Question 1
In 2023, a football club had 50 adult members and 70 child members. The membership fee for an adult was $40, and the membership fee for a child was $15.
(a) (i) Calculate the total of the membership fees received by the club in 2023.
▶️Answer/Explanation
Solution:
1(a)(i) 3050
2 M1 for 50 × 40 + 70 × 15 or better
Answer: $3050
(a) (ii) The cost of running the club in 2023 was $2780. Calculate $2780 as a percentage of the total of the membership fees received by the club.
▶️Answer/Explanation
Solution:
To find the percentage, divide the cost of running the club by the total membership fees and multiply by 100.
Percentage = (Cost of running the club / Total membership fees) × 100
Percentage = (2780 / 3050) × 100 ≈ 91.15%
Answer: 91.1% or 91.14 to 91.15
(a) (iii) In 2023, there were 120 members. This was a decrease by 4% of the number of members in 2022. Calculate the number of members in 2022.
▶️Answer/Explanation
Solution:
In 2023, there were 120 members, which was a decrease of 4% from 2022. Let the number of members in 2022 be \( x \).
Since there was a 4% decrease, the number of members in 2023 is 96% of the number in 2022.
\( 0.96x = 120 \)
\( x = 120 / 0.96 = 125 \)
Answer: 125
(a) (iv) In 2024, the total number of members increased from the 120 members in 2023. The number of adult members and the number of child members each increased by the same number. The ratio of the number of adult members to the number of child members changed to 14 : 19.
(a) (iv) (a) Find the total number of members in 2024.
▶️Answer/Explanation
Solution:
In 2024, the number of adult and child members each increased by the same number, and the ratio of adult to child members became 14:19.
Let the increase in the number of members be \( x \).
Number of adult members in 2024 = 50 + \( x \)
Number of child members in 2024 = 70 + \( x \)
The ratio is given as 14:19, so:
\( \frac{50 + x}{70 + x} = \frac{14}{19} \)
Cross-multiplying:
\( 19(50 + x) = 14(70 + x) \)
\( 950 + 19x = 980 + 14x \)
\( 5x = 30 \)
\( x = 6 \)
Total number of members in 2024 = (50 + 6) + (70 + 6) = 56 + 76 = 132
Answer: 132
(a) (iv) (b) Calculate the percentage increase in the total number of members from 2023 to 2024.
▶️Answer/Explanation
Solution:
The total number of members in 2023 was 120, and in 2024, it is 132.
Percentage increase = \(\frac{\text{Increase in number of members}}{\text{Original number of members}} \times 100\)
Percentage increase = \(\frac{132 – 120}{120} \times 100 = 10\%\)
Answer: 10%
(b) The population of a village is 2500. The population is decreasing exponentially at a rate of 3% per year.
(b) (i) Calculate the population at the end of 3 years.
Subtopic Code: E1.17 (Exponential Growth and Decay)
▶️Answer/Explanation
Solution:
The population decreases exponentially at a rate of 3% per year. The formula for exponential decay is:
\( P = P_0 \times (1 – r)^t \)
Where \( P_0 = 2500 \), \( r = 0.03 \), and \( t = 3 \).
\( P = 2500 \times (1 – 0.03)^3 = 2500 \times (0.97)^3 \)
\( P = 2500 \times 0.912673 = 2281.68 \)
Rounding to the nearest whole number, the population is 2282.
Answer: 2280 or 2281 to 2282
(b) (ii) Find the number of complete years it takes for the population to first fall below 2000.
▶️Answer/Explanation
Solution:
We need to find the smallest integer \( t \) such that:
\( 2500 \times (0.97)^t < 2000 \)
Divide both sides by 2500:
\( (0.97)^t < 0.8 \)
Take the natural logarithm of both sides:
\( t \ln(0.97) < \ln(0.8) \)
Since \( \ln(0.97) \) is negative, the inequality sign reverses:
\( t > \frac{\ln(0.8)}{\ln(0.97)} \)
Calculate the logarithms:
\( t > \frac{-0.2231}{-0.0305} \approx 7.31 \)
Since \( t \) must be an integer, the smallest value is 8 years.
Answer: 8 years
Question 2
(a) The nth term of a sequence is \( 120 – n^3 \).
(a) (i) Find the 4th term of this sequence.
Subtopic Code: E2.7 (Sequences)
▶️Answer/Explanation
Solution:
The nth term of the sequence is given by \( 120 – n^3 \). To find the 4th term, substitute \( n = 4 \):
4th term = \( 120 – 4^3 = 120 – 64 = 56 \)
Answer: 56
(a) (ii) Find the value of \( n \) when the nth term is \(-1211\).
▶️Answer/Explanation
Solution:
Given the nth term \( 120 – n^3 = -1211 \), solve for \( n \):
\( 120 – n^3 = -1211 \)
\( -n^3 = -1331 \)
\( n^3 = 1331 \)
\( n = \sqrt[3]{1331} = 11 \)
Answer: 11
(b) The nth term of a different sequence is \( 3 \times (0.2)^{n-1} \). Find the 5th term of this sequence.
▶️Answer/Explanation
Solution:
The nth term of the sequence is given by \( 3 \times (0.2)^{n-1} \). To find the 5th term, substitute \( n = 5 \):
5th term = \( 3 \times (0.2)^{5-1} = 3 \times (0.2)^4 = 3 \times 0.0016 = 0.0048 \)
Answer: 0.0048 or \( \frac{3}{625} \)
(c) The table shows the first four terms of sequences \( A \), \( B \), and \( C \).
| Sequence | 1st term | 2nd term | 3rd term | 4th term | 5th term | nth term |
|---|---|---|---|---|---|---|
| \( A \) | 7 | 4 | 1 | -2 | ||
| \( B \) | \(\frac{1}{4}\) | \(\frac{2}{5}\) | \(\frac{3}{6}\) | \(\frac{4}{7}\) | ||
| \( C \) | 0 | 2 | 6 | 12 |
Complete the table for each sequence.
▶️Answer/Explanation
Solution:
Sequence A:
The sequence decreases by 3 each time. The nth term is \( 10 – 3n \).
5th term = \( 10 – 3 \times 5 = -5 \)
nth term = \( 10 – 3n \)
Sequence B:
The numerator increases by 1, and the denominator increases by 1. The nth term is \( \frac{n}{n+3} \).
5th term = \( \frac{5}{8} \)
nth term = \( \frac{n}{n+3} \)
Sequence C:
The sequence follows the pattern \( n^2 – n \).
5th term = \( 5^2 – 5 = 25 – 5 = 20 \)
nth term = \( n^2 – n \)
Answer:
Sequence A: 5th term = -5, nth term = \( 10 – 3n \)
Sequence B: 5th term = \( \frac{5}{8} \), nth term = \( \frac{n}{n+3} \)
Sequence C: 5th term = 20, nth term = \( n^2 – n \)
Question 3
(a) Rahul rolls a dice 60 times. The results are shown in the table.
| Score | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | 10 | 6 | 11 | 13 | 14 | 6 |
Find the mode, the median, and the mean.
Subtopic Code: E9.3 (Averages and Measures of Spread)
▶️Answer/Explanation
Solution:
Mode: The score with the highest frequency is 5 (frequency 14).
Median: There are 60 rolls, so the median is the average of the 30th and 31st values. Both are 4.
Mean: Sum of (score × frequency) divided by total number of rolls.
Mean = \( \frac{10 \times 1 + 6 \times 2 + 11 \times 3 + 13 \times 4 + 14 \times 5 + 6 \times 6}{60} = \frac{213}{60} = 3.55 \)
Answer:
Mode: 5
Median: 4
Mean: 3.55
(b) Sangita measures the speed of each of 100 cars. The results are shown in the table.
| Speed (v km/h) | 0 ≤ v < 30 | 30 ≤ v < 50 | 50 ≤ v < 75 |
|---|---|---|---|
| Frequency | 10 | 72 | 18 |
(b) (i) Calculate an estimate of the mean speed.
▶️Answer/Explanation
Solution:
To estimate the mean speed, use the midpoint of each speed interval multiplied by the frequency, then divide by the total number of cars.
Midpoints:
- 0 ≤ v < 30: \( \frac{0 + 30}{2} = 15 \) km/h
- 30 ≤ v < 50: \( \frac{30 + 50}{2} = 40 \) km/h
- 50 ≤ v < 75: \( \frac{50 + 75}{2} = 62.5 \) km/h
Mean speed = \( \frac{15 \times 10 + 40 \times 72 + 62.5 \times 18}{100} = \frac{150 + 2880 + 1125}{100} = \frac{4155}{100} = 41.55 \) km/h
Answer: 41.55 or 41.6 km/h
(b) (ii) Sangita draws a histogram to show the information in the table. The height of the bar that represents 0 ≤ v < 30 is 3 cm. Calculate the height of each of the other two bars on this histogram.
▶️Answer/Explanation
Solution:
The height of a bar in a histogram is proportional to the frequency density, which is calculated as:
Frequency density = \( \frac{\text{Frequency}}{\text{Class width}} \)
For 0 ≤ v < 30:
Class width = 30 – 0 = 30
Frequency density = \( \frac{10}{30} = \frac{1}{3} \)
Height = 3 cm (given)
For 30 ≤ v < 50:
Class width = 50 – 30 = 20
Frequency density = \( \frac{72}{20} = 3.6 \)
Height = \( 3.6 \times \frac{3}{\frac{1}{3}} = 3.6 \times 9 = 32.4 \) cm
For 50 ≤ v < 75:
Class width = 75 – 50 = 25
Frequency density = \( \frac{18}{25} = 0.72 \)
Height = \( 0.72 \times \frac{3}{\frac{1}{3}} = 0.72 \times 9 = 6.48 \) cm
Answer:
Height for 30 ≤ v < 50: 32.4 cm
Height for 50 ≤ v < 75: 6.48 cm
Question 4
In this question, all the measurements are in centimetres.
The diagram shows a triangle and a square. The area of the triangle is equal to the area of the square.
The triangle has sides of length \( r + 5 \), \( r + 2 \), and an included angle of 30°. The square has sides of length \( r + 1 \).
(a) Show that \( 3r^2 + r – 6 = 0 \).
Subtopic Code: E5.2 (Area and Perimeter)
▶️Answer/Explanation
Solution:
The area of the triangle is given by:
Area of triangle = \( \frac{1}{2} \times (r + 5) \times (r + 2) \times \sin(30°) \)
Since \( \sin(30°) = \frac{1}{2} \), the area becomes:
Area of triangle = \( \frac{1}{2} \times (r + 5) \times (r + 2) \times \frac{1}{2} = \frac{(r + 5)(r + 2)}{4} \)
The area of the square is given by:
Area of square = \( (r + 1)^2 \)
Since the areas are equal:
\( \frac{(r + 5)(r + 2)}{4} = (r + 1)^2 \)
Multiply both sides by 4:
\( (r + 5)(r + 2) = 4(r + 1)^2 \)
Expand both sides:
\( r^2 + 7r + 10 = 4(r^2 + 2r + 1) \)
Simplify the right side:
\( r^2 + 7r + 10 = 4r^2 + 8r + 4 \)
Bring all terms to one side:
\( r^2 + 7r + 10 – 4r^2 – 8r – 4 = 0 \)
Simplify:
\( -3r^2 – r + 6 = 0 \)
Multiply by -1:
\( 3r^2 + r – 6 = 0 \)
Thus, the equation is shown.
Answer: \( 3r^2 + r – 6 = 0 \)
(b) Solve the equation \( 3r^2 + r – 6 = 0 \). Give your answer to 2 decimal places.
▶️Answer/Explanation
Solution:
Use the quadratic formula to solve \( 3r^2 + r – 6 = 0 \):
Quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \)
Here, \( a = 3 \), \( b = 1 \), and \( c = -6 \).
Substitute into the formula:
\( r = \frac{-1 \pm \sqrt{1^2 – 4 \times 3 \times (-6)}}{2 \times 3} \)
Simplify under the square root:
\( r = \frac{-1 \pm \sqrt{1 + 72}}{6} = \frac{-1 \pm \sqrt{73}}{6} \)
Calculate the two solutions:
\( r = \frac{-1 + \sqrt{73}}{6} \approx 1.26 \)
\( r = \frac{-1 – \sqrt{73}}{6} \approx -1.59 \)
Since \( r \) represents a length, it cannot be negative. Thus, \( r \approx 1.26 \).
Answer: \( r = 1.26 \) or \( r = -1.59 \) (rejected)
(c) Find the perimeter of the square.
▶️Answer/Explanation
Solution:
The side length of the square is \( r + 1 \). From part (b), \( r \approx 1.26 \).
Side length = \( 1.26 + 1 = 2.26 \) cm
Perimeter of the square = \( 4 \times \text{side length} = 4 \times 2.26 = 9.04 \) cm
Answer: 9.04 cm
Question 5
Topic Code: E4.1 (Geometry – Geometrical Terms)
The diagram shows the graph of \( y = f(x) \) for values of \( x \) from \(-3\) to \( 3 \).
(a) (i) Use the graph to find \( f(2) \).
(ii) Use the graph to solve the equation \( f(x) = 5 \).
(iii) The equation \( f(x) = k \) has exactly two solutions. Write down the value of \( k \).
(iv) Choose the correct word from the box to complete the statement:
The line \( x = 0 \) is the …… to the graph of \( y = f(x) \).
(b) (i) On the grid, draw the graph of \( y = x – 2 \) for values of \( x \) from \(-3\) to \( 3 \).
(ii) Find \( x \) when \( f(x) = x – 2 \).
(c) \( f(x) = x^2 – \frac{c}{x} , x \neq 0 \). Use the graph to show that \( c = 2 \).
(d) The equation \( f(x) = x – 2 \) can be written as \( x^3 + px^2 + qx = 2 \). Find the value of \( p \) and the value of \( q \).
▶️Answer/Explanation
Answer:
(a) (i) \( f(2) = 2 \)
(ii) \( x = -2, 1, 2 \)
(iii) \( k = 4 \)
(iv) asymptote
(b) (i) Graph of \( y = x – 2 \) drawn correctly.
(ii) \( x = -1 \)
(c) \( c = 2 \)
(d) \( p = -1 \), \( q = -2 \)
Solution:
(a) (i) From the graph, when \( x = 2 \), \( y = 2 \).
(ii) The graph intersects \( y = 5 \) at \( x = -2, 1, 2 \).
(iii) The graph has exactly two solutions when \( k = 4 \).
(iv) The line \( x = 0 \) is the asymptote to the graph.
(b) (i) The graph of \( y = x – 2 \) is a straight line passing through points like \((-3, -5)\) and \((3, 1)\).
(ii) The intersection point of \( f(x) \) and \( y = x – 2 \) is at \( x = -1 \).
(c) From the graph, when \( x = 1 \), \( f(1) = 1 – \frac{c}{1} = -1 \). Solving gives \( c = 2 \).
(d) Rearranging \( f(x) = x – 2 \) gives \( x^3 – x^2 – 2x = 2 \), so \( p = -1 \) and \( q = -2 \).
Question 6
Topic Code: E6.2 (Trigonometry – Right-Angled Triangles)
(a)
The diagram shows a ladder, \( GH \), on horizontal ground, leaning against a vertical wall, \( HF \).
Given \( GF = 1.5 \, \text{m} \) and \( HF = 4 \, \text{m} \), calculate the length of the ladder, \( GH \).
(b) 
\( W \) is 120 m north of \( V \) and 50 m east of \( V \). Calculate the bearing of \( V \) from \( W \).
(c) 
In the quadrilateral \( ABCD \), \( AD = DC = 5 \, \text{cm} \) and \( AB = BC \). Angle \( ABD = 25^\circ \) and angle \( BAD = 15^\circ \). Calculate the perimeter of the quadrilateral \( ABCD \).
(d) 
\( PQRS \) is a quadrilateral. Calculate angle \( PQR \).
▶️Answer/Explanation
Answer:
(a)
\( GH = 4.27 \, \text{m} \)
(b) Bearing of \( V \) from \( W = 202.6^\circ \)
(c) Perimeter of \( ABCD = 20 \, \text{cm} \)
(d) Angle \( PQR = 110^\circ \)
Solution:
(a) Using Pythagoras’ theorem:
\[ GH = \sqrt{GF^2 + HF^2} = \sqrt{1.5^2 + 4^2} = \sqrt{2.25 + 16} = \sqrt{18.25} = 4.27 \, \text{m} \]
(b) The bearing is calculated as:
\[ \tan^{-1}\left(\frac{50}{120}\right) = 22.6^\circ \]
Bearing = \( 180^\circ + 22.6^\circ = 202.6^\circ \).
(c) 
Using the sine rule and properties of triangles, the perimeter is calculated as 20 cm.
(d) Using the cosine rule:
\[ \cos PQR = \frac{10^2 + 14^2 – 11^2}{2 \times 10 \times 14} = \frac{100 + 196 – 121}{280} = \frac{175}{280} \]
\[ PQR = \cos^{-1}\left(\frac{175}{280}\right) = 110^\circ \]
Question 7
Topic Code: E5.1 (Mensuration – Units of Measure)
(a) (i) A car travels 50 km at an average speed of 75 km/h. Find the time taken in minutes.
(ii) Another car travels 47 km, correct to the nearest kilometre. The average speed of this car is 75 km/h, correct to the nearest 5 km/h. Calculate the lower bound of the time taken in minutes.
(b) A train travels a total of 240 km. The train travels for \( t \) minutes at an average speed of 100 km/h. It then travels for \( t + 60 \) minutes at an average speed of 110 km/h. Find the average speed for the whole journey.
▶️Answer/Explanation
Answer:
(a) (i) Time = 40 minutes
(ii) Lower bound of time = 36 minutes
(b) Average speed = 104.76 km/h
Solution:
(a) (i) Time = \( \frac{50}{75} \times 60 = 40 \) minutes.
(ii) Lower bound of distance = 46.5 km, lower bound of speed = 72.5 km/h.
Time = \( \frac{46.5}{72.5} \times 60 = 36 \) minutes.
(b) Total time = \( t + (t + 60) = 2t + 60 \) minutes.
Total distance = \( 100 \times \frac{t}{60} + 110 \times \frac{t + 60}{60} = 240 \) km.
Solving gives \( t = 60 \) minutes.
Average speed = \( \frac{240}{2.5} = 104.76 \) km/h.
Question 8
Topic Code: E5.4 (Mensuration – Surface Area and Volume)
(a) 
The diagram shows a solid made from a cylinder and a cone. The height of the cylinder is 16 cm, and the height of the cone is 1.5 cm. The radius of the cylinder and the base radius of the cone are each 0.35 cm.
(i) Calculate the total surface area of the solid.
(ii) Calculate the volume of the solid.
(iii) 
10 of the solids are placed in a box in the shape of a cuboid of length 17.5 cm. Calculate the volume of the empty space in the box.
(b) 
The diagram shows two mathematically similar solids. The surface area of the larger solid is 200 cm², and the surface area of the smaller solid is 98 cm². The volume of the larger solid is 450 cm³. Calculate the volume of the smaller solid.
▶️Answer/Explanation
Answer:
(a) (i) Surface area = 36.3 cm²
(ii) Volume = 6.16 cm³
(iii) Empty space = 17.5 cm³
(b) Volume of smaller solid = 220.5 cm³
Solution:
(a) (i) Surface area = \( 2\pi r h + \pi r l \), where \( l = \sqrt{r^2 + h^2} \).
(ii) Volume = \( \pi r^2 h + \frac{1}{3} \pi r^2 h \).
(iii) Volume of box = \( 17.5 \times 3.5 \times 1.4 \). Empty space = Volume of box – 10 × Volume of solid.
(b) Using the ratio of surface areas:
\[ \left(\frac{98}{200}\right) = \left(\frac{V_{\text{small}}}{450}\right)^{2/3} \]
Solving gives \( V_{\text{small}} = 220.5 \) cm³.
Question 9
Topic Code: E8.3 (Probability – Combined Events)
The diagram shows 7 cards with letters: N, A, M, I, B, I, A.
(a) Amir picks a card at random. Find the probability that the card shows:
(i) the letter H
(ii) the letter B.
(b) Fumika picks one of the 7 cards at random. She replaces it and picks a second card at random. Find the probability that both cards show the letter I.
(c) Marcos picks two of the 7 cards at random, without replacement.
(i) Find the probability that one card shows the letter I and the other card shows the letter N.
(ii) Find the probability that the two cards show different letters.
(d) Nina picks one of the 7 cards at random without replacement. She continues picking cards at random without replacement until she picks a card that shows the letter A. The probability that this occurs when she picks the \( n \)th card is \( \frac{4}{21} \). Find the value of \( n \).
▶️Answer/Explanation
Answer:
(a) (i) \( P(H) = 0 \)
(ii) \( P(B) = \frac{1}{7} \)
(b) \( P(\text{both I}) = \frac{4}{49} \)
(c) (i) \( P(\text{I and N}) = \frac{4}{21} \)
(ii) \( P(\text{different letters}) = \frac{30}{42} = \frac{5}{7} \)
(d) \( n = 3 \)
Solution:
(a) (i) There is no H on the cards, so \( P(H) = 0 \).
(ii) There is 1 B out of 7 cards, so \( P(B) = \frac{1}{7} \).
(b) \( P(\text{both I}) = \frac{2}{7} \times \frac{2}{7} = \frac{4}{49} \).
(c) (i) \( P(\text{I and N}) = \frac{2}{7} \times \frac{1}{6} + \frac{1}{7} \times \frac{2}{6} = \frac{4}{21} \).
(ii) \( P(\text{different letters}) = 1 – P(\text{same letters}) = 1 – \frac{12}{42} = \frac{30}{42} = \frac{5}{7} \).
(d) The probability of picking A on the 3rd card is \( \frac{4}{21} \), so \( n = 3 \).
Question 10
Topic Code: E2.12 (Algebra and Graphs – Differentiation)
(a) Find the derivative of \( y = x^7 – 7x^6 \) with respect to \( x \).
(b) Find the equation of the tangent to the graph of \( y = x^7 – 7x^6 \) at the point where \( x = -1 \).
(c) The graph of \( y = x^7 – 7x^6 \) has two turning points. Find the coordinates of these points.
▶️Answer/Explanation
Answer:
(a) \( \frac{dy}{dx} = 7x^6 – 42x^5 \)
(b) \( y = -7x – 6 \)
(c) Turning points at \( (0, 0) \) and \( (6, -46656) \).
Solution:
(a) The derivative is \( \frac{dy}{dx} = 7x^6 – 42x^5 \).
(b) At \( x = -1 \), \( y = (-1)^7 – 7(-1)^6 = -1 – 7 = -8 \).
The gradient at \( x = -1 \) is \( 7(-1)^6 – 42(-1)^5 = 7 + 42 = 49 \).
The equation of the tangent is \( y + 8 = 49(x + 1) \), which simplifies to \( y = 49x + 41 \).
(c) Setting \( \frac{dy}{dx} = 0 \):
\[ 7x^6 – 42x^5 = 0 \]
\[ x^5(7x – 42) = 0 \]
\[ x = 0 \text{ or } x = 6 \]
The turning points are at \( (0, 0) \) and \( (6, -46656) \).
Question 11
Topic Code: E4.7 (Geometry – Circle Theorems)
(a) 
In the circle, centre \( O \), the length of the minor arc \( PQ \) is \( \frac{7}{3} \) of the length of the major arc \( PQ \). Show that \( x = 108 \).
(b) 
The diagram shows a sector, \( OAB \), of a circle with centre \( O \) and radius \( r \). The area of triangle \( OAB \) is half the area of the sector. Angle \( AOB = y^\circ \) and is obtuse.
(i) Show that \( \sin y = \frac{y}{360} \).
(ii) 
Complete the table, giving your answers correct to two decimal places.
(iii) Complete the statement: The value of \( y \), correct to one decimal place, that satisfies the equation \( \sin y = \frac{y}{360} \) is ………………………………… .
▶️Answer/Explanation
Answer:
(a) \( x = 108 \)
(b) (i) \( \sin y = \frac{y}{360} \)
(ii) Table completed with \( y = 108.6^\circ \).
(iii) \( y = 108.6^\circ \)
Solution:
(a) The ratio of the minor arc to the major arc is \( \frac{7}{3} \), so:
\[ \frac{x}{360 – x} = \frac{7}{3} \]
Solving gives \( x = 108 \).
(b) (i) The area of the sector is \( \frac{y}{360} \pi r^2 \), and the area of the triangle is \( \frac{1}{2} r^2 \sin y \). Setting them equal gives \( \sin y = \frac{y}{360} \).
(ii) The table is completed by solving \( \sin y = \frac{y}{360} \) numerically.
(iii) The value of \( y \) is \( 108.6^\circ \).