Maria invested Rs 80,000 in a business. She would be paid interest at 5% per annum compounded annually. Find (i) the amount standing to her credit at the end of the second year. (ii) The interest for the third year.
Last updated date: 29th Mar 2023
•
Total views: 308.4k
•
Views today: 2.85k
Answer
308.4k+ views
Hint: Here we go through by applying the formula of amount after t year at the rate of r% compound i.e. $A = P{\left[ {1 + \dfrac{r}{{100}}} \right]^t}$ here A=amount, P=principal amount, r=rate and t=time.
Here in this question it is given that
Principal, P= Rs. 80,000
Rate of interest, r=5%
And we have to find the amount at the end of two year so t=2.
(i) The amount credited at the end of the second year, $A = P{\left[ {1 + \dfrac{r}{{100}}} \right]^t}$
Now put the values in the formula to find the amount.
$A = 80000{\left( {1 + \dfrac{5}{{100}}} \right)^2} = 80000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} = 88200$
Hence, the amount standing to her credit at the end of the second year is Rs. 88200.
Now for solving the (ii) part,
(ii) We have to first calculate the total amount in third year and then for finding the interest for third year we will subtract the amount of two years.
Here Principal, P= Rs. 80,000
Rate of interest, r=5%
And we have to find the amount at the end of three year so t=3.
The amount credited at the end of the third year, $A = P{\left[ {1 + \dfrac{r}{{100}}} \right]^t}$
$A = 80000{\left( {1 + \dfrac{5}{{100}}} \right)^3} = 80000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} = 92610$.
Interest of third year=Amount of three years-Amount of two years.
Interest=92610-88200 (as we find the amount for two years above).
=4410
$\therefore $ The interest for the third year is 44410. Answer
Note: Whenever we face such a type of question the key concept for solving the question is you must always remember the formula related to compound interest for solving such a type of question. By putting the given terms in formula you can easily get the answer. And for finding the interest for a specific year just subtract the amount of its previous year from the amount of that year.
Here in this question it is given that
Principal, P= Rs. 80,000
Rate of interest, r=5%
And we have to find the amount at the end of two year so t=2.
(i) The amount credited at the end of the second year, $A = P{\left[ {1 + \dfrac{r}{{100}}} \right]^t}$
Now put the values in the formula to find the amount.
$A = 80000{\left( {1 + \dfrac{5}{{100}}} \right)^2} = 80000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} = 88200$
Hence, the amount standing to her credit at the end of the second year is Rs. 88200.
Now for solving the (ii) part,
(ii) We have to first calculate the total amount in third year and then for finding the interest for third year we will subtract the amount of two years.
Here Principal, P= Rs. 80,000
Rate of interest, r=5%
And we have to find the amount at the end of three year so t=3.
The amount credited at the end of the third year, $A = P{\left[ {1 + \dfrac{r}{{100}}} \right]^t}$
$A = 80000{\left( {1 + \dfrac{5}{{100}}} \right)^3} = 80000 \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} \times \dfrac{{21}}{{20}} = 92610$.
Interest of third year=Amount of three years-Amount of two years.
Interest=92610-88200 (as we find the amount for two years above).
=4410
$\therefore $ The interest for the third year is 44410. Answer
Note: Whenever we face such a type of question the key concept for solving the question is you must always remember the formula related to compound interest for solving such a type of question. By putting the given terms in formula you can easily get the answer. And for finding the interest for a specific year just subtract the amount of its previous year from the amount of that year.
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 ?
