Answer
Hint: First, find the amount for the compound interest compounded yearly by applying the formula $A = P{\left( {1 + \dfrac{r}{{100}}} \right)^t}$ for the first year and then the formula $A = P{\left( {1 + \dfrac{r}{{2 \times 100}}} \right)^{t \times 2}}$ for the next 6 months. Then subtract the principal from the amount to get the interest.Then, find the amount for the compound interest compounded half-yearly by applying the formula $A = P{\left( {1 + \dfrac{r}{{2 \times 100}}} \right)^{t \times 2}}$. Then subtract the principal from the amount to get the interest. After that subtract the values of the interest to find the difference of the interest.
Complete step by step answer:
The formula for compound interest compounded yearly is,$A = P{\left( {1 + \dfrac{r}{{100}}} \right)^t}$The formula for compound interest compounded half-yearly is,$A = P{\left( {1 + \dfrac{r}{{2 \times 100}}} \right)^{t \times 2}}$Given: - Principal, P = Rs. 10000Time, t = 18 months = 1.5 yearsRate, r = 10% p.a.For the compound interest compounded yearly,Calculate the amount for 1 year. Then calculate the amount for $\dfrac{1}{2}$ year.For 1st year,$P = 10000$$t = 1$$r = 10\% $Then,$A = 10000{\left( {1 + \dfrac{{10}}{{100}}} \right)^1}$Cancel out common factors and take LCM,$ \Rightarrow A = 10000\left( {\dfrac{{10 + 1}}{{10}}} \right)$Add the terms and cancel out the common factor,$ \Rightarrow A = 1000 \times 11$Multiply the terms,$ \Rightarrow A = 11000$Now, for $\dfrac{1}{2}$ year,$P = 11000$$r = 10\% $$t = \dfrac{1}{2}$Substitute the values in the formula for compounded half-yearly,$A = 11000{\left( {1 + \dfrac{{10}}{{2 \times 100}}} \right)^{\dfrac{1}{2} \times 2}}$Cancel out the common factors,$ \Rightarrow A = 11000{\left( {1 + \dfrac{1}{{20}}} \right)^1}$Take LCM,$ \Rightarrow A = 11000 \times \dfrac{{20 + 1}}{{20}}$Cancel out the common factors,$ \Rightarrow A = 550 \times 21$Multiply the terms,$ \Rightarrow A = {\text{Rs}}{\text{. }}11550$So, the interest is,$I = A - P$Substitute the value of amount and principal,$ \Rightarrow I = 11550 - 10000$Subtract the term,$ \Rightarrow I = {\text{Rs}}{\text{. }}1550$.....….. (1)For the compound interest compounded half-yearly,$P = 10000$$r = 10\% $$t = \dfrac{3}{2}$Substitute the values in the formula for compounded half-yearly,$A = 10000{\left( {1 + \dfrac{{10}}{{2 \times 100}}} \right)^{\dfrac{3}{2} \times 2}}$Cancel out the common factors,$ \Rightarrow A = 10000{\left( {1 + \dfrac{1}{{20}}} \right)^3}$Take LCM,$ \Rightarrow A = 10000{\left( {\dfrac{{20 + 1}}{{20}}} \right)^3}$Add the terms,$ \Rightarrow A = 10000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}}$Cancel the terms and multiply the remaining terms,$ \Rightarrow A = {\text{Rs}}{\text{. }}11576.25$So, the interest is,$I = A - P$Substitute the value of amount and principal,$ \Rightarrow I = 11576.25 - 10000$Subtract the term,$ \Rightarrow I = {\text{Rs}}{\text{. }}1576.25$..........….. (2)For the difference between two compound interest is,$\therefore 1576.25 - 1550 = {\text{Rs}}{\text{. 2}}6.25$Hence, the difference between the compound interest compounded yearly and half-yearly is Rs. 26.25.
Note: The students might make mistakes in calculating the amount for the 6 months compounded yearly.
Compound interest is the interest calculated on the principal and the interest accumulated over the previous period. It is different from the simple interest where interest is not added to the principal while calculating the interest during the next period.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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