UPSC CISF-AC-EXE Archives

21. In which one of the following kingdoms, would you place an organism wh

In which one of the following kingdoms, would you place an organism which is eukaryotic, multi-cellular and non-photosynthetic ?

Protista

Monera

Fungi

Animalia

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The question asks for a kingdom containing an organism that is eukaryotic, multi-cellular, and non-photosynthetic. Let’s examine the options based on these characteristics.

– Monera: Prokaryotic. Does not fit the ‘eukaryotic’ criterion.
– Protista: Eukaryotic, but mostly unicellular. While some multi-cellular forms exist (like some algae, which are photosynthetic) and some non-photosynthetic forms exist (like amoeba, which are unicellular), being *both* multi-cellular and non-photosynthetic is not a defining characteristic of the kingdom as a whole.
– Fungi: Eukaryotic (fits). Mostly multi-cellular (fits, except for yeasts). Non-photosynthetic (all are heterotrophic, so fits). Fungi contain organisms like mushrooms that are eukaryotic, multicellular, and non-photosynthetic.
– Animalia: Eukaryotic (fits). Multi-cellular (all are, fits perfectly). Non-photosynthetic (all are heterotrophic, fits perfectly). Animalia contain organisms like humans or insects that are eukaryotic, multicellular, and non-photosynthetic.

Both Fungi and Animalia fit the criteria of containing organisms that are eukaryotic, multicellular, and non-photosynthetic. However, all members of the Kingdom Animalia are multicellular, whereas the Kingdom Fungi includes unicellular organisms like yeasts. Given that all three criteria perfectly describe all members of the Kingdom Animalia, and multicellularity is not universal in Fungi, Animalia is a stronger fit as a kingdom primarily characterized by these features among the options, when compared to Fungi. The question asks in which kingdom *an organism* with these properties would be placed, and Animalia universally contains such organisms.

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22. Infectious diseases can spread through various means, such as air (sne

Infectious diseases can spread through various means, such as air (sneezes and coughs), water, food, insects and vectors. Which one of the following diseases does not have a vector ?

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AIDS

Malaria

Dengue

Zika

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A vector in the context of infectious diseases is an organism, typically an arthropod like a mosquito or tick, that transmits a pathogen from one host to another.

AIDS (Acquired Immunodeficiency Syndrome) is caused by the Human Immunodeficiency Virus (HIV). HIV is primarily transmitted through direct contact with infected bodily fluids, such as blood, semen, vaginal fluids, and breast milk. It is not transmitted through insect bites or other vectors.

Malaria is transmitted by the Anopheles mosquito. Dengue is transmitted by the Aedes mosquito. Zika is primarily transmitted by the Aedes mosquito, although it can also be transmitted sexually or from mother to fetus. These diseases rely on a vector organism for transmission. AIDS transmission pathways are sexual contact, parenteral (blood/needles), and mother-to-child (vertical transmission), none of which involve a vector organism in the typical sense.

23. Which one of the following is known as the life-supporting zone of the

Which one of the following is known as the life-supporting zone of the Earth ?

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Ecosystem

Biosphere

Atmosphere

Biome

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The biosphere is the sum of all ecosystems on Earth. It is the narrow layer of the Earth where life exists, encompassing parts of the land (lithosphere), water (hydrosphere), and air (atmosphere).

The biosphere is essentially the life-supporting zone of the Earth because it is the region where living organisms interact with the physical environment.

An ecosystem is a community of living organisms interacting with each other and their physical environment in a specific area. The atmosphere is the layer of gases surrounding the Earth. A biome is a large area characterized by its climate and the plant and animal communities adapted to that climate (e.g., forest, desert, grassland). While these are all components related to the Earth and its life, the biosphere is the overarching term for the entire zone where life is found.

24. The organic component of soil, formed by the decomposition of leaves a

The organic component of soil, formed by the decomposition of leaves and other plant materials that adds to the fertility of the soil is called :

Humus.

Top soil.

B-horizon.

Mineralization.

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Humus is the stable organic component of soil that results from the decomposition of dead plant and animal matter by microorganisms.

Humus is dark in color, rich in nutrients essential for plant growth, and significantly improves soil structure, water retention, and aeration. It is a key indicator of soil fertility.

Top soil (A-horizon) is the uppermost layer of soil, typically the richest in organic matter (including humus) and biological activity. B-horizon (subsoil) is below the topsoil and generally contains less organic matter but may accumulate minerals leached from the A-horizon. Mineralization is a process of decomposition where organic matter is broken down into inorganic substances, releasing nutrients, but it refers to the process, not the organic component itself.

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25. When the square of the sum of two numbers are added to the square of t

When the square of the sum of two numbers are added to the square of their difference, we get 416. The difference between the square of the sum and square of the difference is 384. What are the numbers ?

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18 and 24

12 and 16

8 and 12

10 and 12

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Let the two numbers be ‘a’ and ‘b’. The sum of the numbers is (a + b) and the difference is (a – b). The problem gives two conditions based on the squares of the sum and difference.
Condition 1: The square of the sum added to the square of the difference is 416.
(a + b)² + (a – b)² = 416.
Using the identities (a+b)² = a² + 2ab + b² and (a-b)² = a² – 2ab + b², we get:
(a² + 2ab + b²) + (a² – 2ab + b²) = 416.
2a² + 2b² = 416.
Dividing by 2, we get: a² + b² = 208. (Equation 1)

Condition 2: The difference between the square of the sum and the square of the difference is 384.
(a + b)² – (a – b)² = 384.
Using the identities again:
(a² + 2ab + b²) – (a² – 2ab + b²) = 384.
a² + 2ab + b² – a² + 2ab – b² = 384.
4ab = 384.
Dividing by 4, we get: ab = 96. (Equation 2)

We now have a system of two equations:
1) a² + b² = 208
2) ab = 96
We can check the options given to see which pair of numbers satisfies both equations.
Option C gives the numbers 8 and 12.
Let’s test if a=8 and b=12 satisfy the equations:
Equation 1: 8² + 12² = 64 + 144 = 208. (Satisfied)
Equation 2: 8 * 12 = 96. (Satisfied)
Since the numbers 8 and 12 satisfy both conditions derived from the problem statement, they are the correct numbers. Alternatively, we can solve the system using the identities (a+b)² = a²+b²+2ab and (a-b)² = a²+b²-2ab.
(a+b)² = 208 + 2(96) = 208 + 192 = 400 => a+b = √400 = 20 (assuming positive numbers).
(a-b)² = 208 – 2(96) = 208 – 192 = 16 => a-b = √16 = 4 (assuming a>b).
Solving a+b=20 and a-b=4: Adding gives 2a=24 => a=12. Substituting gives 12+b=20 => b=8. The numbers are 12 and 8.

26. In order to convert a loss of 5% to a profit of 5%, a shopkeeper raise

In order to convert a loss of 5% to a profit of 5%, a shopkeeper raises the price of an item by ₹ 500. What is the cost price of the item ?

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₹ 1,000

₹ 5,000

₹ 10,000

₹ 1,200

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Let the cost price (CP) of the item be ₹ x. Initially, the shopkeeper incurs a loss of 5%. The selling price (SP1) in this case is CP – 5% of CP = x – 0.05x = 0.95x. To convert this to a profit of 5%, the new selling price (SP2) must be CP + 5% of CP = x + 0.05x = 1.05x.

The problem states that the shopkeeper raises the price of the item by ₹ 500 to achieve the desired change. This means the difference between the new selling price (SP2) and the original selling price (SP1) is ₹ 500.
SP2 – SP1 = ₹ 500.
(1.05x) – (0.95x) = 500.

Subtracting the two expressions for the selling prices in terms of CP (x):
(1.05 – 0.95) * x = 500.
0.10 * x = 500.
To find x (the cost price), divide 500 by 0.10:
x = 500 / 0.10 = 500 / (1/10) = 500 * 10 = ₹ 5,000.
The price increase of ₹ 500 represents the change from a 5% loss margin to a 5% profit margin, which is a total change of 10% of the cost price (5% below CP to 5% above CP). So, 10% of CP = 500, which directly leads to CP = 5000.

27. A shopkeeper gives a discount of 10% on his items. ₹ 280 is the marked

A shopkeeper gives a discount of 10% on his items. ₹ 280 is the marked price of an item. On selling this item, the shopkeeper earns a profit of 26%. What is the cost price of the item ?

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₹ 260

₹ 240

₹ 200

₹ 180

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The marked price (MP) of the item is ₹ 280. A discount of 10% is given. The selling price (SP) is calculated after applying the discount on the marked price.
Discount amount = 10% of ₹ 280 = (10/100) * 280 = ₹ 28.
Selling Price (SP) = Marked Price – Discount = ₹ 280 – ₹ 28 = ₹ 252.

The shopkeeper earns a profit of 26% on selling this item. Profit is calculated on the cost price (CP). The selling price is the cost price plus the profit.
SP = CP + Profit = CP + 26% of CP = CP + (26/100) * CP = CP * (1 + 0.26) = 1.26 * CP.

We have the selling price SP = ₹ 252 and the relationship SP = 1.26 * CP.
So, 252 = 1.26 * CP.
To find the cost price, divide the selling price by 1.26:
CP = 252 / 1.26 = 252 / (126/100) = (252 * 100) / 126.
Since 252 is twice 126 (2 * 126 = 252), the calculation simplifies:
CP = (2 * 126 * 100) / 126 = 2 * 100 = ₹ 200.

28. If the 1st of January of a non-leap year is a Monday, then which one o

If the 1st of January of a non-leap year is a Monday, then which one of the following days will the 31st of December of that year be ?

Monday

Tuesday

Sunday

None of the above

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A non-leap year has 365 days. To find the day of the week for the 31st of December given the day of the 1st of January, we need to find the number of odd days between these two dates. The number of days from January 1st to December 31st inclusive is 365.

The number of odd days in a given period is the remainder when the total number of days in the period is divided by 7. A non-leap year has 365 days. Dividing 365 by 7 gives a remainder of 1 (365 = 52 * 7 + 1). Therefore, there is 1 odd day in a non-leap year.

The day of the week repeats every 7 days. The day of the week for the day after the 1st of January will be (Day of 1st Jan + Number of days after 1st Jan) modulo 7, shifted appropriately. The 31st of December is 364 days after the 1st of January (365 total days – 1 day for Jan 1st). 364 days = 52 weeks (364 / 7 = 52). The number of odd days is 364 mod 7 = 0. Thus, the day of the week on December 31st will be the same as the day of the week on January 1st. If January 1st is a Monday, December 31st of a non-leap year will also be a Monday. A leap year has 366 days (52 weeks and 2 days), so the 31st of December in a leap year is one day ahead of the 1st of January.

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29. A mother gives pocket money to her three children A, B and C in the ra

A mother gives pocket money to her three children A, B and C in the ratio 3 : 4 : 5 respectively. Then the father gives ₹ 40 to each child. As a result, the pocket money of A, B and C now has the ratio 5 : 6 : 7 respectively. How much does C get from her mother ?

₹ 80

₹ 100

₹ 120

₹ 60

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The correct answer is B) ₹100.
Let the initial pocket money from the mother for children A, B, and C be 3x, 4x, and 5x respectively.
After the father gives ₹40 to each child, their pocket money becomes:
A: 3x + 40
B: 4x + 40
C: 5x + 40
The new ratio of their pocket money is 5:6:7.
We can set up a proportion using any two children’s new amounts and their corresponding ratio:
(3x + 40) / (4x + 40) = 5 / 6
Cross-multiply:
6(3x + 40) = 5(4x + 40)
18x + 240 = 20x + 200
240 – 200 = 20x – 18x
40 = 2x
x = 20
The question asks for how much C gets from her mother, which is the initial amount for C.
Initial amount for C = 5x = 5 * 20 = ₹100.

– Represent the initial amounts using a common variable based on the ratio.
– Add the fixed amount received from the father to each child’s money.
– Use the new ratio to form an equation and solve for the variable.
– Calculate the initial amount for C using the value of the variable.

We can verify the new amounts and their ratio:
A: 3(20) + 40 = 60 + 40 = 100
B: 4(20) + 40 = 80 + 40 = 120
C: 5(20) + 40 = 100 + 40 = 140
The new ratio is 100:120:140, which simplifies to 10:12:14, and further to 5:6:7, confirming the value of x is correct.

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30. A train of 150 m length crosses a milestone in 10 seconds. Another tra

A train of 150 m length crosses a milestone in 10 seconds. Another train with the same speed crosses the milestone in 12 seconds. What is the length of the second train ?

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180 m

120 m

200 m

100 m

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The correct answer is A) 180 m. When a train crosses a milestone, it travels a distance equal to its own length.
For the first train:
Length (L1) = 150 m
Time taken (t1) = 10 seconds
Speed (S) = Distance / Time = L1 / t1 = 150 m / 10 s = 15 m/s.
The second train has the same speed (S = 15 m/s).
Time taken by the second train (t2) = 12 seconds.
Length of the second train (L2) = Speed × Time = S × t2 = 15 m/s × 12 s = 180 m.

– Speed = Distance / Time
– Distance covered by a train crossing a milestone is its own length.
– Both trains have the same speed.

This is a standard problem involving speed, distance, and time for trains. The key is recognizing that crossing a point object (like a milestone) means the train covers a distance equal to its length.