Find compound interest on rs 10000 at 10% per annum for 2 years and 3 months, compounded annually.

hello guys the question is find the amount and the compound interest on rupees 10000 of for 1.5 years at 10% per annum compounded half-yearly so let's start our amount our principal is equal to 10000 and time period given is equal to 1.5 and the rate of interest that is given is 10% per annum now we have to calculate for half yearly for 6 month so hot six month this rate of interest become half become 5% and time period become double so it is 3 so using the formula a is equal to amount is equal to principal oneplus rate of interest divided by 100 power and so put the values

you will get amount amount is equal to principal that is 10000 X 1 + R is 5% / and raised to the power 3 so it become 10000 multiply 105 divided by 100 x 105 divided by 100 multiply 10 500/100 now they will cancel 21 by 20 by 50 it become 2020 21 20 so the answer is 1157 6.25 so this is the amount so this is the amount is equal to 1157 6.25 now we have to calculate the compound interest so

compound interest is equal to amount that is 1157 6.25 -10000 at is equal to 15/7 6.25 so this is our compound interest and this is our amount thank you

The compound interest on Rs 4000 at 10% per annum for 2 years 3 months compounded annually is A). Rs. 916B). Rs. 900C). Rs. 961D). Rs. 896

Answer

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Hint: First we’ll find the compound interest for 2 years by using its formula. Rest of the 2 months’ interest will be simple because it’s given that the compound interest will be applied annually. In the end, we’ll add both the interest and subtract from the principal amount.Complete step by step solution:
Here, we have given the principal amount (P) as 4000 and interest (r) 10% annually. As we’ll divide the total time duration into two parts because in the time span of 2 years compound interest will be applied and for the rest of the 3 months simple interest will be applied.
The formula for the compound interest is $A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n}$
Where A is the total amount, P is the principal amount, r is the rate of interest and n is the time duration.
According to our question, $P=4000, r=10\%, n =2 $years. On Calculation A we get,
\[ A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} \\
   \Rightarrow A = 4000{\left( {1 + \dfrac{{10}}{{100}}} \right)^2} \\
   \Rightarrow A = P{\left( {1 + \dfrac{1}{{10}}} \right)^2} \\
   \Rightarrow A = 4000{\left( {\dfrac{{11}}{{10}}} \right)^2} \\
   \Rightarrow A = 4000 \times \dfrac{{121}}{{100}} \\
   \Rightarrow A = 4840{\text{ Rs}}{\text{.}} \]
Hence the amount after 2 years will be 4840 and it’ll only work as the principal amount for simple interest. The formula for the simple interest is = $\dfrac{{PRT}}{{100}}$, where P is the principal amount, R is the rate of interest, and T is time spam.
On putting the values, we get,
$ \dfrac{{PRT}}{{100}} \\
   = \dfrac{{4840 \times 10 \times 1}}{{100 \times \times 4}} \\
   = \dfrac{{4840 \times 1 \times 1}}{{10 \times \times 4}} \\
   = \dfrac{{484 \times 1 \times 1}}{{1 \times \times 4}} \\
   = 121 $
So, the total amount after 2 years and 3 months will be Rs 4840 + Rs 121 which is equal to Rs 4961.
Total interest will be = total amount after 2 years 3 months- principal amount
That is, $Rs 4961 – Rs 4000$
And, Rs. 961

Hence, option (c) is the correct option.

Note: Students usually make mistakes in such a type of problem where for some time period compound interest is applied and for some time period, simple interest is applied. It’s always recommended from our side to read the question carefully, especially the interest section. Whether the interest is applied monthly or annually also of which kind, simple of the compound.

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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