This problem involves relative speed. When a train crosses a person, the distance covered is the length of the train. Since the persons are moving in the same direction as the train, the relative speed is the difference between the train's speed and the person's speed. We can set up two equations based on the given information and solve for the train's speed and length.
Let the speed of the train be \(S_T\) kmph and its length be \(L\) meters.
The first person is travelling at \(S_1 = 3\) kmph. The train crosses this person in \(t_1 = 10\) seconds.
Convert speeds to m/s for consistency with time in seconds:
\[ S_T \text{ kmph} = S_T \times \frac{5}{18} \text{ m/s} \]\[ S_1 = 3 \text{ kmph} = 3 \times \frac{5}{18} = \frac{5}{6} \text{ m/s} \]Relative speed of the train with respect to the first person (moving in the same direction):
\[ S_{rel1} = \left(S_T - 3\right) \text{ kmph} = \left(S_T - 3\right) \times \frac{5}{18} \text{ m/s} \]The distance covered by the train to cross the person is its own length \(L\). Using the formula \(\text{Distance} = \text{Speed} \times \text{Time}\):
\[ L = \left(S_T - 3\right) \times \frac{5}{18} \times 10 \quad \text{(Equation 1)} \]The second person is travelling at \(S_2 = 5\) kmph. The train crosses this person in \(t_2 = 11\) seconds.
Convert speed to m/s:
\[ S_2 = 5 \text{ kmph} = 5 \times \frac{5}{18} = \frac{25}{18} \text{ m/s} \]Relative speed of the train with respect to the second person (moving in the same direction):
\[ S_{rel2} = \left(S_T - 5\right) \text{ kmph} = \left(S_T - 5\right) \times \frac{5}{18} \text{ m/s} \]Using the distance formula for the second person:
\[ L = \left(S_T - 5\right) \times \frac{5}{18} \times 11 \quad \text{(Equation 2)} \]Since the length of the train \(L\) is the same in both cases, we can equate Equation 1 and Equation 2:
\[ \left(S_T - 3\right) \times \frac{5}{18} \times 10 = \left(S_T - 5\right) \times \frac{5}{18} \times 11 \]Cancel out the common term \(\frac{5}{18}\) from both sides:
\[ \left(S_T - 3\right) \times 10 = \left(S_T - 5\right) \times 11 \]Expand and solve for \(S_T\):
\[ 10S_T - 30 = 11S_T - 55 \]\[ 11S_T - 10S_T = 55 - 30 \]\[ S_T = 25 \]The speed of the train is 25 kmph.
Correct Option: C) 25 kmph
This problem involves the concept of work and time, specifically relating efficiency (skill) to the time taken to complete a task. If a person is less skilled, they take more time to complete the same amount of work. We can establish a relationship between their efficiencies and then use it to find the time taken by B.
First, let's understand the relationship between A's and B's skill levels. A is 20% less skilled than B.
If B's skill (efficiency) is \(E_B\), then A's skill \(E_A\) is:
\[ E_A = E_B - 0.20 \times E_B = 0.80 \times E_B \]The time taken to complete a work is inversely proportional to the skill (efficiency) of the worker. That is, if a worker is more skilled, they take less time.
Let \(T_A\) be the time taken by A and \(T_B\) be the time taken by B.
So, we can write the relationship as:
\[ \frac{T_A}{T_B} = \frac{E_B}{E_A} \]We are given that A takes \(7 \frac{1}{2}\) hours to complete the work. Convert this mixed fraction to an improper fraction:
\[ T_A = 7 \frac{1}{2} = \frac{(7 \times 2) + 1}{2} = \frac{14 + 1}{2} = \frac{15}{2} \text{ hours} \]Now substitute the values into the inverse proportionality equation:
\[ \frac{\frac{15}{2}}{T_B} = \frac{E_B}{0.80 \times E_B} \]Simplify the right side of the equation:
\[ \frac{\frac{15}{2}}{T_B} = \frac{1}{0.80} \]We know that \[0.80 = \frac{80}{100} = \frac{4}{5}\]. So, \[\frac{1}{0.80} = \frac{1}{\frac{4}{5}} = \frac{5}{4}\].
The equation becomes:
\[ \frac{15}{2T_B} = \frac{5}{4} \]Now, solve for \(T_B\):
Cross-multiply:
\[ 15 \times 4 = 5 \times 2T_B \]\[ 60 = 10T_B \]\[ T_B = \frac{60}{10} \]\[ T_B = 6 \text{ hours} \]C) 6 hours
To find the principal sum, we will use the formula for the difference between compound interest (CI) and simple interest (SI) for 3 years. This formula directly relates the principal, rate, and the difference in interest, making it efficient for such problems.
The formula for the difference between CI and SI for 3 years is:
\[ \text{CI} - \text{SI} = P \left( \frac{R}{100} \right)^2 \left( 3 + \frac{R}{100} \right) \]Let \(P\) be the principal sum. We have:
\[ 15.25 = P \left( \frac{5}{100} \right)^2 \left( 3 + \frac{5}{100} \right) \]A) The calculated principal sum is Rs. 2000, which matches option A.
We need to find the value of a natural number P such that when it is added to the product of four consecutive natural numbers, the result is a perfect square. We can represent the four consecutive natural numbers algebraically and then simplify their product to identify the pattern that leads to a perfect square.
Let the four consecutive natural numbers be \(n\), \(n+1\), \(n+2\), and \(n+3\).
Their product is \(P_n = n(n+1)(n+2)(n+3)\).
Rearrange the terms to group the first with the last, and the two middle terms:
\[ P_n = [n(n+3)][(n+1)(n+2)] \]Expand these products:
\[ n(n+3) = n^2 + 3n \]\[ (n+1)(n+2) = n^2 + 2n + n + 2 = n^2 + 3n + 2 \]Now, substitute these back into the expression for \(P_n\):
\[ P_n = (n^2 + 3n)(n^2 + 3n + 2) \]To simplify, let \(x = n^2 + 3n\). Then the expression becomes:
\[ P_n = x(x+2) \]\[ P_n = x^2 + 2x \]We are given that when a natural number P is added to this product, the result is a perfect square. So, \(P_n + P = \text{perfect square}\).
\[ x^2 + 2x + P = \text{perfect square} \]We know that \((x+1)^2 = x^2 + 2x + 1\). Comparing this with our expression, we can see that if \(P=1\), the expression becomes a perfect square.
\[ x^2 + 2x + 1 = (x+1)^2 \]Since \(x = n^2 + 3n\), which is an integer for any natural number \(n\), \(x+1\) will also be an integer, and \((x+1)^2\) will be a perfect square.
Therefore, the value of P must be 1.
D) 1 is the correct value. When \(P=1\), the product of four consecutive natural numbers plus P becomes a perfect square, specifically \((n^2+3n+1)^2\).
To find the fraction lying between two given fractions, we can convert all fractions (the given ones and the options) into decimal form or find a common denominator to compare them. Comparing decimal values is often simpler.
First, convert the two boundary fractions into decimals:
So, we are looking for a fraction whose decimal value is between \( 0.6 \) and \( 0.666... \).
Now, let's convert each option to its decimal form and check if it falls within the range \( (0.6, 0.666...) \).
\( \frac{2}{5} = 0.4 \)
This is less than \( 0.6 \), so it is not between the given fractions.
\( \frac{31}{50} = 0.62 \)
This value is greater than \( 0.6 \) and less than \( 0.666... \). So, \( 0.6 < 0.62 < 0.666... \). This fraction lies between the given fractions.
\( \frac{1}{15} = 0.0666... \) (approximately \( 0.07 \))
This is less than \( 0.6 \), so it is not between the given fractions.
\( \frac{1}{3} = 0.333... \) (approximately \( 0.33 \))
This is less than \( 0.6 \), so it is not between the given fractions.
B) \( \frac{31}{50} \)
As calculated, \( \frac{31}{50} = 0.62 \), which lies between \( \frac{3}{5} = 0.6 \) and \( \frac{2}{3} \approx 0.666... \).
To find the smallest 5-digit number divisible by 41, we first identify the smallest 5-digit number. Then, we divide this number by 41 to find the remainder. Using this remainder, we can determine how much needs to be added to the smallest 5-digit number to make it perfectly divisible by 41.
Identify the smallest 5-digit number:
The smallest 5-digit number is \(10000\).
Divide the smallest 5-digit number by 41:
We perform the division:
\[ 10000 \div 41 \]Let's do the long division:
\[ \begin{array}{r} 243 \[-3pt] 41\overline{)10000} \[-3pt] -82\phantom{000} \[-3pt] \hline 180\phantom{00} \[-3pt] -164\phantom{00} \[-3pt] \hline 160\phantom{0} \[-3pt] -123\phantom{0} \[-3pt] \hline 37\phantom{0} \end{array} \]So, \(10000 = 41 \times 243 + 37\).
The quotient is \(243\) and the remainder is \(37\).
Find the number to add to make it divisible by 41:
For \(10000\) to be perfectly divisible by \(41\), the remainder should be \(0\). Since the remainder is \(37\), we need to add the difference between the divisor and the remainder to \(10000\).
Number to add \( = \text{Divisor} - \text{Remainder} = 41 - 37 = 4\).
Calculate the smallest 5-digit number divisible by 41:
Smallest 5-digit number divisible by \(41 = 10000 + 4 = 10004\).
C) 10004 — As calculated, adding 4 to the smallest 5-digit number (10000) makes it 10004, which is perfectly divisible by 41. \((10004 = 41 \times 244)\).
The problem asks us to identify the odd one out from a set of pairs of numbers. We need to find a common pattern or relationship between the numbers in each pair and then identify the pair that does not follow this pattern.
Correct Option: A) 42 – 4. In this pair, 42 is not perfectly divisible by 4. For the other options, the first number is perfectly divisible by the second number.
The problem is based on a direct letter coding pattern. We need to identify the code assigned to each letter from the given examples and then apply the same coding scheme to the target word.
MOLEK will be coded as 72495.
Correct Option: C) 72495
The problem involves a coding-decoding pattern. We need to identify the rule by which "ASSIGN" is transformed into "SASING" and then apply the same rule to "KIDNAP" to find the coded word.
So, the pattern is: (1st & 2nd letters swap), (3rd & 4th letters remain), (5th & 6th letters swap).
A) IKNDPA is the correct answer based on the identified pattern.
The question asks about the primary nutrient supplied to plants by urea. To answer this, we need to recall the chemical composition of urea and its role as a fertilizer.
D) Nitrogen — Urea is a widely used nitrogen fertilizer, providing a significant source of nitrogen to plants, which is vital for their growth and development, particularly for vegetative growth and leaf development.