This problem involves work and time. We need to find the individual work rates of A, B, and C, and then combine them to find their combined work rate. Finally, we can calculate the time taken for them to complete the task together.
Determine the ratio of work rates between A and B:
Given that A completes in 2 days as much as B completes in 3 days. This means:
\[ 2 \times \text{Work rate of A} = 3 \times \text{Work rate of B} \]\[ \frac{\text{Work rate of A}}{\text{Work rate of B}} = \frac{3}{2} \]Determine the ratio of work rates between B and C:
Given that B completes in 4 days as much as C completes in 5 days. This means:
\[ 4 \times \text{Work rate of B} = 5 \times \text{Work rate of C} \]\[ \frac{\text{Work rate of B}}{\text{Work rate of C}} = \frac{5}{4} \]Find the combined ratio of work rates for A, B, and C:
We have \(\text{A}:\text{B} = 3:2\) and \(\text{B}:\text{C} = 5:4\). To combine these ratios, we need to make the 'B' part common. Multiply the first ratio by 5 and the second by 2:
So, the ratio of work rates is \(\text{A}:\text{B}:\text{C} = 15:10:8\).
Let their individual daily work rates be \(15x\), \(10x\), and \(8x\) units of work per day, respectively.
Calculate the total work:
A alone can complete the task in 22 days. Since A's daily work rate is \(15x\), the total work is:
\[ \text{Total Work} = \text{A's daily work rate} \times \text{Time taken by A} \]\[ \text{Total Work} = 15x \times 22 = 330x \text{ units} \]Calculate the combined daily work rate of A, B, and C:
Combined daily work rate \( = 15x + 10x + 8x = 33x \) units per day.
Calculate the time taken for A, B, and C to complete the task together:
\[ \text{Time} = \frac{\text{Total Work}}{\text{Combined daily work rate}} \]\[ \text{Time} = \frac{330x}{33x} = 10 \text{ days} \]Correct Option: C) 10 days
To find the median of a given set of data, the first step is to arrange the data in ascending or descending order. Then, identify the number of observations (\(n\)). If \(n\) is odd, the median is the middle value. If \(n\) is even, the median is the average of the two middle values.
Therefore, the median of the given data is 17.
B) 17 — This is the correct median calculated by arranging the data in ascending order and taking the average of the two middle values.
We need to find the amount lent at 10% per annum. We can set up equations based on the simple interest formula for each part of the loan and the total interest received. Let the amount lent at 10% be \(x\) and the amount lent at 8% be \(30,000 - x\). Then, we calculate the simple interest for each part for 3 years and sum them up to equal the total interest received.
Correct Option: D) ₹18,000
To solve this problem, we first need to calculate the total distance covered by the train. Then, we will determine how many segments of 50 m are in that total distance. Finally, we will add 1 to the number of segments to account for both the starting and ending poles.
Given speed of the train = \(45 \text{ km/h}\).
To convert kilometers to meters, multiply by 1000:
\[ \text{Speed} = 45 \times 1000 \text{ m/h} = 45000 \text{ m/h} \]Time taken = \(4 \text{ h}\).
Using the formula: \(\text{Distance} = \text{Speed} \times \text{Time}\)
\[ \text{Distance} = 45000 \text{ m/h} \times 4 \text{ h} = 180000 \text{ m} \]The distance between two consecutive telegraph poles is \(50 \text{ m}\).
Number of gaps = \[\frac{\text{Total Distance}}{\text{Distance between poles}}\]
\[ \text{Number of gaps} = \frac{180000 \text{ m}}{50 \text{ m}} = 3600 \]If there are \(N\) gaps, there will be \(N+1\) poles (including the pole at the start and the pole at the end).
Total number of poles = \(\text{Number of gaps} + 1\)
\[ \text{Total number of poles} = 3600 + 1 = 3601 \]Correct Option: D) 3601
The problem asks us to identify the relationship between the numbers in the given set \((9, 81, 6561)\) and then find an option that follows the same relationship. We need to analyze how the second and third numbers are derived from the first number in the given set.
C) (2, 4, 16)
Thus, option C follows the same relationship as the given set.
The height of a house is h and 5.2 m long ladder is just sufficient to go at the top of the house when its inclination with the horizontal ground is 45°. Find the height of the house.
This problem can be solved using basic trigonometry. We are given the length of the ladder (hypotenuse) and the angle of inclination with the horizontal. We need to find the height of the house (opposite side to the angle). The sine function relates the opposite side, hypotenuse, and the angle.
In a right-angled triangle, the sine of an angle is defined as the ratio of the length of the opposite side to the length of the hypotenuse.
\[ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \]Correct Option: C) \(\frac{13\sqrt{2}}{5}\) m
The problem asks us to identify the number that is different from the rest among the given options. This type of question usually involves finding a common property shared by three of the numbers, while one number does not possess that property. We will examine each number to see if it is a perfect square.
To check if 836 is a perfect square, we can try to find its square root or recall common perfect squares. We know that \(20^2 = 400\), \(30^2 = 900\). So, if 836 is a perfect square, its root would be between 20 and 30. Let's check numbers ending in 6: \(24^2 = 576\), \(26^2 = 676\). We can also perform prime factorization:
\[ 836 = 2 \times 418 = 2 \times 2 \times 209 = 2^2 \times 11 \times 19 \]Since the prime factors 11 and 19 do not appear with an even exponent, 836 is not a perfect square.
Let's find the square root of 1296. We know that \(30^2 = 900\) and \(40^2 = 1600\). The number ends in 6, so its square root might end in 4 or 6. Let's try 36:
\[ 36^2 = 36 \times 36 = 1296 \]Thus, 1296 is a perfect square.
Let's find the square root of 576. We know that \(20^2 = 400\) and \(30^2 = 900\). The number ends in 6, so its square root might end in 4 or 6. Let's try 24:
\[ 24^2 = 24 \times 24 = 576 \]Thus, 576 is a perfect square.
Let's find the square root of 1156. We know that \(30^2 = 900\) and \(40^2 = 1600\). The number ends in 6, so its square root might end in 4 or 6. Let's try 34:
\[ 34^2 = 34 \times 34 = 1156 \]Thus, 1156 is a perfect square.
A) 836 is the number that is different from the rest because it is not a perfect square, while options B, C, and D are all perfect squares.
The question asks to identify the South Indian classical dance form whose name literally translates to 'Story-Play'. We need to examine the etymology of each given option.
Based on the etymological analysis, Kathakali is the only dance form among the options whose name literally means 'Story-Play' and is also a South Indian classical dance.
Correct Option: A) Kathakali — The name "Kathakali" is derived from the Malayalam words "Katha" (story) and "Kali" (play/performance), thus literally meaning 'Story-Play'. It is a prominent classical dance form from Kerala, a South Indian state.
The question asks about the primary commodity stored as 'buffer stock' by the Indian government. This relates to government policies aimed at ensuring food security and price stability.
C) food grains — The Indian government, primarily through the Food Corporation of India (FCI), procures and stores large quantities of food grains (mainly wheat and rice) to maintain a buffer stock. This stock is crucial for ensuring food security, stabilizing prices, and distributing food through the Public Distribution System (PDS) to vulnerable sections of society, especially during times of scarcity or natural calamities.
This problem involves the relationship between distance, speed, and time. We are given a fixed distance and two scenarios with different speeds and corresponding times. We can set up equations based on the formula \(\text{Distance} = \text{Speed} \times \text{Time}\) and solve for the unknown original speed.
Let the original speed of the car be \(x \text{ km/h}\).
The distance to be covered is \(300 \text{ km}\).
Using the formula \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\), the original time taken is \(T_1 = \frac{300}{x} \text{ hours}\).
In the second scenario, the speed of the car is \(5 \text{ km/h}\) more, so the new speed is \((x+5) \text{ km/h}\).
The time taken to cover the same distance at the new speed is \(T_2 = \frac{300}{x+5} \text{ hours}\).
According to the problem, if the speed is \(5 \text{ km/h}\) more, it takes \(2 \text{ hours}\) less. This means the difference between the original time and the new time is \(2 \text{ hours}\).
\[ T_1 - T_2 = 2 \]
\[ \frac{300}{x} - \frac{300}{x+5} = 2 \]
Now, we solve this equation for \(x\):
Take \(300\) common:
\[ 300 \left( \frac{1}{x} - \frac{1}{x+5} \right) = 2 \]
Find a common denominator for the terms inside the parenthesis:
\[ 300 \left( \frac{(x+5) - x}{x(x+5)} \right) = 2 \]
\[ 300 \left( \frac{5}{x^2 + 5x} \right) = 2 \]
\[ \frac{1500}{x^2 + 5x} = 2 \]
Cross-multiply:
\[ 1500 = 2(x^2 + 5x) \]
Divide by \(2\):
\[ 750 = x^2 + 5x \]
Rearrange into a quadratic equation:
\[ x^2 + 5x - 750 = 0 \]
Solve the quadratic equation. We can factorize it or use the quadratic formula. For factorization, we need two numbers that multiply to \(-750\) and add up to \(5\). These numbers are \(30\) and \(-25\).
\[ x^2 + 30x - 25x - 750 = 0 \]
\[ x(x+30) - 25(x+30) = 0 \]
\[ (x-25)(x+30) = 0 \]
This gives two possible values for \(x\): \(x = 25\) or \(x = -30\).
Since speed cannot be negative, we discard \(x = -30\).
Therefore, the original speed of the car is \(25 \text{ km/h}\).
B) 25 km/h is the correct answer. If the original speed is \(25 \text{ km/h}\), the original time is \(\frac{300}{25} = 12 \text{ hours}\). If the speed increases to \(30 \text{ km/h}\), the new time is \(\frac{300}{30} = 10 \text{ hours}\). The difference in time is \(12 - 10 = 2 \text{ hours}\), which matches the condition given in the problem.