The question asks about the atomicity of Helium. Atomicity refers to the number of atoms present in one molecule of an element. To answer this, we need to recall the nature of noble gases, to which Helium belongs.
Correct Option: A) Monoatomic
Helium is a noble gas and exists as single, uncombined atoms. Thus, its atomicity is 1, making it monoatomic.
To find the net percentage loss or gain, we need to calculate the cost price (CP) of each item, then the total cost price and total selling price (SP). The net percentage loss or gain can then be calculated using these totals.
Selling Price (\(SP_1\)) = Rs. 200
Gain = 10%
The formula for selling price with gain is: \(SP = CP \times (1 + \frac{\text{Gain}\%}{100})\)
So, \(200 = CP_1 \times (1 + \frac{10}{100})\)
\(200 = CP_1 \times (1 + 0.1)\)
\(200 = CP_1 \times 1.1\)
\[CP_1 = \frac{200}{1.1} = \frac{2000}{11}\] Rs.
Selling Price (\(SP_2\)) = Rs. 200
Loss = 10%
The formula for selling price with loss is: \(SP = CP \times (1 - \frac{\text{Loss}\%}{100})\)
So, \(200 = CP_2 \times (1 - \frac{10}{100})\)
\(200 = CP_2 \times (1 - 0.1)\)
\(200 = CP_2 \times 0.9\)
\[CP_2 = \frac{200}{0.9} = \frac{2000}{9}\] Rs.
Total Selling Price (\(TSP\)) = \(SP_1 + SP_2 = 200 + 200 = 400\) Rs.
Total Cost Price (\(TCP\)) = \[CP_1 + CP_2 = \frac{2000}{11} + \frac{2000}{9}\]
\[TCP = 2000 \times (\frac{1}{11} + \frac{1}{9})\]
\(TCP = 2000 \times (\frac{9 + 11}{99})\)
\[TCP = 2000 \times \frac{20}{99} = \frac{40000}{99}\] Rs.
Since \(TCP = \frac{40000}{99} \approx 404.04\) Rs. and \(TSP = 400\) Rs., we can see that \(TCP > TSP\), which means there is a net loss.
Net Loss = \(TCP - TSP = \frac{40000}{99} - 400\)
\(Net \text{ Loss} = \frac{40000 - 400 \times 99}{99}\)
\[Net \text{ Loss} = \frac{40000 - 39600}{99} = \frac{400}{99}\] Rs.
Percentage Loss = \((\frac{\text{Net Loss}}{\text{TCP}}) \times 100\)
Percentage Loss = \((\frac{400/99}{40000/99}) \times 100\)
Percentage Loss = \[(\frac{400}{99} \times \frac{99}{40000}) \times 100\]
Percentage Loss = \((\frac{400}{40000}) \times 100\)
Percentage Loss = \((\frac{1}{100}) \times 100 = 1\%\)
Alternatively, for cases where two identical items are sold at the same price, one at x% gain and the other at x% loss, there is always a net loss given by the formula: \((x/10)^2 \%\). In this case, \(x=10\), so the loss is \((10/10)^2 \% = 1^2 \% = 1\%\).
Correct Option: D) 1% loss
This problem can be solved using the principles of set theory, specifically the formula for the union of two sets. We are given percentages of people reading newspaper A, newspaper B, and neither. From this, we can find the percentage of people who read at least one newspaper. Then, using the formula for the union of two sets, we can find the percentage of people who read both newspapers. Finally, we can use the given number of people who read both to find the total number of people surveyed.
Let \(P(A)\) be the percentage of people who read newspaper A, and \(P(B)\) be the percentage of people who read newspaper B.
We are given that 20% read neither A nor B. This means the percentage of people who read at least one newspaper (A or B or both) is \(100\% - 20\% = 80\%\). So, \(P(A \cup B) = 80\%\).
The formula for the union of two sets is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]Substitute the known percentages into the formula to find \(P(A \cap B)\), which is the percentage of people who read both A and B:
\[ 80\% = 50\% + 40\% - P(A \cap B) \]\[ 80\% = 90\% - P(A \cap B) \]\[ P(A \cap B) = 90\% - 80\% \]\[ P(A \cap B) = 10\% \]We are given that the number of persons who read both A and B is 500. This corresponds to the 10% we just calculated.
Let \(T\) be the total number of persons surveyed. Then:
\[ 10\% \text{ of } T = 500 \]\[ 0.10 \times T = 500 \]Solve for \(T\):
\[ T = \frac{500}{0.10} \]\[ T = 5000 \]Correct Option: D) 5000
The question asks for the time taken for one complete oscillation of the density of the medium in a sound wave. This is a fundamental definition in wave physics. We need to identify the term that specifically refers to the duration of one full cycle of a wave.
D) Time period is the correct answer because it is defined as the time taken for one complete oscillation or cycle of a wave or any periodic motion. In the context of a sound wave, this means the time for the density of the medium at a point to go through one full cycle of compression and rarefaction.
The question asks about the definition of an element's atomic number. We need to recall the fundamental components of an atom and how the atomic number is defined in relation to these components.
Therefore, the atomic number is solely determined by the number of protons in the nucleus.
Correct Option: D) Protons
If a sum becomes 4 times in 20 years what is the rate of simple interest given?
This problem involves calculating the rate of simple interest. We are given the principal amount becomes 4 times itself over a certain period. We can use the formula for simple interest to find the rate.
Let the principal amount be \(P\). If the sum becomes 4 times in 20 years, it means the Amount (\(A\)) after 20 years is \(4P\).
The simple interest (\(SI\)) earned is the difference between the Amount and the Principal:
\[ SI = A - P \]\[ SI = 4P - P \]\[ SI = 3P \]The formula for simple interest is:
\[ SI = \frac{P \times R \times T}{100} \]where \(P\) is the Principal, \(R\) is the Rate of interest per annum, and \(T\) is the Time in years.We are given \(SI = 3P\), \(T = 20\) years, and we need to find \(R\). Substitute these values into the formula:
\[ 3P = \frac{P \times R \times 20}{100} \]Now, we can solve for \(R\). Divide both sides by \(P\) (assuming \(P \neq 0\)):
\[ 3 = \frac{R \times 20}{100} \]Simplify the right side:
\[ 3 = \frac{R}{5} \]Multiply both sides by 5 to find \(R\):
\[ R = 3 \times 5 \]\[ R = 15\% \]Correct Option: A) 15%
This is an analogy question based on letter series. We need to find the pattern or relationship between the letters in the first pair (FDA : MKH) and then apply the same pattern to the second pair (QOL : ?).
The pattern is that each letter in the first word is shifted forward by 7 positions in the alphabet to get the corresponding letter in the second word.
Correct Option: D) XVS
The question asks to identify the first Indian footballer to receive the Arjuna Award. This requires knowledge of Indian sports history, specifically the recipients of the Arjuna Award in football.
Correct Option: A) P.K. Banerjee was indeed the first Indian footballer to receive the Arjuna Award. He was honored in 1961, the inaugural year of the award, for his outstanding contributions to Indian football as a player.
To find the number of 3-digit numbers divisible by 3, we first identify the smallest and largest 3-digit numbers. Then, we find the smallest and largest 3-digit numbers that are divisible by 3. Finally, we use the formula for the number of terms in an arithmetic progression.
D) 300 is the correct answer, as calculated by finding the range of 3-digit numbers divisible by 3 and applying the arithmetic progression formula.
The question asks to identify the specific gas that leaked during the Bhopal tragedy. This is a factual recall question related to a significant industrial disaster in India's history.
Correct Option: C) Methyl isocyanate