This is an analogy question where we need to find the relationship between the first pair of words and apply the same relationship to the second pair. The relationship is about what is consumed or experienced by the respective person.
Correct Option: D) Song
The question asks to identify the specific type of downslope winds that flow from high elevations of mountains, plateaus, and hills down to valleys or plains. This is a common concept in physical geography and meteorology, related to local wind phenomena.
A) Katabatic winds are downslope winds that occur when air cools over elevated terrain (like mountains, plateaus, or hills) and becomes denser. This dense, cold air then flows downwards under the influence of gravity into valleys or plains below. This process is common at night or in winter when radiative cooling is significant.
The question asks for the boiling point of liquid nitrogen. This is a standard physical constant that needs to be known. We need to identify the correct value among the given options.
Correct Option: C) –196 °C is the correct boiling point of liquid nitrogen at standard atmospheric pressure. This is a well-established physical constant.
The question asks to identify the incorrectly matched pair of a desert and its location. We need to check the geographical location of each desert listed in the options.
Correct Option: C) Kalahari Desert - India. This pair is NOT matched correctly because the Kalahari Desert is located in Southern Africa, not India.
What is the HCF of \( \frac{2}{3} \), \( \frac{3}{4} \) and \( \frac{5}{6} \)?
\( \frac{1}{72} \)
\( \frac{5}{24} \)
\( \frac{1}{12} \)
\( \frac{5}{12} \)
To find the HCF (Highest Common Factor) of fractions, we use a specific formula: the HCF of fractions is equal to the HCF of their numerators divided by the LCM (Least Common Multiple) of their denominators.
Identify the fractions: The given fractions are \( \frac{2}{3} \), \( \frac{3}{4} \), and \( \frac{5}{6} \).
Extract numerators and denominators:
Calculate the HCF of the numerators:
The numerators are 2, 3, and 5. These are all prime numbers. The only common factor for distinct prime numbers is 1.
\[ \text{HCF}(2, 3, 5) = 1 \]
Calculate the LCM of the denominators:
The denominators are 3, 4, and 6.
The least common multiple is 12.
\[ \text{LCM}(3, 4, 6) = 12 \]
Apply the formula for HCF of fractions:
\[ \text{HCF of fractions} = \frac{\text{HCF of numerators}}{\text{LCM of denominators}} \]
Substitute the calculated values:
\[ \text{HCF} \left( \frac{2}{3}, \frac{3}{4}, \frac{5}{6} \right) = \frac{1}{12} \]
Correct Option: C) \( \frac{1}{12} \)
The question asks to identify a nuclear-powered submarine from the given options. This requires knowledge of the types of submarines operated by the Indian Navy and their propulsion systems.
D) INS Arihant is India's first indigenous nuclear-powered ballistic missile submarine (SSBN). It is powered by a pressurised water reactor and is designed for long endurance underwater operations, making it a nuclear-powered submarine.
A alone can do a task in 39 days, whereas B alone can do it in 52 days. The duo start working together, but A leaves 3 days before the task gets over. For how many days did the two work together?
This problem involves work and time. We need to calculate the individual work rates, combine them, and then account for a scenario where one person leaves before the task is completed. The total work can be assumed as the LCM of the individual days taken to complete the work.
A alone can do the task in 39 days. So, A's 1-day work is:
\[ \text{A's 1-day work} = \frac{1}{39} \]B alone can do the task in 52 days. So, B's 1-day work is:
\[ \text{B's 1-day work} = \frac{1}{52} \]To simplify calculations, let's assume the total work is the Least Common Multiple (LCM) of 39 and 52.
Prime factorization of 39: \(3 \times 13\)
Prime factorization of 52: \(2^2 \times 13\)
LCM(39, 52) = \(2^2 \times 3 \times 13 = 4 \times 3 \times 13 = 12 \times 13 = 156\)
Let the total work be 156 units.
A's efficiency (units per day) = \[\frac{\text{Total Work}}{\text{Days taken by A}} = \frac{156}{39} = 4\] units/day.
B's efficiency (units per day) = \[\frac{\text{Total Work}}{\text{Days taken by B}} = \frac{156}{52} = 3\] units/day.
A leaves 3 days before the task gets over. This means that in the last 3 days, only B worked.
Work done by B in the last 3 days = \(\text{B's efficiency} \times 3 \text{ days} = 3 \text{ units/day} \times 3 \text{ days} = 9\) units.
Total work = 156 units.
Work done by A and B together = Total Work - Work done by B alone in the last 3 days
Work done by A and B together = \(156 - 9 = 147\) units.
Combined efficiency of A and B = A's efficiency + B's efficiency = \(4 + 3 = 7\) units/day.
Days A and B worked together = \[\frac{\text{Work done by A and B together}}{\text{Combined efficiency of A and B}}\]
Days A and B worked together = \(\frac{147}{7} = 21\) days.
D) 21
The duo worked together for 21 days.
To find the minimum possible value for the smallest integer, we need to maximize the other integers given the constraints. We will use the properties of averages and the difference between the largest and smallest numbers.
The average of 6 positive integers is 45. The sum of these integers is given by:
\[ \text{Sum} = \text{Average} \times \text{Number of integers} \]\[ \text{Sum} = 45 \times 6 = 270 \]Let the 6 positive integers be \(x_1, x_2, x_3, x_4, x_5, x_6\) in ascending order. So, \(x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6\).
The difference between the largest and the smallest of these numbers is 18.
\[ x_6 - x_1 = 18 \]\[ x_6 = x_1 + 18 \]To minimize \(x_1\), we need to maximize the other integers. Since the integers are positive, \(x_1 \ge 1\). To maximize \(x_2, x_3, x_4, x_5\), we should make them as large as possible without exceeding \(x_6\). The best way to do this is to set them equal to \(x_6\).
However, the problem states that the integers are positive. If we set \(x_2, x_3, x_4, x_5\) equal to \(x_6\), then \(x_1\) would be minimized. But this is not correct. To minimize \(x_1\), we need to maximize \(x_2, x_3, x_4, x_5, x_6\). Given \(x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6\), to maximize \(x_2, x_3, x_4, x_5\), we should set them equal to \(x_6\).
So, the integers would be \(x_1, x_6, x_6, x_6, x_6, x_6\). This is incorrect because it violates the ascending order \(x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6\). Instead, to minimize \(x_1\), we should make \(x_2, x_3, x_4, x_5\) as large as possible, which means making them equal to \(x_6\).
Let's re-evaluate. To minimize \(x_1\), we must maximize the other five numbers \(x_2, x_3, x_4, x_5, x_6\). Given the ordering \(x_1 \le x_2 \le x_3 \le x_4 \le x_5 \le x_6\), the maximum possible value for \(x_2, x_3, x_4, x_5\) is \(x_6\).
So, to minimize \(x_1\), we should set:
\[ x_2 = x_3 = x_4 = x_5 = x_6 \]Substitute these values into the sum equation:
\[ x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 270 \]\[ x_1 + x_6 + x_6 + x_6 + x_6 + x_6 = 270 \]\[ x_1 + 5x_6 = 270 \]We know \(x_6 = x_1 + 18\). Substitute this into the equation:
\[ x_1 + 5(x_1 + 18) = 270 \]\[ x_1 + 5x_1 + 90 = 270 \]\[ 6x_1 + 90 = 270 \]If \(x_1 = 30\), then \(x_6 = 30 + 18 = 48\).
The numbers would be \(30, 48, 48, 48, 48, 48\). All are positive integers. The sum is \(30 + 5 \times 48 = 30 + 240 = 270\). The average is \(270/6 = 45\). The difference between the largest (48) and smallest (30) is \(48 - 30 = 18\). All conditions are met.
A) 30 is the minimum value possible for the smallest of these 6 integers.
The question asks to identify the incorrect statement among the given options related to Indian geography and polity. We need to evaluate each statement for its factual accuracy.
Based on this evaluation, statement B is factually incorrect.
Correct Option: B) Capital of Mizoram is Kohima.
The question asks to identify the alloy that does NOT contain tin. To answer this, we need to recall the primary compositions of the given alloys: Gun metal, Bell metal, Brass, and Bronze.
Correct Option: C) Brass is an alloy primarily composed of copper and zinc. It does not typically contain tin as a major component.