin{pmatrix} -23 & 14 \\ 13 & -10 \end{pmatrix} \)
-
Find the number of white balls.
Let \(w\) be the number of white balls.
Number of red balls = 8.
Total number of balls = \( 8 + w \).The probability of drawing a white ball is \( P(W) = \frac{w}{8+w} \).
The probability of drawing a red ball is \( P(R) = \frac{8}{8+w} \).
We are given the condition: \( P(W) = \frac{1}{2} P(R) \).
\[ \frac{w}{8+w} = \frac{1}{2} \left( \frac{8}{8+w} \right) \] \[ \frac{w}{8+w} = \frac{4}{8+w} \] Since the denominators are equal and non-zero, the numerators must be equal. \[ w = 4 \]There are 4 white balls in the bag.
Question 12
- The 8th term of an A.P. is 46; the sum of the first 8 terms is 200. Find the:
- First term
- Sum of the first 12 terms
- Points \( X(70^\circ S, 60^\circ E) \) and \( Y(7^\circ S, 60^\circ E) \).
- Illustrate in a diagram
- Calculate the distance between X and Y along the meridian. [Take \(\pi=\frac{22}{7}\); \(R=6400\, \text{km}\)]
Solution to Question 12
-
Arithmetic Progression (A.P.)
Let the first term be \(a\) and the common difference be \(d\).
The formula for the nth term is \(T_n = a + (n-1)d\).
The formula for the sum of n terms is \(S_n = \frac{n}{2}(2a + (n-1)d)\).From the given information:
8th term is 46: \( T_8 = a + (8-1)d \implies a + 7d = 46 \) (Equation 1)
Sum of first 8 terms is 200: \( S_8 = \frac{8}{2}(2a + (8-1)d) = 200 \implies 4(2a + 7d) = 200 \)
\[ 2a + 7d = 50 \quad \text{(Equation 2)} \]-
Find the first term (\(a\))
Subtract Equation 1 from Equation 2: \[ (2a + 7d) - (a + 7d) = 50 - 46 \] \[ a = 4 \]The first term is 4.
-
Find the sum of the first 12 terms (\(S_{12}\))
First, find the common difference \(d\) by substituting \(a=4\) into Equation 1: \[ 4 + 7d = 46 \implies 7d = 42 \implies d = 6 \] Now, use the sum formula for \(n=12\): \[ S_{12} = \frac{12}{2}(2a + (12-1)d) = 6(2a + 11d) \] \[ S_{12} = 6(2(4) + 11(6)) = 6(8 + 66) = 6(74) = 444 \]The sum of the first 12 terms is 444.
-
Find the first term (\(a\))
Question 13
The marks of 20 students in a test are as follows:
15, 11, 17, 25, 13, 15, 16, 22, 24, 27, 20, 22, 15, 16, 15, 19, 22, 24, 22, 11
- Prepare a frequency table using class intervals: 10–12, 13–15, 16–18, etc.
- Calculate the variance of the distribution.
- If the pass mark is 16, find the probability that a student selected at random failed the test.
Solution to Question 13
-
Frequency Table
Class Interval Tally Frequency (f) 10 – 12 || 2 13 – 15 ||||| 5 16 – 18 ||| 3 19 – 21 || 2 22 – 24 ||||| | 6 25 – 27 || 2 Total 20 -
Calculate the variance.
To calculate the variance from grouped data, we use the formula \( \sigma^2 = \frac{\sum fx^2}{\sum f} - \left( \frac{\sum fx}{\sum f} \right)^2 \). We expand our table to include the midpoint (x), fx, x², and fx².
First, calculate the mean \( (\bar{x}) = \frac{\sum fx}{\sum f} = \frac{373}{20} = 18.65 \).Interval f Midpoint (x) fx x² fx² 10–12 2 11 22 121 242 13–15 5 14 70 196 980 16–18 3 17 51 289 867 19–21 2 20 40 400 800 22–24 6 23 138 529 3174 25–27 2 26 52 676 1354 Total ∑f=20 ∑fx=373 ∑fx²=7417
Now calculate the variance: \[ \text{Variance } (\sigma^2) = \frac{\sum fx^2}{\sum f} - (\bar{x})^2 = \frac{7417}{20} - (18.65)^2 \] \[ \sigma^2 = 370.85 - 347.8225 = 23.0275 \]The variance of the distribution is 23.0275.
-
Probability that a student failed.
The pass mark is 16. A student fails if their score is less than 16. We must use the original raw data for this calculation: 15, 11, 17, 25, 13, 15, 16, 22, 24, 27, 20, 22, 15, 16, 15, 19, 22, 24, 22, 11.The scores less than 16 are: 15, 11, 13, 15, 15, 15, 11.
Number of students who failed = 7.
Total number of students = 20.
The probability of a random student failing is: \[ P(\text{failed}) = \frac{\text{Number of students who failed}}{\text{Total number of students}} = \frac{7}{20} \]The probability that a student failed is \( \frac{7}{20} \) or 0.35.
