UPSC CISF-AC-EXE

Let R’s age last year be $R_L$ and V’s age last year be $V_L$.
Last year’s age is one year less than the present age:
$R_L = R_P – 1$
$V_L = V_P – 1$

According to the problem, last year the age of R was three times that of V:
$R_L = 3 V_L$ (Equation 2)

Substitute the expressions for $R_L$ and $V_L$ from the ages at present into Equation 2:
$(R_P – 1) = 3 (V_P – 1)$
$R_P – 1 = 3V_P – 3$

Now substitute the expression for $R_P$ from Equation 1 into the above equation:
$(2 V_P) – 1 = 3V_P – 3$
$2 V_P – 1 = 3V_P – 3$

Rearrange the terms to solve for $V_P$:
$3 – 1 = 3V_P – 2V_P$
$2 = V_P$

So, V’s age at present is 2 years.
Now find R’s age at present using Equation 1:
$R_P = 2 V_P = 2 \times 2 = 4$.

R’s age at present is 4 years.
Let’s check this:
Present ages: R=4, V=2. R is double V (4 = 2*2). Correct.
Last year’s ages: R=3, V=1. R was three times V (3 = 3*1). Correct.

The question asks: How old is R? This refers to R’s age at present.
R is 4 years old.

For y:
y is the smallest integer greater than 2.
When y is divided by 6, the remainder is 2. This can be written as $y \equiv 2 \pmod 6$, or $y – 2$ is divisible by 6.
When y is divided by 8, the remainder is 2. This can be written as $y \equiv 2 \pmod 8$, or $y – 2$ is divisible by 8.
So, $y – 2$ is a common multiple of 6 and 8.
To find the smallest such integer y, $y-2$ must be the smallest positive common multiple of 6 and 8. This is the LCM(6, 8).
LCM(6, 8) = 24.
So, $y – 2$ must be a multiple of 24. $y – 2 = 24j$ for some integer j.
$y = 24j + 2$.
We need the smallest integer y greater than 2.
If j=0, $y = 24(0) + 2 = 2$. This is not greater than 2.
If j=1, $y = 24(1) + 2 = 26$. This is greater than 2 and is the smallest value of y satisfying the condition. So, $y=26$.

The question asks for the value of $x – y$.
$x – y = 27 – 26 = 1$.

The strength of the class is increased by 25%.
New strength = $N + 0.25N = 1.25N$.
The number of additional students is $N_{add} = 0.25N$.

The average weight of the class gets increased by 1%.
Initial average weight = 20 kg.
Increase in average weight = 1% of 20 kg = $0.01 \times 20 = 0.2$ kg.
New average weight = $20 + 0.2 = 20.2$ kg.

The new total weight of the class is $W_{new} = (\text{New strength}) \times (\text{New average weight})$
$W_{new} = (1.25N) \times 20.2$ kg.

The total weight of the class after adding students is the sum of the initial total weight and the total weight of the additional students.
Let the average weight of the additional students be $W_{avg\_add}$.
Total weight of additional students = $N_{add} \times W_{avg\_add} = (0.25N) \times W_{avg\_add}$.

So, $W_{new} = W_{initial} + \text{Total weight of additional students}$.
$(1.25N) \times 20.2 = (N \times 20) + (0.25N \times W_{avg\_add})$.

We can divide the entire equation by N (assuming $N > 0$):
$1.25 \times 20.2 = 20 + 0.25 \times W_{avg\_add}$.

Calculate the left side:
$1.25 \times 20.2 = (5/4) \times 20.2 = 5 \times (20.2 / 4) = 5 \times 5.05 = 25.25$.

The equation becomes:
$25.25 = 20 + 0.25 \times W_{avg\_add}$.

Subtract 20 from both sides:
$25.25 – 20 = 0.25 \times W_{avg\_add}$
$5.25 = 0.25 \times W_{avg\_add}$.

Solve for $W_{avg\_add}$:
$W_{avg\_add} = 5.25 / 0.25 = 5.25 / (1/4) = 5.25 \times 4$.
$W_{avg\_add} = (5 + 0.25) \times 4 = 5 \times 4 + 0.25 \times 4 = 20 + 1 = 21$.

The average weight of the additional students is 21.00 kg.

When the cylinder is tilted by an angle $\theta$ from the vertical, the water surface remains horizontal. Water will start spilling out when the horizontal water surface reaches the rim of the cylinder.
When the cylinder is tilted to the maximum possible angle $\theta_{max}$ without spilling, the horizontal water surface must pass through the lowest point on the base rim and the highest point on the top rim. These two points are diametrically opposite when viewed along the direction perpendicular to the tilt axis.

Consider a vertical cross-section of the cylinder along a diameter perpendicular to the tilt axis. This cross-section is a rectangle of width 2R and height H.
The lowest point on the base rim in this cross-section is at one corner (say, position -R, height 0). The highest point on the top rim on the diametrically opposite side is at the other corner (position +R, height H).
The horizontal water surface, when the cylinder is tilted to $\theta_{max}$ without spilling (and containing half the volume), forms a plane that cuts this cross-section along a line segment from (-R, 0) to (R, H) *in the coordinate system aligned with the cylinder’s base diameter and height*.

Let’s consider the cylinder tilted by angle $\theta$ from the vertical. The base of the cylinder is inclined at angle $\theta$ to the horizontal. The water surface is horizontal. The angle between the base and the water surface is $\theta$.
Consider the line segment on the cylinder wall connecting the lowest point on the base circumference to the highest point on the opposite side of the top circumference. This line segment lies on the cylinder wall and spans a vertical distance H and a horizontal distance 2R (across the diameter).
The angle this line makes with the base is $\alpha$ such that $\tan \alpha = H / (2R)$.
The water surface is horizontal and contains this line segment when the cylinder is tilted to the maximum angle without spilling (for half volume).
The angle the cylinder axis makes with the vertical is $\theta$. The water surface is horizontal. The angle between the cylinder axis and the water surface is $90^\circ – \theta$.
The line segment on the wall connecting the lowest point on the base to the highest point on the top (on the opposite side) lies within the plane of the cylinder wall. The angle this line makes with the cylinder axis is $90^\circ – \alpha$.
The water surface is perpendicular to the vertical, and the cylinder axis makes angle $\theta$ with the vertical. The line segment connecting the points lies on the water surface.
Let’s re-examine the cross-section rectangle (width 2R, height H). The horizontal water surface cuts this rectangle.
The volume of water is $\pi R^2 (H/2)$.
The volume of water in a tilted cylinder below a horizontal plane cutting the cylinder is a wedge. If the plane passes through one side of the base diameter and the opposite side of the top diameter, the volume of the wedge is $\pi R^2 \times (H/2)$, which is exactly half the cylinder volume.
This configuration occurs when the angle $\alpha$ made by the line connecting the lowest point on the base and the highest point on the opposite top rim with the horizontal is $0^\circ$.
The angle of tilt $\theta$ from the vertical corresponds to the angle the cylinder’s side makes with the horizontal.
In the cross-section (rectangle of width 2R and height H), the line from (-R, 0) to (R, H) makes an angle $\beta$ with the *horizontal* axis (representing the diameter direction) such that $\tan \beta = H / (2R)$. This line is the water surface in this cross-section when the cylinder contains half volume and is tilted to the spilling point.
The angle the base makes with the horizontal is $\theta$. The water surface is horizontal. The angle between the water surface and the base is $\theta$.
The angle between the base (horizontal line in the upright cross-section view) and the line segment (-R, 0) to (R, H) is $\beta$. This angle $\beta$ is the maximum tilt angle $\theta_{max}$ of the base from the horizontal.
So, $\tan \theta_{max} = \frac{H}{2R}$.
Given H = 10 cm, R = 5 cm.
$\tan \theta_{max} = \frac{10}{2 \times 5} = \frac{10}{10} = 1$.
$\theta_{max} = \arctan(1) = 45^\circ$.

The cylinder is tilted so that no water goes out. This means the angle of tilt $\theta$ from the vertical must be less than or equal to the maximum angle before spilling.
So, the angle by which the cylinder is tilted is $\theta \leq 45^\circ$.
Option B states “not more than 45°”, which means $\theta \leq 45^\circ$. This is the condition for no water going out.

Consider statement 1: $m^2 n \leq x^2 y$.
We can substitute $y = mn/x$ from the fundamental relationship into the inequality:
$m^2 n \leq x^2 (mn/x)$
$m^2 n \leq xmn$
Since x, m, and n are positive for positive integers x, y (unless x or y is 0 or 1, but the context implies standard HCF/LCM), we can divide both sides by mn:
$m \leq x$.
The HCF (m) of x and y is always a divisor of x. Therefore, for positive integers, $m \leq x$ is always true.
Thus, statement 1 is correct.

Consider statement 2: $mn^2 \leq x^2 y$.
Substitute $y = mn/x$ into the inequality:
$mn^2 \leq x^2 (mn/x)$
$mn^2 \leq xmn$
Since x, m, and n are positive, we can divide both sides by mn:
$n \leq x$.
The LCM (n) of x and y is a multiple of x. For positive integers, this means $n \geq x$. The inequality $n \leq x$ is only true in specific cases (e.g., when y divides x, in which case $n=x$) but is generally false. For example, if x=2 and y=3, LCM(2,3)=6, and $6 \not\leq 2$.
Thus, statement 2 is incorrect in general.

Since only statement 1 is correct, the answer is A.

When going from house to office:
Let the house be the origin (displacement = 0) and the office be at a displacement D (assuming a straight line path for simplicity, though the problem just says “same route” implying the distance is the same).
Time taken ($t_{going}$) = 30 minutes = 0.5 hours.
Displacement while going = D.
Average velocity while going ($v_{going}$) = Displacement / Time = D / 0.5 = 2D. (Taking direction from house to office as positive).

When returning from office to house:
The person travels back to the origin.
Time taken ($t_{returning}$) = 45 minutes = 0.75 hours.
Displacement while returning = -D (from office position D back to house position 0).
Average velocity while returning ($v_{returning}$) = Displacement / Time = -D / 0.75 = -4D/3. (Taking direction from office to house as negative).

The question asks for the ratio of the average velocity while going and the average velocity while returning. The options are positive numbers, implying the ratio of the magnitudes of the average velocities (average speeds for each leg, since distance = |displacement|).
Magnitude of $v_{going}$ = $|2D| = 2D$.
Magnitude of $v_{returning}$ = $|-4D/3| = 4D/3$.

Ratio of magnitudes of average velocities = $|v_{going}| / |v_{returning}|$ = $(2D) / (4D/3)$
$= 2 \times (3 / 4) = 6 / 4 = 3 / 2$.

If the question was strictly about vector velocities, the ratio would be $(2D) / (-4D/3) = -3/2$, which is not an option. The positive options confirm the question is asking for the ratio of speeds or magnitudes of velocities.