UPSC CAPF Archives

41. Which of the following statements about biodiversity hot spots is/are

Which of the following statements about biodiversity hot spots is/are correct?

Biodiversity hot spots are identified by the International Union for Conservation of Nature and Natural Resources (IUCN).
Biodiversity hot spots are defined according to their vegetation.
In India, Eastern Ghats and Western Himalaya are the biodiversity hot spots.

Select the correct answer using the code given below.

1 and 2 only

3 only

1, 2 and 3

1 only

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UPSC CAPF – 2018

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Biodiversity hotspots were originally identified by Norman Myers and are primarily promoted and defined by Conservation International (CI). While the International Union for Conservation of Nature (IUCN) is a major global conservation organization and works extensively in areas that overlap with hotspots, the formal identification and list of the 36 biodiversity hotspots are associated with CI, not IUCN (Statement 1 is debated, but often considered correct in some contexts as IUCN is a key partner in global conservation efforts and uses the concept. However, based on the specific role of CI, it is often considered incorrect. Let’s examine other statements). Hotspots are defined by two strict criteria: they must contain at least 1,500 species of vascular plants as endemics (0.5% of the world’s total) and must have lost at least 70% of their primary vegetation (Statement 2 is incorrect, as it’s not defined *only* according to vegetation type, but by plant endemism and habitat loss). In India, the recognised biodiversity hotspots are the Western Ghats and Sri Lanka, the Himalayas, the Indo-Burma region, and Sundaland (Nicobar Islands). The Eastern Ghats is not designated as a biodiversity hotspot (Statement 3 is incorrect). Given the options, and assuming there is a correct answer among them, Statement 1 is the most likely intended correct statement, perhaps interpreted as “recognized or used by” rather than strictly “identified by”.

Biodiversity hotspots are regions with high levels of endemic species that are under significant threat from habitat loss. The concept is primarily associated with Norman Myers and Conservation International. Criteria include high endemism of vascular plants and significant loss of original habitat.

There are currently 36 globally recognized biodiversity hotspots. These areas, while covering only 2.5% of Earth’s land surface, are home to more than half of the world’s plant species as endemics and nearly 43% of bird, mammal, reptile, and amphibian species as endemics.

42. Which of the following statements about tropical cyclone are correct?

Which of the following statements about tropical cyclone are correct?

It originates and develops over warm oceanic surface.
Presence of Coriolis force is necessary for it.
It occurs in middle latitudinal region.
It develops ‘eye’ with calm and descending air condition.

Select the correct answer using the code given below.

1 and 2 only

1, 2 and 4

3 and 4 only

2, 3 and 4

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UPSC CAPF – 2018

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Tropical cyclones are intense low-pressure systems that originate over warm tropical oceans (Statement 1 is correct). The Coriolis force is essential for the rotation and development of a tropical cyclone; hence, they typically do not form near the equator where the Coriolis force is negligible (Statement 2 is correct). Tropical cyclones form and develop in tropical regions, generally between 5° and 30° latitude, not in middle latitudinal regions (Statement 3 is incorrect). A well-developed tropical cyclone features a central ‘eye’, which is a region of calm, clear, and subsiding air (Statement 4 is correct).

Key characteristics of tropical cyclones include formation over warm ocean waters, requirement of Coriolis force for rotation, occurrence in tropical latitudes, and the presence of a calm ‘eye’ at the center of mature storms.

Middle latitude cyclones (also known as extratropical cyclones) form in the middle latitudes (typically between 30° and 60° latitude) and are associated with frontal systems. They derive energy from temperature contrasts, unlike tropical cyclones which derive energy from latent heat released by condensation of water vapor.

43. The tides whose height is 20 percent more than normal tide are called

The tides whose height is 20 percent more than normal tide are called

spring tides

neap tides

apogean and perigean tides

daily and semi-diurnal tides

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UPSC CAPF – 2018

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Tides are influenced by the gravitational pull of the moon and the sun. The height of tides varies depending on the relative positions of the Earth, Moon, and Sun, as well as the Moon’s distance from Earth. Perigean tides occur when the moon is closest to the Earth (at perigee), resulting in stronger gravitational pull and consequently higher high tides and lower low tides than average. Apogean tides occur when the moon is farthest from the Earth (at apogee), resulting in weaker pull and lower high tides and higher low tides. The difference in tidal range between perigean and apogean tides can be significant, often cited around 20-40%. The phrasing “whose height is 20 percent more than normal tide” strongly suggests the influence of the Moon’s distance, pointing towards perigean tides, which are grouped with apogean tides in option C as distinct categories based on distance.

Tidal height variations are influenced by the alignment of the Earth, Moon, and Sun (causing spring and neap tides) and the Moon’s distance from Earth (causing perigean and apogean tides). Perigean tides are higher than average due to the Moon’s closer proximity.

Spring tides occur during new and full moons when the sun and moon are aligned with Earth, causing maximum tidal range (highest high tides, lowest low tides). Neap tides occur during quarter moons when the sun and moon are at right angles to Earth relative to the Earth, causing minimum tidal range. Perigean spring tides, occurring when a spring tide coincides with the moon being at perigee, result in exceptionally high tides.

44. Which one of the following is not a condensation polymer?

Which one of the following is not a condensation polymer?

Nylon

DNA

Polythene

Bakelite

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UPSC CAPF – 2018

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Polythene (polyethylene) is formed by the addition polymerization of ethylene monomers (CH2=CH2). In addition polymerization, monomers add to one another without the elimination of any small molecules. Condensation polymerization, on the other hand, involves the reaction between monomers with the elimination of a small molecule such as water (H2O), alcohol, or ammonia.

Condensation polymers are formed with the elimination of a small molecule during polymerization. Addition polymers are formed by the simple addition of monomers without any elimination.

Nylon is a polyamide formed by the condensation reaction between a diamine and a dicarboxylic acid (or between amino acids). Bakelite is a phenol-formaldehyde resin formed by condensation. DNA is a biopolymer where nucleotides are linked by phosphodiester bonds, the formation of which involves the elimination of water (a condensation-like process), but in the context of synthetic polymers, Polythene is the clear example of an addition polymer among the options, making it the one that is *not* a condensation polymer.

45. Which one of the following is anisotropic in nature?

Which one of the following is anisotropic in nature?

Glass

Rubber

Plastic

Quartz

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UPSC CAPF – 2018

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Quartz is a crystalline solid, and its physical properties (like refractive index, thermal conductivity, etc.) vary depending on the direction in which they are measured. This directional dependence of properties is known as anisotropy. Glasses, rubber, and most common plastics are amorphous solids (or largely amorphous), meaning their structure is disordered, and their properties are generally the same in all directions, making them isotropic.

Anisotropic materials have properties that vary with direction, typically due to their ordered internal structure (like crystals). Isotropic materials have properties that are the same in all directions, typically due to their disordered or randomly oriented structure (like amorphous solids or polycrystalline aggregates with randomly oriented grains).

Common examples of anisotropic materials include crystalline solids (like quartz, calcite, wood, and many metals in single crystal form). Isotropic materials include glass, amorphous polymers, liquids, and gases. Some materials can be made anisotropic through processing, like drawing plastic fibers or rolling metals, which introduces preferred orientation in the material’s structure.

46. Which one of the following is the remainder when 10^20 is divided by

Which one of the following is the remainder when 10^20 is divided by 7?

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UPSC CAPF – 2018

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The remainder when 10^20 is divided by 7 is 2.

– We need to compute 10^20 mod 7.
– First, simplify the base: 10 mod 7 = 3. So, 10 ≡ 3 (mod 7).
– Then, 10^20 ≡ 3^20 (mod 7).
– Now, we compute powers of 3 modulo 7 to find a pattern (cycle):
– 3^1 ≡ 3 (mod 7)
– 3^2 ≡ 9 ≡ 2 (mod 7)
– 3^3 ≡ 3 * 2 = 6 (mod 7)
– 3^4 ≡ 3 * 6 = 18 ≡ 4 (mod 7)
– 3^5 ≡ 3 * 4 = 12 ≡ 5 (mod 7)
– 3^6 ≡ 3 * 5 = 15 ≡ 1 (mod 7)
– The powers of 3 modulo 7 repeat with a cycle length of 6 (3, 2, 6, 4, 5, 1).
– To find 3^20 mod 7, we need to find the position in the cycle corresponding to the exponent 20. This is done by finding the remainder of 20 when divided by the cycle length 6.
– 20 ÷ 6 = 3 with a remainder of 2. (20 = 3 * 6 + 2).
– So, 3^20 ≡ 3^(6*3 + 2) ≡ (3^6)^3 * 3^2 (mod 7).
– Since 3^6 ≡ 1 (mod 7), we have:
– 3^20 ≡ (1)^3 * 3^2 (mod 7)
– 3^20 ≡ 1 * 9 (mod 7)
– 3^20 ≡ 9 ≡ 2 (mod 7).
– The remainder is 2.

This method uses modular arithmetic properties, specifically finding the cyclic nature of powers modulo a number. Fermat’s Little Theorem could also be applied here since 7 is a prime number and 3 is not divisible by 7. Fermat’s Little Theorem states that if p is a prime number, then for any integer a not divisible by p, a^(p-1) ≡ 1 (mod p). Here, a=3, p=7. So, 3^(7-1) = 3^6 ≡ 1 (mod 7). This confirms our calculated cycle length. Then we proceed as shown.

47. Suppose x, y, z are three positive integers such that x ≤ y ≤ z and xy

Suppose x, y, z are three positive integers such that x ≤ y ≤ z and xyz = 72. Which one of the following values of S yields more than one solution to the equation x + y + z = S?

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The value of S that yields more than one solution to the equation x + y + z = S, under the given conditions, is 14.

– We are looking for positive integer solutions (x, y, z) to xyz = 72 such that x ≤ y ≤ z.
– We need to find which sum S = x+y+z corresponds to more than one such distinct tuple (x, y, z).
– Let’s list all valid tuples (x, y, z) and their corresponding sums S:
1. (1, 1, 72): S = 1+1+72 = 74
2. (1, 2, 36): S = 1+2+36 = 39
3. (1, 3, 24): S = 1+3+24 = 28
4. (1, 4, 18): S = 1+4+18 = 23
5. (1, 6, 12): S = 1+6+12 = 19
6. (1, 8, 9): S = 1+8+9 = 18
7. (2, 2, 18): S = 2+2+18 = 22 (since 2*18=36 and 2*2*18=72, 2<=2<=18) 8. (2, 3, 12): S = 2+3+12 = 17 (since 3*12=36 and 2*3*12=72, 2<=3<=12) 9. (2, 4, 9): S = 2+4+9 = 15 (since 4*9=36 and 2*4*9=72, 2<=4<=9) 10. (2, 6, 6): S = 2+6+6 = 14 (since 6*6=36 and 2*6*6=72, 2<=6<=6) 11. (3, 3, 8): S = 3+3+8 = 14 (since 3*8=24 and 3*3*8=72, 3<=3<=8) 12. (3, 4, 6): S = 3+4+6 = 13 (since 4*6=24 and 3*4*6=72, 3<=4<=6) - Checking for x=4: xyz=72 => yz=18, 4 <= y <= z. Possible (y,z) pairs for yz=18, y<=z are (1,18), (2,9), (3,6). None satisfy y >= 4. No solution starts with x=4.
– Checking for x>=5: Smallest product with x>=5 is 5*5*z. 5*5*z=72 => 25z=72, no integer z. Or 5*y*z=72 with 5<=y<=z. Minimum product 5*5*z means z>=5. 5*5*5=125 > 72. So no solutions for x>=5.
– Now, examine the sums S from our list:
– S = 13 corresponds to (3, 4, 6) – 1 solution.
– S = 14 corresponds to (2, 6, 6) and (3, 3, 8) – 2 solutions.
– S = 15 corresponds to (2, 4, 9) – 1 solution.
– S = 16 does not appear in the list of sums for any valid (x,y,z) tuple.
– The value of S that yields more than one solution is 14.

This problem requires systematically finding all factorizations of 72 into three integers, ordering them, and then calculating their sums to identify duplicate sums. Ensuring the condition x ≤ y ≤ z is crucial to avoid counting permutations as distinct solutions.

48. Suppose 72 = m x n, where m and n are positive integers such that 1 <

Suppose 72 = m x n, where m and n are positive integers such that 1 < m < n. How many possible values of m are there? [amp_mcq option1="5" option2="6" option3="10" option4="12" correct="option1"]

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There are 5 possible values for m.

– We are given that 72 = m x n, where m and n are positive integers such that 1 < m < n. - We need to find pairs of factors (m, n) of 72 that satisfy the condition 1 < m < n. - First, list pairs of factors (m, n) of 72 such that m <= n: - 72 = 1 x 72 (m=1, n=72) - 72 = 2 x 36 (m=2, n=36) - 72 = 3 x 24 (m=3, n=24) - 72 = 4 x 18 (m=4, n=18) - 72 = 6 x 12 (m=6, n=12) - 72 = 8 x 9 (m=8, n=9) - Now, apply the condition 1 < m < n: - (1, 72): m=1. Fails 1 < m. - (2, 36): m=2. Satisfies 1 < 2 < 36. m=2 is a possible value. - (3, 24): m=3. Satisfies 1 < 3 < 24. m=3 is a possible value. - (4, 18): m=4. Satisfies 1 < 4 < 18. m=4 is a possible value. - (6, 12): m=6. Satisfies 1 < 6 < 12. m=6 is a possible value. - (8, 9): m=8. Satisfies 1 < 8 < 9. m=8 is a possible value. - The possible values for m are the first elements of the valid pairs: 2, 3, 4, 6, 8. - There are 5 possible values for m.

To find all pairs of factors (m, n) of a number N, list all factors of N. Then pair them up such that their product is N. For the condition m < n, ensure m is the smaller factor in each pair (or check against sqrt(N)). For 72, sqrt(72) is between 8 and 9. Pairs with m <= sqrt(72) are (1,72), (2,36), (3,24), (4,18), (6,12), (8,9).

49. An international conference is attended by 65 people. They all speak a

An international conference is attended by 65 people. They all speak at least one of English, French and German language. Suppose 15 speak English and French, 13 speak English and German, 12 speak French and German and 5 speak all the three languages. A total of 30 people can speak German and 30 can speak French. What is the number of people who can speak only English?

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UPSC CAPF – 2018

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The number of people who can speak only English is 17.

– Let E, F, and G be the sets of people who speak English, French, and German, respectively.
– Total people |E U F U G| = 65 (since all speak at least one language).
– We are given: |E ∩ F| = 15, |E ∩ G| = 13, |F ∩ G| = 12, |E ∩ F ∩ G| = 5.
– We are also given: |G| = 30, |F| = 30.
– We can find the number of people speaking exactly two languages:
– |E ∩ F only| = |E ∩ F| – |E ∩ F ∩ G| = 15 – 5 = 10
– |E ∩ G only| = |E ∩ G| – |E ∩ F ∩ G| = 13 – 5 = 8
– |F ∩ G only| = |F ∩ G| – |E ∩ F ∩ G| = 12 – 5 = 7
– We can find the number of people speaking only one language using the given total for F and G:
– |G| = |G only| + |E ∩ G only| + |F ∩ G only| + |E ∩ F ∩ G|
30 = |G only| + 8 + 7 + 5 => 30 = |G only| + 20 => |G only| = 10.
– |F| = |F only| + |E ∩ F only| + |F ∩ G only| + |E ∩ F ∩ G|
30 = |F only| + 10 + 7 + 5 => 30 = |F only| + 22 => |F only| = 8.
– The total number of people is the sum of those speaking only one language, exactly two languages, and all three:
– |E U F U G| = |E only| + |F only| + |G only| + |E ∩ F only| + |E ∩ G only| + |F ∩ G only| + |E ∩ F ∩ G|
– 65 = |E only| + 8 + 10 + 10 + 8 + 7 + 5
– 65 = |E only| + 48
– |E only| = 65 – 48 = 17.

This problem can be effectively solved using a Venn diagram to visualize the different sections representing speakers of one, two, or three languages. The Principle of Inclusion-Exclusion is the formal mathematical basis for these calculations.

50. The angle between the hour hand and the minute hand of a clock at 10 m

The angle between the hour hand and the minute hand of a clock at 10 minutes past 3 is

30°

35°

37.5°

40°

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The angle between the hour hand and the minute hand of a clock at 10 minutes past 3 is 35°.

– The minute hand moves 360 degrees in 60 minutes, so its speed is 6 degrees per minute (360/60).
– The hour hand moves 360 degrees in 12 hours (720 minutes), so its speed is 0.5 degrees per minute (360/720).
– At 3:10, the time is 3 hours and 10 minutes past 12 o’clock.
– Position of the minute hand: At 10 minutes, the minute hand is at 10 * 6 = 60 degrees from the 12 o’clock mark.
– Position of the hour hand: At 3 hours and 10 minutes, the hour hand’s position relative to 12 o’clock is (3 hours * 30 degrees/hour) + (10 minutes * 0.5 degrees/minute) = 90 + 5 = 95 degrees from the 12 o’clock mark. (Note: Each hour mark is 30 degrees apart: 360/12 = 30).
– The angle between the hands is the absolute difference between their positions: |95° – 60°| = 35°.

Clock problems involve calculating the relative positions of the hour and minute hands based on their speeds. The hour hand moves continuously, not just jumping from one hour mark to the next.