The question asks to identify the Chairman of ISRO during the launch of Chandrayaan-1. This requires knowledge of key historical figures and events in India's space program.
B) G Madhavan Nair was the Chairman of ISRO when Chandrayaan-1 was launched on October 22, 2008. He held the position from 2003 to 2009, overseeing significant milestones in India's space program, including the Chandrayaan-1 mission.
The question asks to identify the correctly matched pair of a dance form and its associated state. This requires knowledge of various Indian folk and tribal dance forms and their geographical origins.
Correct Option: D) Gaur-Chhattisgarh
When a train passes a bridge, the total distance covered by the train is the sum of its own length and the length of the bridge. Since the train's speed is constant, we can set up two equations based on the formula \(\text{Distance} = \text{Speed} \times \text{Time}\) and solve for the unknown length of the train and its speed.
Let the length of the train be \(L_T\) meters and its speed be \(S\) meters/second.
Case 1: Passing the first bridge
The length of the first bridge is \(L_{B1} = 600\) m.
The time taken to pass the first bridge is \(T_1 = 80\) s.
The total distance covered by the train is \(L_T + L_{B1}\).
Using the formula \(\text{Distance} = \text{Speed} \times \text{Time}\), we get:
Case 2: Passing the second bridge
The length of the second bridge is \(L_{B2} = 200\) m.
The time taken to pass the second bridge is \(T_2 = 40\) s.
The total distance covered by the train is \(L_T + L_{B2}\).
Using the formula \(\text{Distance} = \text{Speed} \times \text{Time}\), we get:
We now have a system of two linear equations with two variables \(L_T\) and \(S\). We can solve for \(S\) from both equations:
From Equation 1:
\[ S = \frac{L_T + 600}{80} \]From Equation 2:
\[ S = \frac{L_T + 200}{40} \]Since the speed \(S\) is constant, we can equate the two expressions for \(S\):
\[ \frac{L_T + 600}{80} = \frac{L_T + 200}{40} \]Now, solve for \(L_T\):
Multiply both sides by 80 to clear the denominators:
\[ L_T + 600 = 2(L_T + 200) \]\[ L_T + 600 = 2L_T + 400 \]Subtract \(L_T\) from both sides:
\[ 600 = L_T + 400 \]Subtract 400 from both sides:
\[ L_T = 600 - 400 \]\[ L_T = 200 \]So, the length of the train is 200 meters.
D) 200 m — The calculations show that the length of the train is 200 m.
To find Madhu's rank from the last, we can use a standard formula that relates the total number of students, her rank from the top, and her rank from the last. The total number of students in a row or class is equal to the sum of the rank from the top and the rank from the bottom, minus one (because the person themselves is counted twice).
The total number of students in a line or class can be calculated using the formula:
\[ \text{Total students} = \text{Rank from top} + \text{Rank from last} - 1 \]Thus, Madhu's rank from the last is 19.
Correct Option: B) 19
The problem involves decoding words based on a given pattern. We need to identify the rule used to code "CUP" and "CAKE" and then apply the same rule to "PACK". This is a letter-to-digit coding problem, often based on alphabetical position or a simple arithmetic operation.
Correct Option: B) 5427
This is a syllogism problem. We need to analyze the given statements and determine which conclusions logically follow. We can use Venn diagrams or logical deduction to represent the relationships between the categories.
B) Only conclusion 2 follows
This problem involves simple interest. We are given the time it takes for a sum to amount to 3 times itself and need to find the time it takes for the same sum to amount to 5 times itself at the same simple interest rate. We will use the simple interest formula to find the rate and then use that rate to find the required time.
Let the original sum (Principal) be \(P\). The amount \(A\) after \(T\) years at a simple interest rate \(R\) is given by the formula:
\[ A = P + \frac{P \times R \times T}{100} \]In the first case, the sum amounts to 3 times the original sum in 15 years. So, \(A_1 = 3P\) and \(T_1 = 15\) years.
Substitute these values into the formula:
\[ 3P = P + \frac{P \times R \times 15}{100} \]Subtract \(P\) from both sides:
\[ 2P = \frac{P \times R \times 15}{100} \]Divide both sides by \(P\) (assuming \(P \neq 0\)):
\[ 2 = \frac{R \times 15}{100} \]Solve for \(R\):
\[ R = \frac{2 \times 100}{15} = \frac{200}{15} = \frac{40}{3}\% \]Now, in the second case, we want to find the time \(T_2\) when the original sum amounts to 5 times itself. So, \(A_2 = 5P\). The rate \(R\) is the same, which is \(\frac{40}{3}\%\).
Substitute these values into the simple interest formula:
\[ 5P = P + \frac{P \times \frac{40}{3} \times T_2}{100} \]Subtract \(P\) from both sides:
\[ 4P = \frac{P \times \frac{40}{3} \times T_2}{100} \]Divide both sides by \(P\):
\[ 4 = \frac{\frac{40}{3} \times T_2}{100} \]Solve for \(T_2\):
\[ T_2 = \frac{4 \times 100}{\frac{40}{3}} = \frac{400 \times 3}{40} = 10 \times 3 = 30 \text{ years} \]D) 30 is the correct answer. The calculations show that it will take 30 years for the original sum to amount to 5 times itself at the same rate of simple interest.
The problem involves calculating the total distance of a journey given the total time and different speeds for the two halves of the journey. We can set up equations for time taken for each half of the journey and sum them up to the total given time. From this, we can find the total distance.
A room is 750 cm long and 525 cm wide. Its floor is to be paved with square tiles. What is the least number of tiles required?
To find the least number of square tiles required to pave a rectangular floor, we need to determine the largest possible size for each square tile. This means the side length of the square tile must be the greatest common divisor (GCD) of the room's length and width. Once the side length of the tile is found, we can calculate the area of the room and the area of one tile, and then divide the room's area by the tile's area to get the total number of tiles.
Length \(L = 750 \text{ cm}\)
Width \(W = 525 \text{ cm}\)
The side length of the largest possible square tile will be the GCD of 750 and 525. We can use the prime factorization method or Euclidean algorithm.
Prime factorization of 750:
\[ 750 = 2 \times 3 \times 5^3 \]\[ 750 = 2 \times 3 \times 5 \times 5 \times 5 \]Prime factorization of 525:
\[ 525 = 3 \times 5^2 \times 7 \]\[ 525 = 3 \times 5 \times 5 \times 7 \]The common prime factors are \(3\) and \(5^2\).
So, \( \text{GCD}(750, 525) = 3 \times 5^2 = 3 \times 25 = 75 \).
Thus, the side length of each square tile is \(75 \text{ cm}\).
Number of tiles along the length \[= \frac{\text{Length}}{\text{Side of tile}} = \frac{750 \text{ cm}}{75 \text{ cm}} = 10\]
Number of tiles along the width \[= \frac{\text{Width}}{\text{Side of tile}} = \frac{525 \text{ cm}}{75 \text{ cm}} = 7\]
Total number of tiles \( = (\text{Tiles along length}) \times (\text{Tiles along width}) = 10 \times 7 = 70 \)
Alternatively, calculate the areas:
Area of the room \( = L \times W = 750 \text{ cm} \times 525 \text{ cm} = 393750 \text{ cm}^2 \)
Area of one square tile \( = (\text{Side})^2 = (75 \text{ cm})^2 = 5625 \text{ cm}^2 \)
Number of tiles \[= \frac{\text{Area of room}}{\text{Area of one tile}} = \frac{393750 \text{ cm}^2}{5625 \text{ cm}^2} = 70\]
D) 70
This problem tests the understanding of how adding a constant to each value in a dataset affects its standard deviation and variance. A key property of standard deviation and variance is that they are measures of spread and are not affected by shifts in the data (i.e., adding or subtracting a constant from each data point).
Variance is the square of the standard deviation.
\[ \text{Variance} = (\text{Standard Deviation})^2 = \sigma^2 \]When a constant is added to or subtracted from each value in a dataset, the mean of the dataset changes by that constant, but the spread or dispersion of the data points around the mean remains the same. Consequently, the standard deviation and variance do not change.
Let the original values be \(x_1, x_2, \dots, x_n\). The new values are \(y_i = x_i + 4\).
The standard deviation for the original set is:
\[ \sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} \]The mean of the new set of values will be \(\bar{y} = \bar{x} + 4\).
The standard deviation for the new set of values (\(\sigma_\text{new}\)) is:
\[ \sigma_\text{new} = \sqrt{\frac{\sum (y_i - \bar{y})^2}{n}} = \sqrt{\frac{\sum ((x_i + 4) - (\bar{x} + 4))^2}{n}} \] \[ \sigma_\text{new} = \sqrt{\frac{\sum (x_i + 4 - \bar{x} - 4)^2}{n}} = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}} = \sigma \]This shows that adding a constant to each value does not change the standard deviation.
Since the standard deviation remains unchanged, the standard deviation of the new set of values is still 3.
Therefore, the variance of the new set of values is:
\[ \text{Variance}_{\text{new}} = (\sigma_\text{new})^2 = (3)^2 = 9 \]Correct Option: D) 9