Tag: maturity value of recurring deposits

Questions Related to maturity value of recurring deposits

An amount of Rs. $20000$ is deposited in a bank for $2$ years and paying an annual interest rate of $5\%$, compounded yearly. Find the maturity value.

maturity value of recurring deposits banking maths

  1. $12050$

  2. $22050$

  3. $32050$

  4. $42050$

Show answer
Correct Option: B
Explanation:

Given, principal Amount (P) $=$ Rs. $20000 $ 
Rate of Interest Amount (r) $= 5\% = 0.05  $
Number of Period (t) $= 2$ years 
Compounded Interest (n) $= 1$ (yearly)
Maturity value $=$ $P\times \left (1+\dfrac{r}{n}\right)^{nt}$
$=$ $20000\times \left (1+\dfrac{0.05}{1}\right)^{2}$
$=$ Rs. $22050$

Sonya deposited Rs. $200$ per month in her bank for six months under the recurring deposit scheme. What will be the maturity value of her deposit, if the rate of interest is $7\%$ per annum and the interest is calculated at the end of every month?

maturity value of recurring deposits banking maths

  1. $1936$

  2. $24936$

  3. $3936$

  4. $4936$

Show answer
Correct Option: D
Explanation:

Given, $P =$ Rs. $200$, $n = 6 $ months, $r = 7\%$
Interest $=$ $\dfrac{Pn(n+1)r}{2400}$

$=\dfrac{200\times6(6+1)7}{100}$
$= 24.5$
Maturity value $= 200 \times  24.5 + 36 =$ Rs. $4936$

Sneha deposits Rs. $2500$ per month in a bank for $48$ months under a recurring deposit scheme. If she is entitled to get Rs. $150000$ as maturity value, find the rate of interest per annum.

maturity value of recurring deposits banking maths

  1. $6\%$

  2. $8\%$

  3. $10\%$

  4. $12\%$

Show answer
Correct Option: D
Explanation:

Let the rate of interest per annum be $r\%$.
$P =$ Rs. $2500, n = 48$ months
Interest $=$ $\dfrac{Pn(n+1)r}{2400}=\dfrac{2500\times48(48+1)r}{2400}= 2540r$
Maturity amount $ = 2500 \times 48 + 2450r = 120000 + 2450r$
$\Rightarrow 150000 - 120000 = 2450r$
$\Rightarrow 30000 = 2450r$
$\Rightarrow r = 12.24\%$

In which account, a depositor choses a fixed amount and deposits it for fixed period every month?

maturity value of recurring deposits banking maths

  1. saving account

  2. fixed account

  3. recurring deposit account

  4. current account

Show answer
Correct Option: C
Explanation:

In recurring deposit account, a depositor chooses a fixed amount and deposits it for fixed period every month.

John deposited Rs. $50000$ in a bank for $1$ year and paying an annual interest rate of $14\%$, compounded quarterly. What is the maturity amount?

maturity value of recurring deposits banking maths

  1. Rs $54548$

  2. Rs $48884$

  3. Rs $84448$

  4. Rs $44888$

Show answer
Correct Option: C
Explanation:

Given, principal Amount $(P)$ $=$ Rs. $50000$  
Rate of Interest Amount $(r)$ $= 14\% = 0.14  $
Number of Period $(t)$ $= 1$ year  
Compounded Interest $(n)$ $= 4$ (quarterly)
Maturity value $=$ $P\times \left (1+\dfrac{r}{n}\right)^{nt}$
$=$ $50000\times \left (1+\dfrac{0.14}{1}\right)^{4}$
$=$ Rs. $84448$

If the account statement states that the interest is compounded annually then n = ?

maturity value of recurring deposits banking maths

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: A
Explanation:

$A=P(1+\cfrac{r}{100})^n$

Here $n=$ no. of interest periods
If interest is compounded annually
$\implies$ no. of interest period $1\times 1=1$
$\implies n=1$

The maturity value of a R.D Account is Rs. $16,176$. If the monthly installment is Rs. $400$ and the rate of interest is $8$ $\%$. Find the time period of this R.D account.

maturity value of recurring deposits banking maths

  1. $2$ years

  2. $3$ years

  3. $4$ years

  4. None of the above

Show answer
Correct Option: B
Explanation:

It is given that

Maturity values $=$ Rs. $16,176$
Monthly installment $=$ Rs. $400$
Rate of interest $=8\%$

Let the time be $x$ months
$\therefore$ Qualifying amount $=\dfrac {400\times (x\times x+1)}{2}=200(x^2+x)$

Now, S.I. $=\dfrac {200(x^2+x)\times 8\times 1}{100\times 12}=\dfrac {2(x^2+x)\times 2}{3}=\dfrac {4}{3}(x^2+x)$

Also, Principal $=$ Rs. $400\times x$

Therefore, $ 400x+\dfrac {4}{3}(x^2+x)=16176$

$\Rightarrow 1200x+4x^2+4x=48528$

$\Rightarrow 4x^2+1204x-48528=0$

$\Rightarrow x^2+301x-12132=0$

$\Rightarrow (x-36)(x+337)=0$

$\Rightarrow x=36$ or $x=-337$

Since the time cannot be negative, we have $x=36$.

The time of RD account is $36$ months or $3$ years.

A man deposited a certain amount in a bank that would repay double the amount after a year. At the
beginning of the second year, the man took out Rs 8000 and deposited the rest in the same bank. Again at the beginning of the third year, he took out Rs 8000 and deposited the rest in the same bank. At the beginning  of the fourth year, he took out Rs 8000 as before but was not left with any balance in the bank. What was  his initial deposit?

maturity value of recurring deposits banking maths

  1. Rs 6000

  2. Rs 9000

  3. Rs 8000

  4. Rs 7000

Show answer
Correct Option: D
Explanation:

$Let\quad the\quad initial\quad deposit\quad be\quad Rs.\quad x,\ amount\quad after\quad 1st\quad year=Rs.\quad 2x\ amount\quad left\quad after\quad withdrawl=Rs.\quad 2x-8000,\ amount\quad after\quad 2nd\quad year=Rs.\quad 4x-16000\ amonut\quad left\quad after\quad withdrawl=Rs.\quad 4x-24000,\ amount\quad after\quad 3rd\quad year=Rs.\quad 8x-48000=Rs.\quad 8000,\ 8x=56000,\ x=Rs.\quad 7000$

The fixed period of recurring deposit account can be 

maturity value of recurring deposits banking maths

  1. $2$ months

  2. $3$ months

  3. $6$ months

  4. All of the above

Show answer
Correct Option: C
Explanation:

The fixed period of recurring deposit  account can be $6$ months or more.

Therefore, options C is the correct answer.

Kiran deposited Rs.$200$ per month for $36$ months in a bank's recurring deposit account. If the bank pays interest at the rate of $11$ $\%$ per annum, find the amount she gets on maturity.

maturity value of recurring deposits banking maths

  1. Rs.$8412$

  2. Rs.$8421$

  3. Rs.$2481$

  4. Rs.$1234$

Show answer
Correct Option: B
Explanation:

Let the monthly installment be $P$.

Given, $P=200$, $n=36$, $r=11\%$
Interest $=\dfrac {Pn(n+1)r}{2400}$$=\dfrac {200\times 36 \times 37\times 11}{2400}$$=1221$
We know, maturity amount $=(Pn+1221)$$=(200\times 36+1221)$$=$ Rs. $8421$