Question

Mr Sharma has a recurring deposit account for 2 years at 6 %interest p.a.. He receives ₹975 as interest on maturity.
(1) find the monthly installment amount.
(2) find the maturity amount.
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THIS QUESTION ANSWER.

Here, n=2×12, R=6 and total interest =₹975
Let the monthly installment be p.
(1) We have,
Interest earned = PRn(n+1) /2400

$975 = \frac{p \times 6 \times 24 \times 25 }{2400?}$
$p = 975 \times \frac{2400}{6 \times 24 \times 25?} = 650$
(2) Maturity amount =
$np(1 + \times \frac{r(n + 1)}{2400} )$
$24 \times 650 \times (1 + \times \frac{6 \times 25}{2400} )$
$15600(1 + \times \frac{6 \times 25}{2400} )$
$15600(1 + \times \frac{150}{2400} )$
$15600(1 + \times \frac{150}{16 \times 150} )$
$15600(1 + \times \frac{1}{16} )$
$15600 \times 1 \times \frac{1}{16}$
= ₹16575 ans

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Answer 1

Answer:

In case of recurring deposits, the compounding happens on quarterly basis. Here, A is the maturity amount in Rs., the recurring deposit amount is 'P' in Rs., 'N' is the compounding frequency, interest rate R in percentage and 't' is the tenure.

Step-by-step explanation: