To find the cost price when the selling price and loss percentage are given, we can use the formula relating these three quantities. A loss of 6% means the selling price is 94% of the cost price.
Let the Cost Price (CP) of the article be \( \text{Rs } x \).
The man incurs a loss of 6%.
The Selling Price (SP) is given as Rs 940.
The formula for loss percentage is:
\[ \text{Loss Percentage} = \frac{\text{Loss}}{\text{Cost Price}} \times 100 \]Also, \( \text{Loss} = \text{Cost Price} - \text{Selling Price} \).
Substitute these into the loss percentage formula:
\[ 6 = \frac{x - 940}{x} \times 100 \]Now, solve for \( x \):
\[ 6x = (x - 940) \times 100 \]\[ 6x = 100x - 94000 \]\[ 94000 = 100x - 6x \]\[ 94000 = 94x \]\[ x = \frac{94000}{94} \]\[ x = 1000 \]Alternatively, if there is a loss of 6%, it means the selling price is \( (100 - 6)\% = 94\% \) of the cost price.
\[ \text{Selling Price} = \frac{100 - \text{Loss Percentage}}{100} \times \text{Cost Price} \]\[ 940 = \frac{94}{100} \times \text{Cost Price} \]\[ \text{Cost Price} = 940 \times \frac{100}{94} \]\[ \text{Cost Price} = 10 \times 100 \]\[ \text{Cost Price} = 1000 \]Thus, the cost price of the article is Rs 1,000.
C) Rs 1,000 is the correct cost price, as calculated.
Find the arithmetic mean of the following data.
| x | 5 | 10 | 15 | 20 | 25 | 30 |
| f | 3 | 5 | 6 | 2 | 1 | 3 |
To find the arithmetic mean of grouped data, we use the formula for the weighted average, where the 'x' values are the variates and 'f' values are their corresponding frequencies. The formula is the sum of the products of each variate and its frequency, divided by the sum of the frequencies.
A) 15.5 — The arithmetic mean calculated using the formula \(\bar{x} = \frac{\sum (x \cdot f)}{\sum f}\) is 15.5.
Refer to the given number, symbol series and answer the question that follows.
(Left) 9 + $ 5 @ $ 2 6 3 % 2 & 8 # * 2 9 ! 4 3 (Right)
How many such symbols are there (from left to right) which are immediately preceded by a symbol and also immediately followed by an even number?
The problem asks us to identify symbols in a given series that satisfy two conditions: they must be immediately preceded by another symbol and immediately followed by an even number. We will scan the series from left to right, checking each symbol against these conditions.
D) 2. There are two such symbols in the series that are immediately preceded by a symbol and immediately followed by an even number.
The question asks for the specific date when the 17th Foundation Day of the National Commission for Protection of Child Rights (NCPCR) was observed. To answer this, one needs to know the establishment date of the NCPCR and then calculate the 17th anniversary.
The 17th Foundation Day of the National Commission for Protection of Child Rights (NCPCR) was celebrated on 1 March 2022. This is a specific event date that needs to be recalled.
Correct Option: A) 1 March 2022
To find the missing number in the series, we need to identify the pattern or the relationship between consecutive numbers. We will calculate the difference between each adjacent pair of numbers to see if there is a constant difference or a progressive pattern.
Therefore, the number that replaces the question mark is \(10.4\).
Correct Option: D) 10.4
The problem asks us to calculate the total power dissipated in an electrical circuit and the resistance of each identical bulb. We are given the voltage of the battery, the total current drawn, and that five identical bulbs are connected in parallel.
The power dissipated in an electrical circuit can be calculated using the formula:
\[ P = V \times I \]where \(P\) is the power, \(V\) is the voltage, and \(I\) is the total current.
Given:
Substituting the values:
\[ P = 20 \text{ V} \times 5 \text{ A} = 100 \text{ W} \]So, the total power dissipated in the circuit is \(100 \text{ W}\).
Using Ohm's Law, the total equivalent resistance \(R_{eq}\) can be found:
\[ R_{eq} = \frac{V}{I} \]Substituting the values:
\[ R_{eq} = \frac{20 \text{ V}}{5 \text{ A}} = 4 \text{ \Omega} \]Since five identical bulbs are connected in parallel, the formula for equivalent resistance in a parallel circuit is:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} \]As all bulbs are identical, let the resistance of each bulb be \(R_{bulb}\). Then \(R_1 = R_2 = R_3 = R_4 = R_5 = R_{bulb}\).
\[ \frac{1}{R_{eq}} = \frac{1}{R_{bulb}} + \frac{1}{R_{bulb}} + \frac{1}{R_{bulb}} + \frac{1}{R_{bulb}} + \frac{1}{R_{bulb}} = \frac{5}{R_{bulb}} \]Therefore,
\[ R_{eq} = \frac{R_{bulb}}{5} \]We found \(R_{eq} = 4 \text{ \Omega}\). So, we can solve for \(R_{bulb}\):
\[ 4 \text{ \Omega} = \frac{R_{bulb}}{5} \] \[ R_{bulb} = 4 \text{ \Omega} \times 5 = 20 \text{ \Omega} \]Thus, the resistance of each bulb is \(20 \text{ \Omega}\).
B) The power dissipated in the circuit is \(100 \text{ W}\) and the resistance of each bulb is \(20 \text{ \Omega}\). This matches our calculations.
The question asks to identify the type of soil that covers approximately 40% of India's total area. This requires knowledge of the distribution and extent of major soil types in India.
Correct Option: D) Alluvial soils cover about 40-43% of the total area of India, making them the most widespread soil type.
The question asks to identify the winner of the third edition of the BBC Indian Sportswoman of the Year Award in 2021. This is a factual question related to current affairs and sports awards.
Correct Option: A) Mirabai Chanu won the third edition of the BBC Indian Sportswoman of the Year Award in 2021. She was recognized for her outstanding performance, particularly her silver medal win at the Tokyo Olympics.
To balance a chemical equation, we must ensure that the number of atoms of each element on the reactant side is equal to the number of atoms of that same element on the product side. This adheres to the Law of Conservation of Mass.
The given reaction is: Zinc + Silver nitrate \(\rightarrow\) Zinc nitrate + Silver
The chemical formulas are:
So, the unbalanced equation is:
\[ \text{Zn (s)} + \text{AgNO}_3 \text{ (aq)} \rightarrow \text{Zn(NO}_3)_2 \text{ (aq)} + \text{Ag (s)} \]
| Element | Reactant Side | Product Side |
|---|---|---|
| Zn | 1 | 1 |
| Ag | 1 | 1 |
| N | 1 | 2 |
| O | 3 | 6 |
On the product side, there are two nitrate groups \(((\text{NO}_3)_2)\), while on the reactant side, there is only one \(((\text{NO}_3))\). To balance this, place a coefficient of 2 in front of \(\text{AgNO}_3\) on the reactant side:
\[ \text{Zn (s)} + 2\text{AgNO}_3 \text{ (aq)} \rightarrow \text{Zn(NO}_3)_2 \text{ (aq)} + \text{Ag (s)} \]
Now, let's recount:
| Element | Reactant Side | Product Side |
|---|---|---|
| Zn | 1 | 1 |
| Ag | 2 | 1 |
| N | 2 | 2 |
| O | 6 | 6 |
Silver (\(\text{Ag}\)) is now unbalanced. There are 2 silver atoms on the reactant side and 1 on the product side. Place a coefficient of 2 in front of \(\text{Ag}\) on the product side:
\[ \text{Zn (s)} + 2\text{AgNO}_3 \text{ (aq)} \rightarrow \text{Zn(NO}_3)_2 \text{ (aq)} + 2\text{Ag (s)} \]
| Element | Reactant Side | Product Side |
|---|---|---|
| Zn | 1 | 1 |
| Ag | 2 | 2 |
| N | 2 | 2 |
| O | 6 | 6 |
All atoms are balanced.
Correct Option: D) Zn (s) + 2AgNO₃ (aq) → Zn(NO₃)₂ (aq) + 2Ag (s)
The problem asks us to find the value of an expression involving \(\tan^2\theta\) and \(\cot^2\theta\), given an equation involving \(\tan\theta\) and \(\cot\theta\). We can use the algebraic identity \((a+b)^2 = a^2 + b^2 + 2ab\) to relate the given equation to the required expression. We also know that \(\tan\theta \cdot \cot\theta = 1\).
We are given the equation:
\[ \tan\theta + \cot\theta = 14 \]Square both sides of the equation:
\[ (\tan\theta + \cot\theta)^2 = (14)^2 \]Apply the algebraic identity \((a+b)^2 = a^2 + b^2 + 2ab\) to the left side, where \(a = \tan\theta\) and \(b = \cot\theta\):
\[ \tan^2\theta + \cot^2\theta + 2 \tan\theta \cot\theta = 196 \]Use the trigonometric identity \(\tan\theta \cdot \cot\theta = 1\):
\[ \tan^2\theta + \cot^2\theta + 2(1) = 196 \]\[ \tan^2\theta + \cot^2\theta + 2 = 196 \]Subtract 2 from both sides to find the value of \(\tan^2\theta + \cot^2\theta\):
\[ \tan^2\theta + \cot^2\theta = 196 - 2 \]\[ \tan^2\theta + \cot^2\theta = 194 \]Now, we need to find the value of \(7\tan^2\theta + 7\cot^2\theta\). Factor out 7:
\[ 7\tan^2\theta + 7\cot^2\theta = 7(\tan^2\theta + \cot^2\theta) \]Substitute the value we found in step 5:
\[ 7(194) \]Perform the multiplication:
\[ 7 \times 194 = 1358 \]Thus, the value of \(7\tan^2\theta + 7\cot^2\theta\) is 1358.
D) 1358 is the correct answer, as calculated through the steps above.