The question describes a body covering equal distances in equal intervals of time along a straight path. This is the definition of uniform motion.
D) Uniform motion. This is correct because the body covers equal distances in equal intervals of time along a straight path, implying constant speed and constant direction (constant velocity).
Study the given table and answer the question that follows.
The table depicts new admissions in the Primary and Middle wings of a school in the years 2021 and 2022.
| Years | New admission in Primary Wing | New admission in Middle Wing |
| 2021 | 25 | 20 |
| 2022 | 12 | 15 |
How many new admissions were there in the Primary wing of the school in the years 2021 and 2022 together?
The problem asks for the total number of new admissions in the Primary wing for the years 2021 and 2022 combined. We need to extract the relevant data from the provided table and sum the values.
Total new admissions in Primary wing (2021 and 2022 together) = (New admissions in Primary wing in 2021) + (New admissions in Primary wing in 2022)
\[ \text{Total admissions} = 25 + 12 = 37 \]
C) 37 is the correct answer, as calculated by summing the new admissions in the Primary wing for 2021 (25) and 2022 (12).
To calculate the concentration of a solution in terms of volume by volume percentage (\(v/v\%\)), we use the formula:
\[ \text{Volume by volume percentage} = \frac{\text{Volume of solute}}{\text{Volume of solution}} \times 100 \% \]First, we need to find the total volume of the solution, which is the sum of the volume of the solute (acid) and the volume of the solvent (water).
Identify the given values:
Calculate the total volume of the solution:
The volume of the solution is the sum of the volume of the solute and the volume of the solvent.
\[ \text{Volume of solution} = \text{Volume of acid} + \text{Volume of water} \] \[ \text{Volume of solution} = 40 \text{ ml} + 120 \text{ ml} = 160 \text{ ml} \]Calculate the volume by volume percentage:
Now, use the formula for volume by volume percentage:
\[ \text{Volume by volume percentage} = \frac{\text{Volume of acid}}{\text{Volume of solution}} \times 100 \% \] \[ \text{Volume by volume percentage} = \frac{40 \text{ ml}}{160 \text{ ml}} \times 100 \% \] \[ \text{Volume by volume percentage} = \frac{1}{4} \times 100 \% \] \[ \text{Volume by volume percentage} = 0.25 \times 100 \% = 25 \% \]B) 25% — The calculation shows that the concentration of the solution in terms of volume by volume percentage is 25%.
To find the combined ratio \(a:b:c\), we need to ensure that the common term, which is 'b' in this case, has the same value in both given ratios. Once the value of 'b' is consistent, we can directly combine the ratios.
We are given two ratios:
The common term in both ratios is 'b'. In the first ratio, \(a:b\), the value corresponding to 'b' is 4. In the second ratio, \(b:c\), the value corresponding to 'b' is also 4.
Since the value of 'b' is already the same in both ratios (which is 4), we can directly combine them.
When the common term has the same value, we can write the combined ratio \(a:b:c\) by taking the respective values:
Therefore, \(a:b:c = 3:4:1\).
C) 3 : 4 : 1 — The common term 'b' has the same value (4) in both given ratios, allowing for direct combination of the values of a, b, and c.
The question asks about the location of a specific joint military exercise conducted in 2024 involving the Indian Army's Kharga Corps and the Indian Air Force. This requires knowledge of recent defense exercises and their geographical locations.
Correct Option: A) Punjab
The joint exercise "Gagan Strike" involving the Indian Army's Kharga Corps and the Indian Air Force was indeed conducted in Punjab in 2024. This exercise focused on validating the operational concepts of the Indian Army's Strike Corps, supported by the Indian Air Force, in a high-intensity conflict scenario.
The question asks to identify the specific civilian award conferred upon Narayan Chakraborty in 2024 for his contributions to Science and Engineering. This requires knowledge of recent national awards, specifically the Padma Awards.
Correct Option: B) Padma Shri
Narayan Chakraborty was awarded the Padma Shri in 2024 for his distinguished service in the field of Science and Engineering. The Padma Shri is the fourth-highest civilian award in the Republic of India, awarded for distinguished service in any field.
To solve this problem, we need to apply two rules: change each consonant to the letter immediately preceding it and change each vowel to the letter immediately succeeding it. We will then form the new word and identify any repeated letters.
Combining the transformed letters, the new group of letters is C, B, M, B, F, Q.
In the new group of letters (C, B, M, B, F, Q), the letter 'B' appears twice.
B) The letter 'B' is repeated twice in the new group of letters (C, B, M, B, F, Q).
The problem requires us to apply specific rules for changing letters in the word KNIGHT based on whether they are consonants or vowels. We then need to count the number of vowels in the newly formed word.
Correct Option: D) Three
The problem requires us to evaluate a given mathematical expression after interchanging specific operators. We need to replace '+' with '−', '−' with '+', '×' with '÷', and '÷' with '×'. After performing these interchanges, we will solve the new expression following the order of operations (BODMAS/PEMDAS).
Thus, the value of the expression after interchanging the operators is 273.
A) 273 is the correct answer, as derived from the step-by-step calculation after interchanging the operators.
This problem requires us to find the mode of a data set given its mean and median. We can use the empirical relationship between mean, median, and mode, which is often applicable for moderately skewed distributions.
Recall the Empirical Formula: The empirical relationship between mean, median, and mode is given by:
\[ \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean} \]Identify the given values:
Substitute the values into the formula:
\[ \text{Mode} = (3 \times 32) - (2 \times 28.5) \]Perform the multiplication:
\[ 3 \times 32 = 96 \] \[ 2 \times 28.5 = 57 \]Perform the subtraction to find the mode:
\[ \text{Mode} = 96 - 57 = 39 \]Correct Option: D) 39