This problem involves setting up and solving linear equations based on the given age relationships at different points in time. We will define variables for the current ages of the father and son and then translate the given conditions into algebraic equations.
Correct Option: A) 3
To find the simple interest, we use the standard formula for simple interest: \[\text{Simple Interest} = \frac{\text{Principal} \times \text{Rate} \times \text{Time}}{100}\]. We are given the principal amount, the rate of interest per annum, and the time period in years. We will substitute these values into the formula to calculate the simple interest.
\[ \text{SI} = \frac{P \times R \times T}{100} \]
\[ \text{SI} = \frac{400 \times 4 \times 6.5}{100} \]
\[ 400 \times 4 \times 6.5 = 1600 \times 6.5 \]
To multiply \(1600 \times 6.5\):
\[ 1600 \times 6.5 = 1600 \times \left(6 + 0.5\right) = 1600 \times 6 + 1600 \times 0.5 \]
\[ = 9600 + 800 = 10400 \]
\[ \text{SI} = \frac{10400}{100} = 104 \]
The simple interest is ₹104.
Correct Option: D) 104
This problem involves work and time, where the efficiency of men and women needs to be compared first. We will calculate the individual work rate of one man and one woman, then determine the combined work rate of 2 women and 4 men, and finally find the total time required to complete the work.
Correct Option: D) 20 days
To find the population at the end of the third year, we need to apply the given percentage changes sequentially to the initial population. First, calculate the population after a 25% increase, then apply a 60% increase to that new population, and finally, apply a 57% decrease to the result of the second year.
Initial Population:
\(P_0 = 15,250\)
Population after the first year (25% increase):
An increase of 25% means the population becomes \(100\% + 25\% = 125\%\) of the previous population.
\[ P_1 = P_0 \times \left(1 + \frac{25}{100}\right) = 15,250 \times 1.25 \] \[ P_1 = 19,062.5 \]Since population must be a whole number, we can consider it as 19,063 for practical purposes, or carry the decimal for exact calculation until the end. For competitive exams, it's usually best to carry the exact value.
Population after the second year (60% increase):
An increase of 60% means the population becomes \(100\% + 60\% = 160\%\) of the population after the first year.
\[ P_2 = P_1 \times \left(1 + \frac{60}{100}\right) = 19,062.5 \times 1.60 \] \[ P_2 = 30,500 \]Population after the third year (57% decrease):
A decrease of 57% means the population becomes \(100\% - 57\% = 43\%\) of the population after the second year.
\[ P_3 = P_2 \times \left(1 - \frac{57}{100}\right) = 30,500 \times 0.43 \] \[ P_3 = 13,115 \]A) 13,115 is the population of the town at the end of the third year, calculated by applying the successive percentage changes.
To solve this problem, we need to simplify the given expression by following the order of operations (PEMDAS/BODMAS). First, simplify the fraction inside the parentheses, then perform the addition and subtraction from left to right.
Simplify the term inside the parentheses: \[\left(\frac{1}{1+\frac{4}{9}}\right)\].
First, evaluate the denominator \( 1+\frac{4}{9} \):
\[ 1+\frac{4}{9} = \frac{9}{9} + \frac{4}{9} = \frac{9+4}{9} = \frac{13}{9} \]Now, substitute this back into the fraction:
\[ \frac{1}{1+\frac{4}{9}} = \frac{1}{\frac{13}{9}} = 1 \times \frac{9}{13} = \frac{9}{13} \]Substitute this simplified term back into the original expression:
\[ \frac{2}{5} + \frac{9}{13} - \frac{4}{5} \]Group the terms with the same denominator to simplify calculations:
\[ \left(\frac{2}{5} - \frac{4}{5}\right) + \frac{9}{13} \]Perform the subtraction:
\[ \frac{2}{5} - \frac{4}{5} = \frac{2-4}{5} = \frac{-2}{5} \]Now, add the remaining terms:
\[ \frac{-2}{5} + \frac{9}{13} \]To add these fractions, find a common denominator, which is the least common multiple (LCM) of 5 and 13. Since 5 and 13 are prime numbers, their LCM is their product, \( 5 \times 13 = 65 \).
\[ \frac{-2}{5} + \frac{9}{13} = \frac{-2 \times 13}{5 \times 13} + \frac{9 \times 5}{13 \times 5} \] \[ = \frac{-26}{65} + \frac{45}{65} \] \[ = \frac{-26 + 45}{65} \] \[ = \frac{19}{65} \]A) \( \frac{19}{65} \) is the correct result obtained after simplifying the expression step-by-step.
We need to find the principal sum given the interest earned in the second year, and the annual compound interest rate. We can work backward from the interest earned in the second year to find the principal amount. The interest for the second year is calculated on the principal plus the interest accumulated in the first year.
Let the principal sum be \(P\).
The rate of interest \(R = 20\%\) per annum.
Interest for the first year (\(I_1\)) will be calculated on the principal \(P\):
\[ I_1 = P \times \frac{R}{100} = P \times \frac{20}{100} = 0.2P \]Amount at the end of the first year (\(A_1\)) will be:
\[ A_1 = P + I_1 = P + 0.2P = 1.2P \]The interest for the second year (\(I_2\)) is calculated on the amount at the end of the first year (\(A_1\)).
\[ I_2 = A_1 \times \frac{R}{100} \]We are given that \(I_2 = \text{₹}5,916\).
Substitute the values into the equation for \(I_2\):
\[ 5916 = (1.2P) \times \frac{20}{100} \] \[ 5916 = 1.2P \times 0.2 \] \[ 5916 = 0.24P \]Now, solve for \(P\):
\[ P = \frac{5916}{0.24} \] \[ P = \frac{5916}{\frac{24}{100}} \] \[ P = \frac{5916 \times 100}{24} \] \[ P = \frac{591600}{24} \]Divide 591600 by 24:
\[ P = 24650 \]So, the principal sum is ₹24,650.
A) ₹24,650 is the correct sum, as calculated above.
The problem involves finding an angle in a triangle using given angles and perpendicular lines. We will use the properties of angles in a right-angled triangle and the sum of angles in a triangle.
Analyze the given information:
Find \(\angle BCD\) in \(\triangle BDC\):
Since \(\triangle BDC\) is a right-angled triangle at \(D\), the sum of its angles is \(180^\circ\).
\[ \angle BCD + \angle BDC + \angle DBC = 180^\circ \]Substitute the known values:
\[ \angle BCD + 90^\circ + 55^\circ = 180^\circ \]\[ \angle BCD + 145^\circ = 180^\circ \]\[ \angle BCD = 180^\circ - 145^\circ \]\[ \angle BCD = 35^\circ \]Find \(\angle ACB\):
From the previous step, we found \(\angle BCD = 35^\circ\). Since \(D\) is on \(AC\), \(\angle BCD\) is the same as \(\angle ACB\).
\[ \angle ACB = 35^\circ \]Find \(\angle AEC\) in \(\triangle AEC\):
We have \(\angle CAE = 56^\circ\) and \(\angle ACE = \angle ACB = 35^\circ\).
The sum of angles in \(\triangle AEC\) is \(180^\circ\).
\[ \angle AEC + \angle CAE + \angle ACE = 180^\circ \]Substitute the known values:
\[ \angle AEC + 56^\circ + 35^\circ = 180^\circ \]\[ \angle AEC + 91^\circ = 180^\circ \]\[ \angle AEC = 180^\circ - 91^\circ \]\[ \angle AEC = 89^\circ \]Find \(\angle AEB\):
Angles \(\angle AEB\) and \(\angle AEC\) form a linear pair on the line \(BC\).
Therefore, their sum is \(180^\circ\).
\[ \angle AEB + \angle AEC = 180^\circ \]Substitute the value of \(\angle AEC\):
\[ \angle AEB + 89^\circ = 180^\circ \]\[ \angle AEB = 180^\circ - 89^\circ \]\[ \angle AEB = 91^\circ \]Correct Option: D) 91°
To find the ratio of the volumes, we need to use the formula for the volume of a sphere. We will calculate the volume of the original sphere and the new sphere (with half the radius) and then find their ratio.
Define the radius of the original sphere:
Let the radius of the original sphere be \(R\).
Calculate the volume of the original sphere:
The formula for the volume of a sphere is \(V = \frac{4}{3} \pi r^3\).
So, the volume of the original sphere, \(V_{\text{original}}\), is:
Define the radius of the new sphere:
The radius of the sphere is halved. So, the radius of the new sphere, \(r_{\text{new}}\), is:
Calculate the volume of the new sphere:
Substitute \(r_{\text{new}}\) into the volume formula:
Find the ratio of the volume of the original sphere to that of the new sphere:
We need to find \(V_{\text{original}} : V_{\text{new}}\):
Cancel out the common term \(\frac{4}{3} \pi R^3\):
\[ \frac{V_{\text{original}}}{V_{\text{new}}} = \frac{1}{\frac{1}{8}} \] \[ \frac{V_{\text{original}}}{V_{\text{new}}} = 8 \]So, the ratio is \(8:1\).
Correct Option: B) 8 : 1
To find the average speed for the whole journey, we need to calculate the total distance traveled and the total time taken. The average speed is then the total distance divided by the total time. The journey is divided into three segments, each with a given distance and speed. We will calculate the time taken for each segment and sum them up.
Correct Option: A) 76.47
The problem involves solving for an unknown variable \(z\) using given exponential equations. We will use the properties of exponents, specifically that if \(a^m = a^n\), then \(m = n\). We will express \(x\) and \(y\) in terms of their base and exponents, substitute them into the third equation, and then solve for \(z\).
Given the first two equations:
\[ x = 73^{0.51} \] \[ y = 73^{0.73} \]Given the third equation:
\[ x^z = y^{10} \]Substitute the expressions for \(x\) and \(y\) from step 1 into the equation from step 2:
\[ (73^{0.51})^z = (73^{0.73})^{10} \]Apply the exponent rule \((a^m)^n = a^{mn}\) to both sides of the equation:
\[ 73^{(0.51 \times z)} = 73^{(0.73 \times 10)} \] \[ 73^{0.51z} = 73^{7.3} \]Since the bases are the same, the exponents must be equal:
\[ 0.51z = 7.3 \]Solve for \(z\):
\[ z = \frac{7.3}{0.51} \]Perform the division:
\[ z \approx 14.3137 \]Rounding to two decimal places, the value of \(z\) is approximately \(14.31\).
Correct Option: C) 14.31