Competency Focused Questions
Q1: What annual instalment will discharge a debt of ₹4600 due in 4 years at 10% simple interest?
i. Set Up the Instalment Logic
Step 1: Understand how simple interest instalments work.
Let each annual instalment be \( x \).
The total debt of \( \text{₹}4600 \) is paid back over 4 years through 4 separate instalments.
Each instalment paid before the end of the final year accumulates simple interest until the end of the 4th year.
Step 2: Calculate the worth of each instalment at the end of the 4th year.
The 1st instalment stays for 3 years: worth = \( x + \frac{x \times 10 \times 3}{100} = x + 0.3x = 1.3x \)
The 2nd instalment stays for 2 years: worth = \( x + \frac{x \times 10 \times 2}{100} = x + 0.2x = 1.2x \)
The 3rd instalment stays for 1 year: worth = \( x + \frac{x \times 10 \times 1}{100} = x + 0.1x = 1.1x \)
The 4th instalment is paid at the very end: worth = \( x \)
ii. Solve for the Instalment Amount (x)
Step 1: Sum up all the future values of the instalments and equate them to the total debt.
\( 1.3x + 1.2x + 1.1x + x = 4600 \)
Step 2: Add the coefficients together.
\( 4.6x = 4600 \)
Step 3: Isolate \( x \) to find the value of each instalment.
\( x = \frac{4600}{4.6} \)
\( x = \frac{46000}{46} \)
\( x = 1000 \)
Answer:a. ₹1000
Q2: The simple interest on a sum of money at 8% per annum for 6 years is half the sum. The sum is:
i. Set Up Simple Interest Equation
Step 1: Identify the given values and define variables.
Let the principal sum of money be \( P \).
Rate of Interest \( (R) = 8\% \text{ per annum} \)
Time period \( (T) = 6 \text{ years} \)
According to the question, Simple Interest \( (\text{SI}) = \frac{P}{2} \).
Step 2: Substitute the variables into the standard Simple Interest formula.
Formula: \( \text{SI} = \frac{P \times R \times T}{100} \)
\( \frac{P}{2} = \frac{P \times 8 \times 6}{100} \)
ii. Analyze the Equation Profile
Step 1: Simplify the algebraic expression.
\( \frac{P}{2} = \frac{48P}{100} \)
\( \frac{P}{2} = 0.48P \)
Step 2: Attempt to isolate the principal variable \( P \).
\( 0.5P = 0.48P \)
\( 0.5P – 0.48P = 0 \)
\( 0.02P = 0 \Rightarrow P = 0 \)
Since the variable \( P \) cancels out during analysis or results in zero, we cannot determine a specific non-zero numerical value for the principal sum from the data provided.
Answer:d. data inadequate
Q3: What is the principal amount which earns ₹132 as compound interest for the second year at 10% per annum?
i. Express Second Year Interest in Terms of Principal (P)
Step 1: Identify given variables and define the initial principal.
Let the initial principal amount be \( P \).
Rate of Interest \( (R) = 10\% \text{ per annum} \)
Step 2: Calculate the interest earned during the first year.
Formula: \( \text{Interest for Year 1 } (I_1) = \frac{P \times R \times 1}{100} \)
\( I_1 = \frac{P \times 10 \times 1}{100} = 0.1P \)
Step 3: Find the principal at the beginning of the second year.
\( \text{Principal for Year 2 } (P_2) = \text{Initial Principal } (P) + I_1 \)
\( P_2 = P + 0.1P = 1.1P \)
Step 4: Formulate the interest expression for the second year.
\( \text{Interest for Year 2 } (I_2) = \frac{P_2 \times R \times 1}{100} \)
\( I_2 = \frac{1.1P \times 10 \times 1}{100} = 0.11P \)
ii. Solve for the Principal Amount (P)
Step 1: Equate the second-year interest expression to the given value.
Given: \( I_2 = \text{₹}132 \)
\( 0.11P = 132 \)
Step 2: Isolate \( P \) to calculate the initial principal amount.
\( P = \frac{132}{0.11} \)
\( P = \frac{13200}{11} \)
\( P = 1200 \)
Answer:c. ₹1200
Q4: In an industrial area, the price of a plot of area 1 acre is ₹ 5 crores. If the price of land increases by 2% quarterly, then what will be the price of the plot after 1 year?
i. Identify Given Values and Compound Framework
Step 1: List all the given parameters from the problem statement.
Initial Price of the plot \( (P) = 5 \text{ crores} \)
Quarterly Increase Rate \( (R) = 2\% \text{ per quarter} \)
Time Duration = \( 1 \text{ year} \)
Step 2: Determine the total number of compounding cycles \( (n) \).
Since the price increases quarterly and there are 4 quarters in 1 year:
Total quarterly periods \( (n) = 4 \)
ii. Formulate the Final Price Expression
Step 1: State the standard exponential growth formula for compounding intervals.
Formula: \( A = P \times \left(1 + \frac{R}{100}\right)^n \)
Step 2: Substitute the initial values into the template framework to construct the final expression.
\( A = 5 \times \left(1 + \frac{2}{100}\right)^4 \text{ crores} \)
Answer:a. \( 5\left(1+\frac{2}{100}\right)^4 \) crores
Q5: If the compound interest on a certain sum for 2 years at 10% p.a. is ₹2100, the simple interest on it at the same rate for 2 years will be
i. Find the Sum of Money (Principal)
Step 1: Identify the given variables from the Compound Interest portion.
Compound Interest \( (\text{CI}) = \text{₹}2100 \)
Rate of Interest \( (R) = 10\% \text{ per annum} \)
Time period \( (n) = 2 \text{ years} \)
Step 2: Set up the equation using the standard Compound Interest formula.
Formula: \( \text{CI} = P \times \left[\left(1 + \frac{R}{100}\right)^n – 1\right] \)
\( 2100 = P \times \left[\left(1 + \frac{10}{100}\right)^2 – 1\right] \)
\( 2100 = P \times \left[\left(1.1\right)^2 – 1\right] \)
Step 3: Expand the decimal expression and solve for Principal \( (P) \).
\( 2100 = P \times \left[1.21 – 1\right] \)
\( 2100 = P \times 0.21 \)
\( P = \frac{2100}{0.21} \)
\( P = \frac{210000}{21} \)
\( P = \text{₹}10000 \)
The initial sum of money is ₹10000.
ii. Calculate the Simple Interest
Step 1: Identify the variables for the Simple Interest calculation.
Principal \( (P) = \text{₹}10000 \)
Rate of Interest \( (R) = 10\% \text{ per annum} \)
Time Period \( (T) = 2 \text{ years} \)
Step 2: Compute the Simple Interest value using the standard interest equation.
Formula: \( \text{SI} = \frac{P \times R \times T}{100} \)
\( \text{SI} = \frac{10000 \times 10 \times 2}{100} \)
\( \text{SI} = 100 \times 20 \)
\( \text{SI} = \text{₹}2000 \)
Answer:c. ₹2000
Q6: Mr Goyal bought a car for ₹6,25,000. Its value depreciates at 20% p.a. How many years it would take for the price of the car to go down by ₹3,69,000?
i. Identify Given Values and Setup Formula
Step 1: List all the given parameters from the problem statement.
Initial Price of the car \( (P) = \text{₹}625000 \)
Total Depreciation amount = \( \text{₹}369000 \)
Rate of Depreciation \( (R) = 20\% \text{ p.a.} \)
Let the required time be \( n \) years.
Step 2: Calculate the final depreciated value (A) of the car.
\( A = \text{Initial Price} – \text{Depreciation Amount} \)
\( A = 625000 – 369000 \)
\( A = \text{₹}256000 \)
ii. Solve for the Number of Years (n)
Step 1: Substitute the values into the standard annual depreciation formula framework.
Formula: \( A = P \times \left(1 – \frac{R}{100}\right)^n \)
\( 256000 = 625000 \times \left(1 – \frac{20}{100}\right)^n \)
Step 2: Isolate the exponential term by dividing both sides by 625000.
\( \frac{256000}{625000} = \left(1 – \frac{1}{5}\right)^n \)
Cancel out the zeros and simplify the base:
\( \frac{256}{625} = \left(\frac{4}{5}\right)^n \)
Step 3: Express the fraction on the left side as a power with base \( \frac{4}{5} \).
We know that \( 4^4 = 256 \) and \( 5^4 = 625 \).
\( \left(\frac{4}{5}\right)^4 = \left(\frac{4}{5}\right)^n \)
Step 4: Equate the exponents since the bases are identical.
\( n = 4 \text{ years} \)
Answer:It would take 4 years for the price of the car to go down by ₹3,69,000.
Q7: The population of a town was decreasing every year due to migration. The present population of the town is 6,31,680. Last year the migration was 4% and the year before last, it was 6%. What was the population 2 years ago?
i. Identify Given Values and Formula Framework
Step 1: List all the given parameters from the problem statement.
Present Population \( (A) = 631680 \)
Migration Rate last year \( (R_2) = 4\% \text{ (Decrease)} \)
Migration Rate year before last \( (R_1) = 6\% \text{ (Decrease)} \)
Let the population 2 years ago be \( P \).
Step 2: State the compound interest template formula for successive population decrease variations.
Formula: \( A = P \times \left(1 – \frac{R_1}{100}\right) \times \left(1 – \frac{R_2}{100}\right) \)
ii. Calculate the Population 2 Years Ago
Step 1: Substitute the given values into the formula framework.
\( 631680 = P \times \left(1 – \frac{6}{100}\right) \times \left(1 – \frac{4}{100}\right) \)
Step 2: Simplify the fraction expressions inside the parentheses.
\( 1 – \frac{6}{100} = \frac{94}{100} \)
\( 1 – \frac{4}{100} = \frac{96}{100} \)
Step 3: Combine and multiply the fraction values together.
\( 631680 = P \times \frac{94}{100} \times \frac{96}{100} \)
\( 631680 = P \times \frac{94 \times 96}{10000} \)
\( 631680 = P \times \frac{9024}{10000} \)
Step 4: Isolate \( P \) to compute the final population value from 2 years ago.
\( P = \frac{631680 \times 10000}{9024} \)
Since \( \frac{631680}{9024} = 70 \):
\( P = 70 \times 10000 \)
\( P = 700000 \)
Answer:The population 2 years ago was 700000
Leave a Comment