Question

give some example of simple interest and compound interest

Answer 1

Problem 1. A sum of Rs. 25000 becomes Rs. 27250 at the end of 3 years when calculated at simple interest. Find the rate of interest.

Solution:

Simple interest = 27250 – 25000 = 2250

Time = 3 years.

SI = PTR / 100 → R = SI * 100 / PT

R = 2250 * 100 / 25000 * 3 → R = 3%.

Problem 2. Find the present worth of Rs. 78000 due in 4 years at 5% interest per year.

Solution:

Amount with interest after 4 years = Rs. 78000

Therefore, simple interest = 78000 – Principal.

Let the principal amount be p.

78000 – p = p*4*5/100 → p=13000

Principal = 78000 – 13000 = Rs. 65000

Problem 3. A certain principal amounts to Rs. 15000 in 2.5 years and to Rs. 16500 in 4 years at the same rate of interest. Find the rate of interest.

Solution:

Amount becomes 15000 in 2.5 years and 16500 in 4 years.

Simple interest for (4-2.5) years = 16500 – 15000

Therefore, SI for 1.5 years = Rs. 1500.

SI for 2.5 years = 1500/1.5 * 2.5 = 2500

Principal amount = 15000 – 2500 = Rs. 12500.

Rate of Interest = 2500 * 100 / 12500 * 2.5 → R = 8%.

Problem 4. Find the compound interest on Rs. 3000 at 5% for 2 years, compounded annually.

Solution:

Amount with CI = 3000 (1+ 5/100)2 = Rs. 3307.5

Therefore, CI = 3307.5 – 3000 = Rs. 307.5

Problem 5. Find the compound interest on Rs. 10000 at 12% rate of interest for 1 year, compounded half-yearly.

Solution:

Amount with CI = 10000 [1+ (12/2 * 100)]2 = Rs. 11236

Therefore, CI = 11236 – 10000 = Rs. 1236

Problem 6. The difference between SI and CI compounded annually on a certain sum of money for 2 years at 8% per annum is Rs. 12.80. Find the principal.

Solution:

Let the principal amount be x.

SI = x * 2 * 8 / 100 = 4x/25

CI = x[1+ 8/100]2 – x → 104x/625

Therefore, 104x/625 – 4x/25 = 12.80

Solving which gives x, Principal = Rs. 2000.

Problem 7. Find the simple interest on Rs. 5000 at a certain rate if the compound interest on the same amount for 2 years is Rs. 253.125.

Solution:

Let the rate of interest be r.

5000[1+ r/100]2 = 5000+253.125

→ [1+r/100]2 = 5253.125/5000

Solving which gives

[1+ r/100]2 = 1681/1600

→ 1+r/100 = 41/40

→ r = 2.5

Therefore, SI = 5000* 2 * 2.5/ 100 = Rs. 250.

Problem 8. A certain amount becomes Rs. 5760 in 2 years and Rs. 6912 in 3 years. What is the principal amount and the rate of interest?

Solution:

SI on Rs. 5760 for 1 year = 6912 – 5760 = Rs. 1152

Therefore, Rate of interest for 1 year = 100*1152/5760*1 = 20%

Let the principal be p.

Then, Principal = p[1+ 20/100]2 = 5760

Solving which gives Principal = Rs. 4000

Problem 9. How long will it take a certain amount to increase by 30% at the rate of 15% simple interest?

Solution:

Let the principal be Rs. x

Simple interest = x*30/100 = 3x/10

T = 100*SI/PR = 100*3x/10 / x*15 = 2%

Alternatively, this can be solved by considering principal amount to be Rs. 100. Then simple interest becomes Rs. 30.

Then, T = 100*30/100*15 = 2%




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Answer 2

1.   Alina takes a loan of $8,000 to buy a used truck at the rate of 9% simple interest .Find the annual interest to be paid for the loan.

SOLUTION  = From the detail given in the problem
Principle=P=$8,000 and R=9%or0.09expressed as decimal

As the annual Interest has to be calculated ,the time period=1
Enter the value in the formula-

I=p×r×t

8,000×0.09×1

=720.00