This problem involves analyzing statements and conclusions based on the principles of syllogism. We can use Venn diagrams to visually represent the relationships described in the statements and then evaluate each conclusion.
Correct Option: A) 4
The question asks about the state government that instituted the Kalidas Samman Award. This is a factual question related to Indian awards and their origins, which requires knowledge of general awareness.
D) The Kalidas Samman Award is a prestigious arts award presented annually by the Government of Madhya Pradesh. It was instituted in 1980.
The question asks to identify which of the given options is NOT an element. To answer this, we need to understand the definition of an element in chemistry and classify each option accordingly.
Correct Option: A) Silica is a compound (\(\text{SiO}_2\)), not an element. It is composed of silicon and oxygen atoms.
The question asks to identify which statement is NOT a main feature of the FAME-2 India Scheme. This requires knowledge of the objectives and key provisions of the FAME-2 scheme.
D) Its target is only private transport vehicles. This statement is incorrect because the FAME-2 scheme also significantly targets public and commercial transport vehicles, such as electric buses and three-wheelers used for public transport, in addition to private vehicles.
This is an analogy question where you need to identify the relationship between the first pair of words and apply the same relationship to the second pair to find the missing word.
Correct Option: C) Temperature. A clock measures time, and similarly, a thermometer measures temperature.
The question asks to identify the youngest Nobel Laureate among the given options as of October 2020. This requires knowledge of the Nobel laureates' ages at the time they received their awards.
Correct Option: B) Malala Yousafzai is the youngest Nobel Laureate. She received the Nobel Peace Prize in 2014 at the age of 17 for her struggle against the suppression of children and young people and for the right of all children to education.
To find the time taken by a train to cross a bridge, we need to calculate the total distance the train travels and then divide it by the train's speed. The total distance will be the sum of the train's length and the bridge's length. The speed must be converted from km/h to m/s to match the units of distance (meters) and time (seconds).
When a train crosses a bridge, the total distance it must travel is its own length plus the length of the bridge.
\[ \text{Total Distance } (D) = L_T + L_B \]\[ D = 110 \text{ m} + 180 \text{ m} \]\[ D = 290 \text{ m} \]To convert km/h to m/s, multiply by \(\frac{5}{18}\).
\[ S = 90 \text{ km/h} \times \frac{5}{18} \text{ m/s per km/h} \]\[ S = 5 \times 5 \text{ m/s} \]\[ S = 25 \text{ m/s} \]Time is calculated using the formula: \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\).
\[ \text{Time } (T) = \frac{D}{S} \]\[ T = \frac{290 \text{ m}}{25 \text{ m/s}} \]\[ T = 11.6 \text{ s} \]Correct Option: A) 11.6 s is the time the train will take to cross the bridge.
The question asks for the year when the Planning Commission of India was dissolved. This is a factual question related to Indian governance and economic planning history. Recall the major governmental changes related to economic policy in recent years.
Correct Option: B) 2014. The decision to dissolve the Planning Commission was announced by Prime Minister Narendra Modi on August 15, 2014, and it was subsequently replaced by NITI Aayog. While NITI Aayog officially came into existence on January 1, 2015, the dissolution of the Planning Commission as an entity was effectively decided and announced in 2014.
What is the remainder when 220 is divided by 3?
To find the remainder when a large power of a number is divided by another number, we can use the concept of modular arithmetic. We look for a pattern in the remainders of successive powers of the base number when divided by the divisor.
Let's find the remainders of the first few powers of \(2\) when divided by \(3\):
We observe a pattern here:
This pattern repeats: \(2^3 = 8\). \(8 \div 3\) gives a remainder of \(2\). So, \(2^3 \equiv 2 \pmod{3}\).
\(2^4 = 16\). \(16 \div 3\) gives a remainder of \(1\). So, \(2^4 \equiv 1 \pmod{3}\).
In general, for any positive integer \(n\):
The given exponent is \(20\). Since \(20\) is an even number, based on our pattern, the remainder when \(2^{20}\) is divided by \(3\) will be \(1\).
Alternatively, using the property that if \(a \equiv b \pmod{m}\), then \(a^k \equiv b^k \pmod{m}\):
We know that \(2^2 \equiv 1 \pmod{3}\).
We can write \(2^{20}\) as \((2^2)^{10}\).
So, \(2^{20} \equiv (1)^{10} \pmod{3}\).
\(2^{20} \equiv 1 \pmod{3}\).
Thus, the remainder is \(1\).
D) 1 — As derived from the pattern of remainders of powers of 2 when divided by 3, \(2^{20}\) leaves a remainder of \(1\).
Which fraction bears the same ratio to \( \frac{1}{27} \) as \( \frac{3}{11} \)does to \( \frac{5}{9} \)?
\( \frac{1}{15} \)
\( \frac{1}{99} \)
\( \frac{1}{27} \)
\( \frac{1}{55} \)
The problem asks us to find a fraction that bears the same ratio to \( \frac{1}{27} \) as \( \frac{3}{11} \) does to \( \frac{5}{9} \). This means we need to set up a proportion where two ratios are equal. We will first calculate the ratio of \( \frac{3}{11} \) to \( \frac{5}{9} \) and then use this ratio to find the unknown fraction.
The ratio of \( A \) to \( B \) is \( \frac{A}{B} \). So, the ratio of \( \frac{3}{11} \) to \( \frac{5}{9} \) is:
\[ \frac{\frac{3}{11}}{\frac{5}{9}} = \frac{3}{11} \times \frac{9}{5} = \frac{3 \times 9}{11 \times 5} = \frac{27}{55} \]Let the unknown fraction be \( x \). The problem states that \( x \) bears the same ratio to \( \frac{1}{27} \) as \( \frac{3}{11} \) does to \( \frac{5}{9} \). This can be written as:
\[ \frac{x}{\frac{1}{27}} = \frac{\frac{3}{11}}{\frac{5}{9}} \]From step 1, we know that \[\frac{\frac{3}{11}}{\frac{5}{9}} = \frac{27}{55}\]. So, the proportion becomes:
\[ \frac{x}{\frac{1}{27}} = \frac{27}{55} \]To find \( x \), multiply both sides of the equation by \( \frac{1}{27} \):
\[ x = \frac{27}{55} \times \frac{1}{27} \]\[ x = \frac{27 \times 1}{55 \times 27} \]\[ x = \frac{1}{55} \]D) \( \frac{1}{55} \) is the correct answer because it satisfies the given ratio condition.