To find the quantity of toor dal sold, we need to calculate 4.5% of the total available quantity, which is 1500 kg. Percentage means "per hundred", so 4.5% can be written as a fraction \(\frac{4.5}{100}\) or a decimal 0.045. We will then multiply this fraction or decimal by the total quantity.
Understand the given values:
Convert the percentage to a decimal or fraction:
To convert a percentage to a decimal, divide by 100.
\[ 4.5\% = \frac{4.5}{100} = 0.045 \]Calculate the quantity sold:
Multiply the total quantity by the decimal equivalent of the percentage.
\[ \text{Quantity sold} = \text{Percentage sold} \times \text{Total quantity} \]\[ \text{Quantity sold} = 0.045 \times 1500 \]Perform the multiplication:
\[ 0.045 \times 1500 = 45 \times 15 \] (by moving the decimal point two places to the right in 0.045 and removing two zeros from 1500, or by multiplying directly)
Let's do the multiplication:
\[ 45 \times 10 = 450 \]
\[ 45 \times 5 = 225 \]
\[ 450 + 225 = 675 \]
So, \(0.045 \times 1500 = 67.5\) kg.
Correct Option: D) 67.5 kg
This question asks us to identify which assumption(s) are implicit in the given statement. An implicit assumption is something that must be true for the statement to make sense or be logically valid. We need to analyze each assumption in relation to the statement: "School students are more influenced by their faculty nowadays."
Correct Option: C) Only I is implicit
To find the average speed of an object, we need to calculate the total distance traveled and divide it by the total time taken. The problem provides two segments of travel with their respective distances and times.
Correct Option: A) 7.6 m s⁻¹ is the calculated average speed, including the correct units.
The problem presents a series of terms, each consisting of a letter, a number, and another letter. To find the next term, we need to identify the pattern in each component (first letter, number, second letter) separately.
The first letters are X, V, T, ...
The pattern is a decrease of 2 in the alphabetical position (\(24 - 2 = 22\), \(22 - 2 = 20\)). Following this pattern, the next letter should be the 18th letter, which is R (\(20 - 2 = 18\)).
The numbers are 24, 22, 20, ...
The pattern is a decrease of 2 (\(24 - 2 = 22\), \(22 - 2 = 20\)). Following this pattern, the next number should be \(20 - 2 = 18\).
The second letters are C, E, G, ...
The pattern is an increase of 2 in the alphabetical position (\(3 + 2 = 5\), \(5 + 2 = 7\)). Following this pattern, the next letter should be the 9th letter, which is I (\(7 + 2 = 9\)).
Combining the next first letter (R), the next number (18), and the next second letter (I), the next term in the series is R18I.
C) R18I — This option correctly follows all three identified patterns for the first letter, the number, and the second letter.
This problem involves calculating the time taken to empty a cistern when two pipes, one filling and one draining, operate alternately. We will first determine the individual rates of filling and draining, then calculate the net change in water level over a 2-hour cycle, and finally account for the last hour when only the draining pipe operates to empty the remaining water.
Calculate the filling rate of Pipe A:
Pipe A fills the cistern in 69 hours. So, in 1 hour, Pipe A fills:
Calculate the draining rate of Pipe B:
Pipe B drains the cistern in 46 hours. So, in 1 hour, Pipe B drains:
Determine the net change in water level over a 2-hour cycle:
The pipes are opened one at a time for an hour each, starting with Pipe B. So, in the first hour, Pipe B drains, and in the second hour, Pipe A fills.
Net change in 2 hours (1st hour B drains, 2nd hour A fills) = Amount drained by B - Amount filled by A
\[ \text{Net change in 2 hours} = \text{Rate}_B - \text{Rate}_A \]\[ = \frac{1}{46} - \frac{1}{69} \]\[ = \frac{3}{138} - \frac{2}{138} \]\[ = \frac{1}{138} \text{ of the cistern drained} \]Calculate the amount of water to be drained before the last cycle:
The cistern is initially full (1 unit). In each 2-hour cycle, \(\frac{1}{138}\) of the cistern is drained. If the cistern were to be completely emptied by this alternating process, the last operation would be draining by Pipe B. To avoid over-draining in the calculation, we consider the state just before the final draining operation. The maximum amount that can be drained by Pipe B in one hour is \(\frac{1}{46}\) of the cistern. So, we need to find how many cycles it takes to drain \(1 - \frac{1}{46}\) of the cistern.
Amount to be drained by cycles = \[1 - \frac{1}{46} = \frac{45}{46}\] of the cistern.
Calculate the number of 2-hour cycles:
Number of cycles = \[\frac{\text{Amount to be drained by cycles}}{\text{Net draining per cycle}}\]
Calculate the time taken for these cycles:
Time for 135 cycles = \(135 \times 2 \text{ hours} = 270 \text{ hours}\).
After 270 hours, the amount of water remaining in the cistern is \(\frac{1}{46}\) of the cistern. The last operation in a cycle is filling by Pipe A. So, after 135 cycles (270 hours), the cistern has \(\frac{1}{46}\) of its capacity filled, and it's Pipe B's turn to operate.
Calculate the time for the final draining:
Pipe B drains \(\frac{1}{46}\) of the cistern in 1 hour.
So, the remaining \(\frac{1}{46}\) of the cistern will be drained by Pipe B in 1 hour.
Calculate the total time:
Total time = Time for cycles + Time for final draining
Convert total hours to days and hours:
Since 1 day = 24 hours:
B) 11 days 7 hours
To find the magnitude of the applied force, we need to first calculate the acceleration of the object. We can do this using the given initial velocity, final velocity, and time duration. Once the acceleration is known, Newton's second law of motion (\(F = ma\)) can be applied to find the force.
Using the first equation of motion:
\[ v = u + at \]Rearrange to solve for \(a\):
\[ a = \frac{v - u}{t} \]Substitute the given values:
\[ a = \frac{6 \text{ m s}^{-1} - 4 \text{ m s}^{-1}}{3 \text{ s}} \]\[ a = \frac{2 \text{ m s}^{-1}}{3 \text{ s}} \]\[ a = 0.666... \text{ m s}^{-2} \]Using Newton's second law of motion:
\[ F = ma \]Substitute the mass and calculated acceleration:
\[ F = 8 \text{ kg} \times 0.666... \text{ m s}^{-2} \]\[ F = 5.333... \text{ N} \]Rounding to two decimal places, the force is \(5.33 \text{ N}\).
A) 5.33 N is the correct magnitude of the applied force, calculated using the change in momentum over time or by finding acceleration first and then applying Newton's second law.
To find the next term in the series, we need to identify the pattern in both the alphabetical and numerical components of each term.
The letters are decreasing by 2 positions in the English alphabet (Z \(\rightarrow\) X \(\rightarrow\) V). Following this pattern, the next letter should be two positions before V. Counting backward from V (V, U, T), the next letter is T.
The numbers are decreasing by 2 each time (26 \(\rightarrow\) 24 \(\rightarrow\) 22). Following this pattern, the next number should be \(22 - 2 = 20\).
The next letter is T and the next number is 20. Therefore, the next term in the series is T20.
Correct Option: A) T20
The problem asks us to calculate the kinetic energy of an object given its mass and speed. We will use the standard formula for kinetic energy to solve this.
Kinetic energy (\(KE\)) is given by the formula:
\[ KE = \frac{1}{2}mv^2 \]Substitute \(m = 10 \text{ kg}\) and \(v = 2 \text{ m/s}\) into the kinetic energy formula:
\[ KE = \frac{1}{2} \times (10 \text{ kg}) \times (2 \text{ m/s})^2 \]First, calculate \(v^2\):
\[ (2 \text{ m/s})^2 = 4 \text{ m}^2/\text{s}^2 \]Then, multiply the terms:\[ KE = \frac{1}{2} \times 10 \text{ kg} \times 4 \text{ m}^2/\text{s}^2 \]\[ KE = 5 \text{ kg} \times 4 \text{ m}^2/\text{s}^2 \]\[ KE = 20 \text{ J} \]The unit of kinetic energy is Joules (\(\text{J}\)).
Correct Option: A) 20 J
If a is positive and a2 + \( \frac{1}{a^2} \) = 7, then a3 + \( \frac{1}{a^3} \) =
The problem asks us to find the value of \(a^3 + \frac{1}{a^3}\) given that \(a^2 + \frac{1}{a^2} = 7\) and \(a\) is positive. We can use algebraic identities to solve this. First, we'll find the value of \(a + \frac{1}{a}\) from the given equation, and then use this to find \(a^3 + \frac{1}{a^3}\).
Find \(a + \frac{1}{a}\):
We know the identity: \[\left(a + \frac{1}{a}\right)^2 = a^2 + 2 \cdot a \cdot \frac{1}{a} + \left(\frac{1}{a}\right)^2 = a^2 + 2 + \frac{1}{a^2}\].
Rearranging this, we get: \[\left(a + \frac{1}{a}\right)^2 = \left(a^2 + \frac{1}{a^2}\right) + 2\].
Given that \(a^2 + \frac{1}{a^2} = 7\), substitute this value:
\[ \left(a + \frac{1}{a}\right)^2 = 7 + 2 \]\[ \left(a + \frac{1}{a}\right)^2 = 9 \]\[ a + \frac{1}{a} = \pm\sqrt{9} \]\[ a + \frac{1}{a} = \pm 3 \]Since \(a\) is positive, \(a + \frac{1}{a}\) must also be positive. Therefore, \(a + \frac{1}{a} = 3\).Find \(a^3 + \frac{1}{a^3}\):
We use the identity: \(x^3 + y^3 = (x+y)(x^2 - xy + y^2)\) or more directly, \((x+y)^3 = x^3 + y^3 + 3xy(x+y)\).
Let \(x = a\) and \(y = \frac{1}{a}\). Then \(xy = a \cdot \frac{1}{a} = 1\).
Substitute these into the identity:
\[ \left(a + \frac{1}{a}\right)^3 = a^3 + \left(\frac{1}{a}\right)^3 + 3 \cdot a \cdot \frac{1}{a} \left(a + \frac{1}{a}\right) \]\[ \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3 \left(a + \frac{1}{a}\right) \]We already found that \(a + \frac{1}{a} = 3\). Substitute this value:\[ (3)^3 = a^3 + \frac{1}{a^3} + 3(3) \]\[ 27 = a^3 + \frac{1}{a^3} + 9 \]Now, solve for \(a^3 + \frac{1}{a^3}\):\[ a^3 + \frac{1}{a^3} = 27 - 9 \]\[ a^3 + \frac{1}{a^3} = 18 \]Correct Option: C) 18
To determine which element has the largest atoms among the given options, we need to understand the periodic trends in atomic size. Atomic radius generally decreases across a period from left to right and increases down a group. We will locate each element on the periodic table and compare their positions.
D) Li — Lithium is in Period 2, Group 1. Among the given options, it is the leftmost element in Period 2 (compared to O and F) and is below Hydrogen in Group 1, making it the largest in terms of atomic size due to periodic trends.