This problem involves relative speed. When a train crosses a person moving in the same direction, the effective speed is the difference between the train's speed and the person's speed. The distance covered by the train to cross the person is equal to the length of the train itself. We can set up equations based on the formula \(\text{Distance} = \text{Speed} \times \text{Time}\) for both scenarios and solve for the train's speed.
The speed of the train is 39 km/h.
What will be the next term in the following series?
126, 132, 142, 148, ___
To find the next term in the series, we need to identify the pattern or rule governing the sequence of numbers. This typically involves looking at the differences between consecutive terms.
First, let's find the difference between the first and second terms:
\[ 132 - 126 = 6 \]Next, find the difference between the second and third terms:
\[ 142 - 132 = 10 \]Then, find the difference between the third and fourth terms:
\[ 148 - 142 = 6 \]The differences are \(6, 10, 6\). This suggests an alternating pattern where the differences are \(+6\), then \(+10\), then \(+6\), and so on.
Following this alternating pattern, the next difference should be \(+10\).
So, to find the next term in the series, we add \(10\) to the last given term:
\[ 148 + 10 = 158 \]B) 158. This is the number obtained by following the established alternating pattern of differences (\(+6, +10, +6, +10, \dots\)).
The question asks to identify the summer capital of British India. This is a factual question related to the administrative history of British India. We need to recall the historical significance of the given options.
C) Shimla was the summer capital of British India. Its pleasant climate and strategic location made it an ideal choice for the British administration to escape the summer heat of the plains.
The question asks about the relationship shown by the second equation of motion. We need to recall the three equations of motion for uniformly accelerated linear motion and identify which one is the second equation and what physical quantities it relates.
B) Time — The second equation of motion, \(s = ut + \frac{1}{2}at^2\), explicitly relates displacement (\(s\), which is a change in position) to time (\(t\)).
This problem involves two right-angled triangles formed by the ladder, the wall, and the ground on each side of the street. We can use the Pythagorean theorem to find the horizontal distance from the base of the ladder to each wall. The total width of the street will be the sum of these two horizontal distances.
Identify the given information:
Calculate the horizontal distance on the first side:
Let \(W_1\) be the horizontal distance from the base of the ladder to the wall on the first side. Using the Pythagorean theorem (\(a^2 + b^2 = c^2\)):
\[ W_1^2 + H_1^2 = L^2 \] \[ W_1^2 + 48^2 = 73^2 \] \[ W_1^2 + 2304 = 5329 \] \[ W_1^2 = 5329 - 2304 \] \[ W_1^2 = 3025 \] \[ W_1 = \sqrt{3025} \] \[ W_1 = 55 \text{ feet} \]Calculate the horizontal distance on the second side:
Let \(W_2\) be the horizontal distance from the base of the ladder to the wall on the second side. Using the Pythagorean theorem:
\[ W_2^2 + H_2^2 = L^2 \] \[ W_2^2 + 55^2 = 73^2 \] \[ W_2^2 + 3025 = 5329 \] \[ W_2^2 = 5329 - 3025 \] \[ W_2^2 = 2304 \] \[ W_2 = \sqrt{2304} \] \[ W_2 = 48 \text{ feet} \]Calculate the total width of the street:
The total width of the street (\(W_{\text{total}}\)) is the sum of the horizontal distances on both sides:
\[ W_{\text{total}} = W_1 + W_2 \] \[ W_{\text{total}} = 55 \text{ feet} + 48 \text{ feet} \] \[ W_{\text{total}} = 103 \text{ feet} \]A) 103 ft. As calculated, the total width of the street is 103 feet.
Consider the following question and decide which statement is sufficient to answer the question.
Question: Find the value of p.
Statement:
1. p : p2 :: p2 : 1000
2. p2 + 2pn + n2 = 21
Either 1 or 2 is sufficient.
Only 1 is sufficient
Neither 1 nor 2 is sufficient.
Only 2 is sufficient
We need to determine which of the given statements is sufficient to find the unique value of 'p'. We will analyze each statement independently.
This is a proportion, which can be written as:
\[ \frac{p}{p^2} = \frac{p^2}{1000} \]Assuming \(p \neq 0\), we can simplify the left side:
\[ \frac{1}{p} = \frac{p^2}{1000} \]Now, cross-multiply:
\[ p^3 = 1000 \]To find \(p\), take the cube root of both sides:
\[ p = \sqrt[3]{1000} \] \[ p = 10 \]This statement gives a unique value for \(p\). Therefore, Statement 1 alone is sufficient.
This equation can be recognized as a perfect square trinomial:
\[ (p + n)^2 = 21 \]Taking the square root of both sides:
\[ p + n = \pm \sqrt{21} \]From this equation, we have two unknowns, \(p\) and \(n\), and only one equation. We cannot find a unique value for \(p\) without knowing \(n\), or another independent equation relating \(p\) and \(n\). Therefore, Statement 2 alone is not sufficient.
Statement 1 alone is sufficient to find the value of \(p\) because it leads to a unique solution \(p=10\). Statement 2 alone is not sufficient as it involves two variables, \(p\) and \(n\), and does not provide a unique value for \(p\).
A) Either 1 or 2 is sufficient. — This is incorrect because Statement 2 is not sufficient.
C) Neither 1 nor 2 is sufficient. — This is incorrect because Statement 1 is sufficient.
D) Only 2 is sufficient. — This is incorrect because Statement 2 is not sufficient, while Statement 1 is.
Three pipes A, B and C can fill an empty tank in 15, 10 and 5 minutes respectively. When the tank is empty, the three pipes are opened simultaneously. What is the part of work done by B and C?
\( \frac{9}{11} \)
\( \frac{12}{11} \)
\( \frac{6}{11} \)
\( \frac{2}{11} \)
This problem involves calculating the work done by pipes filling a tank. We can determine the rate at which each pipe fills the tank (part of the tank filled per minute). Then, we find the combined rate of all pipes to determine the total time taken to fill the tank. Finally, we calculate the work done by pipes B and C by summing their individual contributions over the total filling time.
The combined rate is the sum of individual rates:
\[ \text{Combined Rate} = \frac{1}{15} + \frac{1}{10} + \frac{1}{5} \]To add these fractions, find a common denominator, which is 30:
\[ \text{Combined Rate} = \frac{2}{30} + \frac{3}{30} + \frac{6}{30} = \frac{2+3+6}{30} = \frac{11}{30} \]So, all three pipes together fill \( \frac{11}{30} \) of the tank in 1 minute.
If the pipes fill \( \frac{11}{30} \) of the tank in 1 minute, then the total time to fill the entire tank (1 unit of work) is the reciprocal of the combined rate:
\[ \text{Total Time} = \frac{1}{\text{Combined Rate}} = \frac{1}{\frac{11}{30}} = \frac{30}{11} \text{ minutes} \]The part of work done by B and C is their combined rate multiplied by the total time taken to fill the tank.
First, find the combined rate of B and C:
\[ \text{Rate of B and C} = \frac{1}{10} + \frac{1}{5} \]Find a common denominator, which is 10:
\[ \text{Rate of B and C} = \frac{1}{10} + \frac{2}{10} = \frac{1+2}{10} = \frac{3}{10} \]Now, multiply this combined rate by the total time:
\[ \text{Work done by B and C} = (\text{Rate of B and C}) \times (\text{Total Time}) \]Substituting the values:
\[ \text{Work done by B and C} = \frac{3}{10} \times \frac{30}{11} \]Simplify the expression:
\[ \text{Work done by B and C} = \frac{3 \times 3}{1 \times 11} = \frac{9}{11} \]Correct Option: A) \( \frac{9}{11} \)
The question asks for the international (SI) unit of electric current. We need to recall the SI base units for various physical quantities, specifically focusing on electrical quantities.
The SI base unit for electric current is the Ampere (A).
D) Ampere is the SI unit of electric current. One Ampere is defined as the flow of one Coulomb of charge per second. Mathematically, it can be expressed as:
\[ 1 \text{ A} = 1 \frac{\text{C}}{\text{s}} \]where \(\text{A}\) is Ampere, \(\text{C}\) is Coulomb (unit of charge), and \(\text{s}\) is second (unit of time).
The question asks to identify the leading rubber-producing state in India. This requires knowledge of agricultural geography and crop distribution within India.
A) Kerala is unequivocally the leading producer of natural rubber in India. It accounts for the largest share of rubber production in the country due to its favorable climatic conditions (hot and humid tropical climate, high rainfall) and extensive cultivation. The Rubber Board of India, which oversees the development of the rubber industry, is also headquartered in Kottayam, Kerala, further highlighting the state's importance in rubber production.
We are given information about four numbers: their overall mean, the mean of the three smallest, and the data range. We need to find the mean of the three largest numbers. We will use the definitions of mean and range to set up equations and solve for the individual numbers, or at least the sum of the required numbers.
Let the four numbers be \(a, b, c, d\) in ascending order, so \(a \le b \le c \le d\).
The mean of the four numbers is 22. This means:
\[ \frac{a + b + c + d}{4} = 22 \]\[ a + b + c + d = 22 \times 4 \]\[ a + b + c + d = 88 \quad \text{(Equation 1)} \]The mean of the three smallest numbers (\(a, b, c\)) is 19. This means:
\[ \frac{a + b + c}{3} = 19 \]\[ a + b + c = 19 \times 3 \]\[ a + b + c = 57 \quad \text{(Equation 2)} \]Substitute Equation 2 into Equation 1 to find the largest number, \(d\):
\[ (a + b + c) + d = 88 \]\[ 57 + d = 88 \]\[ d = 88 - 57 \]\[ d = 31 \]The data range is 15. The range is the difference between the largest and the smallest number. So:
\[ \text{Range} = d - a \]\[ 15 = 31 - a \]\[ a = 31 - 15 \]\[ a = 16 \]Now we need to find the mean of the three largest numbers, which are \(b, c, d\). We know \(d = 31\). We also know \(a + b + c = 57\) and \(a = 16\). Substitute \(a\) into \(a + b + c = 57\):
\[ 16 + b + c = 57 \]\[ b + c = 57 - 16 \]\[ b + c = 41 \]The sum of the three largest numbers is \(b + c + d\). We have \(b + c = 41\) and \(d = 31\). So:
\[ b + c + d = 41 + 31 \]\[ b + c + d = 72 \]The mean of the three largest numbers is:
\[ \text{Mean} = \frac{b + c + d}{3} \]\[ \text{Mean} = \frac{72}{3} \]\[ \text{Mean} = 24 \]A) 24