Interest

Q1: Multiple Choice Type:

i. The S.I. on a certain sum in 3 years and at 8% per year is ₹720. The sum is:

Step 1: Given:
Simple Interest (SI) = ₹720
Rate (R) = 8% per annum
Time (T) = 3 years
Step 2: Use the formula for simple interest: \[ SI = \frac{P \times R \times T}{100} \] Substitute values: \[ 720 = \frac{P \times 8 \times 3}{100} = \frac{24P}{100} \]Step 3: Solve for \(P\): \[ 24P = 720 \times 100 = 72000 \\ \Rightarrow P = \frac{72000}{24} = ₹3,000 \]Answer: c. ₹3,000

ii. A sum of money becomes \(\frac{5}{4}\) of itself in 5 years. The rate of interest is:

Step 1: Let principal be \(P\). Amount after 5 years is \(\frac{5}{4}P\).
Step 2: Using simple interest formula: \[ A = P + SI = P + \frac{P \times R \times T}{100} \\ \Rightarrow \frac{5}{4}P = P + \frac{P \times R \times 5}{100} \]Step 3: Simplify: \[ \frac{5}{4}P – P = \frac{5PR}{100} \\ \Rightarrow \frac{1}{4}P = \frac{5PR}{100} \]Cancel \(P\): \[ \frac{1}{4} = \frac{5R}{100}\\ \Rightarrow 5R = 25\\ \Rightarrow R = 5\% \]Answer: b. 5%

iii. \(\frac{3}{5}\) part of certain sum is lent at S.I. and the remaining is lent at C.I. If the rate of interest in both the cases is 20%. On the whole the total interest in 1 year is 1,000 then the sum is:

Step 1: Let sum = \(P\).
SI part = \(\frac{3}{5}P\)
CI part = \(\frac{2}{5}P\)
Rate \(R = 20\%\), Time \(T = 1\) year.
Step 2: Calculate SI on \(\frac{3}{5}P\): \[ SI = \frac{3}{5}P \times \frac{20 \times 1}{100} = \frac{3}{5}P \times 0.20 = \frac{3}{25}P \]Step 3: Calculate CI on \(\frac{2}{5}P\) for 1 year: \[ CI = \frac{2}{5}P \times \left[\left(1 + \frac{20}{100}\right)^1 – 1\right] = \frac{2}{5}P \times (1.20 – 1) = \frac{2}{5}P \times 0.20 = \frac{2}{25}P \]Step 4: Total interest = SI + CI = \( \frac{3}{25}P + \frac{2}{25}P = \frac{5}{25}P = \frac{1}{5}P \)
Given total interest = ₹1,000
Step 5: Solve for \(P\): \[ \frac{1}{5}P = 1000\\ \Rightarrow P = 5000 \]Answer: d. ₹5,000

iv. The amount, of ₹1,000 invested for 2 years at 5% per annum compounded annually is:

Step 1: Given:
Principal (P) = ₹1,000
Rate (R) = 5% per annum
Time (T) = 2 years
Step 2: Use compound interest formula: \[ A = P \times \left(1 + \frac{R}{100}\right)^T = 1000 \times (1.05)^2 = 1000 \times 1.1025 = ₹1,102.50 \]Answer: b. ₹1,102.50

v. If the interest is compounded half-yearly, the time is:

Answer: b. Double

vi. Statement 1: On a certain sum, at the same rate of interest and for the same time period, compound interest is always greater than the simple interest.
Statement 2: In compound interest, the principal remains constant for the whole time period, however the compound interest keeps increasing every year.
Which of the following options is correct?

Step 1: Analyze Statement 1:
Compound Interest and Simple Interest are always equal for 1 year.
So, Statement 1 is FALSE.
Step 2: Analyze Statement 2:
In compound interest, the **principal increases each year** because interest is added to it.
So, the **principal does not remain constant**.
Therefore, Statement 2 is FALSE.
Answer: b. Both the statements are false.

vii. Assertion (A): The simple interest on a certain sum is \(\frac{9}{16}\) of the principal. If the numbers representing the rate of interest in percent and time in years are equal, then the time for which the principal is lent out is \(7\frac{1}{2}\) years.
Reason (R): In simple interest, \(Time=\ \frac{S.I.\ \times 100}{Principal\ \times Rate}\)

Step 1: Given: \[ SI = \frac{9}{16} P, \quad R = T \quad (\text{rate and time equal}) \]Step 2: Using simple interest formula: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow \frac{9}{16} P = \frac{P \times R \times R}{100} \]Cancel \(P\): \[ \frac{9}{16} = \frac{R^2}{100}\\ \Rightarrow R^2 = \frac{9 \times 100}{16} = \frac{900}{16} \] Taking Squareroot on both sides: \[ R = \frac{30}{4} = \frac{30 \div 2}{4 \div 2} = \frac{15}{2} = 7.5 \] Since \(R = T\), time \(T = 7.5\) years.
Conclusion: Both A and R are true, and R does not correctly explains A.
Answer: b. Both A and R are correct, and R is not the correct explanation for A.

viii. Assertion (A): The simple interest on ₹15000 in 2 years at 6% p.a. is ₹1800. Then compound interest on the same sum at the same rate of interest for 2 years will never be less than ₹1800.
Reason (R): For a given principal, rate and time, both simple interest and compound interest are equal for the 1st year.

Step 1: Calculate simple interest: \[ SI = \frac{15000 \times 6 \times 2}{100} = ₹1800 \]Step 2: Compound interest in 1st year equals simple interest.
Since compound interest adds interest on interest, CI for 2 years will be more than SI.
Conclusion: Both A and R are true, but R is not the explanation for A (because A states CI will never be less than SI, which is true due to compounding, but R only states equality for 1st year).
Answer: b. Both A and R are correct, and R is not the correct explanation for A.

ix. Assertion (A) : Compound interest for the 2nd year on ₹8000 at 5% p.a. is ₹820.
Reason (R): \(Compound\ interest\ for\ 2\ years\ =\ Amount\ at\ the\ end\ of\ 2nd\ year\ -\ original\ sum\).

Step 1: Calculate compound interest for 2nd year:
Calculate amount after 1 year: \[ A_1 = 8000 \times \left(1 + \frac{5}{100}\right) = 8000 \times 1.05 = ₹8,400 \]Calculate amount after 2 years: \[ A_2 = 8400 \times 1.05 = ₹8,820 \]Compound interest for 2nd year: \[ CI_2 = A_2 – A_1 = 8820 – 8400 = ₹420 \]Given assertion states \(CI_2 = ₹820\), which is incorrect.
Reason (R) is true.
Answer: d. A is false, but R is true.

x. Assertion (A): On ₹8750 at 8% p.a., the simple interest for 1st year is equal to the simple interest for 4th year.
Reason (R): \(Simple\ interest\ for\ 4th\ year\ =\ Amount\ at\ the\ end\ of\ 4th\ year\ -\ original\ sum\).

Step 1: Simple interest is calculated on principal for every year.
So, SI for 1st year = SI for 4th year = \(\frac{8750 \times 8 \times 1}{100} = ₹700\)
Step 2: Reason R states difference between amount and original sum which is simple interest.
So, Reason R is correct.
Answer: a. Both A and R are correct, and R the correct explanation for A.

Q2: Mohan ₹4,800 to John for \(4\frac{1}{2}\) years and ₹2,500 to Shyam for 6 years and receives a total sum of ₹2.196 as interest. Find the rate per cent per annum, provided it is same in both the cases.

Step 1: Given:
Principal to John, \(P_1 = ₹4,800\)
Time for John, \(T_1 = 4\frac{1}{2} = \frac{9}{2} = 4.5\) years
Principal to Shyam, \(P_2 = ₹2,500\)
Time for Shyam, \(T_2 = 6\) years
Total interest, \(SI_{total} = ₹2,196\)
Rate of interest, \(R = ?\)
Step 2: Simple interest formula: \[ SI = \frac{P \times R \times T}{100} \]Total interest received is sum of interest from John and Shyam: \[ SI_1 + SI_2 = 2196 \\ \frac{4800 \times R \times 4.5}{100} + \frac{2500 \times R \times 6}{100} = 2196 \]Simplify: \[ \frac{4800 \times 4.5}{100} R + \frac{2500 \times 6}{100} R = 2196 \]Calculate constants: \[ \frac{21600}{100} R + \frac{15000}{100} R = 2196 \\ \Rightarrow 216 R + 150 R = 2196 \\ \Rightarrow 366 R = 2196 \]Step 3: Solve for \(R\): \[ R = \frac{2196}{366} = 6\% \]Answer: The rate of interest per annum is 6%.

Q3: John lent ₹2,550 to Mohan at 7.5 per cent per annum. If Mohan discharges the debt after 8 months by giving an old television and ₹1,422.50. find the price of the television.

Step 1: Given:
Principal \(P = ₹2,550\)
Rate of interest \(R = 7.5\%\) per annum
Time \(T = \frac{8}{12} = \frac{2}{3}\) years
Amount paid in cash = ₹1,422.50
Price of television = ?
Step 2: Calculate the simple interest for 8 months: \[ SI = \frac{P \times R \times T}{100} = \frac{2550 \times 7.5 \times \frac{2}{3}}{100} \] Calculate numerator: \[ 2550 \times 7.5 = 19125 \] Multiply by \(\frac{2}{3}\): \[ 19125 \times \frac{2}{3} = 12750 \] Divide by 100: \[ SI = \frac{12750}{100} = ₹127.50 \]Step 3: Calculate total amount to be paid: \[ A = P + SI = 2550 + 127.50 = ₹2677.50 \]Step 4: The debt is discharged by ₹1,422.50 cash and price of television.
Let price of television = \(x\)
So, \[ 1422.50 + x = 2677.50 \\ \Rightarrow x = 2677.50 – 1422.50 = ₹1,255 \]Answer: The price of the television is ₹1,255.

Q4: Divide ₹10,800 into two parts so that if one part is put at 18% per annum S.I. and the other part is put at 20% p.a. S.I. the total annual interest is ₹2,060.

Step 1: Let the first part be \(x\).
Then the second part is \(10800 – x\).
Step 2: Interest from first part at 18% for 1 year: \[ SI_1 = \frac{x \times 18 \times 1}{100} = \frac{18x}{100} = 0.18x \]Interest from second part at 20% for 1 year: \[ SI_2 = \frac{(10800 – x) \times 20 \times 1}{100} = \frac{20(10800 – x)}{100} = 0.20(10800 – x) = 2160 – 0.20x \]Step 3: Total interest is given as ₹2,060: \[ SI_1 + SI_2 = 2060 \\ \Rightarrow 0.18x + 2160 – 0.20x = 2060 \]Simplify: \[ 2160 – 0.02x = 2060 \\ \Rightarrow -0.02x = 2060 – 2160 = -100 \\ \Rightarrow 0.02x = 100 \\ \Rightarrow x = \frac{100}{0.02} = 5000 \]Step 4: Calculate second part: \[ 10800 – 5000 = 5800 \]Answer: ₹5,000 at 18% and ₹5,800 at 20%

Q5: Find the amount and the compound interest on ₹16.000 for 3 years at 5% per annum compounded annually.

Step 1: Given:
Principal \(P = ₹16,000\)
Rate of interest \(R = 5\%\) per annum
Time \(T = 3\) years
Step 2: Use compound interest formula: \[ A = P \times \left(1 + \frac{R}{100} \right)^T = 16000 \times \left(1 + \frac{5}{100} \right)^3 = 16000 \times (1.05)^3 \]Step 3: Calculate \((1.05)^3\): \[ 1.05 \times 1.05 = 1.1025 \quad \text{and} \quad 1.1025 \times 1.05 = 1.157625 \]Step 4: Calculate amount: \[ A = 16000 \times 1.157625 = ₹18,522 \] (Rounded to nearest rupee)
Step 5: Calculate compound interest: \[ CI = A – P = 18522 – 16000 = ₹2,522 \]Answer: The amount is ₹18,522 and the compound interest is ₹2,522.

Q6: Find the amount and the compound interest on ₹20,000 for \(1\frac{1}{2}\) years at 10% per annum compounded half-yearly.

Step 1: Given:
Principal \(P = ₹20,000\)
Rate \(R = 10\%\) per annum
Time \(T = 1\frac{1}{2} = \frac{3}{2} = 1.5\) years
Compounded half-yearly ⇒ \(n = 2\) times in a year
Step 2: Convert to half-yearly terms:
Half-yearly rate = \(\frac{10}{2} = 5\%\)
Number of half-years = \(1.5 \times 2 = 3\)
Step 3: Use the compound interest formula: \[ A = P \times \left(1 + \frac{r}{100} \right)^n = 20000 \times (1.05)^3 \]Step 4: Compute \( (1.05)^3 \): \[ 1.05 \times 1.05 = 1.1025 \quad \text{then} \quad 1.1025 \times 1.05 = 1.157625 \]Step 5: Calculate amount: \[ A = 20000 \times 1.157625 = ₹23,152.50 \]Step 6: Calculate compound interest: \[ CI = A – P = 23152.50 – 20000 = ₹3,152.50 \]Answer: The amount is ₹23,152.50 and the compound interest is ₹3,152.50.

Q7: Find the amount and the compound interest on ₹32,000 for 1 year at 20% per annum compounded half-yearly.

Step 1: Given:
Principal \(P = ₹32,000\)
Rate \(R = 20\%\) per annum
Time \(T = 1\) year
Compounded half-yearly ⇒ \(n = 2\) times in a year
Step 2: Convert rate and time to half-yearly terms:
Rate per half-year = \(\frac{20}{2} = 10\%\)
Number of half-years = \(1 \times 2 = 2\)
Step 3: Use the compound interest formula: \[ A = P \times \left(1 + \frac{r}{100} \right)^n = 32000 \times (1.10)^2 \]Step 4: Compute \( (1.10)^2 \): \[ 1.10 \times 1.10 = 1.21 \]Step 5: Calculate amount: \[ A = 32000 \times 1.21 = ₹38,720 \]Step 6: Calculate compound interest: \[ CI = A – P = 38720 – 32000 = ₹6,720 \]Answer: The amount is ₹38,720 and the compound interest is ₹6,720.

Q8: Find the amount and the compound interest on ₹4,000 in 2 years, if the rate of interest for first year is 10% and for the second year is 15%.

Step 1: Given:
Principal \(P = ₹4,000\)
Rate for 1st year = 10%
Rate for 2nd year = 15%
Time = 2 years
Step 2: Calculate amount after 1st year: \[ A_1 = P \times \left(1 + \frac{10}{100}\right) = 4000 \times 1.10 = ₹4,400 \]Step 3: Use ₹4,400 as new principal for 2nd year at 15%: \[ A_2 = A_1 \times \left(1 + \frac{15}{100}\right) = 4400 \times 1.15 = ₹5,060 \]Step 4: Find compound interest: \[ CI = A_2 – P = 5060 – 4000 = ₹1,060 \]Answer: The amount is ₹5,060 and the compound interest is ₹1,060.

Q9: Find the amount and the compound interest on ₹10.000 in 3 years, if the rates of interest for the successive years are 10%, 15% and 20% respectively.

Step 1: Given:
Principal \(P = ₹10,000\)
Rate for 1st year = 10%
Rate for 2nd year = 15%
Rate for 3rd year = 20%
Step 2: Calculate amount after 1st year: \[ A_1 = P \times \left(1 + \frac{10}{100}\right) = 10000 \times 1.10 = ₹11,000 \]Step 3: Calculate amount after 2nd year: \[ A_2 = A_1 \times \left(1 + \frac{15}{100}\right) = 11000 \times 1.15 = ₹12,650 \]Step 4: Calculate amount after 3rd year: \[ A_3 = A_2 \times \left(1 + \frac{20}{100}\right) = 12650 \times 1.20 = ₹15,180 \]Step 5: Find compound interest: \[ CI = A_3 – P = 15180 – 10000 = ₹5,180 \]Answer: The amount is ₹15,180 and the compound interest is ₹5,180.

Q10: A sum of money lent at simple interest amounts to ₹3,224 in 2 years and ₹4,160 in 5 years. Find the sum and the rate of interest.

Step 1: Let the principal (sum) be ₹\(P\) and rate of interest be \(R\%\).
We are given:
Amount after 2 years = ₹3,224
Amount after 5 years = ₹4,160
Step 2: Difference in time = \(5 – 2 = 3\) years
Difference in amount = \(₹4,160 – ₹3,224 = ₹936\)
So, interest for 3 years = ₹936 ⇒ Interest per year = \(\frac{936}{3} = ₹312\)
Step 3: Now, interest for 2 years = ₹312 × 2 = ₹624
So, principal = Amount after 2 years − interest for 2 years: \[ P = 3224 – 624 = ₹2,600 \]Step 4: Use simple interest formula to find rate: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow 624 = \frac{2600 \times R \times 2}{100} \\ \Rightarrow 624 = \frac{5200R}{100} \\ \Rightarrow 624 = 52R \\ \Rightarrow R = \frac{624}{52} = 12\% \]Answer: The sum is ₹2,600 and the rate of interest is 12% per annum.

Q11: At what rate percent per annum compound interest will ₹5,000 amount to ₹5,832 in 2 years?

Step 1: Given:
Principal \(P = ₹5,000\)
Amount \(A = ₹5,832\)
Time \(T = 2\) years
Rate \(R = ?\)
Step 2: Use compound interest amount formula: \[ A = P \times \left(1 + \frac{R}{100} \right)^T \\ \Rightarrow 5832 = 5000 \times \left(1 + \frac{R}{100} \right)^2 \]Step 3: Divide both sides by 5000: \[ \left(1 + \frac{R}{100} \right)^2 = \frac{5832}{5000} = 1.1664 \]Step 4: Take square root on both sides: \[ 1 + \frac{R}{100} = \sqrt{1.1664} = 1.08 \\ \Rightarrow \frac{R}{100} = 1.08 – 1 = 0.08 \\ \Rightarrow R = 0.08 \times 100 = 8\% \]Answer: The required compound interest rate is 8% per annum.

Q12: ₹16,000 invested at 10% p.a. compounded semi-annually amounts to ₹18,522. Find the time period of investment.

Step 1: Given:
Principal \(P = ₹16,000\)
Amount \(A = ₹18,522\)
Rate \(R = 10\%\) per annum compounded semi-annually ⇒ \(\frac{10}{2} = 5\%\) per half year
Let time = \(n\) half-years
Step 2: Use compound interest formula: \[ A = P \times \left(1 + \frac{r}{100} \right)^n \\ \Rightarrow 18522 = 16000 \times (1.05)^n \]Step 3: Divide both sides by 16000: \[ (1.05)^n = \frac{18522}{16000} = 1.157625 \]Step 4: Use log or known powers: \[ (1.05)^3 = 1.157625 \\ \Rightarrow n = 3 \text{ half-years} \]Step 5: Convert half-years to years: \[ \text{Time} = \frac{3}{2} = 1.5 \text{ years} \]Answer: The time period of investment is 1.5 years.