Tag: maturity value of recurring deposits

Questions Related to maturity value of recurring deposits

Mr Bittu deposits a certain sum of money each month in a recurring deposit account. If the rate of interest is $8\%$ per annum and Mr Bittu gets Rs $8,088$ from the bank after $3$ years. Find the value of his monthly instalment.

  1. $400$

  2. $800$

  3. $200$

  4. $140$


Correct Option: C
Explanation:

Let the monthly installment be $P$

Given, maturity value $=8088$, $n=3$ years $=3\times 12=36$ months, $r=8\%$
Interest $=\dfrac {Pn(n+1)r}{2400}$
$=\dfrac {P\times 36\times 37 \times 8}{2400}$
$=\dfrac {111P}{25}$
We know, amount $=Pn+\dfrac {111P}{25}$
$8088=36P+\dfrac {111P}{25}$
$=\dfrac {1011P}{25}$
$\therefore 8088= \dfrac {1011P}{25}$
$\therefore P=200$
Therefore, monthly installment  is Rs. $200$.

Divya has a recurring deposit in a bank at $ 5\%$ per annum simple interest. If she pay monthly installment of Rs. $2000$ for annually. Find her Maturity value.

  1. $6049.931$

  2. $6249.931$

  3. $6149.931$

  4. $6549.931$


Correct Option: A
Explanation:

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\dfrac{-1}{3}}}$ 
R $=$ Monthly Installment $= 2000 $
i $=$ Rate of Interest $5\%$
n $=$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{2000 \times [(1+5/400)^{n} - 1]}{1-(1+5/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{2000\times\dfrac{5}{400}}{1-(\dfrac{405}{400})^{-1/3}}$
$M = 6049.931$

Sheela has a recurring deposit in a bank at $2\%$ per annum simple interest. If she pay monthly installments Rs.$200$ for annually. Find her Maturity value.

  1. $621.99$

  2. $611.99$

  3. $631.99$

  4. $601.99$


Correct Option: D
Explanation:

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R =$ Monthly Installment $= 200$ 
$i =$ Rate of Interest/ 400 $= \dfrac {2}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{200 \times [(1+2/400)^{n} - 1]}{1-(1+2/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{200\times\dfrac{2}{400}}{1-(\dfrac{402}{400})^{-1/3}}$
$M = 601.99$

______ type of account is operated by salaried persons and petty traders.

  1. fixed deposit

  2. save deposit

  3. recurring deposit

  4. current deposit


Correct Option: C
Explanation:

Recurring deposit type of account is operated by salaried persons and petty traders.

Rita has a recurring deposit in a bank at $4\%$ per annum simple interest. If she pay monthly installment s Rs.$1000$ for annually. Find her Maturity value.

  1. $3019.97$

  2. $4019.97$

  3. $2019.97$

  4. $5019.97$


Correct Option: A
Explanation:

Using the formula,
$M =\dfrac{R \times [(1+i)^{n} - 1]}{1-(1+i)^{\frac{-1}{3}}}$ 
$R $= Monthly Installment $= 1000 $
$i =$ Rate of Interest/ 400 $= \dfrac {4}{400}$
$n =$ Number of Quarters $= 1$
Substitute the values, 
$M =\dfrac{1000 \times [(1+4/400)^{n} - 1]}{1-(1+4/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{1000\times\dfrac{4}{400}}{1-(\dfrac{404}{400})^{-1/3}}$
$M = 3019.97$

Ahmed has a recurring deposit account in a bank.He deposits Rs $5,500$ per month for 2 years.If he gets Rs $1,46,437.5$ at the time of maturity. Find the interest paid by the bank.

  1. $10.5$%

  2. $30$%

  3. $35$%

  4. None of the above


Correct Option: A
Explanation:

Given, maturity value $=146437.5$, monthly installment $=5,500$, $n=2$ years $=2\times 12=24$ months

Interest $=\dfrac {Pn(n+1)r}{2400}$
$=\dfrac {5500\times 24 \times 25 \times r}{2400}$
$=1375r$
Therefore, amount $=(Pn+1375r)$
$\Rightarrow 146437.5=(5500\times 24+1375r)$
$\Rightarrow 146437.5=132,000+1375r$
$\Rightarrow 14437.5=1375r$
$\Rightarrow  r=10.5$

Nithya deposited Rs. $100$ per month for $12$ months in a bank's recurring deposit account. If the bank pays interest at the rate of $10\%$ per annum, find the amount she gets on maturity.

  1. $1205$

  2. $1265$

  3. $1345$

  4. $1450$


Correct Option: B
Explanation:

Given: $P = 100, T(n) = 12$ months, $R = 10\%$
Equivalent principal for $12$ months $=$ $100\times \dfrac{n(n+1)}{2}$
$=$ $100\times\dfrac{12(13)}{2}$
$= 50 \times  12 \times  13$
Interest $=$ $\dfrac{PRT}{100}$
$=$ $\dfrac{50\times 12 \times 13\times 10\times 1}{100\times 12}$
$=$ $65$
Maturity amount $=$ P$\times$T $+$ Interest
$=$ $100 \times 12 + 65$
$=$ Rs. $1265$

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.

  1. $6811.20$

  2. $2000$

  3. $4811.20$

  4. $3811.20$


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.

  1. $2034.904$

  2. $2134.904$

  3. $3134.904$

  4. $4134.904$


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.

  1. $1123.92$

  2. $1223.92$

  3. $1623.92$

  4. $1823.92$


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, 
$M =\dfrac{400 \times [(1+12/400)^{n} - 1]}{1-(1+12/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{400\times\dfrac{12}{400}}{1-(\dfrac{412}{400})^{-1/3}}$
$M = 1223.92$