Q. If a shopkeeper sells a bicycle for $240 after a discount of 20%, what was the marked price?
A.
$280
B.
$300
C.
$320
D.
$350
Show solution
Solution
Let the marked price be x. Selling Price = x - 20% of x = 240. Thus, 0.8x = 240, so x = 300.
Correct Answer:
B
— $300
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Q. If a shopkeeper sells a product for $80 after a discount of 20%, what was the marked price?
A.
$100
B.
$90
C.
$110
D.
$120
Show solution
Solution
Let the marked price be x. Selling Price = x - (20% of x) = 0.80x. Thus, 0.80x = $80, so x = $80 / 0.80 = $100.
Correct Answer:
A
— $100
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Q. If a shopkeeper sells an item at a loss of 10% and the cost price is $200, what is the selling price?
A.
$180
B.
$190
C.
$200
D.
$210
Show solution
Solution
Selling Price = Cost Price - (Loss Percentage * Cost Price) = $200 - (0.10 * $200) = $200 - $20 = $180.
Correct Answer:
A
— $180
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Q. If a shopkeeper sells an item at a loss of 10% and the cost price is $500, what is the selling price?
A.
$450
B.
$500
C.
$550
D.
$400
Show solution
Solution
Selling Price = Cost Price - (Loss Percentage * Cost Price) = $500 - (0.10 * $500) = $500 - $50 = $450.
Correct Answer:
A
— $450
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Q. If a shopkeeper sells an item at a loss of 10% and the selling price is $90, what was the cost price?
A.
$100
B.
$110
C.
$80
D.
$90
Show solution
Solution
Let the cost price be x. Selling Price = Cost Price - (10% of Cost Price) = x - 0.10x = 0.90x. Thus, 0.90x = $90, so x = $90 / 0.90 = $100.
Correct Answer:
A
— $100
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Q. If a shopkeeper sells an item for $120 after applying a discount of 20%, what was the original price?
A.
$140
B.
$150
C.
$160
D.
$180
Show solution
Solution
Let the original price be x. Selling Price = x - 20% of x = 120. Thus, 0.8x = 120, so x = 150.
Correct Answer:
A
— $140
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Q. If a shopkeeper sells an item for $120 after giving a discount of 20%, what was the original price?
A.
$140
B.
$150
C.
$160
D.
$180
Show solution
Solution
Let the original price be x. Then, 120 = x - 20% of x => 120 = x - 0.2x => 120 = 0.8x => x = 150.
Correct Answer:
A
— $140
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Q. If a shopkeeper sells an item for $120 after giving a discount of 20%, what was the marked price?
A.
$140
B.
$150
C.
$160
D.
$170
Show solution
Solution
Let the marked price be x. Selling Price = Marked Price - Discount = x - 0.20x = 0.80x. So, 0.80x = 120. Therefore, x = 120 / 0.80 = 150.
Correct Answer:
B
— $150
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Q. If a shopkeeper sells an item for $240 after a discount of 20%, what was the marked price?
A.
$280
B.
$300
C.
$320
D.
$350
Show solution
Solution
Let the marked price be x. Selling Price = Marked Price - Discount = x - 0.2x = 0.8x. So, 0.8x = 240. Therefore, x = 240/0.8 = $300.
Correct Answer:
B
— $300
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Q. If a shopkeeper sells an item for $240 after applying a discount of 20%, what was the marked price? (2023)
A.
$280
B.
$300
C.
$320
D.
$350
Show solution
Solution
Let the marked price be x. Then, Selling Price = x - (20% of x) = 240 => 0.8x = 240 => x = 300.
Correct Answer:
B
— $300
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Q. If a shopkeeper sells an item for $240 after applying a discount of 20%, what was the original marked price? (2023)
A.
$250
B.
$300
C.
$320
D.
$350
Show solution
Solution
Let the marked price be x. Selling Price = Marked Price - Discount = x - 0.2x = 0.8x. Thus, 0.8x = 240, so x = 300.
Correct Answer:
B
— $300
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Q. If a shopkeeper sells an item for $240 after applying a discount of 20%, what was the original price of the item?
A.
$300
B.
$280
C.
$250
D.
$320
Show solution
Solution
Let the original price be x. Selling Price = x - 20% of x = 0.8x. Thus, 0.8x = 240, so x = 300.
Correct Answer:
A
— $300
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Q. If a shopkeeper sells an item for $80 after a discount of 20%, what was the marked price?
A.
$100
B.
$90
C.
$110
D.
$120
Show solution
Solution
Let the marked price be x. After a 20% discount, the selling price is x - (0.20 * x) = 0.80x. Setting this equal to $80 gives 0.80x = $80, so x = $80 / 0.80 = $100.
Correct Answer:
A
— $100
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Q. If a shopkeeper sells an item for $90 at a loss of 10%, what was the cost price?
A.
$100
B.
$95
C.
$85
D.
$80
Show solution
Solution
Let the cost price be x. Selling price = cost price - loss = x - (0.10 * x) = 0.90x. Setting this equal to $90 gives 0.90x = $90, so x = $100.
Correct Answer:
A
— $100
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Q. If a simple harmonic oscillator has a frequency of 1 Hz, what is the time period?
A.
0.5 s
B.
1 s
C.
2 s
D.
4 s
Show solution
Solution
Time period (T) is the reciprocal of frequency (f). T = 1/f = 1/1 = 1 s.
Correct Answer:
B
— 1 s
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Q. If a simple harmonic oscillator has a maximum displacement of 5 cm, what is the amplitude?
A.
2.5 cm
B.
5 cm
C.
10 cm
D.
0 cm
Show solution
Solution
The amplitude of a simple harmonic oscillator is defined as the maximum displacement from the equilibrium position, which is 5 cm in this case.
Correct Answer:
B
— 5 cm
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Q. If a simple harmonic oscillator has a total energy E, what is the kinetic energy when the displacement is half of the amplitude?
A.
E/4
B.
E/2
C.
3E/4
D.
E
Show solution
Solution
The total energy E is conserved. When the displacement is half the amplitude, the potential energy is (1/2)E, so the kinetic energy is E - (1/2)E = (1/2)E.
Correct Answer:
C
— 3E/4
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Q. If a soap film is formed on a wire frame, what is the effect of adding more soap to the film?
A.
Surface tension increases
B.
Surface tension decreases
C.
Surface tension remains the same
D.
Surface tension becomes zero
Show solution
Solution
Adding more soap decreases the surface tension because soap molecules disrupt the cohesive forces between water molecules.
Correct Answer:
B
— Surface tension decreases
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Q. If a solar panel generates 300 watts of power per hour, how much power will it generate in 5 hours?
A.
1200 watts
B.
1500 watts
C.
1800 watts
D.
2000 watts
Show solution
Solution
300 watts/hour * 5 hours = 1500 watts
Correct Answer:
C
— 1800 watts
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Q. If a solar panel generates 300 watts per hour, how much energy will it generate in 5 hours?
A.
1500 watts
B.
1200 watts
C.
1800 watts
D.
2000 watts
Show solution
Solution
300 watts/hour * 5 hours = 1500 watts
Correct Answer:
A
— 1500 watts
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Q. If a soldier can carry a load of 40 kg and he carries 5 such loads, what is the total weight he carries? (2019)
A.
150 kg
B.
200 kg
C.
250 kg
D.
300 kg
Show solution
Solution
Total Weight = Load × Number of Loads = 40 kg × 5 = 200 kg
Correct Answer:
B
— 200 kg
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Q. If a soldier can carry a load of 40 kg for 10 km, how much weight can he carry for 5 km at the same rate? (2020)
A.
20 kg
B.
30 kg
C.
40 kg
D.
50 kg
Show solution
Solution
Weight carried is proportional to distance. For 5 km, he can carry 20 kg.
Correct Answer:
A
— 20 kg
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Q. If a soldier can carry a load of 40 kg for 5 km, how much weight can he carry for 10 km if the weight carried is directly proportional to the distance? (2020)
A.
20 kg
B.
30 kg
C.
40 kg
D.
50 kg
Show solution
Solution
If distance doubles, weight carried is halved. So, 40 kg / 2 = 20 kg.
Correct Answer:
A
— 20 kg
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Q. If a soldier can carry a load of 40 kg, how many soldiers are needed to carry a total load of 320 kg? (2022)
Show solution
Solution
Number of Soldiers = Total Load / Load per Soldier = 320 kg / 40 kg = 8
Correct Answer:
C
— 8
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Q. If a soldier can march 12 km in 2 hours, what is his speed in m/s? (2020)
A.
2 m/s
B.
3 m/s
C.
4 m/s
D.
5 m/s
Show solution
Solution
Speed = (12 km × 1000 m/km) / (2 h × 3600 s/h) = 3 m/s
Correct Answer:
B
— 3 m/s
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Q. If a soldier can march 4 km in 40 minutes, what is his speed in m/s? (2020)
A.
1.5 m/s
B.
2.0 m/s
C.
2.5 m/s
D.
3.0 m/s
Show solution
Solution
Speed = Distance / Time = 4 km / (40/60) h = 4 km / (2/3) h = 6 km/h = 1.67 m/s (approx 2.0 m/s)
Correct Answer:
B
— 2.0 m/s
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Q. If a soldier can march 4 km in 50 minutes, what is his speed in m/s? (2020)
A.
1.2 m/s
B.
1.5 m/s
C.
1.8 m/s
D.
2.0 m/s
Show solution
Solution
Speed = (4 km × 1000 m/km) / (50 min × 60 s/min) = 1.33 m/s
Correct Answer:
A
— 1.2 m/s
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Q. If a soldier runs 12 km in 1 hour and 30 minutes, what is his speed in km/h? (2023)
A.
7.5 km/h
B.
8 km/h
C.
8.5 km/h
D.
9 km/h
Show solution
Solution
Speed = Distance / Time = 12 km / (1.5 h) = 8 km/h
Correct Answer:
A
— 7.5 km/h
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Q. If a solenoid has n turns per unit length and carries a current I, what is the magnetic field inside the solenoid?
A.
0
B.
μ₀nI
C.
μ₀I/n
D.
μ₀I
Show solution
Solution
The magnetic field inside a solenoid is given by B = μ₀nI.
Correct Answer:
B
— μ₀nI
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Q. If a solid cylinder is rotated about its diameter, what is its moment of inertia?
A.
1/2 MR^2
B.
1/4 MR^2
C.
1/3 MR^2
D.
MR^2
Show solution
Solution
The moment of inertia of a solid cylinder about its diameter is I = 1/4 MR^2.
Correct Answer:
B
— 1/4 MR^2
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