Compound Amount and Interest

Question 1

What will be the compound interest and compound amount of Rs. 2,000 at the interest rate of 12% p.a. in 2 years? Find the compound interest without using formula.

Solution:

Given information:

  • Principal (P1)=Rs. 2,000(P_1) = \text{Rs. } 2,000
  • Rate of interest (R)=12%(R) = 12\% p.a.
  • Time (T)=2(T) = 2 years

Step 1: Calculate interest for the first year

At the end of the first year, simple interest (I1)=P1TR100(I_1) = \frac{P_1TR}{100}

I1=2000×1×12100=Rs. 240I_1 = \frac{2000 \times 1 \times 12}{100} = \text{Rs. } 240

Step 2: Find principal for the second year

Principal for the second year (P2)=(P_2) = Amount at the end of the first year

P2=P1+I1=Rs. (2000+240)=Rs. 2,240P_2 = P_1 + I_1 = \text{Rs. } (2000 + 240) = \text{Rs. } 2,240

Step 3: Calculate interest for the second year

Again, simple interest for the second year (I2)=P2TR100(I_2) = \frac{P_2TR}{100}

I2=2240×1×12100=Rs. 268.8I_2 = \frac{2240 \times 1 \times 12}{100} = \text{Rs. } 268.8

Step 4: Calculate total compound interest

Thus, the compound interest at the end of 2 years:

CI=I1+I2=240+268.8=Rs. 508.8\text{CI} = I_1 + I_2 = 240 + 268.8 = \text{Rs. } 508.8

Step 5: Calculate compound amount

Compound amount (CA)=P1+CI(CA) = P_1 + \text{CI}

CA=2000+508.8=Rs. 2,508.8CA = 2000 + 508.8 = \text{Rs. } 2,508.8

Final Answers:

  • Compound Interest = Rs. 508.8
  • Compound Amount = Rs. 2,508.8

Verification:

We can verify this by checking that the compound amount equals the principal plus all accumulated interest:

2,000+240+268.8=2,508.82,000 + 240 + 268.8 = 2,508.8

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Question 2

Find the compound interest and compound amount of the borrowed amount of Rs. 25,000 which is paid in exactly 3 years at the rate of yearly compound interest rate 12%.

Solution:

Given information:

  • Principal (P)=Rs. 25,000(P) = \text{Rs. } 25,000
  • Rate of interest (R)=12%(R) = 12\% per year
  • Time (T)=3(T) = 3 years

Finding Compound Interest and Compound Amount:

According to the formula,

Compound interest (CI)=P[(1+R100)T1](CI) = P\left[\left(1 + \frac{R}{100}\right)^T - 1\right]

Substitute the given values:

CI=25,000[(1+12100)31]CI = 25,000\left[\left(1 + \frac{12}{100}\right)^3 - 1\right]

Simplify step by step:

CI=25,000[(112100)31]CI = 25,000\left[\left(\frac{112}{100}\right)^3 - 1\right]
CI=25,000[(1.12)31]CI = 25,000\left[(1.12)^3 - 1\right]
CI=25,000[1.4049281]CI = 25,000[1.404928 - 1]
CI=25,000×0.404928CI = 25,000 \times 0.404928
CI=Rs. 10,123.20CI = \text{Rs. } 10,123.20

Finding Compound Amount:

Again, compound amount (CA)=P+CI(CA) = P + CI

CA=Rs. 25,000+Rs. 10,123.20=Rs. 35,123.20CA = \text{Rs. } 25,000 + \text{Rs. } 10,123.20 = \text{Rs. } 35,123.20

Final Answers:

  • Compound Interest = Rs. 10,123.20
  • Compound Amount = Rs. 35,123.20

Verification:

We can verify using the compound amount formula directly:

CA=P(1+R100)T=25,000×(1.12)3=25,000×1.404928=Rs. 35,123.20CA = P\left(1 + \frac{R}{100}\right)^T = 25,000 \times (1.12)^3 = 25,000 \times 1.404928 = \text{Rs. } 35,123.20
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Question 3

A man borrowed Rs. 32,000 from his friend at the rate of simple interest of 12.5% per annum. He lent the whole sum to a shopkeeper at the same rate of compound interest. How much more money will he get in 3 years? Find.

Solution:

Given information:

  • Principal (P)=Rs. 32,000(P) = \text{Rs. } 32,000
  • Rate of interest (R)=12.5%(R) = 12.5\%
  • Time (T)=3(T) = 3 years

We need to find:

  • Simple interest (SI)=?(SI) = ?
  • Compound interest (CI)=?(CI) = ?

Case I: Simple Interest

Using the simple interest formula:

SI=PTR100=32000×3×12.5100=Rs. 12,000SI = \frac{PTR}{100} = \frac{32000 \times 3 \times 12.5}{100} = \text{Rs. } 12,000

Case II: Compound Interest

Using the compound interest formula:

CI=P[(1+R100)T1]CI = P\left[\left(1 + \frac{R}{100}\right)^T - 1\right]

Substitute the given values:

CI=32,000[(1+12.5100)31]CI = 32,000\left[\left(1 + \frac{12.5}{100}\right)^3 - 1\right]

Simplify step by step:

CI=32,000[(112.5100)31]CI = 32,000\left[\left(\frac{112.5}{100}\right)^3 - 1\right]
CI=32,000[(1.125)31]CI = 32,000\left[(1.125)^3 - 1\right]
CI=32,000[1.4238281251]CI = 32,000[1.423828125 - 1]
CI=32,000×0.423828125CI = 32,000 \times 0.423828125
CI=Rs. 13,562.50CI = \text{Rs. } 13,562.50

Finding the difference:

The more money received by the man = CISICI - SI

Extra money=13,562.5012,000=Rs. 1,562.50\text{Extra money} = 13,562.50 - 12,000 = \text{Rs. } 1,562.50

Final Answer:

The man will get Rs. 1,562.50 more money by lending at compound interest instead of simple interest.

Verification:

Simple Interest = Rs. 12,000

Compound Interest = Rs. 13,562.50

Difference = Rs. 13,562.50 - Rs. 12,000 = Rs. 1,562.50 ✓

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Question 4

Sameer decided to invest Rs. 5,000 at the rate of 8% per annum for 2 years. For this, he has two safe alternatives. The first alternative is to get half yearly compound interest and the second alternative is to get yearly compound interest. If you were to suggest him, which alternative would you suggest? Write with reason.

Solution:

Given information:

  • Principal (P)=Rs. 5,000(P) = \text{Rs. } 5,000
  • Rate of interest (R)=8%(R) = 8\%
  • Time (T)=2(T) = 2 years

a) According to the first alternative:

Half yearly compound interest (CI1)=P[(1+R200)2T1](CI_1) = P\left[\left(1 + \frac{R}{200}\right)^{2T} - 1\right]

Substitute the given values:

CI1=5,000[(1+8200)2×21]CI_1 = 5,000\left[\left(1 + \frac{8}{200}\right)^{2 \times 2} - 1\right]

Simplify step by step:

CI1=5,000[(208200)2×21]CI_1 = 5,000\left[\left(\frac{208}{200}\right)^{2 \times 2} - 1\right]
CI1=5,000[(1.04)41]CI_1 = 5,000\left[(1.04)^4 - 1\right]
CI1=5,000[1.169858561]CI_1 = 5,000[1.16985856 - 1]
CI1=5,000×0.16985856CI_1 = 5,000 \times 0.16985856
CI1=Rs. 849.29CI_1 = \text{Rs. } 849.29

b) According to the second alternative:

Yearly compound interest (CI2)=P[(1+R100)T1](CI_2) = P\left[\left(1 + \frac{R}{100}\right)^T - 1\right]

Substitute the given values:

CI2=5000[(1+8100)21]CI_2 = 5000\left[\left(1 + \frac{8}{100}\right)^2 - 1\right]

Simplify step by step:

CI2=5000[(108100)21]CI_2 = 5000\left[\left(\frac{108}{100}\right)^2 - 1\right]
CI2=5000[(1.08)21]CI_2 = 5000\left[(1.08)^2 - 1\right]
CI2=5000[1.16641]CI_2 = 5000[1.1664 - 1]
CI2=5000×0.1664CI_2 = 5000 \times 0.1664 CI2=Rs. 832CI_2 = \text{Rs. } 832

Finding the difference:

The difference between half yearly compound interest and yearly compound interest is given by:

CI1CI2=Rs. 849.29Rs. 832=Rs. 17.29CI_1 - CI_2 = \text{Rs. } 849.29 - \text{Rs. } 832 = \text{Rs. } 17.29

Conclusion:

Since half yearly compound interest is Rs. 17.29 more than yearly compound interest, I would suggest him to invest in the first alternative (half-yearly compounding).

Reason: When interest is compounded more frequently (half-yearly vs yearly), the effective return is higher because interest is calculated and added to the principal more often, leading to a compounding effect on the interest itself.

Summary:

  • Half-yearly compound interest = Rs. 849.29
  • Yearly compound interest = Rs. 832.00
  • Extra benefit with half-yearly = Rs. 17.29
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Question 5

A twelve-grade student invests Rs. 10,000 for 2 years at the rate of yearly compound interest. If the compound interest in 1 year is Rs. 1,200

  1. Find the rate of yearly compound interest.
  2. Find the yearly compound amount at the end of the second year.

Solution:

(a) Finding the rate of yearly compound interest:

Given information:

  • Principal (P)=Rs. 10,000(P) = \text{Rs. } 10,000
  • Compound interest in 1 year = Rs. 1,200
  • Time (T1)=1(T_1) = 1 year

Compound amount at the end of the first year:

CA=P+CI=10,000+1,200=Rs. 11,200CA = P + CI = 10,000 + 1,200 = \text{Rs. } 11,200

Using the compound amount formula:

CA=P(1+R100)TCA = P\left(1 + \frac{R}{100}\right)^T

For the first year (T = 1):

11,200=10,000(1+R100)111,200 = 10,000\left(1 + \frac{R}{100}\right)^1

Simplify step by step:

11,20010,000=1+R100\frac{11,200}{10,000} = 1 + \frac{R}{100}
1.12=1+R1001.12 = 1 + \frac{R}{100}
R100=1.121=0.12\frac{R}{100} = 1.12 - 1 = 0.12
R=0.12×100=12%R = 0.12 \times 100 = 12\%

Thus, the rate of yearly compound interest is 12%.

(b) Finding the compound amount at the end of second year:

Now we know:

  • Principal (P)=Rs. 10,000(P) = \text{Rs. } 10,000
  • Rate (R)=12%(R) = 12\%
  • Time (T)=2(T) = 2 years

Using the compound amount formula:

CA=P(1+R100)TCA = P\left(1 + \frac{R}{100}\right)^T

Substitute the values:

CA=10,000(1+12100)2CA = 10,000\left(1 + \frac{12}{100}\right)^2 CA=10,000(112100)2CA = 10,000\left(\frac{112}{100}\right)^2 CA=10,000(1.12)2CA = 10,000(1.12)^2 CA=10,000×1.2544CA = 10,000 \times 1.2544 CA=Rs. 12,544CA = \text{Rs. } 12,544

Thus, the compound amount at the end of the second year = Rs. 12,544

Final Answers:

  • (a) Rate of yearly compound interest = 12%
  • (b) Compound amount after 2 years = Rs. 12,544

Verification:

Compound interest for 2 years = Rs. 12,544 - Rs. 10,000 = Rs. 2,544

We can verify: First year CI = Rs. 1,200, Second year CI = Rs. 1,344

Total CI = Rs. 1,200 + Rs. 1,344 = Rs. 2,544 ✓

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