To find the median of a data set, we first need to arrange the data in ascending order. If the number of observations is even, the median is the average of the two middle terms. If the number of observations is odd, the median is the middle term. We will apply this method to find the initial median, then the new median after the data change, and finally calculate the mean of these two medians.
Correct Option: A) 55
This problem involves two trains moving towards each other. We can use the concept of relative speed or analyze the distance covered by each train over the same duration until they meet. The key information is the difference in distance covered by the two trains, which is directly related to the difference in their speeds.
Let the speed of the first train be \(S_1 = 20 \text{ km/h}\) and the speed of the second train be \(S_2 = 25 \text{ km/h}\).
Let \(t\) be the time in hours after which the two trains meet. Since they start at the same time and meet at a certain point, the time taken by both trains until they meet is the same.
The distance covered by the first train is \(D_1 = S_1 \times t = 20t \text{ km}\).
The distance covered by the second train is \(D_2 = S_2 \times t = 25t \text{ km}\).
We are given that one train has travelled 80 km more than the other. Since the second train has a higher speed, it will cover more distance. So, the difference in distance is:
\[ D_2 - D_1 = 80 \text{ km} \]Substitute the expressions for \(D_1\) and \(D_2\):
\[ 25t - 20t = 80 \]\[ 5t = 80 \]Solve for \(t\):
\[ t = \frac{80}{5} = 16 \text{ hours} \]The total distance between the two stations is the sum of the distances covered by both trains until they meet:
\[ \text{Total Distance} = D_1 + D_2 \]\[ \text{Total Distance} = 20t + 25t = 45t \]Substitute the value of \(t\):
\[ \text{Total Distance} = 45 \times 16 \]\[ \text{Total Distance} = 720 \text{ km} \]A) 720 km is the correct answer. This is the total distance between the two stations calculated by summing the distances covered by both trains until they meet.
The question asks for the year of institution of the Maulana Abul Kalam Azad (MAKA) Trophy. This is a factual question requiring knowledge of Indian sports history.
Correct Option: A) 1956-57
To convert a decimal number of hours into hours, minutes, and seconds, we need to separate the whole number part (hours) from the decimal part. The decimal part then needs to be converted into minutes by multiplying by 60, and any remaining decimal part of the minutes needs to be converted into seconds by again multiplying by 60.
The given time is 5.26 hours. The whole number part is 5, so we have 5 hours.
The decimal part is 0.26 hours. To convert this to minutes, we multiply by 60 (since 1 hour = 60 minutes).
\[ \text{Minutes} = 0.26 \times 60 \] \[ \text{Minutes} = 15.6 \text{ minutes} \]So, we have 15 whole minutes.
The decimal part of the minutes is 0.6 minutes. To convert this to seconds, we multiply by 60 (since 1 minute = 60 seconds).
\[ \text{Seconds} = 0.6 \times 60 \] \[ \text{Seconds} = 36 \text{ seconds} \]Combining the whole hours, whole minutes, and seconds, we get 5 hours, 15 minutes, and 36 seconds.
C) 5 hours 15 minutes 36 seconds
The question asks to identify the region with which the Kathiawar Peninsula shares geographical and cultural extensions. This requires knowledge of Indian geography and regional cultural connections.
Correct Option: B) Rajasthan is geographically contiguous with the Kathiawar Peninsula and shares significant cultural and historical ties, making it a natural extension.
The problem states that a positive number N, when divided by 5, leaves a remainder of 3. We need to find the unit's place digit of N. We can use the division algorithm to express N and then analyze the possible unit's digits.
When a number N is divided by a divisor D, it can be expressed as:
\[ N = D \times Q + R \]where Q is the quotient and R is the remainder, with \(0 \le R < D\).
Given that N is divided by 5 and leaves a remainder of 3, we have \(D = 5\) and \(R = 3\). So, we can write:
\[ N = 5 \times Q + 3 \]Here, Q is a non-negative integer (since N is a positive number).
Any number multiplied by 5 will have a unit's digit of either 0 or 5. This depends on whether Q is an even or an odd number.
Now, we need to add the remainder 3 to the unit's digit of \(5 \times Q\).
Then the unit's digit of N will be \(0 + 3 = 3\).
Then the unit's digit of N will be \(5 + 3 = 8\).
Therefore, the unit's place digit of N can be either 3 or 8.
C) 3 or 8
The problem asks us to find the area of a new square formed by joining the mid-points of the sides of an initial square, given the area of the initial square. We will first find the side length of the initial square, then determine the side length of the new square using the Pythagorean theorem, and finally calculate its area.
Find the side length of the initial square:
Let the side length of the initial square be \(a\).
The area of a square is given by the formula: \(\text{Area} = \text{side}^2\).
Given that the area of the initial square is \(36 \text{ cm}^2\).
\[ a^2 = 36 \text{ cm}^2 \]\[ a = \sqrt{36} \text{ cm} \]\[ a = 6 \text{ cm} \]So, the side length of the initial square is \(6 \text{ cm}\).
Determine the side length of the new square:
When the mid-points of the sides of a square are joined, they form another square inside the original one. Let the vertices of the initial square be A, B, C, D. Let the mid-points of AB, BC, CD, DA be P, Q, R, S respectively.
The side length of the initial square is \(a = 6 \text{ cm}\). Therefore, the distance from a vertex to a mid-point on an adjacent side is \(a/2\).
Consider the right-angled triangle formed by a vertex of the initial square and two adjacent mid-points. For example, consider the triangle formed by vertex A, mid-point S on AD, and mid-point P on AB. The lengths of the legs of this right-angled triangle are \(AS = a/2\) and \(AP = a/2\).
The side length of the new square (let's call it \(s\)) is the hypotenuse of this right-angled triangle (e.g., PS).
Using the Pythagorean theorem:
\[ s^2 = (AP)^2 + (AS)^2 \]\[ s^2 = \left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2 \]\[ s^2 = \frac{a^2}{4} + \frac{a^2}{4} \]\[ s^2 = \frac{2a^2}{4} \]\[ s^2 = \frac{a^2}{2} \]We know that \(a^2 = 36 \text{ cm}^2\).
Calculate the area of the new square:
The area of the new square is \(s^2\).
\[ \text{Area of new square} = s^2 = \frac{a^2}{2} \]\[ \text{Area of new square} = \frac{36 \text{ cm}^2}{2} \]\[ \text{Area of new square} = 18 \text{ cm}^2 \]Alternatively, a property states that the area of the square formed by joining the mid-points of the sides of a square is half the area of the original square.
Correct Option: A) The area of the new square is \(18 \text{ cm}^2\).
To find the length of the sides of the square room, we first need to calculate the area covered by one tile and then the total area of the floor. Since the room is square, its side length can be found by taking the square root of its total area. It's important to ensure all units are consistent, preferably converting centimeters to meters at an appropriate step.
The side of each square tile is given as 50 cm.
\[ \text{Side of tile} = 50 \text{ cm} \]\[ \text{Area of one tile} = (\text{Side of tile})^2 = (50 \text{ cm})^2 = 2500 \text{ cm}^2 \]There are 784 such tiles required to tile the floor.
\[ \text{Total area of floor} = \text{Number of tiles} \times \text{Area of one tile} \]\[ \text{Total area of floor} = 784 \times 2500 \text{ cm}^2 = 1,960,000 \text{ cm}^2 \]Since \(1 \text{ m} = 100 \text{ cm}\), then \(1 \text{ m}^2 = (100 \text{ cm})^2 = 10,000 \text{ cm}^2\).
\[ \text{Total area of floor in m}^2 = \frac{1,960,000 \text{ cm}^2}{10,000 \text{ cm}^2/\text{m}^2} = 196 \text{ m}^2 \]Let the length of the side of the square room be \(L\). The area of the square room is \(L^2\).
\[ L^2 = 196 \text{ m}^2 \]\[ L = \sqrt{196 \text{ m}^2} \]\[ L = 14 \text{ m} \]Correct Option: D) 14 m
The length of the sides of the room is 14 meters.
The problem states that if the rate of simple interest were 3% higher, the simple interest for two years would be ₹240 more. We can use the formula for simple interest to find the initial sum invested. The key idea is that the increase in simple interest is solely due to the increase in the rate of interest, as the principal and time remain constant.
C) ₹4,000
The problem asks for the greatest number that divides two given numbers, leaving specific remainders. This is a classic application of finding the Highest Common Factor (HCF) or Greatest Common Divisor (GCD). If a number \(N\) divides \(A\) leaving a remainder \(R_1\), then \(N\) must divide \((A - R_1)\) exactly. Similarly, if \(N\) divides \(B\) leaving a remainder \(R_2\), then \(N\) must divide \((B - R_2)\) exactly. Therefore, we need to find the HCF of the numbers after subtracting their respective remainders.
Correct Option: A) 42 is the greatest number that will divide 2400 and 1810 and leave remainders 6 and 4 respectively.