Interest

Q1: Multiple Choice Type:

i. The interest on ₹483 for 2 years at 5% per annum is:

Step 1: Use formula for Simple Interest: \(SI = \frac{P \times R \times T}{100}\)
Here, P = 483, R = 5%, T = 2 years
Step 2: Substitute values: \[ SI = \frac{483 \times 5 \times 2}{100} = \frac{4830}{100} = 48.30 \]Answer: b. ₹48.30

ii. The simple interest on ₹8,490 at 5% and 73 days is:

Step 1: Use formula: \(SI = \frac{P \times R \times T}{100 \times 365}\)
Here, P = 8490, R = 5%, T = 73 days
Step 2: Substitute values: \[ SI = \frac{8490 \times 5 \times 73}{100 \times 365} = \frac{3098850}{36500} = 84.90 \]Answer: b. ₹84.90

iii. A sum of money, put at simple interest doubles itself in 8 years. The same sum will triple itself in:

Step 1: If sum doubles in 8 years, it means interest = Principal in 8 years.
Step 2: To triple, interest must be twice the Principal.
Step 3: So if 1P is gained in 8 years, then 2P will be gained in 8 × 2 = 16 years.
Answer: a. 16 years

iv. ₹5,000 earns ₹500 as simple interest in 2 years. Then the rate of interest is:

Step 1: Use formula: \(SI = \frac{P \times R \times T}{100}\)
Substitute: SI = 500, P = 5000, T = 2
Step 2: Solve for R: \[ 500 = \frac{5000 \times R \times 2}{100} \Rightarrow 500 = \frac{10000R}{100} \Rightarrow 500 = 100R \Rightarrow R = 5 \]Answer: b. 5%

v. ₹7,000 earns ₹1400 as interest at 5% per annum. Then the time in this case is:

Step 1: Use formula: \(SI = \frac{P \times R \times T}{100}\)
Substitute: SI = 1400, P = 7000, R = 5
Step 2: Solve for T: \[ 1400 = \frac{7000 \times 5 \times T}{100} = \frac{35000T}{100} \Rightarrow 1400 = 350T \Rightarrow T = 4 \]Answer: d. 4 years

Q2: Find interest and the amount on:

i. ₹750 in 3 years 4 months at 10% per annum

Step 1: Convert time into years:
3 years 4 months = \(3 + \frac{4}{12} = \frac{40}{12} = \frac{10}{3}\) years
Step 2: Use formula: \(SI = \frac{P \times R \times T}{100}\)
P = 750, R = 10%, T = \(\frac{10}{3}\)
Step 3: Substitute the values: \[ SI = \frac{750 \times 10 \times \frac{10}{3}}{100} = \frac{75000}{300} = 250 \]Step 4: Calculate Amount:
Amount = Principal + Interest = 750 + 250 = ₹1000
Answer: Interest = ₹250, Amount = ₹1000

ii. ₹5,000 at 8% per year from 23rd December 2011 to 29th July 2012

Step 1: Find the time in days:
From 23rd Dec 2011 to 29th July 2012 = 219 days
Step 2: Convert time to years: \[ T = \frac{219}{365} \]Step 3: Use formula: \(SI = \frac{P \times R \times T}{100}\)
P = 5000, R = 8%, T = \( \frac{219}{365} \)
\[ SI = \frac{5000 \times 8 \times 219}{100 \times 365} = \frac{8760000}{36500} = ₹240
Amount = 5000 + 240 = ₹5240 \]Answer: Interest = ₹240, Amount = ₹5240

iii. ₹2,600 in 2 years 3 months at 1% per month

Step 1: Convert time into months:
2 years 3 months = \(2 \times 12 + 3 = 27\) months
Step 2: Use formula: \(SI = \frac{P \times R \times T}{100}\)
P = 2600, R = 1% per month, T = 27 months
\[ SI = \frac{2600 \times 1 \times 27}{100} = ₹702 \]Step 3: Amount = Principal + Interest = 2600 + 702 = ₹3302
Answer: Interest = ₹702, Amount = ₹3302

iv. ₹4,000 in \(1\frac{1}{3}\) years at 2 paisa per rupee per month

Step 1: Convert time to months: \[ 1\frac{1}{3} \text{ years} = \frac{4}{3} \text{ years} = \frac{4}{3} \times 12 = 16 \text{ months} \]Step 2: Rate = 2 paisa per rupee per month = \(2\%\) per month
Step 3: Use formula: \[ SI = \frac{4000 \times 2 \times 16}{100} = \frac{128000}{100} = ₹1280 \]Step 4: Amount = 4000 + 1280 = ₹5280
Answer: Interest = ₹1280, Amount = ₹5280

Q3: Rohit borrowed ₹24,000 at 7.5 percent per year. How much money will he pay at the end of 4 years to clear his debt?

Step 1: Identify the given values:
Principal (P) = ₹24,000
Rate (R) = 7.5% per annum
Time (T) = 4 years
Step 2: Use the formula for Simple Interest: \[ SI = \frac{P \times R \times T}{100} \]Step 3: Substitute the values: \[ SI = \frac{24000 \times 7.5 \times 4}{100} \\ SI = \frac{720000}{100} = ₹7,200 \]Step 4: Total amount to be paid = Principal + Interest \[ \text{Amount} = 24000 + 7200 = ₹31,200 \]Answer: Rohit will pay ₹31,200 at the end of 4 years to clear his debt.

Q4: On what principal will the simple interest be ₹7,008 in 6 years 3 months at 5% year?

Step 1: Identify the given values:
Simple Interest (SI) = ₹7,008
Rate (R) = 5% per annum
Time (T) = 6 years 3 months = \(6 + \frac{3}{12} = \frac{75}{12} = \frac{25}{4}\) years
Step 2: Use the formula: \[ SI = \frac{P \times R \times T}{100} \]Substitute the values: \[ 7008 = \frac{P \times 5 \times \frac{25}{4}}{100} \]Step 3: Simplify the equation: \[ 7008 = \frac{125P}{400} \]Step 4: Multiply both sides by 400: \[ 7008 \times 400 = 125P \\ \Rightarrow 2803200 = 125P \]Step 5: Solve for P: \[ P = \frac{2803200}{125} = ₹22,425.60 \]Answer: The required principal is ₹22,425.60

Q5: Find the principal which will amount to ₹4,000 in 4 years at 6.25% per annum.

Step 1: Identify the given values:
Amount (A) = ₹4,000
Time (T) = 4 years
Rate (R) = 6.25% per annum
Step 2: Use relation between Amount and Principal: \[ \text{Amount} = \text{Principal} + \text{Simple Interest} \\ \Rightarrow SI = A – P \]But we don’t know Principal yet. So let Principal = P
Step 3: Use Simple Interest formula: \[ SI = \frac{P \times R \times T}{100} = \frac{P \times 6.25 \times 4}{100} = \frac{25P}{100} \]Step 4: Use amount relation: \[ \text{Amount} = P + SI = P + \frac{25P}{100} = \frac{125P}{100} \]So, \[ 4000 = \frac{125P}{100} \\ \Rightarrow P = \frac{4000 \times 100}{125} = ₹3,200 \]Answer: The required principal is ₹3,200

Q6:

i. At what rate per cent per annum will ₹630 produce an interest of ₹126 in 4 years?

Step 1: Use formula: \(SI = \frac{P \times R \times T}{100}\)
Given: SI = ₹126, P = ₹630, T = 4 years
Step 2: Substitute the values: \[ 126 = \frac{630 \times R \times 4}{100} \\ \Rightarrow 126 = \frac{2520R}{100} \\ \Rightarrow 12600 = 2520R \\ \Rightarrow R = \frac{12600}{2520} = 5 \]Answer: The required rate is 5% per annum.

ii. At what rate percent per year will a sum double itself in \(6\frac{1}{4}\) years?

Step 1: If the sum doubles, then Interest = Principal (SI = P)
Let Principal = P, then SI = P
T = \(6\frac{1}{4} = \frac{25}{4}\) years
Step 2: Use formula: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow P = \frac{P \times R \times \frac{25}{4}}{100} \]Cancel P from both sides: \[ 1 = \frac{R \times 25}{400} \\ \Rightarrow R = \frac{400}{25} = 16 \]Answer: The required rate is 16% per annum.

iii. Find the rate of interest per year, if the interest charged for 8 months is 0.06 times the money borrowed.

Step 1: Let Principal = ₹100 (for simplicity)
SI = 0.06 × 100 = ₹6
Time = 8 months = \(\frac{8}{12} = \frac{2}{3}\) years
Step 2: Use formula: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow 6 = \frac{100 \times R \times \frac{2}{3}}{100} \\ \Rightarrow 6 = \frac{2R}{3} \\ \Rightarrow R = \frac{6 \times 3}{2} = 9 \]Answer: The rate of interest is 9% per annum.

Q7:

i. In how many years will ₹950 produce ₹399 as simple interest at 7%?

Step 1: Use formula: \(SI = \frac{P \times R \times T}{100}\)
Given: SI = ₹399, P = ₹950, R = 7%
Step 2: Substitute into the formula and solve for T: \[ 399 = \frac{950 \times 7 \times T}{100} \\ \Rightarrow 399 = \frac{6650T}{100} \\ \Rightarrow 39900 = 6650T \\ \Rightarrow T = \frac{39900}{6650} = 6 \]Answer: Time required is 6 years.

ii. Find the time in which 1₹200 win amount to ₹1,536 at 3.5% per year.

Step 1: Use the relation: Amount = Principal + Simple Interest
Given: A = ₹1,536, P = ₹1,200
Step 2: Find Simple Interest: \[ SI = A – P = 1536 – 1200 = ₹336 \]Step 3: Use formula: \(SI = \frac{P \times R \times T}{100}\)
Substitute: SI = 336, P = 1200, R = 3.5%
\[ 336 = \frac{1200 \times 3.5 \times T}{100} \\ \Rightarrow 336 = \frac{4200T}{100} \\ \Rightarrow 33600 = 4200T \\ \Rightarrow T = \frac{33600}{4200} = 8 \]Answer: The required time is 8 years.

Q8: The simple interest on a certain sum of money is \(\frac{3}{8}\) of sum in \(6\frac{1}{4}\) years. Find the rate percent charged.

Step 1: Let the principal be ₹1 (since SI is a fraction of the principal, we can assume this for simplicity).
Then, SI = \(\frac{3}{8}\) × 1 = \(\frac{3}{8}\)
Step 2: Convert time into improper fraction: \[ T = 6\frac{1}{4} = \frac{25}{4} \text{ years} \]Step 3: Use the formula: \(SI = \frac{P \times R \times T}{100}\)
Substitute values: \(\frac{3}{8} = \frac{1 \times R \times \frac{25}{4}}{100}\)
Step 4: Simplify the right-hand side: \[ \frac{3}{8} = \frac{25R}{400} \]Step 5: Cross-multiply to solve for R: \[ 3 \times 400 = 8 \times 25R \\ \Rightarrow 1200 = 200R \\ \Rightarrow R = \frac{1200}{200} = 6 \]Answer: The rate charged is 6% per annum.

Q9: What sum of money borrowed on 24th May will amount to ₹10,210.20 on 17th October of the same year at 5 percent per annum simple interest?

Step 1: Identify the known values:
Amount (A) = ₹10,210.20
Rate (R) = 5% per annum
Time = From 24th May to 17th October
Step 2: Calculate the number of days:
May: 7 days (from 24 to 31)
June: 30 days
July: 31 days
August: 31 days
September: 30 days
October: 17 days
Total = \(7 + 30 + 31 + 31 + 30 + 17 = 146\) days
Step 3: Convert time into years: \[ T = \frac{146}{365} = \frac{2}{5} \text{ years} \]Step 4: Let Principal be P. Use formula: \[ \text{Amount} = P + SI \\ \Rightarrow 10210.20 = P + \frac{P \times 5 \times \frac{2}{5}}{100} \]Step 5: Simplify the interest term: \[ SI = \frac{P \times 5 \times \frac{2}{5}}{100} = \frac{2P}{100} = \frac{P}{50} \\ \Rightarrow 10210.20 = P + \frac{P}{50} \]Step 6: Combine terms: \[ 10210.20 = \frac{50P + P}{50} = \frac{51P}{50} \\ \Rightarrow P = \frac{10210.20 \times 50}{51} \\ \Rightarrow P = \frac{510510}{51} = ₹10,010 \]Answer: The required sum borrowed is ₹10,010

Q10: In what time will the interest on a certain sum of money at 6% be \(\frac{5}{8}\) of itself?

Step 1: Let the principal be ₹1 (since the answer is a fraction of itself, we can assume this for simplicity).
Then, Simple Interest = \(\frac{5}{8} \times 1 = \frac{5}{8}\)
Step 2: Use the formula: \(SI = \frac{P \times R \times T}{100}\)
Substitute values: \(\frac{5}{8} = \frac{1 \times 6 \times T}{100} = \frac{6T}{100}\)
Step 3: Solve the equation: \[ \frac{5}{8} = \frac{6T}{100} \\ \Rightarrow 5 \times 100 = 6T \times 8 \\ \Rightarrow 500 = 48T \\ \Rightarrow T = \frac{500}{48} = \frac{125}{12} = 10\frac{5}{12} \text{ years} \]Answer: The required time is 10 years 5 months.

Q11: Ashok lent out ₹7,000 at 6% and ₹9,500 at 5% Find his total income from the interest in 3 years.

Step 1: Use the formula for Simple Interest: \[ SI = \frac{P \times R \times T}{100} \]Step 2: Calculate interest on ₹7,000 at 6% for 3 years: \[ SI_1 = \frac{7000 \times 6 \times 3}{100} = \frac{126000}{100} = ₹1,260 \]Step 3: Calculate interest on ₹9,500 at 5% for 3 years: \[ SI_2 = \frac{9500 \times 5 \times 3}{100} = \frac{142500}{100} = ₹1,425 \]Step 4: Add both interests: \[ \text{Total Interest} = 1260 + 1425 = ₹2,685 \]Answer: Ashok’s total income from interest in 3 years is ₹2,685.

Q12: Raj borrows ₹8,000; out of which ₹4,500 at 5% and the remaining at 6%. Find the total interest paid by him in 4 years.

Step 1: Divide the borrowed amount:
Total borrowed = ₹8,000
Part 1: ₹4,500 at 5%
Part 2: ₹8,000 − ₹4,500 = ₹3,500 at 6%
Step 2: Use formula: \(SI = \frac{P \times R \times T}{100}\)
Time (T) = 4 years
Step 3: Calculate SI for ₹4,500 at 5%: \[ SI_1 = \frac{4500 \times 5 \times 4}{100} = \frac{90000}{100} = ₹900 \]Step 4: Calculate SI for ₹3,500 at 6%: \[ SI_2 = \frac{3500 \times 6 \times 4}{100} = \frac{84000}{100} = ₹840 \]Step 5: Total interest paid: \[ SI_{total} = SI_1 + SI_2 = 900 + 840 = ₹1,740 \]Answer: Raj will pay ₹1,740 as total interest in 4 years.