UPSC CISF-AC-EXE Archives

31. The kingdom Plantae as laid down by R. Whittaker comprises which of th

The kingdom Plantae as laid down by R. Whittaker comprises which of the following group of plants?

Bryophyta, Pteridophyta, Protista, Gymnosperms and Angiosperms

Thallophyta, Pteridophyta, Fungi, Gymnosperms and Angiosperms

Thallophyta, Bryophyta, Monera, Pteridophyta and Gymnosperms

Thallophyta, Bryophyta, Pteridophyta, Gymnosperms and Angiosperms

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The kingdom Plantae as laid down by R. Whittaker comprises Thallophyta, Bryophyta, Pteridophyta, Gymnosperms and Angiosperms.

– R.H. Whittaker proposed the five-kingdom classification system: Monera, Protista, Fungi, Plantae, and Animalia.
– According to Whittaker’s system, the kingdom Plantae includes multicellular eukaryotic organisms that are autotrophic (perform photosynthesis) and have cell walls primarily made of cellulose.
– The major groups traditionally placed under Plantae in Whittaker’s system are Algae (specifically multicellular forms), Bryophytes, Pteridophytes, Gymnosperms, and Angiosperms.
– The term ‘Thallophyta’ is an older classification group (or a descriptive term) that included organisms with an undifferentiated body structure (thallus), such as algae, fungi, and lichens. In Whittaker’s system, Fungi were placed in a separate kingdom, and Algae were distributed among Protista (unicellular eukaryotes) and Plantae (multicellular eukaryotes).
– Option D lists Thallophyta, Bryophyta, Pteridophyta, Gymnosperms, and Angiosperms. Given the other options include members of Monera, Protista, and Fungi (separate kingdoms), option D is the most accurate representation of the groups typically included in Whittaker’s Plantae. ‘Thallophyta’ in this context likely refers to the multicellular algae forms included in Plantae by Whittaker, such as some green algae (e.g., Spirogyra, Ulothrix) and brown/red algae.

Whittaker’s five-kingdom system was a significant step in biological classification, separating organisms based on cell structure, mode of nutrition, and ecological role. While ‘Thallophyta’ is somewhat an outdated term as a formal taxon in modern cladistics, its inclusion in Option D reflects the historical context and how these groups were often discussed under the umbrella of Plantae in earlier classification frameworks like Whittaker’s, distinguishing them from Embryophytes (Bryophytes, Pteridophytes, Gymnosperms, Angiosperms).

32. A diameter PQ is drawn to a circle whose diameter length is 1 m. A squ

A diameter PQ is drawn to a circle whose diameter length is 1 m. A square is drawn using the diameter PQ as one of its sides. Assuming that $\pi$ is 22/7, what is the area of the part of the square lying outside the circle ?

3/28 sq. m

11/28 sq. m

15/28 sq. m

17/28 sq. m

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The area of the part of the square lying outside the circle is 17/28 sq. m.

– The diameter PQ of the circle has length 1 m. The radius of the circle is $1/2$ m.
– A square is drawn using the diameter PQ as one of its sides. This means the side length of the square is equal to the length of the diameter, which is 1 m.
– The area of the square is side * side = $1^2 = 1$ sq. m.
– The area of the circle is $\pi r^2 = \pi (1/2)^2 = \pi/4$ sq. m.
– Using $\pi = 22/7$, the area of the circle is $(22/7) / 4 = 22/28 = 11/14$ sq. m.
– Let’s interpret “using the diameter PQ as one of its sides” to mean that the line segment PQ forms one boundary of the square. Let PQ lie on the x-axis from (0,0) to (1,0). The square would then occupy the region $0 \le x \le 1$ and $0 \le y \le 1$ (assuming it’s drawn above PQ). The circle with diameter PQ is centered at the midpoint of PQ, which is (0.5, 0), and has radius 0.5. Its equation is $(x-0.5)^2 + y^2 = 0.5^2 = 0.25$.
– The area of the part of the square lying outside the circle is the Area of the Square minus the Area of the region common to both the square and the circle.
– The square is defined by $0 \le x \le 1$ and $0 \le y \le 1$. The circle is defined by $(x-0.5)^2 + y^2 \le 0.25$.
– The part of the circle within the square is where $y \ge 0$ and $(x-0.5)^2 + y^2 \le 0.25$. This describes the upper semi-circle bounded by the diameter PQ (the base of the square).
– The area of this upper semi-circle is half the area of the full circle = (Area of circle) / 2 = $(\pi/4) / 2 = \pi/8$.
– Using $\pi = 22/7$, the area of the semi-circle is $(22/7) / 8 = 22/56 = 11/28$ sq. m.
– The area of the part of the square lying outside this semi-circular region is Area of Square – Area of the semi-circle.
– Area outside = 1 sq. m – 11/28 sq. m = $(28/28) – (11/28) = (28-11)/28 = 17/28$ sq. m.

The crucial part of this problem is correctly interpreting the geometric setup based on the phrase “A square is drawn using the diameter PQ as one of its sides”. The standard interpretation in such geometry problems is that the diameter forms one edge of the square, and the region considered is the area within the square but outside the shape (or part of it) defined by the diameter.

33. R walks a long distance every Sunday. He walks 2 km towards the north

R walks a long distance every Sunday. He walks 2 km towards the north from his house and then turns right; he walks another 2 km and again turns right; next he walks 5 km and turns left; he further walks 2 km and stops. He rests for some time and returns home following a straight route without any turning point. What is the distance R walks after he has rested ?

11 km

7 km

6 km

5 km

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The distance R walks after he has rested is 5 km.

– Let R’s house be the starting point (0,0).
– 1. Walks 2 km towards North: Reaches point (0, 2).
– 2. Turns right (East) and walks 2 km: Reaches point (0+2, 2) = (2, 2).
– 3. Turns right (South) and walks 5 km: Reaches point (2, 2-5) = (2, -3).
– 4. Turns left (East) and walks 2 km: Reaches point (2+2, -3) = (4, -3).
– R stops at the point (4, -3). This point is 4 km East and 3 km South of his house (0,0).
– He returns home following a straight route from (4, -3) to (0,0).
– This is the distance between these two points, which can be calculated using the distance formula or by recognizing it as the hypotenuse of a right triangle with legs of length 4 km and 3 km.
– Distance = $\sqrt{(4-0)^2 + (-3-0)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ km.
– The distance R walks after he has rested is the length of this straight return journey, which is 5 km.

Direction and distance problems can be solved by representing the movements on a coordinate plane. The final position relative to the starting point can then be determined, and the straight-line distance calculated using the Pythagorean theorem.

34. Consider the following number : $3^5 \times 5^5 \times 6^{10} \times 1

Consider the following number : $3^5 \times 5^5 \times 6^{10} \times 10^6 \times 15^{12} \times 12^{15} \times 25^7$
What is the number of consecutive zeros at the end of the number given above ?

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The number of consecutive zeros at the end of the number is 37.

– The number of consecutive zeros at the end of an integer is determined by the number of times 10 is a factor in its prime factorization. Since $10 = 2 \times 5$, we need to count the number of pairs of factors (2, 5). The number of zeros is equal to the minimum of the total number of factors of 2 and the total number of factors of 5 in the prime factorization of the given number.
– The given number is $3^5 \times 5^5 \times 6^{10} \times 10^6 \times 15^{12} \times 12^{15} \times 25^7$.
– Prime factorize each term:
– $3^5 = 3^5$
– $5^5 = 5^5$
– $6^{10} = (2 \times 3)^{10} = 2^{10} \times 3^{10}$
– $10^6 = (2 \times 5)^6 = 2^6 \times 5^6$
– $15^{12} = (3 \times 5)^{12} = 3^{12} \times 5^{12}$
– $12^{15} = (2^2 \times 3)^{15} = (2^2)^{15} \times 3^{15} = 2^{30} \times 3^{15}$
– $25^7 = (5^2)^7 = 5^{14}$
– Combine the prime factors:
– Factors of 2: $2^{10} \times 2^6 \times 2^{30} = 2^{10+6+30} = 2^{46}$. Total power of 2 is 46.
– Factors of 5: $5^5 \times 5^6 \times 5^{12} \times 5^{14} = 5^{5+6+12+14} = 5^{37}$. Total power of 5 is 37.
– (Factors of 3 are $3^5 \times 3^{10} \times 3^{12} \times 3^{15}$, but these do not contribute to zeros).
– The number of factors of 2 is 46. The number of factors of 5 is 37.
– The number of pairs of (2, 5) is $\min(46, 37) = 37$.
– Therefore, there are 37 consecutive zeros at the end of the number.

Counting trailing zeros involves finding the highest power of 10 that divides the number. Since $10 = 2 \times 5$, this is equivalent to finding the highest power of 5 in the prime factorization of the number (as there are always more factors of 2 than 5 in typical integers).

35. There are four large urns numbered 1 to 4. The number of different way

There are four large urns numbered 1 to 4. The number of different ways all the three balls numbered 1 to 3 can be kept inside the four urns is

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The number of different ways all the three balls numbered 1 to 3 can be kept inside the four urns is 64.

– We have 3 distinct balls (numbered 1, 2, and 3).
– We have 4 distinct urns (numbered 1, 2, 3, and 4).
– Each ball can be placed into any one of the four urns, and the placement of one ball does not affect the choices for the other balls.
– For Ball 1, there are 4 possible urns it can be placed in.
– For Ball 2, there are also 4 possible urns it can be placed in, independently of where Ball 1 was placed.
– For Ball 3, there are similarly 4 possible urns it can be placed in, independently of the placement of the other balls.
– The total number of ways to place all three balls is the product of the number of choices for each ball.
– Total ways = (Choices for Ball 1) * (Choices for Ball 2) * (Choices for Ball 3) = 4 * 4 * 4 = $4^3$.
– $4^3 = 64$.

This is a problem involving permutations with repetition, specifically distributing distinct items into distinct bins. If there are $k$ distinct items and $n$ distinct bins, and each item can go into any bin, the total number of ways is $n^k$. Here, items are balls (k=3) and bins are urns (n=4).

36. The first ten letters of the English Alphabet are used in place of the

The first ten letters of the English Alphabet are used in place of the integers 0, 1, 2, …, 9 respectively. Then what is the value of BAG – ADD + FIG ?

GFI

GAD

GJF

GFJ

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The value of BAG – ADD + FIG in the given code is GFJ.

– The first ten letters of the English Alphabet (A, B, C, D, E, F, G, H, I, J) are used in place of integers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
– A=0, B=1, C=2, D=3, E=4, F=5, G=6, H=7, I=8, J=9.
– The question asks for the value of BAG – ADD + FIG. Assuming the letters represent digits in a number based on their position:
– BAG corresponds to the number formed by B(1), A(0), G(6), which is 106.
– ADD corresponds to the number formed by A(0), D(3), D(3), which is 033 or 33.
– FIG corresponds to the number formed by F(5), I(8), G(6), which is 586.
– Perform the calculation: 106 – 33 + 586.
– 106 – 33 = 73.
– 73 + 586 = 659.
– Now convert the result 659 back into letters using the given mapping:
– 6 corresponds to G.
– 5 corresponds to F.
– 9 corresponds to J.
– The resulting sequence of letters is GFJ.

This type of problem requires interpreting letters as numerical digits based on a defined code and then performing standard arithmetic operations. The result is then converted back into the letter code.

37. Two teams named A and B do business together and their shares are in t

Two teams named A and B do business together and their shares are in the ratio of 2 : 1. Team A has got three members A1, A2 and A3 whose shares are in the ratio of 1 : 2 : 3. Team B has got four members B1, B2, B3 and B4 whose shares are in the ratio of 1 : 2 : 3 : 4. If the actual share of B3 is ₹ 2,25,000, then what is the actual share of A2 ?

₹ 5,00,000

₹ 4,00,000

₹ 2,00,000

₹ 1,00,000

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The actual share of A2 is ₹ 5,00,000.

– The shares of Team A and Team B are in the ratio 2:1. Let the total business amount be 3k. Then Team A’s share is 2k and Team B’s share is k.
– Team A’s share (2k) is divided among A1, A2, A3 in the ratio 1:2:3. The total parts in this ratio are 1+2+3=6.
– A2’s share within Team A is (2/6) of Team A’s total share = (1/3) * 2k.
– Team B’s share (k) is divided among B1, B2, B3, B4 in the ratio 1:2:3:4. The total parts in this ratio are 1+2+3+4=10.
– B3’s share within Team B is (3/10) of Team B’s total share = (3/10) * k.
– We are given that the actual share of B3 is ₹ 2,25,000.
– So, (3/10) * k = 2,25,000.
– From this, we can find k: k = (2,25,000 * 10) / 3 = 2,250,000 / 3 = 7,50,000.
– Team B’s total share is ₹ 7,50,000. Team A’s total share is 2k = 2 * 7,50,000 = ₹ 15,00,000.
– Now we find A2’s share, which is (1/3) of Team A’s share.
– A2’s share = (1/3) * 15,00,000 = 5,00,000.

Problems involving nested ratios can be solved by representing the total shares of the larger groups (Teams A and B) and then distributing those shares according to the internal ratios of their members.

38. Container I contains a mixture of 2 liters of oil and 3 liters of wate

Container I contains a mixture of 2 liters of oil and 3 liters of water. Container II contains a mixture of 6 liters of oil and 7 liters of water. 50% of the mixture of container I and 25% of the mixture of container II are transferred to container III. What is the proportion of oil and water in container III ?

4 : 5

5 : 4

13 : 10

10 : 13

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The correct proportion of oil and water in container III is 10 : 13.

– Container I has 2 L oil and 3 L water (total 5 L). 50% of this mixture is transferred, which is 2.5 L. This 2.5 L contains 50% of the oil from container I (0.5 * 2 L = 1 L oil) and 50% of the water from container I (0.5 * 3 L = 1.5 L water).
– Container II has 6 L oil and 7 L water (total 13 L). 25% of this mixture is transferred, which is 13/4 L = 3.25 L. This 3.25 L contains 25% of the oil from container II (0.25 * 6 L = 1.5 L oil) and 25% of the water from container II (0.25 * 7 L = 1.75 L water).
– Container III receives the transferred amounts. Total oil in container III = 1 L (from I) + 1.5 L (from II) = 2.5 L. Total water in container III = 1.5 L (from I) + 1.75 L (from II) = 3.25 L.
– The proportion of oil to water in container III is 2.5 L : 3.25 L.
– To simplify the ratio, we can multiply both parts by 100 to remove decimals: 250 : 325.
– Dividing both by their greatest common divisor, 25: 250/25 = 10 and 325/25 = 13.
– The ratio is 10 : 13.

When a portion of a mixture is taken out, the ratio of components in the taken portion is the same as the ratio in the original mixture. The amount of each component transferred is calculated based on this ratio and the total amount transferred.

39. A rod with circular cross-section of radius 5 mm is stretched such tha

A rod with circular cross-section of radius 5 mm is stretched such that its cross-section remains circular. If after stretching the rod its length becomes four times the original length, then what is the radius of the new cross-section ?

1.5 mm

2.0 mm

2.5 mm

3.0 mm

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Let the original radius of the circular cross-section be $r_1 = 5$ mm.
Let the original length of the rod be $L_1$.
The original volume of the rod (assuming it’s a cylinder) is $V_1 = \pi r_1^2 L_1 = \pi (5)^2 L_1 = 25\pi L_1$.

The rod is stretched such that its cross-section remains circular. This implies the stretched rod is still a cylinder.
After stretching, the new length becomes four times the original length: $L_2 = 4 L_1$.
Let the new radius of the circular cross-section be $r_2$.
The new volume of the rod is $V_2 = \pi r_2^2 L_2 = \pi r_2^2 (4L_1)$.

Assuming the material of the rod is incompressible (a common assumption in such problems unless otherwise stated, meaning the density remains constant), the volume of the rod remains constant during stretching.
Therefore, $V_1 = V_2$.
$25\pi L_1 = \pi r_2^2 (4L_1)$.

We can cancel $\pi$ and $L_1$ from both sides (assuming $L_1 > 0$):
$25 = 4 r_2^2$.

Solve for $r_2^2$:
$r_2^2 = \frac{25}{4}$.

Solve for $r_2$:
$r_2 = \sqrt{\frac{25}{4}} = \frac{\sqrt{25}}{\sqrt{4}} = \frac{5}{2}$.
$r_2 = 2.5$ mm.

The radius of the new cross-section is 2.5 mm.

– The volume of the material remains constant during stretching (assuming incompressibility).
– The shape of the rod is a cylinder, so its volume is given by $\pi r^2 L$.
– Set the initial volume equal to the final volume and solve for the unknown radius.

This problem relates to the concept of conservation of volume under plastic deformation. When a material is stretched, its length increases, but its cross-sectional area must decrease proportionally to maintain constant volume.

40. A circle is divided into four sectors such that the proportions of the

A circle is divided into four sectors such that the proportions of their areas are 1 : 2 : 3 : 4. What is the angle of the largest sector ?

120°

125°

136°

144°

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The area of a sector of a circle is proportional to the angle subtended by the sector at the center. For sectors within the same circle, the ratio of their areas is equal to the ratio of their central angles.

Let the areas of the four sectors be $A_1, A_2, A_3, A_4$, and their corresponding central angles be $\theta_1, \theta_2, \theta_3, \theta_4$.
The proportions of the areas are given as $A_1 : A_2 : A_3 : A_4 = 1 : 2 : 3 : 4$.
Since the areas are proportional to the angles, the ratio of the angles is also $\theta_1 : \theta_2 : \theta_3 : \theta_4 = 1 : 2 : 3 : 4$.

The sum of the angles of the sectors in a circle is $360^\circ$.
So, $\theta_1 + \theta_2 + \theta_3 + \theta_4 = 360^\circ$.

The total proportion is $1 + 2 + 3 + 4 = 10$.
The angles can be found by dividing the total angle ($360^\circ$) according to the proportions:
$\theta_1 = \frac{1}{\text{Total Proportion}} \times 360^\circ = \frac{1}{10} \times 360^\circ = 36^\circ$.
$\theta_2 = \frac{2}{10} \times 360^\circ = \frac{1}{5} \times 360^\circ = 72^\circ$.
$\theta_3 = \frac{3}{10} \times 360^\circ = \frac{3}{10} \times 360^\circ = 108^\circ$.
$\theta_4 = \frac{4}{10} \times 360^\circ = \frac{2}{5} \times 360^\circ = 144^\circ$.

The four angles are $36^\circ, 72^\circ, 108^\circ, 144^\circ$.
Check the sum: $36 + 72 + 108 + 144 = 108 + 108 + 144 = 216 + 144 = 360^\circ$. The sum is correct.

The largest sector corresponds to the largest proportion, which is 4.
The angle of the largest sector is $\theta_4 = 144^\circ$.

– The area of a sector is directly proportional to its central angle.
– The sum of the central angles of sectors forming a complete circle is 360°.
– Proportional distribution can be calculated by dividing the total quantity by the sum of the ratios.

This concept is fundamental to understanding circle graphs (pie charts), where the size of each slice (sector) represents a proportion of the whole.