Question

Kavita has a cumulative time deposit account in a bank. She deposits 600 per month and
gets 6165 at the time of maturity. If the rate of interest be 6% per annum, find the total
time for which the account was held.​

Answer 1

Answer:

The time for which account was held is 39 years .

Step-by-step explanation:

Given as :

The principal deposited into account = p = Rs 600 per month

Amount received after T years = Rs 6165

The rate of interest = r = 6%

Time period = T years

From Cumulative method

Amount = principal × $(1+\dfrac{Rate}{12\times 100})^{12\times time}$

Or, Rs 6165 = Rs 600 × $(1+\dfrac{6}{1200})^{12T}$

Or, $(1+\dfrac{6}{1200})^{12T}$  =  $\dfrac{6165}{600}$

Or, $(1.005)^{12T}$ = 10.27

Taking log with base 10 both side

Log $(1.005)^{12T}$ = Log 10.27

or, 12 T × 0.002166 = 1.011

Or, T = $\dfrac{1.011}{0.0259}$

∴   Time = 39 years

So, The time for which account was held = T = 39 years

Hence,  The time for which account was held is 39 years . Answer

Answer 2

Solution!!

Given that the monthly deposits are Rs 600 and at the time of maturity, Kavita gets Rs 6165. The rate of interest is equal to 6% per annum.

P = Rs 600 per month

Amount = Rs 6165

R = 6% p.a.

n = ?

$\sf Amount=P\times n+P\times \dfrac{n(n+1)}{2\times 12}\times \dfrac{R}{100}$

$\sf 6165=600n+600\times \dfrac{n(n+1)}{2\times 12}\times \dfrac{6}{100}$

Divide both sides of the equation by 15.

$\sf 411=40n+n(n+1)\times \dfrac{1}{10}$

Use the commutative property.

$\sf 411=40n+\dfrac{1}{10}n(n+1)$

Distribute 1/10 n through the parentheses.

$\sf 411=40n+\dfrac{1}{10}n^{2}+\dfrac{1}{10}n$

Calculate the sum.

$\sf 411=\dfrac{401}{10}n+\dfrac{1}{10}n^{2}$

Multiply both sides of the equation by 10.

$\sf 4110=401n+n^{2}$

Move the variables to the left hand side and change their signs.

$\sf -n^{2}-401n+4110=0$

Change the signs on both sides of the equation.

$\sf n^{2}+401n-4110=0$

Do the middle-term splitting.

$\sf n^{2}+411n-10n-4110=0$

Factor the expressions.

$\sf (n+411)(n-10)=0$

Separate into possible cases.

$\sf n+411=0$

$\sf n-10=0$

Solve the equations.

$\sf n=-412$

$\sf n=10$

The number of months cannot be in negative, therefore, the value of n is equal to 10 months.

$\sf n=10\:months$

The amount of time for which the account was held is 10 months.