UPSC CAPF Archives

1. What is the largest value for n (natural number) such that 6^n divides

What is the largest value for n (natural number) such that 6^n divides the product of the first 100 natural numbers?

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UPSC CAPF – 2018

The largest value for n such that 6^n divides the product of the first 100 natural numbers (100!) is 48.

– To find the highest power of a composite number (like 6) that divides a factorial, we need to find the highest power of its prime factors that divide the factorial.
– 6 = 2 * 3. The highest power of 6 in 100! is limited by the highest power of the less frequent prime factor, which is 3.
– The highest power of a prime p dividing n! is given by Legendre’s formula: floor(n/p) + floor(n/p^2) + floor(n/p^3) + …
– For p=3 and n=100:
Power of 3 in 100! = floor(100/3) + floor(100/9) + floor(100/27) + floor(100/81)
= 33 + 11 + 3 + 1 = 48.
– For p=2 and n=100:
Power of 2 in 100! = floor(100/2) + floor(100/4) + floor(100/8) + floor(100/16) + floor(100/32) + floor(100/64)
= 50 + 25 + 12 + 6 + 3 + 1 = 97.
– So, 100! contains 2^97 * 3^48 * …
– For 6^n = (2*3)^n = 2^n * 3^n to divide 100!, we must have n ≤ 97 and n ≤ 48.
– The largest integer n satisfying both conditions is n = 48.

Legendre’s formula is a powerful tool for finding the exponent of a prime in the prime factorization of n!. This method works because every multiple of a prime p contributes at least one factor of p, every multiple of p^2 contributes an additional factor of p, and so on.

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2. Which one of the following ocean currents is not a warm current?

Which one of the following ocean currents is not a warm current?

North Pacific

Falkland

North Equatorial

Canary

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UPSC CAPF – 2018

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The Falkland current is not a warm current; it is a cold current.

– Ocean currents are broadly classified as warm or cold based on their temperature relative to the surrounding water and whether they bring water from warmer or colder regions.
– The North Pacific Current is part of the North Pacific Gyre and is considered a warm current, carrying warm water eastward across the Pacific.
– The Falkland Current (also known as the Malvinas Current) is a cold current flowing northward along the Atlantic coast of Patagonia, originating from the Antarctic Circumpolar Current.
– The North Equatorial Current (e.g., North Atlantic or North Pacific) flows westward in the tropical regions and carries warm water.
– The Canary Current is a cold current flowing southward along the northwestern coast of Africa, part of the North Atlantic Gyre’s eastern boundary.

The question asks for the current that is “not a warm current”. Both the Falkland Current (B) and the Canary Current (D) are classified as cold currents. However, typically in such multiple-choice questions, only one option is intended as correct. Based on common interpretations of this specific question structure in similar exams, the Falkland Current is often identified as the intended ‘not warm’ (cold) current among these options. The Falkland Current’s connection to Antarctic waters makes it a very distinct cold current.

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3. The lowest temperature is observed/ recorded in which one of the follo

The lowest temperature is observed/ recorded in which one of the following layers of the atmosphere?

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Stratosphere

Mesosphere

Thermosphere

Troposphere

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UPSC CAPF – 2018

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The lowest temperature in the atmosphere is observed/recorded in the Mesosphere.

– The atmosphere is divided into several layers based on temperature profiles: Troposphere, Stratosphere, Mesosphere, Thermosphere, and Exosphere.
– In the Troposphere, temperature generally decreases with altitude.
– In the Stratosphere, temperature increases with altitude due to the absorption of ultraviolet radiation by the ozone layer.
– In the Mesosphere, temperature again decreases with altitude. The coldest temperatures in the Earth’s atmosphere, often dropping below -90°C (-130°F), are found at the top of the mesosphere (the mesopause).
– In the Thermosphere, temperature increases sharply with altitude due to the absorption of high-energy solar radiation.

The mesopause, the boundary between the mesosphere and the thermosphere, is typically the coldest region of the entire atmosphere. Meteors often burn up in the mesosphere.

4. Which one of the following pairs is not correctly matched?

Which one of the following pairs is not correctly matched?

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Indira Gandhi Rashtriya Uran Akademi : Fursatganj

National Flying Training Institute : Belgaum

National Institute of Hydrology : Roorkee

National Water Academy (Headquarters) : Khadakwasla

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UPSC CAPF – 2018

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The pair “National Flying Training Institute : Belgaum” is not correctly matched.

– Indira Gandhi Rashtriya Uran Akademi (IGRUA) is located at Fursatganj, Amethi, Uttar Pradesh. This pair is correctly matched.
– National Flying Training Institute (NFTI) is located at Gondia, Maharashtra. It is not located at Belgaum (Belagavi). This pair is incorrectly matched.
– National Institute of Hydrology (NIH) is located at Roorkee, Uttarakhand. This pair is correctly matched.
– National Water Academy (NWA) Headquarters is located at Khadakwasla, Pune, Maharashtra. This pair is correctly matched.

IGRUA is a premier flying training institute in India. NFTI is another significant flying training academy. NIH is a research and development institution in the field of hydrology. NWA provides training to water resource professionals.

5. Which of the following statements are appropriate to Mangrove Sites?

Which of the following statements are appropriate to Mangrove Sites?

1. Mangrove plants require appropriate mix of saline water and freshwater.
2. Mangrove plants require mudflats to enable it to grow and develop.
3. Mangrove plants are found in the inter-tidal zones of sheltered coasts.
4. Mangrove vegetation has been reported in all the coastal States including Andaman and Nicobar Islands.

Select the correct answer using the code given below.

1 and 2 only

2, 3 and 4 only

1, 3 and 4 only

1, 2, 3 and 4

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UPSC CAPF – 2018

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All four statements provided regarding Mangrove Sites are appropriate and factually correct.

– Mangrove plants are adapted to brackish environments, requiring a mix of saline and freshwater typically found in estuaries and coastal areas.
– They grow well on soft, muddy substrates characteristic of mudflats, which provide stability for their root systems.
– Mangrove ecosystems are primarily located in the inter-tidal zones along sheltered coastlines where sediment deposition occurs and wave action is limited.
– Mangrove vegetation, in varying degrees of cover, has been reported in all the coastal States (Gujarat, Maharashtra, Goa, Karnataka, Kerala, Tamil Nadu, Andhra Pradesh, Odisha, West Bengal) and Union Territories (Andaman & Nicobar Islands, Puducherry) of India.

Mangroves play a crucial ecological role in coastal protection, nurseries for marine life, and carbon sequestration. India has significant mangrove cover, notably in the Sundarbans, Gujarat coast, and Andaman & Nicobar Islands. Conservation efforts are ongoing to protect and restore these vital ecosystems.

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6. Which one of the following sites does not have coral reef formation?

Which one of the following sites does not have coral reef formation?

Gulf of Cambay/Khambhat

Gulf of Mannar

Lakshadweep

Andaman and Nicobar Islands

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UPSC CAPF – 2018

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The correct answer is A) Gulf of Cambay/Khambhat.

Coral reefs require specific environmental conditions, including warm, clear, shallow saline water with adequate sunlight penetration.
– Gulf of Mannar, Lakshadweep, and Andaman and Nicobar Islands are well-known regions in India with significant coral reef formations.
– The Gulf of Mannar is part of a Marine National Park known for its rich marine biodiversity including coral reefs.
– Lakshadweep consists of atolls, which are ring-shaped coral reefs surrounding a lagoon.
– The Andaman and Nicobar Islands are surrounded by fringing and barrier reefs.
– The Gulf of Cambay (Khambhat) is an estuary formed by the confluence of major rivers like the Narmada, Tapti, Mahi, and Sabarmati. The large influx of fresh water and high sediment load from these rivers lead to turbid, brackish conditions which are generally unfavorable for the growth and survival of most coral species.

While some limited coral growth might occur in marginal conditions, extensive reef development is not found in the Gulf of Cambay due to the high freshwater input and turbidity. The other mentioned locations represent India’s four major coral reef areas.

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7. If a cubical container of length, breadth and height each of 10 cm can

If a cubical container of length, breadth and height each of 10 cm can contain exactly 1 litre of water, then a spherical container of radius 10.5 cm can contain

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not more than 4 litres of water

more than 4 litres but less than 4.5 litres of water

more than 4.5 litres but less than 5 litres of water

more than 5 litres of water

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UPSC CAPF – 2018

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The correct answer is C) more than 4.5 litres but less than 5 litres of water.

A cubical container of 10 cm side length has a volume of 10 cm * 10 cm * 10 cm = 1000 cm³.
Given that this container holds exactly 1 litre, we know 1 litre = 1000 cm³.
A spherical container has a radius of 10.5 cm.
The volume of a sphere is given by the formula V = (4/3)πr³.
Using π ≈ 22/7:
V = (4/3) * (22/7) * (10.5)³
V = (4/3) * (22/7) * (10.5 * 10.5 * 10.5)
V = (4/3) * (22/7) * (1157.625)
V = (88/21) * 1157.625
V = 4.1904… * 1157.625
V ≈ 4851 cm³
To convert this volume to litres, divide by 1000:
Volume in litres = 4851 cm³ / 1000 cm³/litre = 4.851 litres.
Now compare 4.851 litres with the options:
A) not more than 4 litres (4.851 is more than 4) – Incorrect
B) more than 4 litres but less than 4.5 litres (4.851 is not less than 4.5) – Incorrect
C) more than 4.5 litres but less than 5 litres (4.5 < 4.851 < 5) - Correct D) more than 5 litres (4.851 is not more than 5) - Incorrect

The value of π used can affect the precision slightly. Using a more precise value like 3.14159 would give V = (4/3) * 3.14159 * (10.5)³ ≈ 4849.05 cm³, which is 4.84905 litres. This value also falls within the range of more than 4.5 litres but less than 5 litres.

8. A king ordered to make a crown from 8 kg of gold and 2 kg of silver. T

A king ordered to make a crown from 8 kg of gold and 2 kg of silver. The goldsmith took away some amount of gold and replaced it by an equal amount of silver and the crown when made, weighed 10 kg. The king knows that under water gold loses $\frac{1}{20}$th of its weight, while silver loses $\frac{1}{10}$th. When the crown was weighed under water, it was 9.25 kg. How much gold was stolen by the goldsmith?

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1 kg

2 kg

3 kg

4 kg

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UPSC CAPF – 2018

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The correct answer is C) 3 kg.

Initial composition: 8 kg gold, 2 kg silver, total 10 kg.
Let ‘x’ kg of gold be stolen and replaced by ‘x’ kg of silver.
Final composition: (8-x) kg gold, (2+x) kg silver. Total weight is (8-x) + (2+x) = 10 kg.
Under water, gold loses 1/20th of its weight, so its apparent weight is (1 – 1/20) = 19/20 of its actual weight.
Under water, silver loses 1/10th of its weight, so its apparent weight is (1 – 1/10) = 9/10 of its actual weight.
The crown weighs 9.25 kg under water.
Apparent weight of gold = (19/20) * (8-x)
Apparent weight of silver = (9/10) * (2+x)
Total apparent weight = (19/20) * (8-x) + (9/10) * (2+x) = 9.25
Multiply by 20 to clear the denominators:
19 * (8-x) + 18 * (2+x) = 9.25 * 20
152 – 19x + 36 + 18x = 185
188 – x = 185
x = 188 – 185
x = 3 kg.
The goldsmith stole 3 kg of gold.

This problem is an application of Archimedes’ principle. The weight loss under water is equal to the weight of the water displaced, which is proportional to the volume of the object. Different densities of gold and silver cause different weight losses for the same weight. Pure gold is much denser than silver, so 1 kg of gold occupies less volume than 1 kg of silver and displaces less water, thus losing less weight. Replacing gold with silver increases the overall volume for the same weight, leading to a greater weight loss under water. The initial weight loss for the pure crown would have been (1/20)*8 + (1/10)*2 = 0.4 + 0.2 = 0.6 kg, making its underwater weight 10 – 0.6 = 9.4 kg. The actual underwater weight is 9.25 kg, a loss of 0.75 kg (10 – 9.25). The increased loss of 0.15 kg (0.75 – 0.6) is due to the replacement of ‘x’ kg of gold with ‘x’ kg of silver. The difference in weight loss per kg is (1/10) – (1/20) = 1/20. So, the increased loss is (1/20)*x = 0.15 kg. x = 0.15 * 20 = 3 kg.

9. Suppose a, b, c, d and e are five consecutive odd numbers in ascending

Suppose a, b, c, d and e are five consecutive odd numbers in ascending order. Consider the following statements:

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1. Their average is (a+4).
2. Their average is (b+2).
3. Their average is (e-4).

Which of the statements given above is/are correct?

1 only

2 and 3 only

1 and 3 only

1, 2 and 3

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UPSC CAPF – 2018

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The correct answer is D) 1, 2 and 3.

Let the five consecutive odd numbers in ascending order be a, a+2, a+4, a+6, and a+8. Thus, b=a+2, c=a+4, d=a+6, e=a+8.
The average of these five numbers is (a + (a+2) + (a+4) + (a+6) + (a+8)) / 5 = (5a + 20) / 5 = a + 4.
Now let’s check the statements:
1. Their average is (a+4). This is correct.
2. Their average is (b+2). Since b = a+2, b+2 = (a+2)+2 = a+4. This is correct.
3. Their average is (e-4). Since e = a+8, e-4 = (a+8)-4 = a+4. This is correct.
All three statements are correct.

For any sequence of consecutive numbers (arithmetic progression), the average is equal to the median. In this case, c is the middle number, and c = a+4. The average is also the average of the first and last term: (a+e)/2 = (a + a+8)/2 = (2a+8)/2 = a+4. Similarly, the average of the second and fourth term: (b+d)/2 = (a+2 + a+6)/2 = (2a+8)/2 = a+4.

10. If the number 2² x 5⁴ x 4⁶ x 10⁸ x 6¹⁰ x 15¹² x 8¹⁴ x 20¹⁶ x 10¹⁸ x 25

If the number 2² x 5⁴ x 4⁶ x 10⁸ x 6¹⁰ x 15¹² x 8¹⁴ x 20¹⁶ x 10¹⁸ x 25²⁰ is divisible by 10ⁿ, then which one of the following is the maximum value of n?

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UPSC CAPF – 2018

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The correct answer is D) 98.

A number is divisible by 10ⁿ if its prime factorization contains at least n factors of 2 and n factors of 5. The maximum value of n is determined by the minimum exponent of 2 and 5 in the prime factorization of the given number.
The given number is N = 2² x 5⁴ x 4⁶ x 10⁸ x 6¹⁰ x 15¹² x 8¹⁴ x 20¹⁶ x 10¹⁸ x 25²⁰.
Let’s express all terms in terms of prime factors 2, 3, 5:
N = 2² * 5⁴ * (2²)⁶ * (2*5)⁸ * (2*3)¹⁰ * (3*5)¹² * (2³)¹⁴ * (2²*5)¹⁶ * (2*5)¹⁸ * (5²)²⁰
N = 2² * 5⁴ * 2¹² * 2⁸ * 5⁸ * 2¹⁰ * 3¹⁰ * 3¹² * 5¹² * 2⁴² * 2³² * 5¹⁶ * 2¹⁸ * 5¹⁸ * 5⁴⁰

Total power of 2: 2 + 12 + 8 + 10 + 42 + 32 + 18 = 124. So, 2¹²⁴.
Total power of 5: 4 + 8 + 12 + 16 + 18 + 40 = 98. So, 5⁹⁸.
Total power of 3: 10 + 12 = 22. So, 3²².

The number is 2¹²⁴ * 3²² * 5⁹⁸.
For this number to be divisible by 10ⁿ = 2ⁿ * 5ⁿ, the maximum possible value of n is min(124, 98) = 98.

Divisibility by 10ⁿ is equivalent to divisibility by (2*5)ⁿ, which means divisibility by 2ⁿ and 5ⁿ simultaneously. The number of trailing zeros in an integer is equal to the exponent of 10 in its prime factorization, which is the minimum of the exponents of 2 and 5.

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