What will be the compound interest on Rs 10000 at the rate of 12% per annum for a year compounded half yearly?

Sum

Find the compound interest on Rs.10000 for 2 years at 8% per annum compounded half-yearly.

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Solution

Principal P = ₹ 10,000

Rate of interest R = 8% p.a. compounded half-yearly

Duration T = 2 years

A = P`(1 + (("R"/2))/100)^"2T"`

= `10000(1 + (8/2)/100)^4`

= `10000(1 + 4/100)^4`

= 10000(1.04)4

= 11698.58

I = A – P

= 11648.58 – 10000

= 1698.58

∴ Compound interest is  ₹ 1698.58.

Concept: Simple and Compound Interest (Entrance Exam)

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Chapter 9: Commercial Mathematics - Exercise 9.3 [Page 130]

Q 5Q 4Q 6

APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board

Chapter 9 Commercial Mathematics
Exercise 9.3 | Q 5 | Page 130

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Solution:

Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly)

Time(n) = 1\ \frac{1}{2} years = 3 years (compounded half yearly)

Amount (A) = P\left(1+\frac{R}{100}\right)^n

= 10000\left(1+\frac{5}{100}\right)^3

= 10000\left(1+\frac{1}{20}\right)^3

= 10000\left(\frac{21}{20}\right)^3

= 10000\times\frac{21}{20}\times\frac{21}{20}\times\frac{21}{20}

= Rs. 11,576.25

Compound Interest (C.I.) = A – P

= Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25

If it is compounded annually, then

Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10%, Time (n) = 1\ \frac{1}{2} years.

Amount (A) for 1 year = P\left(1+\frac{R}{100}\right)^n

= 10000\left(1+\frac{10}{100}\right)^1

= 10000\left(1+\frac{1}{10}\right)^1

= 10000\left(\frac{11}{10}\right)^1

= 10000\times\frac{11}{10}

= Rs. 11,000

Interest for \frac{1}{2} year = \frac{11000\times1\times10}{2\times100}=RS.\ 550

\therefore Total amount = Rs. 11,000 + Rs. 550 = Rs. 11,550

Now, C.I. = A – P = Rs. 11,550 – Rs. 10,000

= Rs. 1,550

Yes, interest Rs. 1,576.25 is more than Rs. 1,550.

Solution:

What is known: Principal, Time Period, and Rate of Interest

What is unknown: Amount and Compound Interest (C.I.)

Reasoning:

A = P[1 + (r/100)]n

P = ₹ 10,000

n = \(1{\Large\frac{1}{2}}\) years

R = 10% p.a. compounded annually and half-yearly

where , A = Amount, P = Principal, n = Time period and R = Rate percent

For calculation of C.I. compounded half-yearly, we will take the Interest rate as 5% and n = 3

A = P[1 + (r/100)]n

A = 10000[1 + (5/100)]3

A = 10000[1 + (1/20)]3

A = 10000 × (21/20)3

A = 10000 × (21/20) × (21/20) × (21/20)

A = 10000 × (9261/8000)

A = 5 × (9261/4)

A = 11576.25

Interest earned at 10% p.a. compounded half-yearly = A - P

= ₹ 11576.25 - ₹ 10000 = ₹ 1576.25

Now, let's find the interest when compounded annually at the same rate of interest.

Hence, for 1 year R = 10% and n = 1

A = P[1 + (r/100)]n

A = 10000[1 + (10/100)]1

A = 10000[1 + (1/10)]

A = 10000 × (11/10)

A = 11000

Now, for the remaining 1/2 year P = 11000, R = 5%

A = P[1 + (r/100)]n

A = 11000[1 + (5/100)]

A = 11000[(105/100)]

A = 11000 × 1.05

A = 11550

Thus, amount at the end of \(1{\Large\frac{1}{2}}\)when compounded annually = ₹ 11550

Thus, compound interest = ₹ 11550 - ₹ 10000 = ₹ 1550

Therefore, the interest will be less when compounded annually at the same rate.

☛ Check: NCERT Solutions for Class 8 Maths Chapter 8

Video Solution:

Find the amount and the compound interest on ₹ 10,000 for \(1{\Large\frac{1}{2}}\) years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

NCERT Solutions Class 8 Maths Chapter 8 Exercise 8.3 Question 8

Summary:

The amount and the compound interest on ₹ 10,000 for \(1{\Large\frac{1}{2}}\) years at 10% per annum, compounded half-yearly is  ₹ 11576.25 and  ₹ 1576.25 respectively. The interest will be less when compounded annually at the same rate.

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