# CLASS-10BANKING - CALCULATION ON RECURRING DEPOSITS

Calculation of Interest on Recurring Deposit

Suppose \$ P per month is deposited each month for ‘n’ months at R% per annum. Then, \$ P deposited in the ‘n’th month will earn interest for 1 month, that deposited in (n – 1)th month will earn interest for 2 months, and so on, while the sum deposited in the first month will earn interest for ‘n’ months.

Thus we have –

Equivalent principal for one month = \$ [P X (1 + 2 + 3 + ……………. + n)]

n(n + 1)

= \$ [P X -----------]

2

Thus, the interest can be calculated using the formula

n(n + 1)          1            R

S.I. =  \$ [P X ----------- X ------- X -------]

2              12          100

Maturity value (M.V) = (P X n) + I

There are some examples are given below for your better understanding -

Example.1) Richard deposited \$ 200 per month for 36 months in a bank’s recurring deposit account. If the bank pays interest at 12% per annum, find the amount she gets on maturity.

Ans.)  Here P = \$ 200, n = 36 months, R = 12%  p.a

Amount deposited in 36 months = \$ (200 X 36) = \$ 7200

n(n + 1)          1            R

S.I. =  \$ [P X ----------- X ------- X -------]

2              12          100

1                              1           12

=  \$ [200 X ------- X 36 X (36 + 1) X ------- X -------]

2                             12          100

1

=  \$ [100 X 36 X 37 X -------]

100

=  \$ (36 X 37)  =  \$ 1332

The amount received by Richard on maturity = \$ (1332 + 7200)

= \$ 8532  (Ans.)

Example.2) Michel has a recurring deposit account in a bank for 2 years at 6% per annum simple interest. If he gets \$1200 as interest at the time of maturity find –

(i) the monthly instalment

(ii) the amount of maturity

Ans.) It is given that SI = \$1200, R = 6% p.a. and n = 24 months

(i) Let the monthly installment be \$P.

1                      1           R

Then, SI = [P X ------- n (n + 1) X ------- X -------]

2                     12          100

1                         1           6

=>  [P X ------- 24 (24 + 1) X ------- X ------] = 1200

2                        12         100

1                     1           6

=>  [P X ------ 24 X 25 X ------- X -------]  =  1200

2                    12          100

3P

=>  ------- = 1200

2

2

=>   P  =  1200 X ------  =  \$ 800

3

Hence the monthly installment is \$ 800

(ii) Amount of maturity = \$ (800 X 24) + Interest

= \$ 19200 + \$ 1200 = \$ 20400    (Ans.)

Example.3) Mr. Stephen opened a recurring deposit account in a bank and deposited \$1000 per month

1

for 1 ----- years. If he received \$18684 at the time of maturity, find

2

the rate of interest per annum.

Ans.) Here P = \$1000, n = 18 months

As per the given condition, total money deposited = \$(1000 X 18)

= \$18000

Total money received = \$22000

SI = \$(18684 – 18000) = \$4000

1                      1           R

Then, SI = [P X ------- n (n + 1) X ------- X -------]

2                     12          100

1                             1          R

Or,  684 =  [1000 X ------ X 18 X (18 + 1) X ------ X ------]

2                            12        100

1

Or,    684  =  (5 X 18 X 19 X ------ X R)

12

684 X 12

Or,    R  =  --------------- = 4.8%

5 X 18 X 19

Hence, the rate of interest is 4.8% per annum.     (Ans.)

Example.4) Mr. Ronald deposits a certain sum of money each month in a recurring deposit account of a bank. If the rate of interest is 8% per annum and Ronald gets \$8088 from the bank after 3 years, find the value of his monthly installment.

Ans.)  Let, the monthly installment be \$P

Total money deposited = \$36P

Total money received = \$8088

Here, n = 36 months,

R = 8% p.a

SI = \$(8088 – 36P)

1                      1             R

Then, SI = \$[P X ------- n (n + 1) X -------- X --------]

2                      12           100

1                            1            8

Or, (8088 – 36P) = \$ [P X ------X 36 X (36 + 1) X ------- X -------]

2                            12         100

1                        1            8

Or, (8088 – 36P) = \$ [P X ------- X 36 X 37 X ------- X -------]

2                        12         100

111P

Or,  (8088 – 36P) = \$ ---------

25

Or,   111P = \$[25 X (8088 – 36P)]

Or,   111P = \$(202200 – 900P)

Or,   111P + 900P = \$202200

Or,   1011P =  \$202200

202200

Or,    P = \$ ----------- = \$200

1011

Hence the value of each monthly installment is \$200.       (Ans.)

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