Secondary 3 Combined Science Semestral Assessment 2 (End of Year) Paper 1

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TuitionGoWhere Practice Paper - Combined Science Secondary 3

TuitionGoWhere Secondary School (AI)

Subject: Combined Science (Physics/Chemistry)
Level: Secondary 3
Paper: SA2 Version 1
Duration: 1 hour 30 minutes
Total Marks: 60

Name: ________________________
Class: ________________________
Date: ________________________

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Instructions to Candidates

1. Write your name, class, and date in the spaces provided above.
2. Answer all questions in the spaces provided.
3. The number of marks is given in brackets [ ] at the end of each question or part question.
4. The total marks for this paper is 60.
5. You may use a calculator.
6. Where appropriate, take $g = 10 \text{ N/kg}$.
7. Show all working for calculation questions.

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Section A: Multiple Choice Questions [15 marks]

Answer all questions. For each question, choose the correct option and write the letter (A, B, C, or D) in the box provided.

1. A ball is thrown vertically upwards. Which statement correctly describes the energy conversion as the ball rises? [1]

• A. Kinetic energy is converted to gravitational potential energy and thermal energy.
• B. Gravitational potential energy is converted to kinetic energy.
• C. Kinetic energy is converted to gravitational potential energy only.
• D. Total energy decreases as the ball rises.

Answer: □

2. A force of 20 N is applied to push a box 5 m across a horizontal floor. The work done against friction is 30 J. What is the net work done on the box? [1]

• A. 70 J
• B. 100 J
• C. 30 J
• D. 130 J

Answer: □

3. Which of the following is a vector quantity? [1]

• A. Energy
• B. Power
• C. Work
• D. Force

Answer: □

4. A car of mass 1200 kg accelerates from rest to 20 m/s in 10 s. What is the average power developed by the engine? [1]

• A. 24 kW
• B. 48 kW
• C. 240 kW
• D. 480 kW

Answer: □

5. The diagram shows a simple pendulum swinging from position P to Q to R. At which position(s) does the bob have maximum kinetic energy? [1]

Image pending generation: diagram for Q5.

• A. P only
• B. Q only
• C. R only
• D. P and R

Answer: □

6. A 500 g object falls from a height of 10 m. Ignoring air resistance, what is its kinetic energy just before it hits the ground? [1]

• A. 5 J
• B. 50 J
• C. 500 J
• D. 5000 J

Answer: □

7. Which energy conversion occurs in a hydroelectric power station? [1]

• A. Electrical → Gravitational potential → Kinetic
• B. Gravitational potential → Kinetic → Electrical
• C. Chemical → Thermal → Electrical
• D. Nuclear → Thermal → Electrical

Answer: □

8. A machine lifts a load of 200 N through a height of 3 m. The effort applied is 80 N and moves through a distance of 10 m. What is the efficiency of the machine? [1]

• A. 25%
• B. 50%
• C. 75%
• D. 80%

Answer: □

9. A student runs up a flight of stairs of vertical height 4 m in 5 s. If the student's mass is 60 kg, what is the average power developed? [1]

• A. 240 W
• B. 480 W
• C. 2400 W
• D. 4800 W

Answer: □

10. A spring is compressed by a force of 10 N. The compression is 0.05 m. What is the elastic potential energy stored in the spring? [1]

• A. 0.25 J
• B. 0.5 J
• C. 2.5 J
• D. 5 J

Answer: □

11. Which of the following statements about work is correct? [1]

• A. Work is done when a force acts on an object, regardless of displacement.
• B. Work is a vector quantity.
• C. No work is done if the force is perpendicular to the displacement.
• D. Work done against friction is always positive.

Answer: □

12. A 2 kg block slides down a frictionless inclined plane of height 3 m. What is its speed at the bottom? [1]

• A. 5.5 m/s
• B. 7.7 m/s
• C. 10 m/s
• D. 15 m/s

Answer: □

13. The rate of doing work is defined as: [1]

• A. Energy
• B. Force
• C. Power
• D. Momentum

Answer: □

14. A weightlifter holds a 100 kg barbell stationary above his head for 10 s. How much work does he do on the barbell during this time? [1]

• A. 0 J
• B. 1000 J
• C. 10000 J
• D. 100000 J

Answer: □

15. A roller coaster car starts from rest at a height of 50 m. Assuming no energy losses, what is its speed at a height of 20 m? [1]

• A. 17 m/s
• B. 20 m/s
• C. 24 m/s
• D. 31 m/s

Answer: □

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Section B: Structured Questions [30 marks]

Answer all questions in the spaces provided.

16. A toy car of mass 0.2 kg is released from rest at point A on a frictionless track as shown. The car passes through a vertical loop of radius 0.3 m and reaches point C at a height of 0.4 m above the ground. [5]

Image pending generation: diagram for Q16.

(a) Calculate the gravitational potential energy of the car at point A. [1]

Answer: _________________________________________________________________ [1]

(b) Calculate the speed of the car at point B (the top of the loop). [2]

Answer: _________________________________________________________________ [2]

(c) Calculate the centripetal force acting on the car at point B. [1]

Answer: _________________________________________________________________ [1]

(d) Explain whether the car will maintain contact with the track at point B. [1]

Answer: _________________________________________________________________ [1]

17. A crane lifts a concrete block of mass 500 kg vertically upwards at a constant speed of 2 m/s. [5]

(a) State the net force acting on the block while it moves at constant speed. [1]

Answer: _________________________________________________________________ [1]

(b) Calculate the tension in the lifting cable. [1]

Answer: _________________________________________________________________ [1]

(c) Calculate the power output of the crane motor. [2]

Answer: _________________________________________________________________ [2]

(d) The crane motor has an efficiency of 80%. Calculate the electrical power input to the motor. [1]

Answer: _________________________________________________________________ [1]

18. A spring with spring constant $k = 200 \text{ N/m}$ is compressed by 0.1 m. A block of mass 0.5 kg is placed against the spring on a horizontal surface with coefficient of kinetic friction $\mu_k = 0.2$. The spring is released, pushing the block. [6]

(a) Calculate the elastic potential energy stored in the spring when compressed. [1]

Answer: _________________________________________________________________ [1]

(b) Calculate the work done against friction as the block moves 0.5 m after release. [2]

Answer: _________________________________________________________________ [2]

(c) Calculate the kinetic energy of the block after it has moved 0.5 m. [2]

Answer: _________________________________________________________________ [2]

(d) How far will the block travel before coming to rest? [1]

Answer: _________________________________________________________________ [1]

19. A hydroelectric dam stores water at a height of 80 m above the turbines. Water flows at a rate of 500 kg/s. [5]

(a) Calculate the gravitational potential energy lost by the water each second. [2]

Answer: _________________________________________________________________ [2]

(b) If the electrical power output is 300 kW, calculate the efficiency of the energy conversion. [2]

Answer: _________________________________________________________________ [2]

(c) State one reason why the efficiency is less than 100%. [1]

Answer: _________________________________________________________________ [1]

20. A student investigates the relationship between the height of a ramp and the speed of a trolley at the bottom. The trolley is released from rest at different heights. [4]

Image pending generation: experimental_setup for Q20.

(a) State the energy conversion that takes place as the trolley moves down the ramp. [1]

Answer: _________________________________________________________________ [1]

(b) The student plots a graph of $v^2$ against $h$. State the expected shape of the graph and explain your answer. [2]

Answer: _________________________________________________________________ [2]

(c) The student finds that the gradient of the graph is 18 m/s². Suggest a reason why this value is less than the theoretical value of $2g$. [1]

Answer: _________________________________________________________________ [1]

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Section C: Longer Structured Questions [15 marks]

Answer all questions in the spaces provided.

21. A roller coaster is designed with a vertical loop of radius 15 m. The coaster car (including passengers) has a total mass of 800 kg. The car starts from rest at a height of 50 m above the bottom of the loop. Assume negligible friction and air resistance. [8]

Image pending generation: diagram for Q21.

(a) Calculate the total mechanical energy of the system at point A. [1]

Answer: _________________________________________________________________ [1]

(b) Calculate the speed of the car at point B (bottom of the hill, before the loop). [2]

Answer: _________________________________________________________________ [2]

(c) Calculate the speed of the car at point C (top of the loop). [2]

Answer: _________________________________________________________________ [2]

(d) Calculate the normal reaction force exerted by the track on the car at point C. [2]

Answer: _________________________________________________________________ [2]

(e) The design is modified so that the car enters the loop at a lower speed. State the minimum speed required at point C for the car to maintain contact with the track, and calculate the corresponding minimum height of point A. [1]

Answer: _________________________________________________________________ [1]

22. A block of mass 3 kg is pulled up a rough inclined plane at 30° to the horizontal by a constant force of 40 N acting parallel to the plane. The block moves 5 m up the plane starting from rest. The coefficient of kinetic friction is 0.15. [7]

Image pending generation: diagram for Q22.

(a) Draw and label all forces acting on the block on the diagram above. [1]

Answer: (Forces drawn on diagram) [1]

(b) Calculate the work done by the applied force. [1]

Answer: _________________________________________________________________ [1]

(c) Calculate the work done against friction. [2]

Answer: _________________________________________________________________ [2]

(d) Calculate the gain in gravitational potential energy of the block. [1]

Answer: _________________________________________________________________ [1]

(e) Using the work-energy theorem, calculate the final speed of the block. [2]

Answer: _________________________________________________________________ [2]

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End of Paper

Total Marks: 60

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TuitionGoWhere Practice Paper - Combined Science Secondary 3 SA2 Version 1 - Answer Key

Total Marks: 60

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Section A: Multiple Choice Questions [15 marks]

1. Answer: A [1]

Explanation: As the ball rises, kinetic energy decreases while gravitational potential energy increases. Some energy is also converted to thermal energy due to air resistance (though often neglected in idealised problems, the most complete correct statement includes thermal energy). The principle of conservation of energy states total energy remains constant, but the conversion involves KE → GPE + thermal.

Marking note: Award 1 mark for A. Option C is incomplete as it ignores air resistance/thermal energy.

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2. Answer: A [1]

Working:

• Work done by applied force = $F \times s = 20 \times 5 = 100 \text{ J}$
• Work done against friction = 30 J (energy lost)
• Net work done on box = Work by applied force - Work against friction = $100 - 30 = 70 \text{ J}$

Alternative: Net work = Change in kinetic energy = 70 J.

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3. Answer: D [1]

Explanation: Force is a vector quantity (has magnitude and direction). Energy, power, and work are scalar quantities (magnitude only).

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4. Answer: Answer: A [1]

Working:

• Final kinetic energy = $\frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 20^2 = 240,000 \text{ J}$
• Time = 10 s
• Average power = $\frac{\text{Work done}}{\text{Time}} = \frac{240,000}{10} = 24,000 \text{ W} = 24 \text{ kW}$

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5. Answer: B [1]

Explanation: At the lowest point Q, gravitational potential energy is minimum, so kinetic energy is maximum (by conservation of energy). At P and R (highest points), speed is zero, so KE = 0.

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6. Answer: B [1]

Working:

• Loss in GPE = Gain in KE (conservation of energy)
• $mgh = 0.5 \times 10 \times 10 = 50 \text{ J}$
• KE just before impact = 50 J

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7. Answer: B [1]

Explanation: Water at height has gravitational potential energy → flows down gaining kinetic energy → turns turbines → generates electrical energy.

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8. Answer: C [1]

Working:

• Useful work output = Load × distance lifted = $200 \times 3 = 600 \text{ J}$
• Work input = Effort × distance moved by effort = $80 \times 10 = 800 \text{ J}$
• Efficiency = $\frac{\text{Useful output}}{\text{Input}} \times 100\% = \frac{600}{800} \times 100\% = 75\%$

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9. Answer: B [1]

Working:

• Work done = Gain in GPE = $mgh = 60 \times 10 \times 4 = 2400 \text{ J}$
• Power = $\frac{\text{Work}}{\text{Time}} = \frac{2400}{5} = 480 \text{ W}$

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10. Answer: A [1]

Working:

• For a spring, average force during compression = $\frac{0 + 10}{2} = 5 \text{ N}$ (if linear)
• Elastic PE = Average force × compression = $5 \times 0.05 = 0.25 \text{ J}$
• Or using formula: $EPE = \frac{1}{2}Fx = \frac{1}{2} \times 10 \times 0.05 = 0.25 \text{ J}$

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11. Answer: C [1]

Explanation: Work done = $F \cdot s \cdot \cos\theta$. If force is perpendicular to displacement ($\theta = 90^\circ$), $\cos 90^\circ = 0$, so no work is done.

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12. Answer: B [1]

Working:

• Loss in GPE = Gain in KE
• $mgh = \frac{1}{2}mv^2$
• $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 3} = \sqrt{60} \approx 7.7 \text{ m/s}$

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13. Answer: C [1]

Explanation: Power is defined as the rate of doing work, or rate of energy transfer. $P = \frac{W}{t}$.

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14. Answer: A [1]

Explanation: Work done = Force × displacement in direction of force. The barbell is stationary (displacement = 0), so work done = 0 J. The weightlifter exerts a force but does no work on the barbell while holding it stationary.

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15. Answer: C [1]

Working:

• Loss in GPE = Gain in KE
• $mg(h_1 - h_2) = \frac{1}{2}mv^2$
• $v = \sqrt{2g(h_1 - h_2)} = \sqrt{2 \times 10 \times (50 - 20)} = \sqrt{600} \approx 24.5 \text{ m/s} \approx 24 \text{ m/s}$

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Section B: Structured Questions [30 marks]

16. [5 marks]

(a) GPE at A = $mgh_A = 0.2 \times 10 \times 1.0 = 2.0 \text{ J}$ [1]

(b) At point B (top of loop), height = 0.6 m

• Total energy at A = Total energy at B (conservation)
• $mgh_A = mgh_B + \frac{1}{2}mv_B^2$
• $2.0 = (0.2 \times 10 \times 0.6) + \frac{1}{2}(0.2)v_B^2$
• $2.0 = 1.2 + 0.1v_B^2$
• $0.1v_B^2 = 0.8$
• $v_B^2 = 8$
• $v_B = \sqrt{8} = 2.83 \text{ m/s}$ [2]

Mark breakdown: 1 mark for correct energy equation, 1 mark for correct final answer with unit.

(c) Centripetal force at B = $\frac{mv_B^2}{R} = \frac{0.2 \times 8}{0.3} = \frac{1.6}{0.3} = 5.33 \text{ N}$ [1]

(d) Weight of car = $mg = 0.2 \times 10 = 2 \text{ N}$

• At top of loop, centripetal force required = 5.33 N downwards
• Forces providing centripetal force: Weight (2 N down) + Normal reaction (N down)
• $N + 2 = 5.33 \Rightarrow N = 3.33 \text{ N}$
• Since $N > 0$, the track pushes down on the car, so the car maintains contact with the track. [1]

Marking note: Must state N > 0 or normal reaction is positive to show contact maintained.

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17. [5 marks]

(a) Net force = 0 N [1] Explanation: Constant speed means zero acceleration. By Newton's First Law, net force = 0.

(b) Tension = Weight = $mg = 500 \times 10 = 5000 \text{ N}$ [1] Explanation: Net force = 0, so Tension - Weight = 0 → Tension = Weight.

(c) Power = Force × velocity = Tension × speed = $5000 \times 2 = 10,000 \text{ W} = 10 \text{ kW}$ [2] Mark breakdown: 1 mark for correct formula/substitution, 1 mark for correct answer with unit.

(d) Efficiency = $\frac{\text{Output power}}{\text{Input power}} \times 100\%$

• $0.80 = \frac{10,000}{\text{Input power}}$
• Input power = $\frac{10,000}{0.80} = 12,500 \text{ W} = 12.5 \text{ kW}$ [1]

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18. [6 marks]

(a) Elastic PE = $\frac{1}{2}kx^2 = \frac{1}{2} \times 200 \times (0.1)^2 = 100 \times 0.01 = 1.0 \text{ J}$ [1]

(b) Friction force = $\mu_k N = \mu_k mg = 0.2 \times 0.5 \times 10 = 1.0 \text{ N}$

• Work against friction = friction force × distance = $1.0 \times 0.5 = 0.5 \text{ J}$ [2] Mark breakdown: 1 mark for friction force calculation, 1 mark for work done.

(c) Initial EPE = 1.0 J

• Work done against friction in 0.5 m = 0.5 J
• KE after 0.5 m = Initial EPE - Work against friction = $1.0 - 0.5 = 0.5 \text{ J}$ [2] Mark breakdown: 1 mark for correct energy balance equation, 1 mark for correct answer.

(d) Total distance before rest: All initial EPE converted to work against friction

• $EPE_{initial} = f \times d_{total}$
• $1.0 = 1.0 \times d_{total}$
• $d_{total} = 1.0 \text{ m}$ [1]

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19. [5 marks]

(a) GPE lost per second = $mgh$ per second = (mass flow rate) × $g$ × $h$

• $= 500 \times 10 \times 80 = 400,000 \text{ J/s} = 400 \text{ kW}$ [2] Mark breakdown: 1 mark for correct formula/approach, 1 mark for correct answer with unit.

(b) Efficiency = $\frac{\text{Useful output}}{\text{Input}} \times 100\% = \frac{300}{400} \times 100\% = 75\%$ [2] Mark breakdown: 1 mark for correct formula, 1 mark for correct answer.

(c) Any one valid reason, e.g.:

• Energy lost as heat due to friction in turbines/generators
• Sound energy produced
• Kinetic energy of water leaving turbines not fully recovered
• Electrical resistance losses in generators/transmission [1]

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20. [4 marks]

(a) Gravitational potential energy → Kinetic energy (+ some thermal energy due to friction) [1]

(b) Graph of $v^2$ against $h$ is a straight line passing through the origin. [1] Explanation: From conservation of energy: $mgh = \frac{1}{2}mv^2 \Rightarrow v^2 = 2gh$. This is of the form $y = mx$ where $y = v^2$, $x = h$, gradient = $2g$. Since $g$ is constant, the graph is linear through origin. [1]

(c) The gradient (18 m/s²) is less than $2g$ (20 m/s²) because energy is lost to friction/air resistance (or work done against friction), so not all GPE is converted to KE. The actual KE gained is less than the GPE lost. [1]

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Section C: Longer Structured Questions [15 marks]

21. [8 marks]

(a) Total mechanical energy at A = GPE at A (since KE = 0 at rest)

• $E = mgh_A = 800 \times 10 \times 50 = 400,000 \text{ J} = 400 \text{ kJ}$ [1]

(b) At point B (h = 0), all energy is KE:

• $400,000 = \frac{1}{2} \times 800 \times v_B^2$
• $v_B^2 = \frac{400,000 \times 2}{800} = 1000$
• $v_B = \sqrt{1000} = 31.6 \text{ m/s}$ [2] Mark breakdown: 1 mark for energy conservation equation, 1 mark for correct answer.

(c) At point C (top of loop, h = 30 m):

• $E_{total} = GPE_C + KE_C$
• $400,000 = (800 \times 10 \times 30) + \frac{1}{2}(800)v_C^2$
• $400,000 = 240,000 + 400v_C^2$
• $400v_C^2 = 160,000$
• $v_C^2 = 400$
• $v_C = 20 \text{ m/s}$ [2] Mark breakdown: 1 mark for correct energy equation, 1 mark for correct answer.

(d) At point C (top of loop):

• Centripetal force required = $\frac{mv_C^2}{R} = \frac{800 \times 400}{15} = \frac{320,000}{15} = 21,333 \text{ N}$
• Forces downwards: Weight ($mg = 8000 \text{ N}$) + Normal reaction ($N$)
• $N + 8000 = 21,333$
• $N = 13,333 \text{ N}$ [2] Mark breakdown: 1 mark for centripetal force calculation, 1 mark for normal reaction calculation.

(e) Minimum speed at C for contact: $N = 0$, so $mg = \frac{mv_{min}^2}{R}$

• $v_{min}^2 = gR = 10 \times 15 = 150$
• $v_{min} = \sqrt{150} = 12.2 \text{ m/s}$
• Minimum height $h_{min}$: $mgh_{min} = mg(2R) + \frac{1}{2}mv_{min}^2$
• $h_{min} = 2R + \frac{v_{min}^2}{2g} = 30 + \frac{150}{20} = 30 + 7.5 = 37.5 \text{ m}$ [1] Marking note: 1 mark for both minimum speed and minimum height correctly stated.

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22. [7 marks]

(a) Forces on diagram [1]:

• Weight $mg$ vertically downwards
• Normal reaction $N$ perpendicular to plane (upwards)
• Applied force $F = 40 \text{ N}$ up the plane
• Friction $f = \mu_k N$ down the plane (opposing motion)

(b) Work done by applied force = $F \times s = 40 \times 5 = 200 \text{ J}$ [1]

(c) Normal reaction $N = mg\cos\theta = 3 \times 10 \times \cos 30^\circ = 30 \times 0.866 = 25.98 \text{ N}$

• Friction $f = \mu_k N = 0.15 \times 25.98 = 3.897 \text{ N}$
• Work against friction = $f \times s = 3.897 \times 5 = 19.485 \text{ J} \approx 19.5 \text{ J}$ [2] Mark breakdown: 1 mark for normal reaction/friction calculation, 1 mark for work done.

(d) Vertical height gained = $s \sin\theta = 5 \times \sin 30^\circ = 5 \times 0.5 = 2.5 \text{ m}$

• Gain in GPE = $mgh = 3 \times 10 \times 2.5 = 75 \text{ J}$ [1]

(e) Work-energy theorem: Net work done = Change in KE

• Net work = Work by applied force - Work against friction - Gain in GPE
• $= 200 - 19.5 - 75 = 105.5 \text{ J}$
• $\frac{1}{2}mv^2 = 105.5$
• $\frac{1}{2} \times 3 \times v^2 = 105.5$
• $1.5v^2 = 105.5$
• $v^2 = 70.33$
• $v = 8.39 \text{ m/s}$ [2] Mark breakdown: 1 mark for correct work-energy equation setup, 1 mark for correct final answer.

Alternative method: Find net force, then acceleration, then $v^2 = u^2 + 2as$. Both methods accepted.

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End of Answer Key

Total Marks: 60