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Q. A customer buys a laptop for $800 after a discount of 20%. What was the marked price?

A. $1000
B. $900
C. $850
D. $800

Solution

Let the marked price be x. After a 20% discount, the selling price is 0.80x. Setting this equal to $800 gives 0.80x = $800, so x = $800 / 0.80 = $1000.

Correct Answer: A — $1000

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Q. A customer buys two items for $120 each. If the shopkeeper gives a discount of 10% on the total bill, what is the final amount paid by the customer? (2023)

A. $216
B. $220
C. $240
D. $250

Solution

Total cost before discount = 2 * $120 = $240. Discount = 10% of $240 = $24. Final amount = $240 - $24 = $216.

Correct Answer: A — $216

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Q. A customer buys two items for $240 after a discount of 10% on each. What was the total marked price of the items?

A. $250
B. $260
C. $270
D. $280

Solution

Let the total marked price be x. After a 10% discount, the selling price is 0.90x. Setting this equal to $240 gives 0.90x = $240, so x = $240 / 0.90 = $266.67, which rounds to $280.

Correct Answer: D — $280

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Q. A cyclist accelerates at 3 m/s² for 4 seconds. What is the final velocity if the initial velocity is 2 m/s? (2023)

A. 10 m/s
B. 11 m/s
C. 14 m/s
D. 20 m/s

Solution

Using v = u + at, we have v = 2 m/s + (3 m/s² * 4 s) = 2 + 12 = 14 m/s.

Correct Answer: B — 11 m/s

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Q. A cyclist accelerates from rest at a rate of 1 m/s². How far will he travel in 10 seconds?

A. 50 m
B. 100 m
C. 150 m
D. 200 m

Solution

Using the formula: distance = initial velocity * time + 0.5 * acceleration * time². Distance = 0 + 0.5 * 1 * 10² = 50 m.

Correct Answer: B — 100 m

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Q. A cyclist accelerates from rest at a rate of 1 m/s². How far will he travel in 8 seconds?

A. 32 m
B. 40 m
C. 48 m
D. 64 m

Solution

Using the formula: distance = initial velocity * time + 0.5 * acceleration * time². Distance = 0 + 0.5 * 1 * 8² = 32 m.

Correct Answer: C — 48 m

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Q. A cyclist accelerates from rest at a rate of 2 m/s². How far will he travel in 10 seconds?

A. 100 m
B. 50 m
C. 200 m
D. 150 m

Solution

Using the formula: distance = ut + 0.5at². Here, u = 0, a = 2 m/s², t = 10 s. Distance = 0 + 0.5 * 2 * 10² = 100 m.

Correct Answer: A — 100 m

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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the acceleration? (2023)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Acceleration = (final velocity - initial velocity) / time = (10 m/s - 0) / 5 s = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the average power output if the cyclist has a mass of 70 kg?

A. 140 W
B. 280 W
C. 560 W
D. 700 W

Solution

Kinetic Energy = 0.5 × m × v² = 0.5 × 70 kg × (10 m/s)² = 3500 J. Power = Work / Time = 3500 J / 5 s = 700 W.

Correct Answer: C — 560 W

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Q. A cyclist accelerates from rest to a speed of 10 m/s in 5 seconds. What is the cyclist's acceleration? (2021)

A. 1 m/s²
B. 2 m/s²
C. 3 m/s²
D. 4 m/s²

Solution

Acceleration a = (final velocity - initial velocity) / time = (10 m/s - 0) / 5 s = 2 m/s².

Correct Answer: B — 2 m/s²

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Q. A cyclist accelerates from rest to a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the work done by the cyclist?

A. 800 J
B. 900 J
C. 1000 J
D. 1200 J

Solution

Work done = Change in Kinetic Energy = 0.5 × mass × (final velocity² - initial velocity²) = 0.5 × 75 kg × (15 m/s)² = 843.75 J.

Correct Answer: C — 1000 J

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Q. A cyclist accelerates from rest to a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the kinetic energy at that speed?

A. 500 J
B. 750 J
C. 1000 J
D. 1250 J

Solution

Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 75 kg × (15 m/s)² = 8437.5 J.

Correct Answer: C — 1000 J

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Q. A cyclist covers 15 km in 30 minutes. What is the speed in km/h? (2023)

A. 20 km/h
B. 25 km/h
C. 30 km/h
D. 35 km/h

Solution

Speed = Distance/Time = 15 km/(30/60) h = 30 km/h

Correct Answer: B — 25 km/h

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Q. A cyclist covers a distance of 30 km at a speed of 15 km/h. How long does it take to complete the journey? (2022)

A. 1 hour
B. 2 hours
C. 2.5 hours
D. 3 hours

Solution

Time = Distance / Speed = 30 km / 15 km/h = 2 hours.

Correct Answer: B — 2 hours

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Q. A cyclist covers a distance of 90 km in 4.5 hours. What is the average speed of the cyclist? (2020)

A. 15 km/h
B. 20 km/h
C. 25 km/h
D. 30 km/h

Solution

Average Speed = Total Distance / Total Time = 90 km / 4.5 h = 20 km/h

Correct Answer: C — 25 km/h

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Q. A cyclist does 1500 J of work to climb a hill. If he takes 5 minutes, what is his average power output?

A. 5 W
B. 10 W
C. 15 W
D. 20 W

Solution

Power = Work / Time = 1500 J / (5 * 60 s) = 5 W.

Correct Answer: B — 10 W

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Q. A cyclist does 300 J of work to climb a hill. If the cyclist's mass is 75 kg, what is the height of the hill? (2023)

A. 0.4 m
B. 0.5 m
C. 0.6 m
D. 0.8 m

Solution

Potential Energy = mass × g × height; 300 J = 75 kg × 10 m/s² × height; height = 300 J / (75 kg × 10 m/s²) = 0.4 m

Correct Answer: B — 0.5 m

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Q. A cyclist does 300 J of work to climb a hill. If the height of the hill is 5 m, what is the mass of the cyclist? (g = 9.8 m/s²)

A. 6.12 kg
B. 10 kg
C. 15 kg
D. 20 kg

Solution

Using Work = mgh; m = Work / (gh) = 300 J / (9.8 m/s² * 5 m) = 6.12 kg.

Correct Answer: A — 6.12 kg

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Q. A cyclist does 300 J of work to climb a hill. If the height of the hill is 5 m, what is the effective weight of the cyclist?

A. 30 kg
B. 60 kg
C. 90 kg
D. 120 kg

Solution

Weight = Work / Height = 300 J / 5 m = 60 N; mass = Weight / g = 60 N / 9.8 m/s² ≈ 6.12 kg.

Correct Answer: B — 60 kg

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Q. A cyclist does 300 J of work to climb a hill. If the hill is 5 m high, what is the weight of the cyclist? (2020)

A. 60 kg
B. 30 kg
C. 50 kg
D. 40 kg

Solution

Work = m × g × h. 300 J = m × 9.8 m/s² × 5 m. m = 300 J / (9.8 m/s² × 5 m) = 6.12 kg.

Correct Answer: A — 60 kg

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Q. A cyclist does 300 J of work to climb a hill. If the hill is 5 m high, what is the effective weight of the cyclist? (2022)

A. 30 kg
B. 60 kg
C. 90 kg
D. 120 kg

Solution

Weight = Work done / height = 300 J / 5 m = 60 kg

Correct Answer: B — 60 kg

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Q. A cyclist does 600 J of work to climb a hill. If he takes 30 seconds, what is his average power output? (2022)

A. 10 W
B. 20 W
C. 30 W
D. 40 W

Solution

Power = Work / Time = 600 J / 30 s = 20 W.

Correct Answer: B — 20 W

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Q. A cyclist exerts a force of 100 N to maintain a speed of 5 m/s. What is the power output of the cyclist?

A. 200 W
B. 500 W
C. 1000 W
D. 250 W

Solution

Power can be calculated using P = F * v. Here, F = 100 N and v = 5 m/s. Thus, P = 100 N * 5 m/s = 500 W.

Correct Answer: B — 500 W

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Q. A cyclist is moving around a circular track of radius 100 m. If he completes one lap in 40 seconds, what is his average speed?

A. 5 m/s
B. 10 m/s
C. 15 m/s
D. 20 m/s

Solution

Circumference = 2πr = 2π(100 m). Average speed = distance/time = (2π(100 m))/40 s ≈ 15.7 m/s.

Correct Answer: B — 10 m/s

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Q. A cyclist is moving at 15 m/s and a pedestrian at 5 m/s in the same direction. How fast does the cyclist appear to be moving to the pedestrian?

A. 10 m/s
B. 15 m/s
C. 20 m/s
D. 5 m/s

Solution

Relative speed = Speed of cyclist - Speed of pedestrian = 15 m/s - 5 m/s = 10 m/s.

Correct Answer: A — 10 m/s

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Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the speed of the cyclist relative to the pedestrian?

A. 10 m/s
B. 15 m/s
C. 5 m/s
D. 20 m/s

Solution

Relative speed = Speed of cyclist - Speed of pedestrian = 15 m/s - 5 m/s = 10 m/s.

Correct Answer: A — 10 m/s

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Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the relative speed of the pedestrian with respect to the cyclist?

A. 10 m/s
B. 5 m/s
C. 20 m/s
D. 15 m/s

Solution

Relative speed = Speed of pedestrian - Speed of cyclist = 5 m/s - 15 m/s = -10 m/s (10 m/s behind).

Correct Answer: A — 10 m/s

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Q. A cyclist is moving at 15 m/s and passes a stationary observer. If the observer starts moving at 5 m/s in the same direction, what is the speed of the cyclist relative to the observer?

A. 10 m/s
B. 15 m/s
C. 20 m/s
D. 5 m/s

Solution

Relative speed = Speed of cyclist - Speed of observer = 15 m/s - 5 m/s = 10 m/s.

Correct Answer: A — 10 m/s

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Q. A cyclist is moving at 15 m/s towards the east while a car is moving at 25 m/s towards the west. What is the relative speed of the cyclist with respect to the car?

A. 10 m/s
B. 15 m/s
C. 40 m/s
D. 25 m/s

Solution

Relative speed = Speed of cyclist + Speed of car = 15 m/s + 25 m/s = 40 m/s.

Correct Answer: C — 40 m/s

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Q. A cyclist is moving at 15 m/s while a car is moving at 25 m/s in the same direction. What is the speed of the cyclist relative to the car?

A. 10 m/s
B. 15 m/s
C. 25 m/s
D. 40 m/s

Solution

Relative speed = Speed of car - Speed of cyclist = 25 m/s - 15 m/s = 10 m/s.

Correct Answer: A — 10 m/s

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Major Competitive Exams

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