Question

A man deposits Rs.50 at the beginning of each month into a Savings Bank which gives simple interest at 3% p.a. Find the amount
accumulated at the end of the year.​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The amount accumulated at the end of the year is Rs.609.75

Step-by-step explanation:

Assumption,
Let the amount deposited by the man be (P)

Given,

A man deposits (P) = Rs. 50 at the beginning of the month.

Which eventually gives the interest of (r) 3% p.a.

To find,

The amount of money(T) accumulated at the end of the year.

Calculation,

First, we try to calculate the amount of money deposited by the man at the end of the year. (Let it be y)

y = Amount deposited for every month(P) × number of months(n)

y = P × n

⇒ y = 50 × 12

⇒ y = 600 rupees.

The period of recurring deposit N is:

$N = \frac{1}{12} [\frac{n(n + 1)}{2} ]$

$\implies N = \frac{1}{24} \times 12 \times 13$

⇒ N = 13/2 years.

Interest I is:

$I = \frac{PNr}{100}$

$\implies I = \frac{50 \times \frac{13}{2} \times 3}{100}$

⇒ I = 9.75 rupees

Therefore, the total amount deposited is(T):

T = y + I

⇒ T = 600 + 9.75

⇒ T = 609.75 rupees.

Therefore, the amount accumulated at the end of the year is Rs.609.75

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