The problem asks us to identify the odd one out from a given set of four words. We need to find a common characteristic shared by three of the words and identify the one that does not share that characteristic.
Correct Option: A) Book is the odd one out because Chalk, Pen, and Marker are all writing instruments, whereas a Book is an item on which one writes or reads from, not an instrument for writing itself.
To find the volume of a cylinder, we need its radius and height. The height is given. We can find the radius using the given curved surface area and height. Once the radius is known, we can calculate the volume.
The curved surface area (CSA) of a cylinder is given by the formula:
\[ \text{CSA} = 2 \pi r h \]Substitute the given values into the CSA formula:
\[ 528 = 2 \times \frac{22}{7} \times r \times 14 \]\[ 528 = 2 \times 22 \times r \times \frac{14}{7} \]\[ 528 = 2 \times 22 \times r \times 2 \]\[ 528 = 88 r \]\[ r = \frac{528}{88} \]\[ r = 6 \text{ cm} \]The volume (\(V\)) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]Substitute the calculated radius and given height into the volume formula:
\[ V = \frac{22}{7} \times (6)^2 \times 14 \]\[ V = \frac{22}{7} \times 36 \times 14 \]\[ V = 22 \times 36 \times \frac{14}{7} \]\[ V = 22 \times 36 \times 2 \]\[ V = 22 \times 72 \]\[ V = 1584 \text{ cm}^3 \]The question asks to identify a multi-barrel rocket system developed by DRDO (Defence Research and Development Organisation) from the given options. We need to recall the types of missile/rocket systems developed by DRDO and their classifications.
B) Pinaka is a multi-barrel rocket launcher (MBRL) system developed by DRDO for the Indian Army. It can fire a salvo of 12 rockets in 44 seconds, engaging targets up to 40-75 km depending on the variant, and is designed to neutralize large areas with rapid, high-volume fire.
The question asks about the mass of a person on the Moon, given their mass on Earth. This requires understanding the fundamental difference between mass and weight and how they are affected by gravitational fields.
C) 60 kg — Mass is an intrinsic property of an object and represents the amount of matter it contains. It does not change with location, even if the gravitational field changes. Therefore, if a person's mass is 60 kg on Earth, it will remain 60 kg on the Moon.
Who was honoured with the 55th Jnanpith Award for the year 2019?
A Achuthan Namboothiri
Krishna Sobti
Shobha Rao
Chitra Mudgal
The question asks to identify the recipient of the 55th Jnanpith Award for the year 2019. This requires knowledge of major literary awards and their recipients, which is a common topic in general awareness sections of competitive exams.
Correct Option: A) A Achuthan Namboothiri
Akkitham Achuthan Namboothiri, often referred to as Akkitham, was a celebrated Malayalam poet and essayist. He was honored with the 55th Jnanpith Award in 2019 for his outstanding contribution to Indian literature.
The question asks for the district in Gujarat where Dandi, the destination of Mahatma Gandhi's famous Salt March, is located. This requires knowledge of Indian history and geography, specifically related to the Dandi March.
Correct Option: A) Navsari
Dandi, the historic site where Mahatma Gandhi broke the salt law, is situated in the Navsari district of Gujarat. The Salt March, also known as the Dandi March or Salt Satyagraha, began on March 12, 1930, from Sabarmati Ashram in Ahmedabad and concluded on April 6, 1930, at Dandi, Navsari district.
The question asks to identify the mountain pass that connects Srinagar and Leh. This requires knowledge of the geographical locations of major mountain passes in India and the regions they connect.
Correct Option: D) Zoji La is a high mountain pass in the Himalayas, located in the Kargil district of Ladakh. It connects Srinagar to Leh, providing a vital link between the Kashmir Valley and Ladakh. It is part of the National Highway 1 (NH1).
The value of \( \frac{6}{5} \) of (\( \frac{1}{3} \) + \( \frac{2}{4} \)) - \( \frac{5}{6} \) of (\( \frac{3}{10} \) + \( \frac{12}{20} \)) - \( \frac{1}{6} \) is:
\( \frac{1}{3} \)
\( \frac{3}{4} \)
\( \frac{5}{6} \)
\( \frac{1}{12} \)
To solve this problem, we need to follow the order of operations (BODMAS/PEMDAS). First, simplify the expressions inside the parentheses, then perform multiplication ('of' means multiplication), and finally, perform addition and subtraction from left to right.
First, calculate the sum inside the first parenthesis: \[\frac{1}{3} + \frac{2}{4}\]
Find a common denominator for 3 and 4, which is 12.
\[ \frac{1}{3} + \frac{2}{4} = \frac{1 \times 4}{3 \times 4} + \frac{2 \times 3}{4 \times 3} = \frac{4}{12} + \frac{6}{12} = \frac{4+6}{12} = \frac{10}{12} = \frac{5}{6} \]Now, multiply this by \( \frac{6}{5} \):
\[ \frac{6}{5} \text{ of } \left( \frac{1}{3} + \frac{2}{4} \right) = \frac{6}{5} \times \frac{5}{6} = 1 \]So, the first term is 1.
Next, calculate the sum inside the second parenthesis: \[\frac{3}{10} + \frac{12}{20}\]
Simplify \( \frac{12}{20} \) to \( \frac{3}{5} \). The common denominator for 10 and 5 is 10.
\[ \frac{3}{10} + \frac{12}{20} = \frac{3}{10} + \frac{3 \times 2}{5 \times 2} = \frac{3}{10} + \frac{6}{10} = \frac{3+6}{10} = \frac{9}{10} \]Now, multiply this by \( \frac{5}{6} \):
\[ \frac{5}{6} \text{ of } \left( \frac{3}{10} + \frac{12}{20} \right) = \frac{5}{6} \times \frac{9}{10} = \frac{5 \times 9}{6 \times 10} = \frac{45}{60} \]Simplify \( \frac{45}{60} \) by dividing both numerator and denominator by 15:
\[ \frac{45 \div 15}{60 \div 15} = \frac{3}{4} \]So, the second term is \( \frac{3}{4} \).
Substitute the simplified terms back into the original expression:
\[ 1 - \frac{3}{4} - \frac{1}{6} \]Find a common denominator for 1, 4, and 6, which is 12.
\[ \frac{12}{12} - \frac{3 \times 3}{4 \times 3} - \frac{1 \times 2}{6 \times 2} = \frac{12}{12} - \frac{9}{12} - \frac{2}{12} \]Now, perform the subtraction:
\[ \frac{12 - 9 - 2}{12} = \frac{3 - 2}{12} = \frac{1}{12} \]D) The final calculated value is \( \frac{1}{12} \).
The problem provides three algebraic equations involving variables \(a\) and \(b\). We need to find the value of the expression \(a^2 + ab + b^2\). We can use the algebraic identities for the difference of cubes and difference of squares to relate the given equations and solve for the required expression.
Recall the algebraic identity for the difference of cubes:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]We are given \(a^3 - b^3 = 625\). So, we can write:
\[ 625 = (a - b)(a^2 + ab + b^2) \quad \text{(Equation 1)} \]Recall the algebraic identity for the difference of squares:
\[ a^2 - b^2 = (a - b)(a + b) \]We are given \(a^2 - b^2 = 25\) and \(a + b = 5\). Substitute these values into the identity:
\[ 25 = (a - b)(5) \]Now, solve for \((a - b)\) from the above equation:
\[ a - b = \frac{25}{5} \]\[ a - b = 5 \quad \text{(Equation 2)} \]Substitute the value of \((a - b)\) from Equation 2 into Equation 1:
\[ 625 = (5)(a^2 + ab + b^2) \]Finally, solve for \(a^2 + ab + b^2\):
\[ a^2 + ab + b^2 = \frac{625}{5} \]\[ a^2 + ab + b^2 = 125 \]A) 125
The sum of the angles in any triangle is always 180°. We are given the three angles of a triangle in terms of 'x'. We can set up an equation by summing these angles and equating them to 180° to find the value of 'x'. Once 'x' is known, we can calculate the measure of each angle and identify the greatest one.
The three angles of the triangle are:
The sum of the angles in a triangle is \(180^\circ\). So, we have:
\[ 2x + (3x - 8) + (5x - 12) = 180 \]Combine like terms:
\[ (2x + 3x + 5x) + (-8 - 12) = 180 \]\[ 10x - 20 = 180 \]\[ 10x = 180 + 20 \]\[ 10x = 200 \]\[ x = \frac{200}{10} \]\[ x = 20 \]Substitute \(x = 20\) into the expressions for each angle:
Comparing the three angles \(40^\circ\), \(52^\circ\), and \(88^\circ\), the greatest angle is \(88^\circ\).
Correct Option: D) 88° is the greatest angle of the triangle.