UPSC CAPF

The total distance walked would be the sum of the lengths of the segments OA, Arc AB, and BO.
Distance OA = radius = 100 m.
Distance BO = radius = 100 m.
Distance = 100 + Length of Arc AB + 100 = 200 + Length of Arc AB.

The length of an arc AB is given by $L = r \times \theta$, where $r$ is the radius and $\theta$ is the central angle AOB in radians. We need to determine the angle AOB from the context or options.
The options for total distance are 703, 723, 743, 823 m.
This implies the length of Arc AB is approximately:
703 – 200 = 503 m
723 – 200 = 523 m
743 – 200 = 543 m
823 – 200 = 623 m

Let’s check if any common central angle results in an arc length close to these values when the radius is 100m.
If the central angle is $\theta$ radians, Arc length = $100\theta$.
If $\theta = 503/100 = 5.03$ rad $\approx 288$ deg.
If $\theta = 523/100 = 5.23$ rad $\approx 300$ deg (using $\pi \approx 3.1416$, $5.23 \times 180/\pi \approx 299.57$ deg).
If $\theta = 543/100 = 5.43$ rad $\approx 311$ deg.
If $\theta = 623/100 = 6.23$ rad $\approx 357$ deg.

A central angle of 300 degrees is a plausible value in geometry problems (e.g., a circle minus a 60-degree sector).
If the central angle AOB is 300 degrees, which is $300 \times \frac{\pi}{180} = \frac{5\pi}{3}$ radians.
Arc length AB = $100 \times \frac{5\pi}{3} = \frac{500\pi}{3}$ metres.
Using $\pi \approx 3.14159$: Arc length $\approx \frac{500 \times 3.14159}{3} \approx \frac{1570.795}{3} \approx 523.598$ metres.
Total distance = $100 + 523.598 + 100 = 723.598$ metres.
Rounded to the nearest metre, this is 724 m.

Let’s try a different approximation for $\pi$, like $\pi \approx 22/7$.
Arc length $\approx \frac{500}{3} \times \frac{22}{7} = \frac{11000}{21} \approx 523.81$ metres.
Total distance = $100 + 523.81 + 100 = 723.81$ metres. Rounded to 724m.

If the arc length was exactly 523m, the total distance would be 723m. This matches Option B. The angle corresponding to an arc length of 523m with radius 100m is 5.23 radians (approx 299.57 degrees).
Given the options, it is highly probable that the intended path is O -> A -> Arc AB -> B -> O and the arc length AB is precisely 523m, leading to a total distance of 723m. This would mean the angle AOB is $5.23$ radians, or the question expects calculation precision that rounds $723.something$ down to 723. Assuming the intended total distance is exactly 723 based on the options, the arc length is 523.

The time changes from 3 hours and 12 minutes to 6 hours.
First, calculate the duration of this time change.
From 3:12 to 4:00 is 48 minutes (60 – 12).
From 4:00 to 6:00 is 2 hours.
Total duration = 2 hours and 48 minutes.

Convert the total duration into minutes:
2 hours = 2 * 60 = 120 minutes.
Total duration in minutes = 120 minutes + 48 minutes = 168 minutes.

Alternatively, calculate the total time in minutes from 12:00 AM/PM.
3 hours 12 minutes = 3 * 60 + 12 = 180 + 12 = 192 minutes past 12.
6 hours 0 minutes = 6 * 60 + 0 = 360 minutes past 12.
Duration = 360 minutes – 192 minutes = 168 minutes.

The hour hand moves 0.5 degrees for every minute.
Total degrees moved by the hour hand = Duration in minutes * Speed per minute
Total degrees = 168 minutes * 0.5 degrees/minute = 168 * (1/2) = 84 degrees.

Therefore, the set of Cows is entirely contained within the set of Animals.
The set of Horses is also entirely contained within the set of Animals.
The set of Cows and the set of Horses are mutually exclusive; they do not overlap.

This relationship is best represented by a Venn diagram consisting of a large circle representing “Animals”, containing two smaller, non-overlapping circles representing “Cows” and “Horses”.
Based on common representations of Venn diagrams in multiple-choice options, option B typically represents this structure: one large circle enclosing two disjoint smaller circles.

Family A: Monthly income ₹ 20,000
Miscellaneous: 20% of ₹ 20,000 = 0.20 * 20000 = ₹ 4,000
Entertainment: 30% of ₹ 20,000 = 0.30 * 20000 = ₹ 6,000
Food: 50% of ₹ 20,000 = 0.50 * 20000 = ₹ 10,000
Total expenditure for Family A = 4000 + 6000 + 10000 = ₹ 20,000.

Family B: Monthly income ₹ 1,00,000
Miscellaneous: 70% of ₹ 1,00,000 = 0.70 * 100000 = ₹ 70,000
Entertainment: 20% of ₹ 1,00,000 = 0.20 * 100000 = ₹ 20,000
Food: 10% of ₹ 1,00,000 = 0.10 * 100000 = ₹ 10,000
Total expenditure for Family B = 70000 + 20000 + 10000 = ₹ 1,00,000.

Now let’s evaluate each statement:
A) Family A spends as much on miscellaneous as family B spends on entertainment.
Family A Misc: ₹ 4,000
Family B Ent: ₹ 20,000
₹ 4,000 ≠ ₹ 20,000. Statement A is false.

B) The food expense of family B is equal to the total expense of family A.
Family B Food: ₹ 10,000
Total expense of family A: ₹ 20,000
₹ 10,000 ≠ ₹ 20,000. Statement B is false.

C) Families A and B spend equally on food.
Family A Food: ₹ 10,000
Family B Food: ₹ 10,000
₹ 10,000 = ₹ 10,000. Statement C is true.

D) Families A and B spend equally on entertainment.
Family A Ent: ₹ 6,000
Family B Ent: ₹ 20,000
₹ 6,000 ≠ ₹ 20,000. Statement D is false.

The only true statement is C.

To find the number of students in each category, we calculate the proportion of the total angle for each category and multiply it by the total number of students.
Number of students who ‘support’ = (Angle of Support Sector / Total Angle) * Total Students
Number of students who ‘support’ = (180 / 360) * 300 = (1/2) * 300 = 150.

Number of students who are ‘neutral’ = (Angle of Neutral Sector / Total Angle) * Total Students
Number of students who are ‘neutral’ = (108 / 360) * 300.
Simplifying the fraction 108/360: Divide by 36: 108/36 = 3, 360/36 = 10.
Number of students who are ‘neutral’ = (3/10) * 300 = 3 * 30 = 90.

Number of students who ‘oppose’ = (Angle of Oppose Sector / Total Angle) * Total Students
Number of students who ‘oppose’ = (72 / 360) * 300.
Simplifying the fraction 72/360: Divide by 72: 72/72 = 1, 360/72 = 5. (Alternatively, divide by 36: 72/36 = 2, 360/36 = 10, so 2/10 = 1/5)
Number of students who ‘oppose’ = (1/5) * 300 = 60.

The numbers for ‘support’, ‘neutral’, and ‘oppose’ are 150, 90, and 60, respectively.
Let’s check the sum: 150 + 90 + 60 = 300. This matches the total number of students.
Comparing with the options:
A) 150, 90, 60 – Matches our calculated numbers.
B) 120, 100, 80
C) 80, 100, 120
D) 60, 90, 150
The correct option is A.

From (4), we can find the cost of one pineapple:
$P = \frac{48}{3} = ₹ 16$.

Now we can use the chain of equivalences to relate the cost of a mango to the cost of a pineapple.
From (3), $4A = 9P$. Substitute the value of P:
$4A = 9 \times 16 = 144$.
So, the cost of 4 apples is ₹ 144. The cost of one apple is $A = \frac{144}{4} = ₹ 36$.

From (2), $5O = 4A$. Substitute the value of 4A:
$5O = 144$.
So, the cost of 5 oranges is ₹ 144. The cost of one orange is $O = \frac{144}{5} = ₹ 28.8$.

From (1), $9M = 5O$. Substitute the value of 5O:
$9M = 144$.
So, the cost of 9 mangoes is ₹ 144. The cost of one mango is $M = \frac{144}{9} = ₹ 16$.

The calculated cost of one mango is ₹ 16. However, ₹ 16 is not among the given options (A) ₹ 9, (B) ₹ 12, (C) ₹ 18, (D) ₹ 27.
Assuming there is a typo in the question and one of the options is correct, let’s consider the possibility that 3 pineapples cost ₹ 36 instead of ₹ 48.
If $3P = 36$, then $P = \frac{36}{3} = ₹ 12$.
Using the equivalences:
$4A = 9P = 9 \times 12 = 108 \implies A = \frac{108}{4} = ₹ 27$.
$5O = 4A = 108 \implies O = \frac{108}{5} = ₹ 21.6$.
$9M = 5O = 108 \implies M = \frac{108}{9} = ₹ 12$.
If 3 pineapples cost ₹ 36, then one mango costs ₹ 12, which is Option B. This suggests a likely typo in the original question’s value for the cost of pineapples. Based on the presence of option B, it is highly probable that the intended cost of 3 pineapples was ₹ 36.

Alternatively, we can look at the ratios: $M/O = 5/9$, $O/A = 4/5$, $A/P = 9/4$.
The ratio $M/P = (M/O) \times (O/A) \times (A/P) = (5/9) \times (4/5) \times (9/4) = \frac{5 \times 4 \times 9}{9 \times 5 \times 4} = 1$.
So, $M/P = 1$, which means $M = P$.
The cost of 1 mango is equal to the cost of 1 pineapple.
From $3P = 48$, the cost of 1 pineapple is $P = ₹ 16$. Thus, $M = ₹ 16$.
This confirms the result ₹ 16 from the question as written. Since this is not an option, we rely on the likely intended question that leads to one of the options. Assuming the typo $3P = 36$, we get $P = 12$, and since $M=P$, $M=12$.

Given that Option B is provided as a choice, the intended answer is likely ₹ 12, based on a probable typo in the question statement regarding the cost of pineapples.