Question

Q. 8. Find the amount and the
compound interest on Rs. 10,000 for 1
years at 10% per annum, compounded half
yearly. Would this interest be more than
the interest he would get if it was
compounded annually ?
​

Answer 1

SOLUTION:-

Given,

Principal= Rs.10,000

Rate = 10% per annum compounded half- yearly

=) 10/2 % per half yearly

=) 5% per half yearly

Time = 1 year

Half- yearly = 2 half years

[As 1 year= 2 half years]

A/q,

$Amount = P(1 + \frac{R}{100}) {}^{n} \\ \\ = > 10000(1 + \frac{5}{100} ) {}^{2} \\ \\ = > 10000(1 + \frac{1}{20} ) {}^{2} \\ \\ = > 10000( \frac{21}{20}) {}^{2} \\ \\ = > 10000 \times \frac{21}{20} \times \frac{21}{20} \\ \\ = > 25 \times 21 \times 21 \\ \\ = > Rs.11025$

Now,

Amount= Principal+ Interest

11025 = 10000+ Interest

Interest = (11025-10000)Rs.

Interest= Rs.1025

Interest when compound Annually

P=Rs.10000

R= 10%

T= 1 year

So,

$Amount = P(1 + \frac{R}{100} ) {}^{2} \\ \\ = > 10000(1 + \frac{10}{100} ) {}^{1} \\ \\ = > 10000(1 + \frac{1}{10} ) \\ \\ = > 10000 \times \frac{11}{10} \\ \\ = > 1000 \times 11 \\ \\ = > Rs.11000$

Now,

Amount = Principal+ Interest

11000= 10000 + Interest

Interest= 11000- 10000

Interest= Rs.1000

Hope it helps ☺️

Answer 2

HERE IS YOUR ANSWER

GIVEN :-

PRINCIPAL (P) = 10000

RATE (R) = 10%

TIME (N) = 1 YEAR

AMOUNT = A

COMPOUND INTEREST = CI

◆COMPOUNDED HALF YEARLY

A = P(1+R/100 × 1/2)^N×2

A = 10000(1+10/100 × 1/2)^2

= 10000(1+5/100)^2

= 10000(1+1/20)^2

= 10000(21/20)^2

= 10000×21/20×21/20

= 25×21×21

= 11025

CI = A - P

= 11025 - 10000

= 1025

◆COMPOUNDED ANNUALLY

A = P(1+R/100)^N

A = 10000(1+10/100)^1

= 10000(1+1/10)^1

= 10000×11/10×11/10

= 100×11×11

= 12100

CI = A - P

= 12100 - 10000

= 2100

NO, THE INTEREST IN COMPOUNDED HALF YEARLY IS LESS THAN THE INTEREST OS COMPOUNDED ANUALLY BY ₹75.