To find the cost price of the jacket, we need to work backward from the marked price. First, calculate the selling price after the discount. Then, use the profit percentage to determine the cost price.
The marked price (MP) is Rs 1,050. A discount of 10% is given.
\[ \text{Discount Amount} = 10\% \text{ of } \text{MP} = 0.10 \times 1050 = \text{Rs } 105 \] \[ \text{Selling Price (SP)} = \text{MP} - \text{Discount Amount} = 1050 - 105 = \text{Rs } 945 \]The shopkeeper earns a profit of 25% on the selling price of Rs 945.
The formula relating Selling Price, Cost Price, and Profit Percentage is:
\[ \text{SP} = \text{CP} \times \left(1 + \frac{\text{Profit Percentage}}{100}\right) \]We need to find CP, so rearrange the formula:
\[ \text{CP} = \frac{\text{SP}}{\left(1 + \frac{\text{Profit Percentage}}{100}\right)} \]Substitute the values:
\[ \text{CP} = \frac{945}{\left(1 + \frac{25}{100}\right)} = \frac{945}{\left(1 + 0.25\right)} = \frac{945}{1.25} \] \[ \text{CP} = 756 \]So, the cost price of the jacket is Rs 756.
D) Rs 756 is the correct cost price of the jacket, calculated by first determining the selling price after discount and then working backward using the profit percentage.
This is a syllogism problem. We will use Venn diagrams to represent the given statements and then check which conclusions logically follow from these diagrams.
C) Only conclusion II follows.
The question asks to identify the animal group where blood passes through the heart only once during a complete circulatory cycle through the body. This refers to the type of circulatory system, specifically whether it's single or double circulation.
D) Fishes — Fishes possess a two-chambered heart and exhibit single circulation. In this system, the heart pumps deoxygenated blood to the gills, where it gets oxygenated. This oxygenated blood then flows directly to the rest of the body tissues before returning to the heart. Thus, blood passes through the heart only once during one complete cycle of passage through the body.
The problem asks us to find a number that completes an analogy based on a specific mathematical relationship between pairs of numbers. We need to identify the pattern connecting the first and second numbers, and the third and fourth numbers, and then apply that same pattern to the fifth number to find the sixth.
Let the first number be \(N_1 = 13\) and the second number be \(N_2 = 179\).
We need to find a relationship between 13 and 179. Let's try squaring the first number and see if it's close to the second number.
\[ 13^2 = 169 \]The difference between \(N_2\) and \(13^2\) is:
\[ 179 - 169 = 10 \]This difference, 10, is \(13 - 3\). So, the pattern could be \(N_1^2 + (N_1 - 3)\).
Let's verify this pattern:
\[ 13^2 + (13 - 3) = 169 + 10 = 179 \]This pattern holds for the first pair.
Let the first number be \(N_3 = 19\) and the second number be \(N_4 = 371\).
Let's apply the same pattern: \(N_3^2 + (N_3 - 3)\).
\[ 19^2 = 361 \]Now, calculate \(N_3 - 3\):
\[ 19 - 3 = 16 \]Add these two values:
\[ 361 + 16 = 377 \]However, the given second number is 371. This means the initial pattern \(N^2 + (N-3)\) is incorrect.
Let's re-examine the first pair: \(13 : 179\). We had \(13^2 = 169\). The difference was \(179 - 169 = 10\). What if the added number is a constant or related to the number in a different way?
Consider the pattern \(N^2 - K\) or \(N^2 + K\).
For \(13 : 179\): \(13^2 = 169\). \(179 - 169 = 10\). So, \(13^2 + 10 = 179\).
For \(19 : 371\): \(19^2 = 361\). \(371 - 361 = 10\). So, \(19^2 + 10 = 371\).
This is a consistent pattern! The relationship is \(N : N^2 + 10\).
Let the first number be \(N_5 = 26\).
Using the identified pattern, the corresponding second number will be \(N_5^2 + 10\).
\[ 26^2 = 676 \]Now, add 10 to this value:
\[ 676 + 10 = 686 \]So, the missing number is 686.
D) 686
The question asks for the location of Shri Lal Bahadur Shastri Rashtriya Sanskrit Vidyapeetha. This requires knowledge of the locations of prominent educational institutions in India, particularly those related to Sanskrit studies.
Correct Option: C) New Delhi
The resistance of a wire depends on its material's resistivity, length, and cross-sectional area. We will calculate the initial resistance of the silver wire and the final resistance of the chromium wire using the given parameters and then find the ratio of the final resistance to the initial resistance to determine how the resistance changes.
The resistance \(R\) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]where \(\rho\) is the resistivity of the material, \(L\) is the length, and \(A\) is the cross-sectional area.
For the silver wire:
So, the initial resistance \(R_1\) is:
\[ R_1 = \rho_{\text{silver}} \frac{L}{A} = 1.6 \times 10^{-8} \frac{L}{A} \quad \text{(Equation 1)} \]For the chromium wire:
So, the final resistance \(R_2\) is:
\[ R_2 = \rho_{\text{chromium}} \frac{L_{\text{chromium}}}{A_{\text{chromium}}} = 12.9 \times 10^{-8} \frac{\frac{L}{3}}{3A} \] \[ R_2 = 12.9 \times 10^{-8} \frac{L}{9A} \quad \text{(Equation 2)} \]To determine how the resistance changes, we calculate the ratio \(\frac{R_2}{R_1}\):
\[ \frac{R_2}{R_1} = \frac{12.9 \times 10^{-8} \frac{L}{9A}}{1.6 \times 10^{-8} \frac{L}{A}} \]Cancel out the common terms \((10^{-8}, L, A)\):
\[ \frac{R_2}{R_1} = \frac{12.9 \times \frac{1}{9}}{1.6} = \frac{12.9}{9 \times 1.6} \] \[ \frac{R_2}{R_1} = \frac{12.9}{14.4} \approx 0.8958 \]The ratio is approximately 0.89. This means the new resistance \(R_2\) is about 0.89 times the original resistance \(R_1\). Since the value is less than 1, the resistance decreases.
Correct Option: A) decrease to 0.89 times of itself
The question asks to identify India's first Unicorn sports enterprise as of January 2022. A Unicorn enterprise is a privately held startup company valued at over $1 billion. This requires knowledge of recent business and sports news in India.
Correct Option: C) Chennai Super Kings
The question asks to identify the river among the given options that flows from east to west. This requires knowledge of the major river systems in India and their general flow directions.
Correct Option: A) Narmada
The Narmada River originates near Amarkantak in Madhya Pradesh and flows westward through a rift valley between the Vindhya Range to the north and the Satpura Range to the south, eventually emptying into the Arabian Sea through the Gulf of Khambhat. This makes it a west-flowing river.
The question asks for the safest method for the disposal of hazardous waste. We need to evaluate each option based on its environmental impact and effectiveness in neutralizing or minimizing the harm from hazardous materials.
Correct Option: C) Incineration
Incineration is a waste treatment process that involves the combustion of organic substances contained in waste materials. For hazardous waste, high-temperature incineration can effectively destroy many toxic organic compounds, reducing the volume of waste and neutralizing its harmful properties. While it can produce air pollutants (like dioxins and furans if not properly managed) and ash that may still be hazardous, modern incinerators with advanced emission control systems are designed to minimize these risks, making it one of the most effective methods for destroying certain types of hazardous waste. It significantly reduces the volume and toxicity of the waste, preventing long-term environmental contamination that other methods might cause.
The question asks about the trend of metallic properties in the modern periodic table. Metallic character is related to an element's tendency to lose electrons and form positive ions. We need to analyze how this tendency changes across a period (left to right) and down a group (top to bottom).
A) Metallic character decreases across a period and increases down a group. This statement accurately describes the trends in metallic properties in the modern periodic table, as explained in the step-by-step analysis.