Are you looking for Compound Interest Questions with solutions to practice for competitive exams? You’ve come to the right place!
Compound Interest (CI) is one of the most important and high-scoring topics in quantitative aptitude. It is a crucial topic asked across major competitive entrance and recruitment examinations, including SSC, Banking, UPSC, Railways, and Campus Placement Tests.
In this comprehensive guide, you will find carefully selected compound interest questions with step-by-step solutions. Whether you are preparing for competitive exams or campus placements, these practice sets will help you solve questions faster and with absolute accuracy.
Let’s dive in and elevate your quantitative problem-solving skills!
Step 1: Identify the given values: Principal P = 5000, Rate R = 10% per annum, and Time T = 2 years.
Step 2: Apply the compound amount formula:
A = P \left(1 + \frac{R}{100}\right)^TStep 3: Substitute the values into the formula:
A = 5000 \left(1 + \frac{10}{100}\right)^2 = 5000 \times (1.1)^2 = 5000 \times 1.21 = 6050Step 4: Calculate the compound interest by subtracting the principal from the total amount:
CI = A - P = 6050 - 5000 = 1050Thus, the compound interest is ₹1050, making option B the correct answer.
Step 1: Identify the given values and adjust for half-yearly compounding. Principal P = 8000, Annual Rate = 20% per annum, and Time T = 1 year.
Step 2: Since interest is compounded half-yearly, the rate per half-year R is half of the annual rate, and the number of periods n is twice the time in years:
R = \frac{20\%}{2} = 10\% \text{ per half-year}
n = 1 \times 2 = 2 \text{ half-years}
Step 3: Apply the compound amount formula with the adjusted rate and periods:
A = P \left(1 + \frac{R}{100}\right)^nStep 4: Substitute the values into the formula:
A = 8000 \left(1 + \frac{10}{100}\right)^2 = 8000 \times (1.1)^2 = 8000 \times 1.21 = 9680Step 5: Calculate the compound interest by subtracting the principal from the total amount:
CI = A - P = 9680 - 8000 = 1680Thus, the compound interest is ₹1680, making option B the correct answer.
Step 1: Identify the given values: Principal P = 10000, Rate R = 10% per annum, and Time T = 1\frac{1}{2} years (1 year and 6 months).
Step 2: When time is in the form of a fraction a\frac{b}{c} years, the amount formula for annual compounding is:
A = P \left(1 + \frac{R}{100}\right)^a \left(1 + \frac{\frac{b}{c} \times R}{100}\right)Step 3: Substitute the values into the formula where a = 1 and \frac{b}{c} = \frac{1}{2}:
A = 10000 \left(1 + \frac{10}{100}\right)^1 \left(1 + \frac{\frac{1}{2} \times 10}{100}\right)Step 4: Simplify the terms inside the brackets:
A = 10000 \times (1.1) \times (1.05) = 10000 \times 1.155 = 11550Step 5: Calculate the compound interest by subtracting the principal from the total amount:
CI = A - P = 11550 - 10000 = 1550Thus, the compound interest is ₹1550, making option C the correct answer.
Step 1: Identify the relationship between compound interest (CI) and simple interest (SI) for 2 years. The direct formula for the difference is:
CI - SI = P \left(\frac{R}{100}\right)^2Step 2: Substitute the given values into the formula where CI - SI = 25 and R = 5\%:
25 = P \left(\frac{5}{100}\right)^2Step 3: Simplify the fraction inside the bracket:
25 = P \left(\frac{1}{20}\right)^2 = P \times \frac{1}{400}Step 4: Solve for the principal P by cross-multiplying:
P = 25 \times 400 = 10000Thus, the sum is ₹10000, making option C the correct answer.
Step 1: Identify the formula for the difference between compound interest (CI) and simple interest (SI) for 3 years:
CI - SI = P \left(\frac{R}{100}\right)^2 \left(\frac{300 + R}{100}\right)Step 2: Substitute the given values into the formula where CI - SI = 310 and R = 10\%:
310 = P \left(\frac{10}{100}\right)^2 \left(\frac{300 + 10}{100}\right)Step 3: Simplify the fractions and terms inside the expression:
310 = P \times \frac{1}{100} \times \frac{310}{100} = P \times \frac{310}{10000}Step 4: Solve for the principal P:
P = \frac{310 \times 10000}{310} = 10000Thus, the sum is ₹10000, making option C the correct answer.
Step 1: Let the principal be P and the annual rate of interest be R%. The amount after 2 years (A_2) and 3 years (A_3) are given by:
A_2 = P \left(1 + \frac{R}{100}\right)^2 = 6050
A_3 = P \left(1 + \frac{R}{100}\right)^3 = 6655
Step 2: Divide the amount after 3 years by the amount after 2 years to find the growth factor for 1 year:
\frac{A_3}{A_2} = \frac{P \left(1 + \frac{R}{100}\right)^3}{P \left(1 + \frac{R}{100}\right)^2} = \frac{6655}{6050}Step 3: Simplify the ratio:
1 + \frac{R}{100} = \frac{6655}{6050} = \frac{11}{10} = 1.1Step 4: Solve for the rate R:
\frac{R}{100} = 1.1 - 1 = 0.1
R = 0.1 \times 100 = 10\%
Thus, the rate of interest per annum is 10%, making option C the correct answer.
Step 1: Understand the property of compound interest where a sum grows geometrically. If a sum becomes x times in t years, it will become x^n times in n \times t years.
Step 2: Express the target multiple (27 times) in terms of the base multiple (3 times):
27 = 3^3Step 3: Here, the power n = 3. Multiply this power by the given initial time (t = 3 years):
\text{Total Time} = 3 \times 3 = 9 \text{ years}Thus, the sum will become 27 times in 9 years, making option B the correct answer.
Step 1: Identify the given values: Principal P = 1000, Amount A = 1331, and Time T = 3 years.
Step 2: Apply the compound amount formula:
A = P \left(1 + \frac{R}{100}\right)^TStep 3: Substitute the known values into the equation:
1331 = 1000 \left(1 + \frac{R}{100}\right)^3Step 4: Divide both sides by 1000 to isolate the term with the rate:
\frac{1331}{1000} = \left(1 + \frac{R}{100}\right)^3
\left(\frac{11}{10}\right)^3 = \left(1 + \frac{R}{100}\right)^3
Step 5: Taking the cube root on both sides and solving for R:
\frac{11}{10} = 1 + \frac{R}{100}
1 + \frac{1}{10} = 1 + \frac{R}{100}
\frac{R}{100} = \frac{1}{10}
R = 10\%
Thus, the rate of interest is 10% per annum, making option C the correct answer.
Step 1: When the rates of interest are different for successive years (say R_1, R_2, R_3), the total amount formula is:
A = P \left(1 + \frac{R_1}{100}\right)\left(1 + \frac{R_2}{100}\right)\left(1 + \frac{R_3}{100}\right)Step 2: Substitute the given values into the formula where P = 10000, R_1 = 4\%, R_2 = 5\%, and R_3 = 6\%:
A = 10000 \left(1 + \frac{4}{100}\right)\left(1 + \frac{5}{100}\right)\left(1 + \frac{6}{100}\right)Step 3: Convert percentages to decimals and multiply:
A = 10000 \times 1.04 \times 1.05 \times 1.06
A = 10000 \times 1.15752 = 11575.20
Step 4: Calculate the compound interest by subtracting the principal from the total amount:
CI = A - P = 11575.20 - 10000 = 1575.20Thus, the compound interest is ₹1575.20, making option A the correct answer.
Step 1: Let the principal be P and the annual rate of interest R = 10\%.
Step 2: The total amount after 1 year is:
A_1 = P \left(1 + \frac{10}{100}\right) = 1.1PStep 3: The total amount after 2 years is:
A_2 = P \left(1 + \frac{10}{100}\right)^2 = 1.21PStep 4: The compound interest for the 2nd year is the difference between the amount after 2 years and the amount after 1 year:
\text{CI for 2nd year} = A_2 - A_1 = 1.21P - 1.1P = 0.11PStep 5: Equate this expression to the given compound interest for the 2nd year and solve for P:
0.11P = 132
P = \frac{132}{0.11} = 1200
Thus, the principal is ₹1200, making option C the correct answer.
Step 1: Identify the given values and adjust for quarterly compounding. Principal P = 16000, Annual Rate = 20% per annum, and Time = 9 months (\frac{9}{12} = \frac{3}{4} years).
Step 2: Since interest is compounded quarterly, the rate per quarter R is one-fourth of the annual rate, and the number of periods n is four times the time in years:
R = \frac{20\%}{4} = 5\% \text{ per quarter}
n = \frac{3}{4} \times 4 = 3 \text{ quarters}
Step 3: Apply the compound amount formula with the adjusted rate and periods:
A = P \left(1 + \frac{R}{100}\right)^nStep 4: Substitute the values into the formula:
A = 16000 \left(1 + \frac{5}{100}\right)^3 = 16000 \times (1.05)^3 = 16000 \times 1.157625 = 18522Step 5: Calculate the compound interest by subtracting the principal from the total amount:
CI = A - P = 18522 - 16000 = 2522Thus, the compound interest is ₹2522, making option B the correct answer.
Step 1: Let the population 2 years ago be P. The annual growth rate is R = 5\% and time t = 2 years.
Step 2: Apply the compound growth formula to relate the past population to the current population:
\text{Current Population} = P \left(1 + \frac{R}{100}\right)^tStep 3: Substitute the known values into the equation:
44100 = P \left(1 + \frac{5}{100}\right)^2Step 4: Simplify the expression:
44100 = P \times (1.05)^2 = P \times 1.1025Step 5: Solve for P:
P = \frac{44100}{1.1025} = 40000Thus, the population 2 years ago was 40000, making option B the correct answer.
Step 1: Let the principal be P and the amount after t years be A_t = P\left(1 + \frac{R}{100}\right)^t. Given that the amount after 2 years (A_2) is ₹4,500 and after 4 years (A_4) is ₹6,750.
Step 2: Use the geometric property of compound interest amounts over equal time intervals. The ratio of amounts separated by a fixed time interval (4 - 2 = 2 years) is constant:
\frac{A_4}{A_2} = \frac{6750}{4500} = 1.5Step 3: This growth factor represents the compounding multiplier over a 2-year period:
\left(1 + \frac{R}{100}\right)^2 = 1.5Step 4: Relate the principal to the amount after 2 years using the amount formula:
A_2 = P \left(1 + \frac{R}{100}\right)^2Step 5: Substitute the known values into the equation:
4500 = P \times 1.5
P = \frac{4500}{1.5} = 3000
Thus, the principal is ₹3,000, making option C the correct answer.
Step 1: Understand the concept of compound interest installments. If a principal P is paid back in 2 equal annual installments x at an annual rate R, the relation is given by:
P = \frac{x}{1 + \frac{R}{100}} + \frac{x}{\left(1 + \frac{R}{100}\right)^2}Step 2: Substitute the given values into the formula where P = 2100 and R = 10\%:
2100 = \frac{x}{1 + \frac{10}{100}} + \frac{x}{\left(1 + \frac{10}{100}\right)^2}Step 3: Simplify the denominators:
2100 = \frac{x}{1.1} + \frac{x}{(1.1)^2} = \frac{x}{1.1} + \frac{x}{1.21}Step 4: Take the common denominator and add the fractions:
2100 = \frac{1.1x + x}{1.21} = \frac{2.1x}{1.21}Step 5: Solve for the installment x:
x = \frac{2100 \times 1.21}{2.1} = \frac{2541}{2.1} = 1210Thus, the value of each installment is ₹1210, making option B the correct answer.
Step 1: Understand the formula for 3 equal annual installments x to pay off a principal P at an annual rate R:
P = \frac{x}{1 + \frac{R}{100}} + \frac{x}{\left(1 + \frac{R}{100}\right)^2} + \frac{x}{\left(1 + \frac{R}{100}\right)^3}Step 2: Substitute the given values into the formula where P = 6620 and R = 10\%:
6620 = \frac{x}{1 + \frac{10}{100}} + \frac{x}{\left(1 + \frac{10}{100}\right)^2} + \frac{x}{\left(1 + \frac{10}{100}\right)^3}Step 3: Simplify the denominators using decimals:
6620 = \frac{x}{1.1} + \frac{x}{(1.1)^2} + \frac{x}{(1.1)^3} = \frac{x}{1.1} + \frac{x}{1.21} + \frac{x}{1.331}Step 4: Multiply the entire equation by 1.331 to clear the fractions:
6620 \times 1.331 = 1.21x + 1.1x + x
8811.22 = 3.31x
Step 5: Solve for the installment x:
x = \frac{8811.22}{3.31} = 2662Thus, the value of each installment is ₹2662, making option C the correct answer.
Step 1: Let the two parts be P_1 and P_2. Given that the total sum is ₹2100:
P_1 + P_2 = 2100Step 2: According to the problem, the amount of the first part for 2 years equals the amount of the second part for 3 years at R = 10\%:
P_1 \left(1 + \frac{10}{100}\right)^2 = P_2 \left(1 + \frac{10}{100}\right)^3Step 3: Simplify the equation by dividing both sides by \left(1 + \frac{10}{100}\right)^2:
P_1 = P_2 \left(1 + \frac{10}{100}\right) = 1.1P_2Step 4: Substitute P_1 = 1.1P_2 into the sum equation:
1.1P_2 + P_2 = 2100
2.1P_2 = 2100 \implies P_2 = \frac{2100}{2.1} = 1000
Step 5: Find the value of P_1:
P_1 = 1.1 \times 1000 = 1100Thus, the two parts are ₹1100 and ₹1000, making option B the correct answer.
Step 1: Find the difference between compound interest (CI) and simple interest (SI) for 2 years:
\text{Difference} = CI - SI = 410 - 400 = 10Step 2: Use the direct shortcut formula relating the 2-year CI, SI, and rate of interest R:
\frac{CI - SI}{SI} = \frac{R}{200}Step 3: Substitute the known values into the equation to find R:
\frac{10}{400} = \frac{R}{200}
R = \frac{10 \times 200}{400} = 5\%
Step 4: Now, use the simple interest formula for 2 years to find the principal P:
SI = \frac{P \times R \times T}{100}
400 = \frac{P \times 5 \times 2}{100} = \frac{10P}{100} = \frac{P}{10}
Step 5: Solve for P:
P = 400 \times 10 = 4000Thus, the sum is ₹4000 and the rate is 5% per annum, making option B the correct answer.
Step 1: Identify the given values: Principal P = 4000, Amount A = 5324, and Annual Rate R = 10%.
Step 2: Apply the compound amount formula:
A = P \left(1 + \frac{R}{100}\right)^TStep 3: Substitute the known values into the equation:
5324 = 4000 \left(1 + \frac{10}{100}\right)^TStep 4: Divide both sides by 4000 and simplify the fraction:
\frac{5324}{4000} = (1.1)^T \implies \frac{1331}{1000} = (1.1)^T
\left(\frac{11}{10}\right)^3 = (1.1)^T \implies (1.1)^3 = (1.1)^T
Step 5: Equating the powers, we get T = 3 years.
Thus, the time required is 3 years, making option B the correct answer.
Step 1: Identify the formula for depreciation where value decreases over time:
A = P \left(1 - \frac{R}{100}\right)^TStep 2: Given that the present value (A) is ₹72,900, the rate of depreciation (R) is 10%, and the time (T) is 3 years, substitute these values into the formula to find the past value (P):
72900 = P \left(1 - \frac{10}{100}\right)^3Step 3: Simplify the term inside the bracket:
72900 = P \times (0.9)^3 = P \times 0.729Step 4: Solve for the past value P:
P = \frac{72900}{0.729} = 100000Thus, the value of the machine 3 years ago was ₹1,00,000, making option B the correct answer.
Step 1: Understand that the interest accrued in 1 year (from the 3rd year to the 4th year) acts as simple interest on the amount at the end of the 3rd year.
Step 2: Calculate the interest for 1 year:
\text{Interest for 1 year} = ₹840 - ₹800 = ₹40Step 3: This ₹40 is the interest on the amount after 3 years (₹800) for 1 year at the rate R:
40 = \frac{800 \times R \times 1}{100}Step 4: Solve for the rate R:
40 = 8R \implies R = \frac{40}{8} = 5\%Thus, the rate of interest per annum is 5%, making option B the correct answer.
Step 1: Identify the relationship between compound interest (CI), principal P, and the annual rate R for 2 years:
CI = P \left[\left(1 + \frac{R}{100}\right)^T - 1\right]Step 2: Substitute the given values where CI = 102, R = 4\%, and T = 2 years into the formula:
102 = P \left[\left(1 + \frac{4}{100}\right)^2 - 1\right]Step 3: Simplify the term inside the bracket:
102 = P \left[(1.04)^2 - 1\right] = P [1.0816 - 1] = P \times 0.0816Step 4: Solve for the principal P:
P = \frac{102}{0.0816} = 1250Step 5: Now, calculate the simple interest (SI) on this principal for 2 years at 4% per annum using the simple interest formula:
SI = \frac{P \times R \times T}{100} = \frac{1250 \times 4 \times 2}{100} = \frac{10000}{100} = 100Thus, the simple interest is ₹100, making option C the correct answer.
Step 1: Let the amount after 2 years be A_2 = 1352 and after 3 years be A_3 = 1406.08. The growth factor for 1 year is the ratio of A_3 to A_2:
1 + \frac{R}{100} = \frac{A_3}{A_2} = \frac{1406.08}{1352} = 1.04Step 2: Solve for the rate of interest R:
\frac{R}{100} = 1.04 - 1 = 0.04
R = 0.04 \times 100 = 4\%
Step 3: Use the amount formula for 2 years to find the principal P:
A_2 = P \left(1 + \frac{R}{100}\right)^2Step 4: Substitute the known values into the equation:
1352 = P \left(1 + \frac{4}{100}\right)^2 = P \times (1.04)^2 = P \times 1.0816Step 5: Solve for P:
P = \frac{1352}{1.0816} = 1250Thus, the sum is ₹1250 and the rate is 4% per annum, making option B the correct answer.
Step 1: Identify the given values: Principal P = 8000, Annual Rate R = 15%, and Time T = 2 years 4 months.
Step 2: Convert the time into years in mixed fraction form (a\frac{b}{c}):
T = 2 \text{ years} + \frac{4}{12} \text{ years} = 2\frac{1}{3} \text{ years} \quad \left(a = 2, \frac{b}{c} = \frac{1}{3}\right)Step 3: Apply the amount formula for annual compounding with a fractional time period:
A = P \left(1 + \frac{R}{100}\right)^a \left(1 + \frac{\frac{b}{c} \times R}{100}\right)Step 4: Substitute the values into the formula:
A = 8000 \left(1 + \frac{15}{100}\right)^2 \left(1 + \frac{\frac{1}{3} \times 15}{100}\right)
A = 8000 \times (1.15)^2 \times (1 + \frac{5}{100})
A = 8000 \times 1.3225 \times 1.05 = 8000 \times 1.388625 = 11109
Step 5: Calculate the compound interest by subtracting the principal from the total amount:
CI = A - P = 11109 - 8000 = 3109Thus, the compound interest is ₹3109, making option B the correct answer.
Step 1: Let the amount after 2 years be A_2 = 6760 and after 3 years be A_3 = 7030.40. The growth factor for 1 year is the ratio of A_3 to A_2:
1 + \frac{R}{100} = \frac{A_3}{A_2} = \frac{7030.40}{6760} = 1.04Step 2: Solve for the rate of interest R:
\frac{R}{100} = 1.04 - 1 = 0.04
R = 0.04 \times 100 = 4\%
Step 3: Use the amount formula for 2 years to find the principal P:
A_2 = P \left(1 + \frac{R}{100}\right)^2Step 4: Substitute the known values into the equation:
6760 = P \left(1 + \frac{4}{100}\right)^2 = P \times (1.04)^2 = P \times 1.0816Step 5: Solve for P:
P = \frac{6760}{1.0816} = 6250Thus, the sum is ₹6,250 and the rate is 4% per annum, making option B the correct answer.
Step 1: Use the direct formula for the difference between compound interest (CI) and simple interest (SI) for 3 years:
\text{Difference} = P \left(\frac{R}{100}\right)^2 \left(3 + \frac{R}{100}\right)Step 2: Substitute the known values into the equation where Difference = ₹620 and Rate R = 10\%:
620 = P \left(\frac{10}{100}\right)^2 \left(3 + \frac{10}{100}\right)Step 3: Simplify the terms inside the parentheses:
620 = P \left(\frac{1}{10}\right)^2 \left(3 + 0.1\right)
620 = P \times \frac{1}{100} \times 3.1 = P \times 0.031
Step 4: Solve for the principal P:
P = \frac{620}{0.031} = 20000Thus, the principal is ₹20,000, making option C the correct answer.
Step 1: Identify the given values: Principal P = 1000, Amount A = 1728, and Time T = 3 years.
Step 2: Apply the compound amount formula:
A = P \left(1 + \frac{R}{100}\right)^TStep 3: Substitute the known values into the equation:
1728 = 1000 \left(1 + \frac{R}{100}\right)^3Step 4: Divide both sides by 1000 to isolate the rate expression:
\frac{1728}{1000} = \left(1 + \frac{R}{100}\right)^3
\left(\frac{12}{10}\right)^3 = \left(1 + \frac{R}{100}\right)^3 \implies \frac{12}{10} = 1 + \frac{R}{100}
Step 5: Solve for the rate R:
1.2 = 1 + \frac{R}{100} \implies \frac{R}{100} = 0.2
R = 0.2 \times 100 = 20\%
Thus, the rate of interest is 20% per annum, making option D the correct answer.
Step 1: Identify the given values and adjust for half-yearly compounding. Principal P = 10,000, Annual Rate = 20%, and Time = 1 year.
Step 2: Since interest is compounded half-yearly, the rate per half-year R is half of the annual rate, and the number of periods n is twice the time in years:
R = \frac{20\%}{2} = 10\% \text{ per half-year}
n = 1 \text{ year} \times 2 = 2 \text{ half-years}
Step 3: Apply the compound amount formula with the adjusted rate and periods:
A = P \left(1 + \frac{R}{100}\right)^nStep 4: Substitute the values into the formula:
A = 10000 \left(1 + \frac{10}{100}\right)^2 = 10000 \times (1.1)^2 = 10000 \times 1.21 = 12100Step 5: Calculate the compound interest by subtracting the principal from the total amount:
CI = A - P = 12100 - 10000 = 2100Thus, the compound interest is ₹2,100, making option B the correct answer.
Step 1: Use the direct formula for the difference between compound interest (CI) and simple interest (SI) for 2 years:
\text{Difference} = P \left(\frac{R}{100}\right)^2Step 2: Substitute the known values into the equation where Difference = ₹25 and Rate R = 5\%:
25 = P \left(\frac{5}{100}\right)^2Step 3: Simplify the fraction inside the parentheses:
25 = P \left(\frac{1}{20}\right)^2 = P \times \frac{1}{400}Step 4: Solve for the principal P:
P = 25 \times 400 = 10000Thus, the sum is ₹10,000, making option C the correct answer.
Step 1: First, use the given simple interest (SI) information to find the principal P. Given SI = ₹1,200, Rate R = 5\%, and Time T = 3 years:
SI = \frac{P \times R \times T}{100}Step 2: Substitute the known values into the simple interest formula:
1200 = \frac{P \times 5 \times 3}{100} = \frac{15P}{100}Step 3: Solve for the principal P:
P = \frac{1200 \times 100}{15} = 8000Step 4: Now, calculate the compound interest (CI) on this principal (P = 8000) for 3 years at 5% per annum using the amount formula:
A = P \left(1 + \frac{R}{100}\right)^T
A = 8000 \left(1 + \frac{5}{100}\right)^3 = 8000 \times (1.05)^3
A = 8000 \times 1.157625 = 9261
Step 5: Calculate the compound interest by subtracting the principal from the total amount:
CI = A - P = 9261 - 8000 = 1261Thus, the compound interest is ₹1,261, making option B the correct answer.
Step 1: Use the simple interest for 2 years to find the interest for 1 year. Since simple interest is equal every year:
\text{SI for 1 year} = \frac{160}{2} = ₹80Step 2: For the first year, simple interest and compound interest are equal. Therefore, the interest for the 1st year is ₹80.
Step 3: For the 2nd year, compound interest includes the interest on the principal (₹80) plus the interest on the first year’s interest at 4%:
\text{Interest on 1st year's interest} = \frac{80 \times 4 \times 1}{100} = ₹3.20Step 4: Add the 1st year’s interest and the 2nd year’s total interest to get the total compound interest for 2 years:
CI = 80 + (80 + 3.20) = 80 + 83.20 = 163.20Thus, the compound interest is ₹163.20, making option B the correct answer.
Step 1: Find the simple interest for 1 year since simple interest is equal every year:
\text{SI for 1 year} = \frac{1000}{2} = ₹500Step 2: Find the difference between compound interest (CI) and simple interest (SI) for the 2-year period:
\text{Difference} = CI - SI = 1050 - 1000 = ₹50Step 3: This difference is the interest accrued on the 1st year’s simple interest for 1 year. Calculate the rate R:
R = \frac{\text{Difference}}{\text{SI for 1 year}} \times 100 = \frac{50}{500} \times 100 = 10\%Step 4: Use the simple interest formula for 1 year to find the principal P:
500 = \frac{P \times 10 \times 1}{100} = \frac{P}{10}
P = 500 \times 10 = 5000
Thus, the principal is ₹5,000 and the rate is 10% per annum, making option B the correct answer.
Step 1: Use the formula for the difference between compound interest (CI) and simple interest (SI) for 3 years to find the principal P:
\text{Difference} = P \left(\frac{R}{100}\right)^2 \left(3 + \frac{R}{100}\right)Step 2: Substitute the known values where Difference = ₹93 and Rate R = 10\%:
93 = P \left(\frac{10}{100}\right)^2 \left(3 + \frac{10}{100}\right)
93 = P \times \frac{1}{100} \times 3.1 = P \times 0.031
Step 3: Solve for the principal P:
P = \frac{93}{0.031} = 3000Step 4: Now, calculate the simple interest on this principal (P = 3000) for 2 years at 10% per annum:
SI = \frac{3000 \times 10 \times 2}{100} = 600Thus, the simple interest for 2 years is ₹600, making option A the correct answer.
Step 1: Let the amount after 2 years be A_2 = 6050 and after 3 years be A_3 = 6655. The growth factor for 1 year is the ratio of A_3 to A_2:
1 + \frac{R}{100} = \frac{A_3}{A_2} = \frac{6655}{6050} = 1.1Step 2: Solve for the rate of interest R:
\frac{R}{100} = 1.1 - 1 = 0.1
R = 0.1 \times 100 = 10\%
Step 3: Use the amount formula for 2 years to find the principal P:
A_2 = P \left(1 + \frac{R}{100}\right)^2Step 4: Substitute the known values into the equation:
6050 = P \left(1 + \frac{10}{100}\right)^2 = P \times (1.1)^2 = P \times 1.21Step 5: Solve for P:
P = \frac{6050}{1.21} = 5000Thus, the sum is ₹5,000 and the rate is 10% per annum, making option B the correct answer.
Step 1: Use the compound amount formula A = P \left(1 + \frac{R}{100}\right)^T. Given that the sum becomes 4 times in 2 years:
4P = P \left(1 + \frac{R}{100}\right)^2Step 2: Simplify by dividing both sides by P:
\left(1 + \frac{R}{100}\right)^2 = 4Step 3: Let the time required to become 64 times be t years:
64P = P \left(1 + \frac{R}{100}\right)^t \implies \left(1 + \frac{R}{100}\right)^t = 64Step 4: Express 64 as a power of 4: 64 = 4^3 = \left(\left(1 + \frac{R}{100}\right)^2\right)^3 = \left(1 + \frac{R}{100}\right)^6
Step 5: Equating the powers, we get t = 6 years.
Thus, the sum will become 64 times of itself in 6 years, making option B the correct answer.
Step 1: Identify the given values: Principal P = 10,000, first-year rate R_1 = 10\%, and second-year rate R_2 = 20\%.
Step 2: Apply the formula for successive compound interest rates to find the total amount A:
A = P \left(1 + \frac{R_1}{100}\right) \left(1 + \frac{R_2}{100}\right)Step 3: Substitute the known values into the equation:
A = 10000 \left(1 + \frac{10}{100}\right) \left(1 + \frac{20}{100}\right)
A = 10000 \times 1.1 \times 1.2 = 10000 \times 1.32 = 13200
Step 4: Calculate the compound interest by subtracting the principal from the total amount:
CI = A - P = 13200 - 10000 = 3200Thus, the compound interest is ₹3,200, making option C the correct answer.
Step 1: Understand that the interest accrued in 1 year (from the 3rd year to the 4th year) acts as simple interest on the amount at the end of the 3rd year.
Step 2: Calculate the interest for 1 year:
\text{Interest for 1 year} = ₹2880 - ₹2400 = ₹480Step 3: This ₹480 is the interest on the amount after 3 years (₹2400) for 1 year at the rate R:
480 = \frac{2400 \times R \times 1}{100}Step 4: Solve for the rate R:
480 = 24R \implies R = \frac{480}{24} = 20\%Thus, the rate of interest per annum is 20%, making option C the correct answer.
Step 1: Identify the given values: Principal P = ₹16,000, Annual Rate = 20%, and Time = 9 months.
Step 2: Adjust the rate and time periods for quarterly compounding (compounded 4 times a year):
\text{Quarterly Rate } (R) = \frac{20\%}{4} = 5\% \text{ per quarter}
\text{Number of quarters } (n) = 9 \text{ months} = \frac{9}{12} \text{ years} \times 4 = 3 \text{ quarters}
Step 3: Apply the compound amount formula:
A = P \left(1 + \frac{R}{100}\right)^nStep 4: Substitute the known values into the formula:
A = 16000 \left(1 + \frac{5}{100}\right)^3 = 16000 \left(\frac{21}{20}\right)^3
A = 16000 \times \frac{9261}{8000} = 2 \times 9261 = 18522
Step 5: Calculate the compound interest by subtracting the principal from the total amount:
CI = A - P = 18522 - 16000 = 2522Thus, the compound interest is ₹2,522, making option C the correct answer.
Step 1: Use the direct formula for the difference between compound interest (CI) and simple interest (SI) for 2 years:
\text{Difference} = P \left(\frac{R}{100}\right)^2Step 2: Substitute the known values into the equation where Difference = ₹32 and Rate R = 8\%:
32 = P \left(\frac{8}{100}\right)^2Step 3: Simplify the fraction inside the parentheses:
32 = P \left(\frac{2}{25}\right)^2 = P \times \frac{4}{625}Step 4: Solve for the principal P:
P = \frac{32 \times 625}{4} = 8 \times 625 = 5000Thus, the sum is ₹5,000, making option B the correct answer.
Step 1: Understand the concept of compound interest installments. If x is the value of each equal annual installment and the rate is R, the present value of the loan is equal to the sum of the present values of each installment:
P = \frac{x}{1 + \frac{R}{100}} + \frac{x}{\left(1 + \frac{R}{100}\right)^2}Step 2: Substitute the known values where Principal P = 2100 and Rate R = 10\%:
2100 = \frac{x}{1 + \frac{10}{100}} + \frac{x}{\left(1 + \frac{10}{100}\right)^2}Step 3: Simplify the fractions:
2100 = \frac{x}{1.1} + \frac{x}{(1.1)^2} = \frac{x}{1.1} + \frac{x}{1.21}Step 4: Take x common and solve the algebraic expression:
2100 = x \left(\frac{10}{11} + \frac{100}{121}\right) = x \left(\frac{110 + 100}{121}\right) = x \left(\frac{210}{121}\right)Step 5: Solve for the installment x:
x = \frac{2100 \times 121}{210} = 10 \times 121 = 1210Thus, the value of each installment is ₹1,210, making option B the correct answer.
Step 1: Set up the present value formula for 3 equal annual installments where x is the installment value and R is the annual rate:
P = \frac{x}{1 + \frac{R}{100}} + \frac{x}{\left(1 + \frac{R}{100}\right)^2} + \frac{x}{\left(1 + \frac{R}{100}\right)^3}Step 2: Substitute the known values where Principal P = 3310 and Rate R = 10\%:
3310 = \frac{x}{1.1} + \frac{x}{(1.1)^2} + \frac{x}{(1.1)^3}Step 3: Convert the decimals to fractions or powers of \frac{11}{10}:
3310 = x \left(\frac{10}{11} + \frac{100}{121} + \frac{1000}{1331}\right)Step 4: Take the common denominator (1331) and sum the terms inside the parentheses:
3310 = x \left(\frac{1210 + 1100 + 1000}{1331}\right) = x \left(\frac{3310}{1331}\right)Step 5: Solve for the installment x:
x = \frac{3310 \times 1331}{3310} = 1331Thus, the value of each installment is ₹1,331, making option C the correct answer.
Step 1: Understand that population growth and depreciation problems follow the exact same multiplicative framework as successive compound interest:
P_{\text{final}} = P_{\text{initial}} \left(1 \pm \frac{R_1}{100}\right) \left(1 \pm \frac{R_2}{100}\right) \left(1 \pm \frac{R_3}{100}\right)Step 2: Substitute the given percentage changes (increase is positive, decrease is negative) into the formula where final population = 29,700:
29700 = P \left(1 + \frac{10}{100}\right) \left(1 - \frac{10}{100}\right) \left(1 + \frac{20}{100}\right)Step 3: Simplify the decimal multipliers for each year:
29700 = P \times (1.1) \times (0.9) \times (1.2)Step 4: Multiply the growth factors together:
29700 = P \times 1.188Step 5: Solve for the initial population P:
P = \frac{29700}{1.188} = 25000Thus, the initial population was 25,000, making option C the correct answer.
Step 1: Set up the compound amount formula A = P \left(1 + \frac{R}{100}\right)^T. Given that the amount A becomes \frac{27}{8}P in 3 years:
\frac{27}{8}P = P \left(1 + \frac{R}{100}\right)^3Step 2: Divide both sides by the principal P to isolate the rate expression:
\left(1 + \frac{R}{100}\right)^3 = \frac{27}{8}Step 3: Express \frac{27}{8} as a cube: \frac{27}{8} = \left(\frac{3}{2}\right)^3. Therefore:
\left(1 + \frac{R}{100}\right)^3 = \left(\frac{3}{2}\right)^3Step 4: Take the cube root on both sides:
1 + \frac{R}{100} = \frac{3}{2} = 1.5Step 5: Solve for the rate R:
\frac{R}{100} = 1.5 - 1 = 0.5
R = 0.5 \times 100 = 50\%
Thus, the rate of interest is 50% per annum, making option C the correct answer.
Step 1: Identify the given values: Principal P = ₹16,000, Amount A = ₹18,522, and Annual Rate = 10%.
Step 2: Adjust the rate for half-yearly compounding (R'):
R' = \frac{10\%}{2} = 5\% \text{ per half-year}Step 3: Set up the compound amount formula where n is the number of half-year periods:
18522 = 16000 \left(1 + \frac{5}{100}\right)^nStep 4: Divide both sides by 16000 and simplify the fraction:
\frac{18522}{16000} = \left(\frac{21}{20}\right)^n \implies \frac{9261}{8000} = \left(\frac{21}{20}\right)^n
\left(\frac{21}{20}\right)^3 = \left(\frac{21}{20}\right)^n \implies n = 3 \text{ half-years}
Step 5: Convert the number of half-years into years:
T = \frac{3}{2} = 1\frac{1}{2} \text{ years}Thus, the time period is 1 \frac{1}{2} years, making option B the correct answer.
Step 1: Let the principal be P and the growth factor for 1 year be X = 1 + \frac{R}{100}. Write the equations for the amounts after 2 years and 4 years:
A_2 = P X^2 = 2400 A_4 = P X^4 = 2880
Step 2: Divide the 4-year amount equation by the 2-year amount equation to find X^2:
\frac{A_4}{A_2} = \frac{P X^4}{P X^2} = X^2 = \frac{2880}{2400} X^2 = \frac{288}{240} = 1.2
Step 3: Substitute X^2 = 1.2 back into the 2-year amount equation (P X^2 = 2400):
P \times 1.2 = 2400Step 4: Solve for the principal P:
P = \frac{2400}{1.2} = 2000Thus, the sum is ₹2,000, making option B the correct answer.
Step 1: Recall the formulas for the difference between compound interest (CI) and simple interest (SI) for 2 years and 3 years:
\text{Difference for 2 years} = P \left(\frac{R}{100}\right)^2 \text{Difference for 3 years} = P \left(\frac{R}{100}\right)^2 \left(3 + \frac{R}{100}\right)
Step 2: Find the ratio of the 3-year difference to the 2-year difference by dividing the two expressions:
\frac{\text{Diff}_3}{\text{Diff}_2} = \frac{P \left(\frac{R}{100}\right)^2 \left(3 + \frac{R}{100}\right)}{P \left(\frac{R}{100}\right)^2} = 3 + \frac{R}{100}Step 3: Equate this ratio to the given value (\frac{31}{10} = 3.1):
3 + \frac{R}{100} = 3.1Step 4: Solve for the rate R:
\frac{R}{100} = 3.1 - 3 = 0.1 R = 0.1 \times 100 = 10\%
Thus, the rate of interest is 10% per annum, making option C the correct answer.
Step 1: Understand the formula for the total sum P when given equal annual installments x and rate R:
P = \frac{x}{1 + \frac{R}{100}} + \frac{x}{\left(1 + \frac{R}{100}\right)^2}Step 2: Substitute the known values where installment x = 6760 and Rate R = 4\%:
P = \frac{6760}{1 + \frac{4}{100}} + \frac{6760}{\left(1 + \frac{4}{100}\right)^2}Step 3: Simplify the fractions 1 + \frac{4}{100} = \frac{26}{25}:
P = \frac{6760}{\frac{26}{25}} + \frac{6760}{\left(\frac{26}{25}\right)^2} = 6760 \times \frac{25}{26} + 6760 \times \frac{625}{676}Step 4: Perform the division and multiplication:
P = 260 \times 25 + 10 \times 625 = 6500 + 6250 = 12500Thus, the sum borrowed is ₹12,500, making option B the correct answer.
Step 1: Calculate the simple interest (SI) for 1.5 years at 10% per annum on P = ₹8,000:
SI = \frac{P \times R \times T}{100} = \frac{8000 \times 10 \times 1.5}{100} = 1200Step 2: Calculate the compound interest (CI) for 1.5 years at 10% per annum, compounded half-yearly. Adjust the rate and time periods:
\text{Half-yearly Rate } (R') = \frac{10\%}{2} = 5\% \text{ per half-year}
\text{Number of half-years } (n) = 1.5 \text{ years} \times 2 = 3 \text{ half-years}
Step 3: Apply the compound amount formula:
A = P \left(1 + \frac{R'}{100}\right)^n = 8000 \left(1 + \frac{5}{100}\right)^3 = 8000 \times (1.05)^3
A = 8000 \times 1.157625 = 9261
Step 4: Find the compound interest by subtracting the principal from the total amount:
CI = A - P = 9261 - 8000 = 1261Step 5: Find the difference between compound interest and simple interest:
\text{Difference} = CI - SI = 1261 - 1200 = 61Thus, the difference is ₹61, making option A the correct answer (with the correct answer successfully rotated to A as requested!).
Step 1: Set up the compound growth relation. Given that the sum becomes 8 times in 3 years:
8P = P \left(1 + \frac{R}{100}\right)^3 \implies \left(1 + \frac{R}{100}\right)^3 = 8Step 2: Express 8 as a power of 2, or find the growth factor for 1 year by taking the cube root on both sides:
1 + \frac{R}{100} = 2Step 3: Let the time required to become 16 times be t years:
16P = P \left(1 + \frac{R}{100}\right)^t \implies \left(1 + \frac{R}{100}\right)^t = 16Step 4: Substitute the 1-year growth factor \left(1 + \frac{R}{100}\right) = 2 into the equation:
2^t = 16 = 2^4Step 5: Equate the exponents to find t:
t = 4 \text{ years}Thus, the sum will become 16 times of itself in 4 years, making option B the correct answer.
Step 1: Set up the compound growth relation. Given that the sum becomes 2 times in 5 years:
2P = P \left(1 + \frac{R}{100}\right)^5 \implies \left(1 + \frac{R}{100}\right)^5 = 2Step 2: Let the time required to become 8 times be t years:
8P = P \left(1 + \frac{R}{100}\right)^t \implies \left(1 + \frac{R}{100}\right)^t = 8Step 3: Express 8 as a power of 2: 8 = 2^3. Substitute 2 = \left(1 + \frac{R}{100}\right)^5 into the equation:
\left(1 + \frac{R}{100}\right)^t = \left(\left(1 + \frac{R}{100}\right)^5\right)^3 = \left(1 + \frac{R}{100}\right)^{15}Step 4: Equate the exponents to find t:
t = 15 \text{ years}Thus, the sum will become 8 times of itself in 15 years, making option C the correct answer (with the correct answer rotated to C!).
Step 1: Calculate the total amount due at the end of the 1st year including 10% interest on the principal ₹10,000:
\text{Amount after 1 year} = 10000 \times \left(1 + \frac{10}{100}\right) = 10000 \times 1.1 = ₹11,000Step 2: Subtract the first year’s repayment of ₹3,000 to find the principal balance for the 2nd year:
\text{Balance for 2nd year} = 11000 - 3000 = ₹8,000Step 3: Calculate the total amount due at the end of the 2nd year including 10% interest on the remaining ₹8,000:
\text{Amount after 2 years} = 8000 \times 1.1 = ₹8,800Step 4: Subtract the second year’s repayment of ₹3,000 to find the final remaining amount to clear the debt:
\text{Final remaining amount} = 8800 - 3000 = ₹5,800Thus, the remaining amount to clear the debt is ₹5,800.00, making option D the correct answer (with the correct answer successfully rotated to D!).