Solution:
(a) Here, Principal (P) = Rs. 10800, Time (n) = 3 years Rate of interest (R) = 12\ \frac{1}{2}\%=\frac{25}{2\%}
Amount(A) = P\left(1+\frac{R}{100}\right)^n
= 10800\left(1+\frac{1}{8}\right)^3=10800\left(\frac{9}{8}\right)^3
= 10800\times\frac{9}{8}\times\frac{9}{8}\times\frac{9}{8}
= Rs. 15,377.34
Compound Interest (C.I.) = A – P
= Rs. 10800 – Rs. 15377.34 = Rs. 4,577.34
(b) Here, Principal (P) = Rs. 18,000, Time (n) = 2\ \frac{1}{2} years, Rate of interest (R) = 10% p.a.
Amount(A) = P\left(1+\frac{R}{100}\right)^n
= 18000\left(1+\frac{10}{100}\right)^2=18000\left(1+\frac{1}{10}\right)^2
= 18000\left(\frac{11}{10}\right)^2=18000\times\frac{11}{10}\times\frac{11}{10}
= Rs. 21,780
Interest for \frac{1}{2} years on Rs. 21,780 at rate of 10% = \frac{21780\times10\times1}{100}= Rs. 1089
Total amount for 2\ \frac{1}{2} years.
= Rs. 21,780 + Rs. 1089 = Rs. 22,869
Compound Interest (C.I.) = A – P
= Rs. 22869 – Rs. 18000 = Rs. 4,869
(c) Here, Principal (P) = Rs. 62500, Time (n) = 1\ \frac{1}{2}=\frac{3}{2} years = 3 years
Rate of interest (R) = 8% = 4% (compounded half yearly)
Amount (A) = P\left(1+\frac{R}{100}\right)^n
= 62500\left(1+\frac{4}{100}\right)^2
= 62500\left(1+\frac{1}{25}\right)^3
= 62500\left(\frac{26}{25}\right)^3
= 62500 \times\frac{26}{25}\times\frac{26}{25}\times\frac{26}{25}
= Rs. 70,304
Compound Interest (C.I.) = A – P
= Rs. 70304 – Rs. 62500 = Rs. 7,804
(d) Here, Principal (P) = Rs. 8000, Time (n) = 1 years = 2 years (compounded half yearly)
Rate of interest (R) = 9% = \frac{9}{2}\% (compounded half yearly)
Amount (A) = P\left(1+\frac{R}{100}\right)^n
= 8000\left(1+\frac{9}{2\times100}\right)^2
= 8000\left(1+\frac{9}{200}\right)^2
= 800\left(\frac{209}{200}\right)^2
= 8000\times\frac{209}{200}\times\frac{209}{200}
= Rs. 8,736.20
Compound Interest (C.I.) = A – P
= Rs. 8736.20 – Rs. 8000
= Rs. 736.20
(e) Here, Principal (P) = Rs. 10,000, Time (n) = 1 years = 2 years (compounded half yearly)
Rate of interest (R) = 8% = 4% (compounded half yearly)
Amount (A) = P\left(1+\frac{R}{100}\right)^n
= 10000\left(1+\frac{4}{100}\right)^2
= 10000\left(1+\frac{1}{25}\right)^2
= 10000\left(\frac{26}{25}\right)^2
= 10000\times\frac{26}{25}\times\frac{26}{25}
= Rs. 10,816
Compound Interest (C.I.) = A – P
= Rs. 10,816 – Rs. 10,000 = Rs. 816
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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