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# How To Solve Compound Interest Questions Quickly

**Solve Compound Interest Problems Quickly**

Go through the page completely to know How To Solve Compound Interest Quickly.

### Definition of Compound Interest

- Compound interest is the interest calculated on the original principal and on the accumulated past interest of a deposit or loan.
- Compound interest calculated by multiplying the original principal amount one plus the annual interest rate raised to the number of compound periods minus one.
- Basic Formula for the Compound Interest ,

**A= P(1 +\mathbf{\frac{r}{n}}) ^{nt}**

Here, A = Amount

P = Principal

r = Interest rate(decimal)

n = number of times interest is compounded per unit ‘t’

t = total time

**Type 1: Problems on ****Compound Interest ****(Yearly, Quarterly, and Half-yearly)**

**Question 1 . If a sum of Rs.100 is invested for 10% p.a at CI then the sum of the amount will be Rs.121 in**

**Options: **

**A. 2 years**

**B.1 year**

**C.1.5 years**

**D. 3 years**

**Solution: **We know that, **Amount** =**P**(1+ \mathbf{\frac{r}{100}})^{t}

121 = 100 (1 + \frac{10}{100})^t

(\frac{121}{100}) = (\frac{11}{10})^t

(\frac{11}{10})^2= (\frac{11}{10})^t

t = 2 years

**Correct option: A**

**Question 2. Find the CI on Rs. 10,000 in 2 years at 2 % per annum, the interest being compounded half-yearly.**

**Options: **

**A. 408**

**B. 406.04**

**C. 409.03**

**D. 405.50**

**Solution: **According to the question

P =10000, r = 2%, t = 2

We know that,

**Amount** = **P** \mathbf{(1+ \frac{{\frac{r}{2}}}{100})^{2t}}^{}

A = 10000 \mathbf{(1+ \frac{{\frac{2}{2}}}{100})^{2*2}}

A = 10000 \mathbf{(1+ \frac{1}{100})^{4}}

A = 10000 (1+ 0.01)^{4}

A = 10000 (1.01)^{4}

A = 10000 (1.04)

A = 10406.04

Now, CI = A – P

CI = 10406.04 – 10000

CI = 406.04

**Correct option: B**

**Question 3. Find the CI on Rs. 2000 in 9 months at 12% p.a. if the interest is calculated quarterly. **

**Options: **

**A. 251.01**

**B. 251**

**C. 304**

**D.185.45**

**Solution: **According to the question,

P = 2000, r = 6%, t = 9 months (3 quarter)

We know that,

Amount = P (1+ \frac{{\frac{r}{4}}}{100})^{4t}

A = 2000 (1+ \frac{{\frac{12}{4}}}{100})^{4*\frac{9}{12}}

A = 2000 (1+ \frac{{\frac{12}{4}}}{100})^{4*\frac{3}{4}}

A = 2000 (1+ \frac{3}{100})^{3}

A = 2000 (1+0.03)^{3}

A = 2000 (1.03)^{ 3}

A = 2000 (1.09)

A = 2185.45

Now, CI = A – P

CI = 2185.45– 2000 = 185.45

**Correct option: D**

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### Type 2: When Rate of Interest, Time period, Principal are given

**Question 1. The CI on Rs. 20,000 at 6% per annum is Rs. 2472. Find the period (in years):**

**Options:**** **

**A. 2**

**B. 4**

**C. 5**

**D. 3**

**Solution: **Amount = 20000 + 2472 = Rs. 22472

Let the time = n years

So, 20000 (1+ \frac{6}{100})^{n} = 22472

(\frac{106}{100})^{n} = (\frac{22472}{20000}) = (\frac{11296}{10000}) = (\frac{106}{100})^{2}

Therefore, n = 2 years

**Correct option: A**

**Question 2. The principal amount is put on CI for two years at 40%. It gets 964 more if the interest is payable half yearly. Calculate the sum.**

**Options:**

**A. Rs 8485****B. Rs 8485.91****C. Rs 8480****D. Rs 8455.91**

**Solution: **Let us assume the Principal as Rs. 100

When compounded annually

A = 100 (1+ \frac{40}{100})^{2}

A= 196

When compounded half yearly

A = 100 (1+ \frac{\frac{40}{2}}{100})^{4}

A = 100 (1+ \frac{20}{100})^{4}

A = 207.36

Difference, 207.36 – 196 = 11.36

If difference is 11.36, then Principal = Rs 100

If difference is 964, then Principal = ( \frac{100}{11.36}) × 964

P = Rs 8485.91

**Correct option: B**

**Question 3. In what time the CI on Rs 800 at 30% pa will amount to Rs.1352 if calculated annually?**

**Options:**

**A. 3 years****B. 1.6 years****C. 2 years****D. 5 years**

**Solution: **We know that,

Amount = P (1+ \frac{r}{100})^{t}

1352 = 800 (1+ \frac{30}{100})^{t}

(\frac{1352}{800})= (1.3)^{t}

1.69 = (1.3)^{t}

1.69 = (1.3)^{2}

Therefore, t = 2 years

**Correct option: C**

### Type 3: When difference of compound and simple interest is given

**Question 1. If the difference between compound interest and simple interest on a certain principal amount is Rs 500 at 10% p.a. for 2 years. Then calculate the sum. **

**Options:**

**A. 45000****B. 50000****C. 55000****D. 5000**

**Solution: **Given the difference between SI and CI = 500

Time = 2 years

When the difference between SI and CI is of **two years** then,

**Difference = P \mathbf{ \frac{(R)^{2}}{(100)^{2}}}**

Where P = principal amount, R = rate of interest

According to the question:

500 = P \frac{(10)^{2}}{(100)^{2}}

P = 50,000

**Correct option: B**

**Question 2. The difference between CI and SI on an amount of Rs. 15,000 for 2 years is Rs. 96. Find the rate of interest?**

**Options:**

**A. 8%****B. 5%****C. 4%****D. 3%**

**Solution: Difference =** **P × **\mathbf{ \frac{(R)^{2}}{(100)^{2}}}

96 =15000 **×** \frac{(R)^{2}}{10000}

\frac{96 ×10000 }{15000} =R^{2}

R^{2} = 64

R = 8%

**Correct option: A**

**Question 3. The difference between CI and SI calculated annually on a certain amount of money for two years at 4% pa is Re. 1. Find the sum:**

**Options:**

**A. 600****B. 700****C. 650****D. 625**

**Solution:** When the difference between SI and CI is of two years then,

**Difference =P × \mathbf{ \frac{(R)^{2}}{(100)^2}} **

Where P = principal amount, R = rate of interest

1 = P **×** \frac{(4)^{2}}{(100)^2}

P = 625

**Correct option: D**

### Type 4: Problem related to sum of money becomes x time in ‘a’ years and y times in ‘b’ years.

**Question 1. A sum of money borrowed under CI gets double in 5 years. When will it become eight times of itself if the rate of interest remains same?**

**Options:**

**A. 15 years****B. 20 years****C. 10 years****D. 5 years**

**Solution: **We know that,

**Correct option: A**

**Question 2. If a certain amount of sum becomes 9 times in 2 years at compound interest, then find out the rate of interest?**

**Options:**

**A. 250%****B. 100%****C. 200%****D. 20%**

**Solution: **We know that,

r = 100 ((\frac{A}{P})^\frac{1}{t} – 1)

Therefore,

r = 100((\frac{9P}{P})^\frac{1}{2} – 1)

r =100 (9^\frac{1}{2} – 1)

r = 100 (3-1)= 100*2

r = 200%

**Correct option: C**

**Question 3. At what rate percentage will certain amount of money become 8 times in three years? **

**Options:**

**A. 113. 79%****B. 100%****C. 110. 79%****D. 130%**

**Solution:** We know that

r = 100 ((\frac{A}{P})^\frac{1}{t} – 1)

r = 100 ((8)^\frac{1}{3} – 1)

r = 100 * (2-1)

r = 100%

**Correct option: B**

### Type 5: How To Solve Compound Interest Quickly (When rates are different for different years)

**Question 1. Ravi took an amount of Rs.20000 as loan at CI charging 5%, 10% and 20% for the 1st year, 2nd year, and 3rd year respectively. Find out the total interest to be paid by Ravi at the end of the 3rd year?**

**Options:**

**A. Rs. 7270**

**B. Rs. 2270**

**C. Rs. 7720**

**D. Rs. 7027**

**Solution: **According to the question,

P = 20000

R = 5%, 10%, and 20%

T= 1, 2, and 3 years

Amount = 20000 (1+\frac{5}{100}) ( 1+\frac{10}{100})(1+\frac{20}{100})

Amount = 20000 (\frac{105}{100}) (\frac{110}{100})(\frac{120}{100})

Amount = (20000) (1.05) (1.1) ( 1.2)

Amount = 27720

We know that, CI = A – P

CI = 27720 – 20000

CI = Rs. 7720

**Correct option: C**

**Question 2. If the rate of CI for the 1 ^{st} year is 8%, 2^{nd} year is 10%, and for 3^{rd} year is 15% then find the amount and the CI on Rs 10000 in three years.**

**Options:**

**A. 3626**

**B. 6236**

**C. 3666**

**D. 3662**

**Solution: **We know that, Amount = P (1 + \frac{r_{1}}{100}) (1 + \frac{r_{2}}{100}) (1 + \frac{r_{3}}{100})

Amount = 10000 (1+\frac{8}{100}) (1+\frac{10}{100}) (1+\frac{15}{100})

Amount = 10000 (1.08) (1.1) (1.15)

Amount = 13662

Now, CI = amount –principal

CI = 13662 – 10000

CI = 3662

**Correct option: D**

**Question 3. Karan bought a bike of Rs. 60000 by paying cash. He borrowed the cash from his friend at rate of interest 5% for the 1st year and 15% for the 2 ^{nd} year. Find out the total amount he has to pay after 2 years to his friend.**

**Options:**

**A. Rs. 72450**

**B. Rs. 74250**

**C. Rs. 72540**

**D. Rs. 75420**

**Solution **We know that, Amount = P (1+\frac{r_{1}}{100})(1+\frac{r_{2}}{100})

Amount after two years = 60000 (1+\frac{5}{100}(1+\frac{15}{100}

Amount after two years = (60000)(1.05)(1.15)

Amount after two years = Rs. 72450

**Correct Option: A**

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