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Questions Related to banking

Mr. Sen has a savings bank account with a Post Office. Calculate the interest canted by Mr. Sen during the year $2010$ at $6.5 \%$ per annum payable for the month of December if the entries during the year in his passbook are as given below:
Date Particulars Withdrwals (Rs.) Deposits(Rs.)
$2.1.10$ By Cash $250.00$
$9.1.10$ By Cheque $825.00$
$13.3.10$ To Cash $325.00$
$24.7.10$ By Cash $1,237.00$
$6.10.10$ To Cheque $250.00$
$22.12.10$ By Cheque $958.00$
  1. Rs. $78.35$

  2. Rs. $81.71$

  3. Rs. $72.58$

  4. Rs. $82.89$


Correct Option: B
Explanation:

In bank, the minimum balance From $10^{th}$ date to last day of month.

Then Product of month January, Ferbruay $= (250+825)\times 2=$ $1075\times2=2150$
Product of month march, April, may, JuneJuly $=(1075-325)\times5=$ $750\times5=3750$ 
Product of month August, September $=(750+1237)\times 2=$ $1987\times2=3974$
Product of month October, November and December $=(1987-250)\times 3=$ $1737\times3=5211$ 
Then total product for year $2010=2150+3750+3974+5211=15085$.Rs
Then interest Payble at $6.5\%$ p.a for the month of December $=$ $\dfrac{15085\times 6.5}{100\times 12}=$ Rs. $81.71$  

If the interest is calculated at $6\%$ p.a. and is compounded at the end of March and September every year, The interest earned up to $31^{st}$ March and then after completing all the entries, find the amount that the account holder would have received had he closed the account on $20^{th}$ Oct. the same year. A page from the passbook of a Saving Book account in a particular year is given below:
Date Particulars Debit (Rs.) Credit (Rs.) Balance (Rs.)
Jan. $13$ By Cash $5,000.00$ $5,000.00$
Feb. $13$ To self $500.00$
March $24$ By Cheque $2,000.00$
March $31$ By Interest
May $20$ By Cash $800.00$
July $7$ To Cheque $1,400.00$
July $18$ By Cash $1,600.00$
Sept. $15$ To Cheque $3,200.00$
Sept. $26$ By Cheque $2,350.00$
  1. Rs. $4517.86$

  2. Rs. $3890.1$

  3. Rs. $4329.39$

  4. Rs. $6898.10$


Correct Option: D
Explanation:

In bank gives interest on minimum balance between 10 Th to last date of month in S.B A\c

Date       Particulars          Debit(Rs)      Credit (Rs)     Balance (Rs )
Jan 13      By cash                                    5000.00       5000.00
Feb 13      To self                  500.00                             4500.00
March 24  By cheque                              2000.00        6500.00
March 31  By interest                                     45.00       6545.00
May 20     By cash                                        800.00      7345.00
July 7        To cheque         1400.00                               5945.00
July 18      BY cash                                        1600.00     7545.00
Sept 15     To cheque          3200.00                              4345.00
Sept 16     By cheque                                   2350.00     6695.00
Sept 30    By interest                                      203.10      6898.10
Oct 20      To A\C closed    6898.10                                  NIL  

Then product for the month Feb and March=$4500\times2=Rs 9000$
Then interest =$\dfrac{9000\times 6}{12\times 100}= 45$ Rs
Then product for month April and May=$6545\times 2=Rs 13090$
  Or product for the month June=Rs 7345
Product for the month July =Rs 5945
Product for the month August =Rs 7545
product for the month Sept=Rs 6695
Then total product up to month Sept=13090+7345+5945+7545+6695=40620 Rs
Then interest up to Sept =$\dfrac{40620\times 6}{12\times 100}= 203.10$Rs
The a|c closed on 20 Th Oct Then no interest paid for the month Oct
Then amount paid Rs. 6898.10.
    

How much interest will you earn if your saving bank account has Rs. $10,000,000$ for $20$ days at $3\%$ per annum is deposited.

  1. $1,043.83$

  2. $1,243.83$

  3. $1,643.83$

  4. $1,603.83$


Correct Option: C
Explanation:

Using the formula,
Saving account interest $=$ Principal or amount in the account $\times$ Number of days $\times$ Daily Interest Rate
At $3\%$ daily interest rate $=$ $\dfrac{3%}{365}$
$=$ $\dfrac{10,000,000\times20\times3}{100\times365}$
$= 1,643.83$

Mark invests Rs. $6500$ in a savings account his annual interest rate is $7\%$ compounded annually. What is the approximate balance of his savings account after $2\dfrac{1}{2}$?

  1. $6500$

  2. $5500$

  3. $7700$

  4. $8200$


Correct Option: C
Explanation:

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Where,
$A =$ total amount
$P =$ principal or amount of money deposited,
$r =$ annual interest rate
$n =$ number of times compounded per year
$t =$ time in years
Given: $P =$ Rs. $6500, r = 7\%, n = 1$ and $t =$ $2\dfrac{1}{2}$ years
$\Rightarrow A = 6500\left (1+\dfrac{0.07}{1}\right)^{1\times 2.5}$
$\Rightarrow A = 6500\times 1.07^{2.5}$
$\Rightarrow A = 6500\times 1.184294$
$\Rightarrow A =$ Rs. $7697.91$ $\text{approx}$ $7700$

How much interest will you earn if your saving bank account has Rs. $30,000,000$ for $30$ days at $12\%$ per annum is deposited.

  1. $195,890.411$

  2. $295,890.411$

  3. $395,890.411$

  4. $495,890.411$


Correct Option: B
Explanation:

Using the formula,
Saving account interest $=$ Principal or amount in the account $\times$ Number of days $\times$ Daily Interest Rate
At $12\% $ daily interest rate $=$ $\dfrac{12%}{365}$
$=$ $\dfrac{30,000,000\times30\times12}{100\times365}$
$= 295,890.411$

If you have a bank account whose principal is Rs. $5000$, and your bank compounds the interest twice a year at an interest rate of $12\%,$ how much money do you have in your account at the year's end?

  1. $4272$

  2. $5272$

  3. $6272$

  4. $7272$


Correct Option: C
Explanation:

Given: $P = 5000, r = 12\%, n = 2$ years
We know the formula $A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
Substituting the given values int he formula, we get

$A = 5000\left [\left (1+\dfrac{12}{100}\right)^2\right]$
$A =$ Rs. $6272$

Mari deposited Rs. $20000$ in a savings bank account. She would be paid interest at $12\%$ per annum compounded annually. Find the interest to her credit at the end of second year.

  1. $1088$

  2. $3088$

  3. $5088$

  4. $7088$


Correct Option: C
Explanation:

Given, $P = $ Rs. $20000$, $r = 12\%$, $n = 1$, $t = 2$ years
$A =$ $P\left (1+\dfrac{r}{n}\right)^{nt}$
$A =$ $20000\times (1+0.12)^{1\times 2}$
$A = 20000 \times  1.2544$
$A = Rs. 25088$
Amount $=$ Principal $+$ Interest
Interest $= A - P$
$= 25088 - 20000$
$=$ Rs. $5088$

What will a deposit of Rs. $4,500$ at $10\%$ in a savings account compounded yearly interest be worth if left in the bank for $9$ years?

  1. $2110.77$

  2. $4110.77$

  3. $6110.77$

  4. $8110.77$


Correct Option: C
Explanation:

Given, $P =$ Rs. $4500$, $r = 10\%$, $n = 1$, $t = 9$ years
$A =$ $P(1+0.1)^{1.9}$
$A = 4500 \times 2.357948$
$A =$ Rs. $10610.77$
Amount $=$ Principal $+$ Interest
Interest $= A - P$
$= 10610.77 - 4500$
$=$ Rs. $6110.77$

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, 
$M =\dfrac{2000 \times [(1+5/400)^{n} - 1]}{1-(1+5/400)^{\frac{-1}{3}}}$ 
$M = \dfrac{2000\times\dfrac{5}{400}}{1-(\dfrac{405}{400})^{-1/3}}$
$M = 6049.931$