Find the amount which Ram will get on Rs4096, if he gives it for 18 months at $12\dfrac{1}{2}$% per annum, interest being compounded half yearly.
Last updated date: 17th Mar 2023
•
Total views: 305.7k
•
Views today: 7.85k
Answer
305.7k+ views
Hint- Ram has given Rs4096 to someone and we have to tell the amount which he will be receiving after 18 months if he lends at some interest rate which is compounded half yearly, thus the principal amount here is Rs4096.
The principal amount P = Rs4096
The duration of time for which he is lending this money is 18 months at $12\dfrac{1}{2}$% per annum and it is compounded half yearly.
Now the interest he will be receiving compounded half yearly in 18 months will be equal to the interest compounded yearly in $18 \times 2 = 36$months.
Here the tenure of interest is 36 months. Now 12 months makes 1 year so 1 month comprises of $\dfrac{1}{{12}}$years and thus 36 months will make $\dfrac{1}{{12}} \times 36 = 3$years. So tenure is of T= $3$years.
Now the rate of interest is given as $12\dfrac{1}{2}$% per annum compounded half yearly.
So we have to change the rate of interest per annum compounded yearly.
Hence that will be $\dfrac{{\dfrac{{25}}{2}}}{2} = \dfrac{{25}}{4}$% per year compounded yearly.
Now the amount that he will be receiving, ${\text{Amount = P}}{\left( {1 + \dfrac{r}{{100}}} \right)^T}$where P is principal amount, R is
Rate of interest, T is the duration of loan given.
${\text{Amount = 4096}}{\left( {1 + \dfrac{{25}}{{4 \times 100}}} \right)^3}$
$
\Rightarrow 4096 \times {\left( {1.0625} \right)^3} \\
\Rightarrow 4096 \times 1.9946 \\
\Rightarrow 4913{\text{ Rs}} \\
$
Thus he will be receiving 4913Rs after 18 months if given at a rate of $12\dfrac{1}{2}$% per annum compounded half yearly.
Note – Whenever we face such a type of problem the key concept that we need to recall is that the amount received is dependent upon the principal amount, rate of interest and the tenure of loan provided, moreover the tenure of loan is always taken in years and not in months while making the calculations.
The principal amount P = Rs4096
The duration of time for which he is lending this money is 18 months at $12\dfrac{1}{2}$% per annum and it is compounded half yearly.
Now the interest he will be receiving compounded half yearly in 18 months will be equal to the interest compounded yearly in $18 \times 2 = 36$months.
Here the tenure of interest is 36 months. Now 12 months makes 1 year so 1 month comprises of $\dfrac{1}{{12}}$years and thus 36 months will make $\dfrac{1}{{12}} \times 36 = 3$years. So tenure is of T= $3$years.
Now the rate of interest is given as $12\dfrac{1}{2}$% per annum compounded half yearly.
So we have to change the rate of interest per annum compounded yearly.
Hence that will be $\dfrac{{\dfrac{{25}}{2}}}{2} = \dfrac{{25}}{4}$% per year compounded yearly.
Now the amount that he will be receiving, ${\text{Amount = P}}{\left( {1 + \dfrac{r}{{100}}} \right)^T}$where P is principal amount, R is
Rate of interest, T is the duration of loan given.
${\text{Amount = 4096}}{\left( {1 + \dfrac{{25}}{{4 \times 100}}} \right)^3}$
$
\Rightarrow 4096 \times {\left( {1.0625} \right)^3} \\
\Rightarrow 4096 \times 1.9946 \\
\Rightarrow 4913{\text{ Rs}} \\
$
Thus he will be receiving 4913Rs after 18 months if given at a rate of $12\dfrac{1}{2}$% per annum compounded half yearly.
Note – Whenever we face such a type of problem the key concept that we need to recall is that the amount received is dependent upon the principal amount, rate of interest and the tenure of loan provided, moreover the tenure of loan is always taken in years and not in months while making the calculations.
Recently Updated Pages
If abc are pthqth and rth terms of a GP then left fraccb class 11 maths JEE_Main

If the pthqth and rth term of a GP are abc respectively class 11 maths JEE_Main

If abcdare any four consecutive coefficients of any class 11 maths JEE_Main

If A1A2 are the two AMs between two numbers a and b class 11 maths JEE_Main

If pthqthrth and sth terms of an AP be in GP then p class 11 maths JEE_Main

One root of the equation cos x x + frac12 0 lies in class 11 maths JEE_Main

Trending doubts
What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?
