The problem asks us to find the principal amount given the final amount, the time period, and the compound interest rate. We can use the compound interest formula to relate these quantities and solve for the principal.
Identify the given values:
Recall the compound interest formula:
\[ A = P \left(1 + \frac{R}{100}\right)^n \]Substitute the given values into the formula:
\[ 936 = P \left(1 + \frac{20}{100}\right)^2 \]Simplify the term inside the parenthesis:
\[ 936 = P \left(1 + \frac{1}{5}\right)^2 \]\[ 936 = P \left(\frac{5+1}{5}\right)^2 \]\[ 936 = P \left(\frac{6}{5}\right)^2 \]Calculate the square of the fraction:
\[ 936 = P \left(\frac{36}{25}\right) \]Solve for the Principal (\(P\)):
\[ P = 936 \times \frac{25}{36} \]Perform the division and multiplication:
\[ P = \frac{936}{36} \times 25 \]\[ P = 26 \times 25 \]\[ P = 650 \]B) 650 is the principal amount. The calculation shows that if the principal is Rs 650, it will grow to Rs 936 in 2 years at a 20% compound interest rate.
To find the day of the week for a future date, we need to calculate the total number of odd days between the given start date and the target date. Odd days are the remainder when the total number of days is divided by 7. Each full year adds 1 odd day (for a normal year) or 2 odd days (for a leap year). We then count the odd days for the months in the partial year.
The period starts on 12 February 2018 and ends on 19 April 2019. First, let's consider the full year from 12 February 2018 to 12 February 2019.
The year 2018 is a normal year (not divisible by 4). A normal year has 365 days.
Number of odd days in a normal year \( = 365 \div 7 \)
\[ 365 = 52 \times 7 + 1 \]Therefore, a normal year has 1 odd day.So, from 12 February 2018 to 12 February 2019, there is 1 odd day.Now we need to count the days from 13 February 2019 to 19 April 2019.
Total number of remaining days \( = 16 + 31 + 19 = 66 \) days.
Number of odd days in these 66 days \( = 66 \div 7 \)
\[ 66 = 9 \times 7 + 3 \]So, there are 3 odd days for this period.Total odd days \( = (\text{odd days from 2018 to 2019}) + (\text{odd days from Feb 2019 to Apr 2019}) \)
Total odd days \( = 1 + 3 = 4 \) odd days.
Given that 12 February 2018 was a Monday. We need to add 4 odd days to Monday.
Monday + 1 day = Tuesday
Monday + 2 days = Wednesday
Monday + 3 days = Thursday
Monday + 4 days = Friday
So, 19 April 2019 will be a Friday.
C) Friday
To calculate the efficiency of a system, we need to determine the ratio of the useful work output to the total energy input. This ratio is then typically expressed as a percentage.
Efficiency is defined as the ratio of useful work output to the total energy input, often multiplied by 100 to express it as a percentage.
\[ \text{Efficiency} = \frac{\text{Useful Work Output}}{\text{Total Energy Input}} \times 100\% \]Correct Option: B) The efficiency of the car is calculated as the ratio of useful work output to total energy input, which is \(500 \, \text{J} / 2000 \, \text{J} = 0.25\). Multiplying by 100% gives 25%.
If \( \frac{0.7}{1-6c} \) = -0.2, then c =?
The problem requires solving a linear equation for the variable \(c\). We will use algebraic manipulation to isolate \(c\) on one side of the equation.
Given the equation:
\[ \frac{0.7}{1-6c} = -0.2 \]Multiply both sides by \((1-6c)\) to eliminate the denominator:
\[ 0.7 = -0.2(1-6c) \]Distribute -0.2 on the right side:
\[ 0.7 = -0.2 + (-0.2)(-6c) \]\[ 0.7 = -0.2 + 1.2c \]Add 0.2 to both sides to isolate the term with \(c\):
\[ 0.7 + 0.2 = 1.2c \]\[ 0.9 = 1.2c \]Divide both sides by 1.2 to solve for \(c\):
\[ c = \frac{0.9}{1.2} \]Simplify the fraction. We can multiply the numerator and denominator by 10 to remove decimals:
\[ c = \frac{9}{12} \]Reduce the fraction by dividing both numerator and denominator by their greatest common divisor, which is 3:
\[ c = \frac{9 \div 3}{12 \div 3} \]\[ c = \frac{3}{4} \]Convert the fraction to a decimal:
\[ c = 0.75 \]Correct Option: D) 0.75
The problem presents an alphabetical sequence with a number followed by a letter. We need to identify the pattern for both the numerical and alphabetical parts separately to predict the next term.
Correct Option: A) 13K — This option correctly follows both the numerical and alphabetical patterns identified.
The question asks to identify the germ cell responsible for storing food for the embryo. This requires knowledge of gamete types and their roles in reproduction.
C) Female gamete. The female gamete (egg or ovum) is typically larger than the male gamete and contains a substantial amount of cytoplasm rich in stored nutrients (like yolk) that provide nourishment for the early development of the embryo after fertilization.
This problem involves setting up and solving a system of linear equations based on the given age relationships. We will define variables for the present ages of Neelotpal and Bharti, form equations from the given statements, and then solve for Neelotpal's present age.
Correct Option: A) 40 years
A and B together can complete a work in 35 days. If A works alone and completes \( \frac{4}{7} \) of the work, then leaves the remaining work for B, the total time taken to complete the work is 114 days. Then, how many days will A, who is more efficient than B, take to complete the entire work alone?
This problem involves work and time. We are given the combined work rate of A and B, and a scenario where they work separately for parts of the work. We need to find the time A takes to complete the entire work alone, given that A is more efficient than B.
Let the time taken by A to complete the work alone be \(x\) days and by B be \(y\) days.
The work rate of A is \( \frac{1}{x} \) per day, and the work rate of B is \( \frac{1}{y} \) per day.
Given that A and B together can complete a work in 35 days:
\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{35} \quad \text{(Equation 1)} \]In the second scenario, A completes \( \frac{4}{7} \) of the work. Let \(t_A\) be the time A works.
\[ t_A \times \frac{1}{x} = \frac{4}{7} \implies t_A = \frac{4x}{7} \]The remaining work is \[1 - \frac{4}{7} = \frac{3}{7}\]. B completes this remaining work. Let \(t_B\) be the time B works.
\[ t_B \times \frac{1}{y} = \frac{3}{7} \implies t_B = \frac{3y}{7} \]The total time taken to complete the work in this scenario is 114 days.
\[ t_A + t_B = 114 \] \[ \frac{4x}{7} + \frac{3y}{7} = 114 \] \[ 4x + 3y = 114 \times 7 \] \[ 4x + 3y = 798 \quad \text{(Equation 2)} \]From Equation 1, we can express \(y\) in terms of \(x\):
\[ \frac{1}{y} = \frac{1}{35} - \frac{1}{x} = \frac{x - 35}{35x} \] \[ y = \frac{35x}{x - 35} \]Substitute the expression for \(y\) into Equation 2:
\[ 4x + 3 \left( \frac{35x}{x - 35} \right) = 798 \] \[ 4x + \frac{105x}{x - 35} = 798 \]Multiply by \( (x - 35) \) to clear the denominator:
\[ 4x(x - 35) + 105x = 798(x - 35) \] \[ 4x^2 - 140x + 105x = 798x - 27930 \] \[ 4x^2 - 35x = 798x - 27930 \] \[ 4x^2 - 35x - 798x + 27930 = 0 \] \[ 4x^2 - 833x + 27930 = 0 \]Solve this quadratic equation for \(x\) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \(a=4\), \(b=-833\), \(c=27930\).
\[ x = \frac{833 \pm \sqrt{(-833)^2 - 4(4)(27930)}}{2(4)} \] \[ x = \frac{833 \pm \sqrt{693889 - 446880}}{8} \] \[ x = \frac{833 \pm \sqrt{247009}}{8} \] \[ x = \frac{833 \pm 497}{8} \]Two possible values for \(x\):
\[ x_1 = \frac{833 + 497}{8} = \frac{1330}{8} = 166.25 \] \[ x_2 = \frac{833 - 497}{8} = \frac{336}{8} = 42 \]Now, we need to check which value of \(x\) is valid. We are given that A is more efficient than B, which means A takes less time than B to complete the work alone (\(x < y\)).
If \(x = 166.25\):
\[ y = \frac{35 \times 166.25}{166.25 - 35} = \frac{5818.75}{131.25} \approx 44.33 \]In this case, \(x > y\), which contradicts the condition that A is more efficient than B.
If \(x = 42\):
\[ y = \frac{35 \times 42}{42 - 35} = \frac{1470}{7} = 210 \]In this case, \(x = 42\) and \(y = 210\). Since \(42 < 210\), A takes less time than B, which means A is more efficient than B. This satisfies the condition.
Therefore, A will take 42 days to complete the entire work alone.
D) 42
The question asks to identify the non-metal that exists in a liquid state at room temperature. We need to recall the common states of matter for elements, especially non-metals, at standard room temperature (approximately 20-25°C).
Correct Option: A) Bromine is a non-metal that is liquid at room temperature. It is a reddish-brown volatile liquid with a strong, unpleasant odor.
The question asks which type of reproduction leads to greater diversity. We need to understand the fundamental mechanisms of each reproduction type and how they contribute to genetic variation.
Comparing these, sexual reproduction inherently introduces genetic variation through the combination of two distinct genetic sets and the processes of meiosis, leading to significantly more diversity than any form of asexual reproduction.
Correct Option: A) Sexual reproduction involves the fusion of gametes from two parents and processes like meiosis (crossing over and independent assortment), which lead to a recombination of genetic material. This results in offspring that are genetically different from both parents and from each other, thus creating significant genetic diversity within a population.