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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