Math, asked by tanu4535, 4 months ago

Mr. Kumar has a recurring deposit account in a bank for 4 years at 10% p.a. rate of interest.
If he gets + 21,560 as interest at the time of maturity, find :
(i) the monthly instalment paid by Mr. Kumar.
(ii) the amount of maturity of this recurring deposit account.

Answers

Answered by ashokshuka272
6

Answer:

1 :- 2000

2:- 80325

Step-by-step explanation:

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Answered by Afreenakbar
0

Answer:

I) The monthly instalment paid by Mr. Kumar is approximately Rs. 4,951.76.

ii) The amount of maturity of Mr. Kumar's recurring deposit account is approximately Rs. 3,09,960.27.

Step-by-step explanation:

We can solve this problem by using the formula for calculating the maturity value of a recurring deposit:

m = p \times  \frac{{(1 +  \frac{r}{n} })^{nt}  - 1}{ \frac{r}{n} }

Where:

M = maturity value

P = monthly instalment

r = rate of interest per annum

n = number of times interest is compounded in a year

t = time period in years

Given that Mr. Kumar has a recurring deposit account for 4 years at a rate of interest of 10% p.a. and the interest earned at maturity is Rs. 21,560. We can calculate the monthly instalment and maturity value as follows:

(i) Monthly instalment paid by Mr. Kumar:

Using the above formula, we can calculate the monthly instalment as follows:

m = p \times  \frac{{(1 +  \frac{r}{n} })^{nt}  - 1}{ \frac{r}{n} }

21560= p \times  \frac{{(1 +  \frac{0.1}{12} })^{124}  - 1}{ \frac{0.1}{12} }

21560= p \times  \frac{{(1 + 0.00833 })^{48}  - 1}{ 0.00833 }

21560= p \times 4.3558

p =  \frac{21560}{4.3558}

p = 4951.76

Therefore, the monthly instalment paid by Mr. Kumar is approximately Rs. 4,951.76.

(ii) Amount of maturity of this recurring deposit account:

Using the above formula, we can calculate the maturity value as follows:

m = p \times  \frac{{(1 +  \frac{r}{n} })^{nt}  - 1}{ \frac{r}{n} }

m = 4951.76 \times  \frac{{(1 + 0.00833 })^{48}  - 1}{ 0.00833}

m = 4951.76 \times  \frac{{(1 +  \frac{r}{n} })^{nt}  - 1}{ \frac{r}{n} }

m ≈ 3,09,960.27

Therefore, the amount of maturity of Mr. Kumar's recurring deposit account is approximately Rs. 3,09,960.27.

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