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Questions Related to maths

Mohan deposited Rs.$80$ per month in a cumulative deposit account for six years. Find the amount payable to him on maturity, if the rate of interest is $6\%$ per annum.

maturity value of recurring deposits banking maths

  1. $6811.20$

  2. $2000$

  3. $4811.20$

  4. $3811.20$

Show answer
Correct Option: A
Explanation:

Let the monthly installment be $P$

Given, monthly installment Rs.$=80$, $r=6\%$, $n=6\times 12=72$
We know, Interest $=\dfrac {Pn(n+1r)}{2400}$
$=\dfrac {80 \times 72 \times 73\times 6}{2400}$
$=\dfrac {5256}{5}$
Required amount $=Pn+{5256}{5}$
$=80 \times 72 +\dfrac {5256}{5}$
$=28800+\dfrac {5256}{5}$
$=6811.20$

Priya has a recurring deposit in a bank at $10\%$ per annum simple interest. If she pay monthly installment s Rs.$700$ for annually. Find her Maturity value.

maturity value of recurring deposits banking maths

  1. $2034.904$

  2. $2134.904$

  3. $3134.904$

  4. $4134.904$

Show answer
Correct Option: B
Explanation:

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R =$ Monthly Installment $= 700 $
$i =$ Rate of Interest/ 400 $= \dfrac {10}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{700 \times [(1+10/400)^{n} - 1]}{1-(1+12/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{700\times\dfrac{10}{400}}{1-(\dfrac{410}{400})^{-1/3}}$
$M = 2134.904$

Meena has a recurring deposit in a bank at $12\%$ per annum simple interest. If she pay monthly installment s Rs.$400$ for annually. Find her Maturity value.

maturity value of recurring deposits banking maths

  1. $1123.92$

  2. $1223.92$

  3. $1623.92$

  4. $1823.92$

Show answer
Correct Option: B
Explanation:

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R =$ Monthly Installment $= 400$ 
$i =$ Rate of Interest/ 400 $= \dfrac {12}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{400 \times [(1+12/400)^{n} - 1]}{1-(1+12/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{400\times\dfrac{12}{400}}{1-(\dfrac{412}{400})^{-1/3}}$
$M = 1223.92$

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.