The question asks for the new deadline set by the Ministry of Corporate Affairs (MCA) for private companies to dematerialise their securities, as extended in February 2025. This is a factual question requiring knowledge of recent regulatory updates.
Correct Option: C) 30 June 2025 — The Ministry of Corporate Affairs (MCA) extended the deadline for private companies (other than small companies) to dematerialise their securities to 30 June 2025. This extension was announced in February 2025, providing companies with additional time to comply with the dematerialisation requirements.
The question asks about the nature of India's industrial development on the eve of independence, specifically as a result of British colonial policies. To answer this, we need to recall the characteristics of industrialization in British India, which was largely designed to serve colonial interests rather than foster comprehensive domestic growth.
A) Limited industrial activity, mainly concentrated in a few cities like Calcutta and Bombay
This option accurately describes the industrial landscape of India on the eve of independence. British policies led to a stunted and skewed industrial structure. Modern industries were few and far between, primarily concentrated in port cities like Calcutta, Bombay, and to a lesser extent, Madras, which served as centers for trade and administration. These industries included textiles (cotton in Bombay, jute in Calcutta) and some basic heavy industries like iron and steel (Jamshedpur).
We are given three numbers in a specific ratio and their Least Common Multiple (LCM). We need to find their Highest Common Factor (HCF). The key concept here is that if numbers are in a given ratio, they can be expressed as multiples of their HCF. We can then use the relationship between the numbers, their HCF, and their LCM.
Let the three numbers be \(N_1\), \(N_2\), and \(N_3\).
Given that the numbers are in the ratio 2 : 1 : 5. Let their HCF be \(H\). Then the numbers can be written as:
\[ N_1 = 2H \]\[ N_2 = 1H = H \]\[ N_3 = 5H \]Now, we need to find the LCM of these three numbers. The LCM of \(2H\), \(H\), and \(5H\) can be found by taking the LCM of the numerical parts (2, 1, 5) and multiplying it by \(H\).
The LCM of 2, 1, and 5 is 10 (since 2 and 5 are prime numbers, and 1 is a factor of all numbers, LCM(2, 1, 5) = 2 * 5 = 10).
Therefore, the LCM of \(N_1\), \(N_2\), and \(N_3\) is \(10H\).
We are given that the LCM of the numbers is 5380.
So, we can set up the equation:
\[ 10H = 5380 \]Now, solve for \(H\):
\[ H = \frac{5380}{10} \]\[ H = 538 \]Thus, the HCF of the three numbers is 538.
A) 538 — The HCF of the three numbers is 538, as calculated by expressing the numbers in terms of HCF and then using the given LCM.
The question asks about an authority strengthened in 2024 to enforce penalties and green clearances under India's environmental regulations. This points to a recent legislative or administrative change concerning environmental governance. We need to identify which of the given bodies has seen such an enhancement in its powers.
Correct Option: D) Central Pollution Control Board
The Central Pollution Control Board (CPCB) was indeed strengthened in 2024 to enforce penalties and green clearances under India's environmental regulations. This move was aimed at enhancing the effectiveness of environmental governance and ensuring stricter compliance with environmental norms across various sectors. The CPCB, established under the Water (Prevention and Control of Pollution) Act, 1974, and also entrusted with powers and functions under the Air (Prevention and Control of Pollution) Act, 1981, is a crucial body for pollution control and environmental protection in India.
This problem can be solved by setting up a system of two linear equations with two variables, representing the cost of one bag and one pen. We will then solve these equations simultaneously to find the individual costs. Finally, we will calculate the difference between the cost of 8 bags and 8 pens.
Define variables:
Let \(B\) be the cost of one bag (in Rs.).
Let \(P\) be the cost of one pen (in Rs.).
Formulate equations from the given information:
From the first statement: "2 bags and 11 pens together cost Rs.505"
\[ 2B + 11P = 505 \quad \text{(Equation 1)} \]From the second statement: "10 bags and 10 pens together cost Rs.1310"
\[ 10B + 10P = 1310 \quad \text{(Equation 2)} \]Solve the system of equations:
We can simplify Equation 2 by dividing by 10:
\[ B + P = 131 \quad \text{(Equation 3)} \]From Equation 3, we can express \(B\) in terms of \(P\):
\[ B = 131 - P \]Substitute this expression for \(B\) into Equation 1:
\[ 2(131 - P) + 11P = 505 \]\[ 262 - 2P + 11P = 505 \]\[ 262 + 9P = 505 \]\[ 9P = 505 - 262 \]\[ 9P = 243 \]\[ P = \frac{243}{9} \]\[ P = 27 \]Now, substitute the value of \(P\) back into Equation 3 to find \(B\):
\[ B + 27 = 131 \]\[ B = 131 - 27 \]\[ B = 104 \]So, the cost of one bag is Rs.104 and the cost of one pen is Rs.27.
Calculate the required difference:
We need to find the cost of 8 bags minus the cost of 8 pens, which is \(8B - 8P\).
\[ 8B - 8P = 8(104) - 8(27) \]\[ 8B - 8P = 832 - 216 \]\[ 8B - 8P = 616 \]C) Rs.616 is the correct answer. The cost of 8 bags exceeds the cost of 8 pens by Rs.616.
The problem presents a pattern involving two letters and a number. We need to identify the logical rule governing the transformation from the first pair (NW -19) to the second (PZ -10), and from the third pair (RD -12) to the fourth (TG -3). Once the rule is established, we apply it to WU -2 to find the corresponding pair.
The consistent rule is:
Correct Option: C) YX 7
The problem asks us to find the average of the remaining numbers after two specific numbers are discarded from a set. We will first calculate the total sum of the original 36 numbers. Then, we will subtract the two discarded numbers from this sum to find the sum of the remaining numbers. Finally, we will divide this new sum by the new count of numbers to find the average.
Calculate the total sum of the original 36 numbers:
The average of 36 numbers is 45. The formula for average is:
So, the sum of the 36 numbers is:
\[ \text{Sum} = \text{Average} \times \text{Count} \]\[ \text{Sum} = 45 \times 36 \]\[ \text{Sum} = 1620 \]Identify the numbers to be discarded:
The two numbers to be discarded are 37 and 53.
Calculate the sum of the discarded numbers:
\[ \text{Sum of discarded numbers} = 37 + 53 \]\[ \text{Sum of discarded numbers} = 90 \]Calculate the sum of the remaining numbers:
Subtract the sum of the discarded numbers from the total sum:
Calculate the count of the remaining numbers:
Initially, there were 36 numbers. After discarding 2 numbers, the new count is:
Calculate the average of the remaining numbers:
Divide the sum of the remaining numbers by their count:
A) 45 is the correct answer. The average of the remaining numbers is 45.
This is a logical reasoning problem involving the arrangement of boxes. We need to deduce the exact order of the seven boxes based on the given clues and then determine the number of boxes above E.
This means B is at the very bottom position (Position 7, if we count from top to bottom, or Position 1 if we count from bottom to top. Let's assume Position 1 is the topmost and Position 7 is the bottommost for clarity).
Order: _ _ _ _ _ _ B (7)
Since B is at position 7, there are two boxes above B, and then G is placed.
Order: _ _ _ G _ _ B
This means G is at position 4.
Order: _ _ _ G (4) _ _ B (7)
This implies that A and F are consecutive, with A immediately above F. Also, it means there is no box above A. So, A must be at the very top (Position 1) and F at Position 2.
Order: A (1) F (2) _ G (4) _ _ B (7)
This tells us the relative order of C, E, and D is C > E > D. We have three empty slots remaining: Position 3, Position 5, and Position 6.
Let's fill these slots with C, E, and D in that order.
Position 3: C
Position 5: E
Position 6: D
1. A
2. F
3. C
4. G
5. E
6. D
7. B
Looking at the final arrangement, the boxes above E (Position 5) are A, F, C, and G.
There are 4 boxes above E.
Correct Option: D) 4
This problem requires the application of the empirical relationship between the mean, median, and mode for a moderately skewed distribution. The formula states that Mode \(= 3 \times\) Median \(- 2 \times\) Mean. We are given the difference between the mode and the mean, and we need to find the difference between the mode and the median.
The empirical relationship between mode, median, and mean is given by:
\[ \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \]We want to find the relationship between Mode and Median. Let's rearrange the formula to isolate terms involving Mode and Median.
\[ \text{Mode} - \text{Median} = 2 \times \text{Median} - 2 \times \text{Mean} \]\[ \text{Mode} - \text{Median} = 2 (\text{Median} - \text{Mean}) \]From the original empirical formula, we can write:
\[ 3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean} \]\[ \text{Median} = \frac{\text{Mode} + 2 \times \text{Mean}}{3} \]Now, substitute the expression for Median into the equation we want to solve: \( \text{Mode} - \text{Median} \).
\[ \text{Mode} - \text{Median} = \text{Mode} - \left( \frac{\text{Mode} + 2 \times \text{Mean}}{3} \right) \]\[ \text{Mode} - \text{Median} = \frac{3 \times \text{Mode} - (\text{Mode} + 2 \times \text{Mean})}{3} \]\[ \text{Mode} - \text{Median} = \frac{3 \times \text{Mode} - \text{Mode} - 2 \times \text{Mean}}{3} \]\[ \text{Mode} - \text{Median} = \frac{2 \times \text{Mode} - 2 \times \text{Mean}}{3} \]\[ \text{Mode} - \text{Median} = \frac{2}{3} (\text{Mode} - \text{Mean}) \]We are given that the mode exceeds its mean by 37.8, which means:
\[ \text{Mode} - \text{Mean} = 37.8 \]Substitute this value into the derived equation:
\[ \text{Mode} - \text{Median} = \frac{2}{3} (37.8) \]\[ \text{Mode} - \text{Median} = 2 \times 12.6 \]\[ \text{Mode} - \text{Median} = 25.2 \]Thus, the mode exceeds the median by 25.2.
D) 25.2 is the correct answer, as calculated using the empirical formula.