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When interest is not compounded annually but with a periodicity of "n", i.e. interest is incurred "n" times annually:
A = $$P\left(1+\frac{R}{n\cdot100}\right)^{n\cdot T}$$
For example, when the periodicity is 2, i.e. "n" = 2, the interest is compounded half-yearly:
A=$$P\left(1+\dfrac{R}{200}\right)^{2T}$$
and when the periodicity is 4, i.e. "n" = 4, the interest is compounded quarterly:
$$A = P\left(1+\dfrac{R}{400}\right)^{4T}$$
Bank A offers 6% interest rate per annum compounded half-yearly. Bank B and Bank C offer simple interest but the annual interest rate offered by Bank C is twice that of Bank B. Raju invests a certain amount in Bank B for a certain period and Rupa invests ₹ 10,000 in Bank C for twice that period. The interest that would accrue to Raju during that period is equal to the interest that would have accrued had he invested the same amount in Bank A for one year. The interest accrued, in INR, to Rupa is
Bank A: 6% p.a. 1/2 yearly (CI)
Bank B: x% p.a (SI)
Bank C: 2x% p.a (SI)
Let Raju invest Rs P in bank B for t years. Hence, Rupa invests Rs 10,000 in bank C for 2t years.
Now,
$$P\left(\frac{x}{100}\right)t\ =\ P\left(1+\frac{3}{100}\right)^2-P$$
$$\left(\frac{x}{100}\right)t\ =\ 1.0609-1$$
$$\left(\frac{x}{100}\right)t\ =\ 0.0609$$
We need to calculate
SI = $$10000\times\ 2t\times\ \left(\frac{2x}{100}\right)=40000\left(\frac{x}{100}\right)t=40000\times\ 0.0609=2436$$
Anil invests Rs. 22000 for 6 years in a certain scheme with 4% interest per annum, compounded half-yearly. Sunil invests in the same scheme for 5 years, and then reinvests the entire amount received at the end of 5 years for one year at 10% simple interest. If the amounts received by both at the end of 6 years are same, then the initial investment made by Sunil, in rupees, is
Correct Answer: 20808
Anil invested 22000 for 6 years at 4% interest compounded half-yearly
=> Amount =$$22000\left(1.02\right)^{12}$$
Let Sunil invest 'S' rupees for 5 years at 4% C.I. half-yearly and 10% S.I. for 1 additional year
=> Amount = $$S\left(1.02\right)^{10}\left(1.1\right)$$
Given that the both amounts are equal
=> $$22000\left(1.02\right)^{12}=S\left(1.02\right)^{10}\left(1.1\right)$$
=> $$S=\dfrac{22000\left(1.02\right)^2}{1.1}=20808$$
Anil borrows Rs 2 lakhs at an interest rate of 8% per annum, compounded half-yearly. He repays Rs 10320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year. Then, the total interest, in rupees, paid over the three years is nearest to
It is given that Anil borrows Rs 2 lakhs at an interest rate of 8% per annum, compounded half-yearly. It is also known that he repays Rs 10320 at the end of the first year and closes the loan by paying the outstanding amount at the end of the third year.
The total amount at the end of the first year is: $$200000\times\ \frac{104}{100}\times\ \frac{104}{100}=216320$$
He repays 10320 rupees at the end of the first year, which implies the amount that remains unpaid at the end of the first year is 206000 rupees.
This unpaid amount will accrue interest for another two years.
Hence, the final amount at the end of three years is $$206000\times\ \frac{104}{100}\times\ \frac{104}{100}\times\ \frac{104}{100}\times\ \frac{104}{100}=240990.86$$
Hence, the accrued interest in these two years is (240990.86-206000) = 34990.86 rupees.
Hence, the total interest accrued over the three years = (34990.86+16320) = 51311 rupees.
The correct option is B
Anil invests Rs 22000 for 6 years in a scheme with 4% interest per annum, compounded half-yearly. Separately, Sunil invests a certain amount in the same scheme for 5 years, and then reinvests the entire amount he receives at the end of 5 years, for one year at 10% simple interest. If the amounts received by both at the end of 6 years are equal, then the initial investment, in rupees, made by Sunil is
Let's take the amount invested by Sunil to be X.
The amount received by Anil at the end of 6 years would be $$22000\left(1+\frac{4}{2\times\ 100}\right)^{6\times\ 2}=22000\left(1.02\right)^{12}$$
The amount received by Sunil at the end of 5 years would be $$X\left(1.02\right)^{10}$$
In the 6th year, Sunil invests this at a simple interest of 10%, giving him an interest of $$X\left(1.02\right)^{10}\times\ 0.1$$
Giving the total amount with him at the end of 6 years to be $$X\left(1.02\right)^{10}\times\ \left(1+0.1\right)$$
Equating the final amount with Sunil and Anil, we get:
$$X\left(1.02\right)^{10}\times\ \left(1.1\right)=22000\left(1.02\right)^{12}$$
$$X=\frac{22000\left(1.02\right)^2}{1.1}=20808$$
Therefore, Option D is the correct answer.
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