Solving the Pramod and Sarayu Distance Problem

This problem requires us to determine the distance between Pramod and Sarayu based on several relative positioning statements involving Nalini and Ravali. We can solve this using a coordinate system and basic geometry.

Step 1: Setting Up the Coordinate System

Let's place Nalini (N) at the origin of a 2D Cartesian coordinate system. We'll assume East corresponds to the positive x-axis and North corresponds to the positive y-axis.

• Nalini (N): Coordinates (0, 0)

Step 2: Locating Ravali and Sarayu

We are given:

• Ravali (R) is 70 m east of Nalini (N).
• Sarayu (S) is west of Ravali (R) at a distance of 40 m.

From this, we can deduce the coordinates:

• Ravali (R): Since R is 70 m east of N(0,0), its coordinates are (70, 0).
• Sarayu (S): Since S is 40 m west of R(70,0), its coordinates are ($70 - 40$, 0) = (30, 0).

The distance between Nalini (N) and Sarayu (S) is 30 m.

Step 3: Analyzing Pramod's Position

We are given information about Pramod (P):

• The distance between Nalini (N) and Pramod (P) is 50 m. This means P lies on a circle centered at N(0,0) with a radius of 50 m. The equation for this circle is: $$x_P^2 + y_P^2 = 50^2 = 2500$$
• Pramod (P) is located west, northwest, or north of Sarayu (S). Sarayu is at (30, 0). This implies Pramod's x-coordinate ($x_P$) must be less than or equal to Sarayu's x-coordinate ($x_P \leq 30$).
• Sarayu (S) is to the south of Pramod (P). This implies Pramod (P) is north of Sarayu (S), meaning Pramod's y-coordinate ($y_P$) must be positive ($y_P > 0$).

Step 4: Calculating the Distance Between Pramod and Sarayu

We need to find the distance between P($x_P, y_P$) and S(30, 0). The distance formula is:

$$Distance(P, S) = \sqrt{(x_P - 30)^2 + (y_P - 0)^2}$$

We are given options for this distance. Let's test the option that matches the provided answer, which is 40 m.

Assume $Distance(P, S) = 40$ m. Squaring both sides gives:

$$(x_P - 30)^2 + y_P^2 = 40^2 = 1600$$

Now we have a system of two equations:

1. $x_P^2 + y_P^2 = 2500$
2. $(x_P - 30)^2 + y_P^2 = 1600$

Step 5: Solving the Equations

Let's solve this system. Expand the second equation:

$$x_P^2 - 60x_P + 900 + y_P^2 = 1600$$

Substitute $y_P^2$ from the first equation ($y_P^2 = 2500 - x_P^2$) into the expanded second equation:

$$x_P^2 - 60x_P + 900 + (2500 - x_P^2) = 1600$$

Simplify the equation:

$$-60x_P + 3400 = 1600$$

Rearrange to solve for $x_P$:

$$60x_P = 3400 - 1600$$ $$60x_P = 1800$$ $$x_P = \frac{1800}{60} = 30$$

Now, substitute $x_P = 30$ back into the first equation to find $y_P$:

$$30^2 + y_P^2 = 2500$$ $$900 + y_P^2 = 2500$$ $$y_P^2 = 2500 - 900 = 1600$$ $$y_P = \sqrt{1600} = 40$$

(We take the positive root $y_P = 40$ because Pramod is north of Sarayu).

Step 6: Verifying the Solution

Pramod's coordinates are (30, 40).

Let's check if these coordinates satisfy all conditions:

• Distance N-P: $\sqrt{(30-0)^2 + (40-0)^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50$ m. (Matches)
• P relative to S(30, 0): $x_P = 30$ and $y_P = 40$. P is directly north of S ($x_P=30$, $y_P>0$). This satisfies the condition "west or northwest or north".
• S relative to P: S(30,0) is south of P(30,40). (Matches)
• Distance P-S: $\sqrt{(30-30)^2 + (40-0)^2} = \sqrt{0^2 + 40^2} = \sqrt{1600} = 40$ m. (Matches the calculated distance and the option)

Therefore, the distance between Pramod and Sarayu is 40 m.