To find the temperature on Thursday in New Delhi, we will use the given mean temperature information and the relationship between Monday's and Thursday's temperatures. Let's denote the temperatures for Monday, Tuesday, Wednesday, and Thursday as \(M\), \(Tu\), \(W\), and \(Th\) respectively.

Temperature Mean Calculation

We are given two pieces of information about the mean temperatures over specific periods:

1. The mean temperature from Monday to Wednesday was 41°C.
2. The mean temperature from Tuesday to Thursday was 43°C.

From the first statement, we can write the equation for the sum of temperatures for Monday, Tuesday, and Wednesday:

\[ \frac{M + Tu + W}{3} = 41 \] Multiplying both sides by 3, we get:

\[ M + Tu + W = 41 \times 3 \] \[ M + Tu + W = 123 \quad \text{(Equation 1)} \]

From the second statement, we can write the equation for the sum of temperatures for Tuesday, Wednesday, and Thursday:

\[ \frac{Tu + W + Th}{3} = 43 \] Multiplying both sides by 3, we get:

\[ Tu + W + Th = 43 \times 3 \] \[ Tu + W + Th = 129 \quad \text{(Equation 2)} \]

Temperature Difference Analysis

Now, let's find the difference between the total temperatures of the two periods. We can subtract Equation 1 from Equation 2:

\[ (Tu + W + Th) - (M + Tu + W) = 129 - 123 \] Notice that \(Tu\) and \(W\) cancel out on the left side, leaving us with:

\[ Th - M = 6 \quad \text{(Equation 3)} \] This equation tells us that the temperature on Thursday was 6°C higher than the temperature on Monday.

Percentage Increase Calculation

The problem states that the temperature on Thursday was 15% higher than that of Monday. We can express this relationship mathematically:

\[ Th = M + 0.15M \] \[ Th = (1 + 0.15)M \] \[ Th = 1.15M \quad \text{(Equation 4)} \]

Solving for Thursday's Temperature

Now we have a system of two equations with two variables (\(M\) and \(Th\)):

1. \(Th - M = 6\) 2. \(Th = 1.15M\)

Substitute the expression for \(Th\) from Equation 4 into Equation 3:

\[ 1.15M - M = 6 \] \[ 0.15M = 6 \] To find \(M\), divide both sides by 0.15:

\[ M = \frac{6}{0.15} \] To simplify the division, we can multiply the numerator and denominator by 100:

\[ M = \frac{6 \times 100}{0.15 \times 100} \] \[ M = \frac{600}{15} \] \[ M = 40 \] So, the temperature on Monday (\(M\)) was 40°C.

Now that we have the temperature on Monday, we can find the temperature on Thursday using Equation 4:

\[ Th = 1.15M \] \[ Th = 1.15 \times 40 \] \[ Th = 46 \]

Therefore, the temperature in °C on Thursday was 46°C.

Summary of Temperatures:

The calculation confirms that Thursday's temperature (46°C) is 15% higher than Monday's temperature (40°C), as \(40 \times 0.15 = 6\), and \(40 + 6 = 46\).