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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Present value = Rs.9000
Interest rate = 10 % per annum
Time = 2 years 4 month = (2 + 1/3) years = (7/2) years
Amount (A) = P (1 + R/100)n × [1 + (1/3 × R)/100]
[Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 9000 (1 + 10/100)2 × [1 + (1/3 × 10)/100]
⇒ A = 9000 (1 + 1/10)2 × [1 + 1/30]
⇒ A = 9000 (11/10)2 × [31/30]
⇒ A = 9000 × 121/100 × 31/30
⇒ A = 9 × 121 × 31/3
⇒ A = 3 × 121 × 31
⇒ A = 11253
∴ Amount = Rs.11253
∴ Compound interest = Rs.(11253 – 9000)
= Rs.2253
Given:
Present value= ₹ 9000
Interest rate= 10 % per annum
Time=2 years 6 months = (2 + ½) years= 5/2 years
To find the amount we have the formula,
Amount (A) = P (1+(R/100))n
Where P is present value, r is rate of interest, n is time in years.
Now substituting the values in above formula we get,
∴ A = 9000 (1 + 10/100)2 [1 + (1/3 × 10)/100]
⇒ A = 9000 (1+1/10)2 (1+1/30)
⇒ A = 9000 (11/10)2 (31/30)
⇒ A = 9000 × 121/100 × 31/30 = 9 × 121 × 31/3
⇒ A = ₹ 11253
∴ Compound interest = A – P
= 11253 – 6000 = ₹ 2253