The question asks for the city where the Hindustan Socialist Republican Association (HSRA) was founded in 1928. This requires knowledge of the history of revolutionary organizations in India.
Correct Option: A) The Hindustan Socialist Republican Association (HSRA) was founded in 1928 at the Feroz Shah Kotla ground in Delhi. This was a crucial meeting where the Hindustan Republican Association (HRA) was reorganized and given a new name and a socialist orientation by revolutionaries like Bhagat Singh, Chandrashekhar Azad, Sukhdev, and others.
This question asks us to identify the letter-cluster that does not follow the same pattern as the others, based on the English alphabetical order. We need to find the positional values of each letter in the alphabet and then look for a consistent mathematical or sequential relationship within each cluster.
From the analysis, Options B, C, and D all follow the same pattern of differences: +8, +3, +5 between consecutive letters' alphabetical positions. Option A does not follow this pattern.
A) ENPU is the letter-cluster that does not belong to the group. The differences between its consecutive letters are +9, +2, +5, which is different from the +8, +3, +5 pattern observed in the other options.
This problem involves symbolic substitution and then evaluating the resulting mathematical expression using the order of operations (BODMAS/PEMDAS).
The original equation is: \(14 + 1 \div 7 \times 32 - 8 = ?\)
Applying the substitutions, the new equation becomes:
\[ 14 - 1 \times 7 + 32 \div 8 = ? \]BODMAS/PEMDAS stands for Brackets, Orders (powers/roots), Division and Multiplication (from left to right), Addition and Subtraction (from left to right).
Correct Option: B) 11
To find the greatest of four consecutive odd numbers whose average is given, we can represent the numbers algebraically. Since they are consecutive odd numbers, they differ by 2. The average of these numbers is their sum divided by the count of numbers.
Let the four consecutive odd numbers be represented as:
\[ x, x+2, x+4, \text{ and } x+6 \]
where \(x\) is the smallest odd number.
The average of these four numbers is given as 36. The formula for average is:
\[ \text{Average} = \frac{\text{Sum of numbers}}{\text{Count of numbers}} \]
Substitute the numbers and the given average into the formula:
\[ 36 = \frac{x + (x+2) + (x+4) + (x+6)}{4} \]
Simplify the sum in the numerator:
\[ 36 = \frac{4x + 12}{4} \]
Multiply both sides by 4:
\[ 36 \times 4 = 4x + 12 \] \[ 144 = 4x + 12 \]
Subtract 12 from both sides:
\[ 144 - 12 = 4x \] \[ 132 = 4x \]
Divide by 4 to find the value of \(x\):
\[ x = \frac{132}{4} \] \[ x = 33 \]
So, the smallest odd number is 33. The four consecutive odd numbers are:
The greatest of these numbers is 39.
Alternatively, for an even number of consecutive terms, their average is the average of the two middle terms. Here, the average is 36. The two middle odd numbers must be 35 and 37, as their average is \((35+37)/2 = 72/2 = 36\). Since the numbers are consecutive odd numbers, the number before 35 is 33, and the number after 37 is 39. So the four numbers are 33, 35, 37, 39. The greatest is 39.
B) 39 โ As calculated, the four consecutive odd numbers are 33, 35, 37, and 39. The greatest among them is 39.
The question asks for the best description of a double displacement reaction. We need to understand the fundamental characteristics of this type of chemical reaction and differentiate it from other common reaction types.
B) Ions in two compounds switch places to form two new compounds. This statement perfectly describes a double displacement reaction. For example, when silver nitrate (\(\text{AgNO}_3\)) reacts with sodium chloride (\(\text{NaCl}\)), the silver ion (\(\text{Ag}^+\)) switches places with the sodium ion (\(\text{Na}^+\)), and the nitrate ion (\(\text{NO}_3^-\)) switches places with the chloride ion (\(\text{Cl}^-\)), resulting in the formation of silver chloride (\(\text{AgCl}\)) and sodium nitrate (\(\text{NaNO}_3\)). The general form is:
\[ \text{AB} + \text{CD} \rightarrow \text{AD} + \text{CB} \]This is a coding-decoding problem where we need to find the code for a specific word by comparing common words and their codes across different statements. We will identify common words and their corresponding codes to deduce the unique code for 'hat'.
Statement 1: 'hat tie watch' is coded as 'zm ph mt'
Statement 2: 'belt bow tie' is coded as 'la oe zm'
Statement 3: 'bow watch clip' is coded as 'oe mt dy'
Compare Statement 1 and Statement 2. The common word is 'tie'. The common code is 'zm'.
So, 'tie' = 'zm'
Compare Statement 1 and Statement 3. The common word is 'watch'. The common code is 'mt'.
So, 'watch' = 'mt'
From Statement 1, we have 'hat tie watch' = 'zm ph mt'.
We know 'tie' = 'zm' and 'watch' = 'mt'.
Substituting these, we get 'hat' + 'zm' + 'mt' = 'zm ph mt'.
Therefore, the code for 'hat' must be 'ph'.
Correct Option: C) ph
To find the numbers that leave no remainder when divided by 8, 12, and 18, we first need to find the Least Common Multiple (LCM) of these three numbers. Any number divisible by 8, 12, and 18 must be a multiple of their LCM. Then, we identify the multiples of this LCM that fall within the given range (250 and 450) and sum them up.
Correct Option: A) 1080
If the mean of the following data is 40, then the value of X is:
| xi | 10 | 25 | 70 | 85 |
| fi | 21 | X | 22 | 35 |
To find the missing frequency \(X\), we will use the formula for the mean of grouped data. The mean is given by the sum of the products of each data point and its frequency, divided by the sum of the frequencies.
First, let's list the given data:
| \(x_i\) | \(f_i\) | \(f_i x_i\) |
|---|---|---|
| 10 | 21 | \(10 \times 21 = 210\) |
| 25 | \(X\) | \(25 \times X = 25X\) |
| 70 | 22 | \(70 \times 22 = 1540\) |
| 85 | 35 | \(85 \times 35 = 2975\) |
Calculate the sum of frequencies (\(\Sigma f_i\)):
\[ \Sigma f_i = 21 + X + 22 + 35 = 78 + X \]Calculate the sum of the products of \(f_i\) and \(x_i\) (\(\Sigma f_i x_i\)):
\[ \Sigma f_i x_i = 210 + 25X + 1540 + 2975 = 4725 + 25X \]The formula for the mean (\(\bar{x}\)) is:
\[ \bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \]We are given that the mean (\(\bar{x}\)) is 40. Substitute the values into the formula:
\[ 40 = \frac{4725 + 25X}{78 + X} \]Now, solve for \(X\):
\[ 40(78 + X) = 4725 + 25X \]\[ 3120 + 40X = 4725 + 25X \]\[ 40X - 25X = 4725 - 3120 \]\[ 15X = 1605 \]\[ X = \frac{1605}{15} \]\[ X = 107 \]B) The value of \(X\) is 107.
If the interest earned during the 2nd year on a certain sum is โน3,726, and the rate of interest is 15% per annum compounded annually, then the sum is:
The problem asks us to find the principal sum given the compound interest earned in the second year and the annual interest rate. We can solve this by understanding how compound interest accrues year by year. The interest for the second year is calculated on the principal plus the interest earned in the first year.
A) โน21,600 is the correct sum.
To find the number of resistors required, we first need to determine the total equivalent resistance needed for the circuit using Ohm's Law. Then, knowing that the resistors are connected in parallel, we can use the formula for equivalent resistance in parallel to find the number of individual resistors.
Calculate the total equivalent resistance (\(R_{eq}\)) of the circuit:
Given: Voltage \(V = 220 \, \text{V}\), Current \(I = 5 \, \text{A}\).
Using Ohm's Law, \(V = I R_{eq}\), we can find \(R_{eq}\):
\[ R_{eq} = \frac{V}{I} = \frac{220 \, \text{V}}{5 \, \text{A}} = 44 \, \Omega \]Determine the number of resistors (\(n\)) connected in parallel:
When \(n\) identical resistors, each with resistance \(R\), are connected in parallel, their equivalent resistance \(R_{eq}\) is given by:
\[ \frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{R} + \dots + \frac{1}{R} \quad (\text{n times}) \]\[ \frac{1}{R_{eq}} = \frac{n}{R} \]\[ R_{eq} = \frac{R}{n} \]Solve for \(n\):
We know \(R_{eq} = 44 \, \Omega\) and the resistance of each individual resistor \(R = 176 \, \Omega\).
Substitute these values into the formula:
\[ 44 \, \Omega = \frac{176 \, \Omega}{n} \]\[ n = \frac{176 \, \Omega}{44 \, \Omega} \]\[ n = 4 \]Therefore, 4 resistors are required.
Correct Option: A) 4