Interest

Q1: Multiple Choice Type

i. ₹5,000 is put in a bank at 5% simple interest. The amount at the end of 2 years will be:

Step 1: Use formula: \(SI = \frac{P \times R \times T}{100}\) \[ SI = \frac{5000 \times 5 \times 2}{100} = ₹500 \]Step 2: Amount = Principal + Interest = 5000 + 500 = ₹5,500
Answer: c. ₹5,500

ii. A sum of money triples itself in 20 years. The rate of interest is:

Step 1: If sum triples, Interest = 2 × Principal
Let P = ₹1, then SI = ₹2, T = 20 years
\[ SI = \frac{P \times R \times T}{100} \Rightarrow 2 = \frac{1 \times R \times 20}{100} \\ \Rightarrow 2 = \frac{20R}{100} \Rightarrow R = \frac{200}{20} = 10 \]Answer: b. 10%

iii. A sum of money earns simple interest equal to 0.5 times the sum in 10 years. The rate of interest per annum is:

Step 1: Let Principal = ₹1, then SI = 0.5 × 1 = ₹0.5
Time = 10 years
\[ SI = \frac{P \times R \times T}{100} \Rightarrow 0.5 = \frac{1 \times R \times 10}{100} \\ \Rightarrow 0.5 = \frac{10R}{100} \\ \Rightarrow R = \frac{0.5 \times 100}{10} = 5 \]Answer: c. 5%

iv. A sum of ₹600 put at 5% S.I. amounts to ₹720 in:

Step 1: SI = 720 − 600 = ₹120
\[ SI = \frac{P \times R \times T}{100} \Rightarrow 120 = \frac{600 \times 5 \times T}{100} \\ \Rightarrow 120 = 30T \Rightarrow T = \frac{120}{30} = 4 \]Answer: b. 4 years

v. Manoj invested ₹8,000 for 10 years at 10% p.a. simple interest. The amount at the end of 2 years will be:

Step 1: Only 2 years are considered here. \[ SI = \frac{8000 \times 10 \times 2}{100} = ₹1600 \\ \Rightarrow A = 8000 + 1600 = ₹9600 \]Answer: a. ₹9,600

Q2: If ₹3,750 amounts to ₹4,620 in 3 years at simple interest. Find:

i. The rate of interest.

Step 1: Given: Principal = ₹3,750, Amount = ₹4,620, Time = 3 years
Step 2: Find the Simple Interest: \[ SI = Amount – Principal = 4620 – 3750 = ₹870 \]Step 3: Use formula: \(SI = \frac{P \times R \times T}{100}\)
Substitute: \[ 870 = \frac{3750 \times R \times 3}{100} \\ \Rightarrow 870 = \frac{11250R}{100} \\ \Rightarrow 87000 = 11250R \\ \Rightarrow R = \frac{87000}{11250} = \frac{87000 \div 750}{11250 \div 750} = \frac {116}{15} = 7\frac{11}{15}\% \]Answer: The rate of interest is \(7\frac{11}{15}\) per annum.

ii. the amount of ₹7,500 in \(5\frac{1}{2}\) years at the same rate of interest.

Step 1: Use P = ₹7,500, R = \(7\frac{11}{15} = \frac{116}{15}\)%, T = \(5\frac{1}{2} = \frac{11}{2}\) years
Step 2: Use formula: \[ SI = \frac{7500 \times \frac{116}{15} \times \frac{11}{2}}{100} = \frac{7500 \times 7.73 \times 11}{200} = \frac{957000}{3000} = 3190 \]Step 3: Find the total amount: \[ A = Principal + Interest = 7500 + 3190 = ₹10,690 \]Answer: The amount is ₹10,690

Q3: A sum of money, lent out at simple interest doubles itself in 8 years. Find:

i. The rate of interest.

Step 1: Let the principal be ₹1 (for easy calculation).
Then, as it doubles, Amount = ₹2 ⇒ Interest = ₹2 − ₹1 = ₹1
Time = 8 years
Step 2: Use the formula: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow 1 = \frac{1 \times R \times 8}{100} \\ \Rightarrow 1 = \frac{8R}{100} \\ \Rightarrow R = \frac{100}{8} = 12.5 \]Answer: The rate of interest is 12.5% per annum.

ii. in how many years will the sum become triple (three times) of itself at the same rate percent?

Step 1: Let Principal = ₹1 again.
To triple ⇒ Amount = ₹3 ⇒ Interest = ₹3 − ₹1 = ₹2
Step 2: Use the formula: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow 2 = \frac{1 \times 12.5 \times T}{100} \\ \Rightarrow 2 = \frac{12.5T}{100} \\ \Rightarrow T = \frac{200}{12.5} = 16 \]Answer: The sum will triple in 16 years at the same rate.

Q4: Rupees 4,000 amounts to ₹5,000 in 8 years; in what time will ₹2,100 amount to ₹2,800 at the same rate?

Step 1: First, find the rate of interest using the first case:
Principal (P₁) = ₹4,000
Amount (A₁) = ₹5,000 ⇒ Simple Interest (SI₁) = 5000 − 4000 = ₹1,000
Time (T₁) = 8 years
Step 2: Use the SI formula to find Rate: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow 1000 = \frac{4000 \times R \times 8}{100} \\ \Rightarrow 1000 = \frac{32000R}{100} \\ \Rightarrow R = \frac{1000 \times 100}{32000} = \frac{100000}{32000} = \frac{100000 \div 4000}{32000 \div 4000} = \frac{25}{8} \]Rate of interest is \(\frac{25}{8}\)% per annum.
Step 3: Now use this rate to find time for second case:
Principal (P₂) = ₹2,100
Amount (A₂) = ₹2,800 ⇒ SI₂ = 2800 − 2100 = ₹700
Rate (R) = \(\frac{25}{8}\)%
Step 4: Use the SI formula again to find T: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow 700 = \frac{2100 \times \frac{25}{8} \times T}{100} \\ \Rightarrow 700 = \frac{525T}{8} \\ \Rightarrow T = \frac{700 \times 8}{525} = \frac{5600 \div 175}{525 \div 175} = \frac{32}{3} = 10\frac{2}{3}\text{ years} \]Step 5: Convert fraction to months: \[ \frac{2}{3} \times 12 = 8 \text{ months} \]Answer: The required time is 10 years 8 months.

Q5: What sum of money lent at 6.5% per annum will produce the same interest in 4 years as ₹7,500 produce in 6 years at 5% per annum?

Step 1: First, calculate the interest produced by ₹7,500 at 5% for 6 years:
Use formula: \(SI = \frac{P \times R \times T}{100}\) \[ SI = \frac{7500 \times 5 \times 6}{100} = \frac{225000}{100} = ₹2,250 \]Step 2: Let the required principal be ₹P. It earns ₹2,250 at 6.5% in 4 years.
Step 3: Use SI formula again: \[ 2250 = \frac{P \times 6.5 \times 4}{100} \\ \Rightarrow 2250 = \frac{26P}{100} \\ \Rightarrow 225000 = 26P \\ \Rightarrow P = \frac{225000}{26} = ₹8,653.85 \]Answer: The required sum is ₹8,653.85

Q6: A certain sum amounts to ₹3,825 in 4 years and to ₹4,050 in 6 years. Find the rate percent and the sum.

Step 1: Find the simple interest for 2 years (difference of amount between 6 years and 4 years): \[ SI = 4050 – 3825 = ₹225 \]Step 2: Time = 6 − 4 = 2 years
Step 3: Use formula \(SI = \frac{P \times R \times T}{100}\), we already know: \[ 225 = \frac{P \times R \times 2}{100} \\ \Rightarrow \frac{PR}{50} = 225 \Rightarrow PR = 225 \times 50 = 11250 \quad \text{(Equation ①)} \]Step 4: Total amount after 4 years is ₹3,825. Let the principal be P.
Then SI for 4 years = ₹3825 − P
Now use SI formula again: \[ SI = \frac{P \times R \times 4}{100} = 3825 – P \quad \text{(Equation ②)} \]Step 5: Substitute R from Equation ①: \(R = \frac{11250}{P}\) into Equation ②: \[ \frac{P \times \frac{11250}{P} \times 4}{100} = 3825 – P \\ \Rightarrow \frac{11250 \times 4}{100} = 3825 – P \\ \Rightarrow \frac{45000}{100} = 3825 – P \\ \Rightarrow 450 = 3825 – P \\ \Rightarrow P = 3825 – 450 = ₹3,375 \]Step 6: Substitute P into Equation ① to find R: \[ PR = 11250 \Rightarrow 3375 \times R = 11250 \\ \Rightarrow R = \frac{11250}{3375} = 3.33\% \]Answer: The sum is ₹3,375 and the rate of interest is 3.33% per annum.

Q7: At what rate per cent of simple interest will the interest on ₹3,750 be one-fifth of itself in 4 years? What will it amount to in 15 years?

Part 1: Find the rate percent

Step 1: Given: Principal = ₹3,750
SI = One-fifth of itself = \(\frac{1}{5} \times 3750 = ₹750\)
Time = 4 years
Step 2: Use the simple interest formula: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow 750 = \frac{3750 \times R \times 4}{100} \\ \Rightarrow 750 = \frac{15000R}{100} \\ \Rightarrow 75000 = 15000R \\ \Rightarrow R = \frac{75000}{15000} = 5 \]Answer (Part 1): The rate of interest is 5% per annum.

Part 2: Find the amount in 15 years

Step 1: Use same P = ₹3,750, R = 5%, T = 15 years
Step 2: Calculate the interest: \[ SI = \frac{3750 \times 5 \times 15}{100} = \frac{281250}{100} = ₹2,812.50 \]Step 3: Now calculate the total amount: \[ A = P + SI = 3750 + 2812.50 = ₹6,562.50 \]Answer (Part 2): The amount in 15 years is ₹6,562.50

Q8: On what date will ₹1,950 lent on 5th January 2011 amount to ₹2,125.50 at 5 percent per annum simple interest?

Step 1: Principal (P) = ₹1,950
Amount (A) = ₹2,125.50
Rate (R) = 5% per annum
Step 2: Find the Simple Interest: \[ SI = A – P = 2125.50 – 1950 = ₹175.50 \]Step 3: Use the formula \(SI = \frac{P \times R \times T}{100}\) to find Time (T): \[ 175.50 = \frac{1950 \times 5 \times T}{100} \\ \Rightarrow 17550 = 9750T \\ \Rightarrow T = \frac{17550}{9750} = \frac{17550 \div 195}{9750 \div 195} = \frac{9}{5} = 1 \frac{4}{5} \text{ years} \]Step 4: Convert fraction years to months: \[ \frac{4}{5} \text{ year} = \frac{4}{5} \times 12 = \frac{48}{5} = 9 \frac{3}{5} \text{ months} \] Step 5: Convert fraction months to days:
\[ \frac{3}{5} \text{ months} = \frac{3}{5} \times 30 = 18 \text{ days} \]Step 5: Add 1 year, 9 months, and 18 days to the date 5th January 2011:
Add 1 year ⇒ 5th January 2012
Add 9 months ⇒ 5th October 2012
Add 18 days ⇒ 23rd October 2012
Answer: The required date is 23rd October 2012.

Q9: If the interest on ₹2,400 is more than the interest on ₹2,000 by ₹60 in 3 years at the same rate per cent, find the rate.

Step 1: Let the rate of interest be \(R\%\).
Time = 3 years (same for both)
Step 2: Write interest formulas for both:
SI₁ (on ₹2,400) = \(\frac{2400 \times R \times 3}{100} = \frac{7200R}{100}\)
SI₂ (on ₹2,000) = \(\frac{2000 \times R \times 3}{100} = \frac{6000R}{100}\)
Step 3: According to the question: \[ SI₁ – SI₂ = ₹60 \\ \Rightarrow \frac{7200R}{100} – \frac{6000R}{100} = 60 \\ \Rightarrow \frac{(7200 – 6000)R}{100} = 60 \\ \Rightarrow \frac{1200R}{100} = 60 \\ \Rightarrow 12R = 60 \\ \Rightarrow R = \frac{60}{12} = 5 \]Answer: The rate of interest is 5% per annum.

Q10: Divide ₹15,600 into two parts such that the interest on one at 5 percent for 5 years may be equal to that on the other at \(4\frac{1}{2}\) per cent for 6 years.

Step 1: Let the two parts be ₹x and ₹(15600 − x)
1st part: ₹x at 5% for 5 years
2nd part: ₹(15600 − x) at 4.5% for 6 years
Step 2: Write expressions for both interests: \[ SI_1 = \frac{x \times 5 \times 5}{100} = \frac{25x}{100} \\ SI_2 = \frac{(15600 – x) \times 4.5 \times 6}{100} = \frac{(15600 – x) \times 27}{100} \]Step 3: Set both interests equal: \[ \frac{25x}{100} = \frac{(15600 – x) \times 27}{100} \\ \Rightarrow 25x = 27(15600 – x) \]Step 4: Solve the equation: \[ 25x = 27 \times 15600 – 27x \\ \Rightarrow 25x + 27x = 421200 \\ \Rightarrow 52x = 421200 \\ \Rightarrow x = \frac{421200}{52} = 8100 \]Step 5: The two parts are:
1st part = ₹8,100
2nd part = ₹15,600 − ₹8,100 = ₹7,500
Answer: The required parts are ₹8,100 and ₹7,500.

Q11: Simple interest on a certain sum is \(\frac{16}{25}\) of the sum. Find the rate of interest and time, if both are numerically equal.

Step 1: Let the Principal = ₹P
Simple Interest = \(\frac{16}{25} \times P\)
Let rate = time = x (numerically equal)
Step 2: Use the formula: \[ SI = \frac{P \times R \times T}{100} \\ \Rightarrow \frac{16}{25} \times P = \frac{P \times x \times x}{100} = \frac{Px^2}{100} \]Step 3: Cancel P from both sides (P ≠ 0): \[ \frac{16}{25} = \frac{x^2}{100} \\ \Rightarrow x^2 = \frac{16}{25} \times 100 = \frac{1600}{25} = 64 \\ \Rightarrow x = \sqrt{64} = 8 \]Answer: Rate = 8% and Time = 8 years

Q12: Divide ₹9,000 into two parts in such a way that S.I. on one part at 16% p.a. and in 2 years is equal to the S.I. on tie other part at 6% p.a. and in 3 years.

Step 1: Let the first part be ₹x.
Then, the second part = ₹(9000 − x)
Step 2: Use the SI formula on both parts:
S.I. on 1st part = \(\frac{x \times 16 \times 2}{100} = \frac{32x}{100}\)
S.I. on 2nd part = \(\frac{(9000 – x) \times 6 \times 3}{100} = \frac{18(9000 – x)}{100}\)
Step 3: Set the two simple interests equal: \[ \frac{32x}{100} = \frac{18(9000 – x)}{100} \\ \Rightarrow 32x = 18(9000 – x) \\ \Rightarrow 32x = 162000 – 18x \\ \Rightarrow 32x + 18x = 162000 \\ \Rightarrow 50x = 162000 \\ \Rightarrow x = \frac{162000}{50} = 3240 \]Step 4: Second part = 9000 − 3240 = ₹5760
Answer: The required parts are ₹3,240 and ₹5,760.