The question asks to identify a famous ruler of the Satavahana dynasty from the given options. This requires knowledge of ancient Indian history, specifically the rulers of major dynasties.
Correct Option: A) Gautamiputra Satakarni was indeed the most famous and powerful ruler of the Satavahana dynasty. He is credited with reviving the Satavahana power after a period of decline and expanding their empire significantly. His military victories and administrative prowess are highlighted in various historical records.
To find the average score of the entire class, we need to calculate the total score of all students and divide it by the total number of students. The total score can be found by summing the total scores of the two groups of students.
Total score = Number of students × Average score
\[ \text{Total score of first group} = 12 \times 70 = 840 \]Total score = Number of students × Average score
\[ \text{Total score of second group} = 18 \times 62 \]\[ 18 \times 62 = 18 \times (60 + 2) = (18 \times 60) + (18 \times 2) = 1080 + 36 = 1116 \]Total score of class = Total score of first group + Total score of second group
\[ \text{Total score of class} = 840 + 1116 = 1956 \]Average score = Total score of class / Total number of students
\[ \text{Average score of class} = \frac{1956}{30} \]\[ \text{Average score of class} = \frac{195.6}{3} = 65.2 \]Correct Option: C) 65.2 is the average score of the class.
The problem involves calculating percentage profit. We are given two scenarios: first, a loss when the article is sold at 25% of its original selling price, and second, the profit when sold at the original selling price. We will use the relationship between Cost Price (CP), Selling Price (SP), and profit/loss percentages to find the required profit.
Let the original selling price of the article be \(SP_1 = \text{₹}x\).
In the first scenario, the article is sold at 25% of this price. Let this new selling price be \(SP_2\).
\[ SP_2 = 25\% \text{ of } SP_1 = \frac{25}{100} \times x = \frac{x}{4} \]When sold at \(SP_2\), there is a loss of 50%. Let \(CP\) be the cost price of the article.
The formula for loss percentage is:
\[ \text{Loss Percentage} = \frac{CP - SP}{CP} \times 100 \]Substitute the given values into the loss percentage formula:
\[ 50 = \frac{CP - \frac{x}{4}}{CP} \times 100 \]Divide both sides by 100:
\[ \frac{50}{100} = \frac{CP - \frac{x}{4}}{CP} \]\[ \frac{1}{2} = \frac{CP - \frac{x}{4}}{CP} \]Cross-multiply to solve for \(CP\):
\[ CP = 2 \left( CP - \frac{x}{4} \right) \]\[ CP = 2CP - \frac{2x}{4} \]\[ CP = 2CP - \frac{x}{2} \]Rearrange the terms to find \(CP\):
\[ 2CP - CP = \frac{x}{2} \]\[ CP = \frac{x}{2} \]Now, we need to find the percentage profit when the article is sold for \(\text{₹}x\) (which is \(SP_1\)).
The formula for profit percentage is:
\[ \text{Profit Percentage} = \frac{SP - CP}{CP} \times 100 \]Substitute \(SP = x\) and \(CP = \frac{x}{2}\) into the profit percentage formula:
\[ \text{Profit Percentage} = \frac{x - \frac{x}{2}}{\frac{x}{2}} \times 100 \]Simplify the numerator:
\[ x - \frac{x}{2} = \frac{2x - x}{2} = \frac{x}{2} \]Substitute this back into the profit percentage formula:
\[ \text{Profit Percentage} = \frac{\frac{x}{2}}{\frac{x}{2}} \times 100 \]\[ \text{Profit Percentage} = 1 \times 100 \]\[ \text{Profit Percentage} = 100\% \]A) 100% — As calculated, when the article is sold for ₹x, the profit percentage is 100%.
The problem asks us to find a relationship between two sets of letters based on their positions in the English alphabet. We need to identify the pattern in the given pairs (ZRVY to TLPS and WOSV to QIMP) and then apply the same pattern to UMQT to find its related set of letters.
The question asks to identify the type of soil that is deficient in nitrogen and phosphorus but rich in iron and aluminium. We need to recall the characteristic properties and chemical composition of different soil types found in India.
D) Laterite soil is formed by intense leaching due to heavy rainfall and high temperatures. This process removes soluble silica, leaving behind a high concentration of iron and aluminium oxides. Consequently, it is deficient in essential nutrients like nitrogen, phosphorus, potash, and lime.
The problem involves finding the volume of a cuboid given information about its edge lengths relative to a cube. We will use the given volume of the cube to find its edge length. Then, we will use the condition that the sum of the lengths of the edges of the cube is equal to the sum of the lengths of the edges of the cuboid. This will allow us to determine the dimensions of the cuboid and subsequently calculate its volume.
Find the edge length of the cube:
The volume of a cube is given by \(V_{\text{cube}} = a^3\), where \(a\) is the length of its edge.
Given \(V_{\text{cube}} = 512 \text{ cm}^3\).
\[ a^3 = 512 \] \[ a = \sqrt[3]{512} \] \[ a = 8 \text{ cm} \]Calculate the sum of the lengths of the edges of the cube:
A cube has 12 edges, all of equal length.
Sum of edge lengths of cube \( = 12a = 12 \times 8 = 96 \text{ cm}\).
Express the dimensions of the cuboid:
Let the length, breadth, and height of the cuboid be \(l\), \(b\), and \(h\) respectively. They are in the ratio 4 : 3 : 1.
So, we can write \(l = 4x\), \(b = 3x\), and \(h = x\) for some constant \(x\).
Calculate the sum of the lengths of the edges of the cuboid:
A cuboid has 12 edges. There are 4 edges of length \(l\), 4 edges of length \(b\), and 4 edges of length \(h\).
Sum of edge lengths of cuboid \( = 4l + 4b + 4h = 4(l + b + h)\).
Substitute the expressions for \(l, b, h\):
Sum of edge lengths of cuboid \( = 4(4x + 3x + x) = 4(8x) = 32x\).
Equate the sum of edge lengths of the cube and cuboid:
Given that the sum of the lengths of the edges of the cube is equal to the sum of the lengths of the edges of the cuboid.
\[ 96 = 32x \] \[ x = \frac{96}{32} \] \[ x = 3 \text{ cm} \]Determine the dimensions of the cuboid:
\(l = 4x = 4 \times 3 = 12 \text{ cm}\)
\(b = 3x = 3 \times 3 = 9 \text{ cm}\)
\(h = x = 3 \text{ cm}\)
Calculate the volume of the cuboid:
The volume of a cuboid is given by \(V_{\text{cuboid}} = l \times b \times h\).
\[ V_{\text{cuboid}} = 12 \times 9 \times 3 \] \[ V_{\text{cuboid}} = 108 \times 3 \] \[ V_{\text{cuboid}} = 324 \text{ cm}^3 \]Correct Option: C) 324
The question asks about the authority responsible for dissolving the Lok Sabha before its full term in the Indian parliamentary system. This involves understanding the roles of key constitutional functionaries like the President, Prime Minister, Speaker, and Council of Ministers.
Correct Option: C) The Prime Minister advises the President to dissolve it.
This option correctly identifies the constitutional mechanism. The Prime Minister, as the head of the Council of Ministers, advises the President, who then formally dissolves the Lok Sabha. This reflects the parliamentary system where the executive (Council of Ministers) is accountable to the legislature and can recommend its dissolution.
The question asks about the number of seats secured by the Jharkhand Mukti Morcha (JMM) in the 2024 Jharkhand Assembly elections and the total number of constituencies. This requires knowledge of the specific election results for the 2024 Jharkhand Assembly elections.
Based on the official results of the 2024 Jharkhand Assembly elections, the total number of constituencies is 81. The Jharkhand Mukti Morcha (JMM) secured 34 seats.
Therefore, JMM secured 34 out of 81 seats.
C) 34 out of 81 — This option correctly states that the Jharkhand Mukti Morcha (JMM) won 34 seats out of a total of 81 constituencies in the 2024 Jharkhand Assembly elections.
This is a coding-decoding problem where letters are assigned numerical codes. We need to identify the pattern by comparing the given coded words and then apply it to find the code for the letter 'S'.
D) 4 — As determined by comparing the unique letters and unique digits in the given coded words, 'S' corresponds to '4'.
To solve this problem, we need to apply the given rules to each digit of the number 5637482. We will identify odd and even digits, apply the respective operations (add 1 to odd, subtract 2 from even), and then form the new number. Finally, we will count how many digits appear more than once in this new number.
C) 2 — The digits 4 and 6 each appear more than once in the new number 6448260. Therefore, there are 2 such digits.