Tag: banking

Questions Related to banking

$A$ started a business with a capital of $Rs.10000$. Four month later, $B$ joined him and $2$ more moth letter , $C$ joined them in the partnership. If at the end of the year, all of them got an equal shares of profit, what was $C's$ investment? 

  1. $Rs.15000$

  2. $Rs.20000$

  3. $Rs.18000$

  4. $Rs.22000$


Correct Option: A

A person pays $ $400$ every year as loan installments to a bank. If every year bank increases the installment amount by $10$%, the find the total amount he pays in installments in $4$ years. 

  1. $ $1324 $

  2. $ $1456.4 $

  3. $ $1856.4 $

  4. $ $884 $


Correct Option: C
Explanation:

Given  person pays $400$ every as loan installments to bank

And every year bank increases installments $10\%$ every year 
Then after one year installments $=$ $400\times \dfrac{110}{100}=$440$
And after two year installments$440\times \dfrac{110}{100}=$484$
And after three year installments$484\times \dfrac{110}{100}=$532.40$
Then person paid installments in $4$ years $=$ $400+440+484+532.40=1856.40$

A man borrows Rs. $200$ at $ 5$% compound interest. At the end of each year he pays back Rs. $50$. At the end of $4 $ years he owes

  1. Rs. $27.59$

  2. Rs. $28.10$

  3. Rs. $27.81$

  4. Rs. $28.14$


Correct Option: A
Explanation:

Amount of one year

$\displaystyle=200\left[1+\frac{5}{100}\right]$

$\displaystyle=200\times\frac{21}{20}=210$

$\therefore$ At the end of year he pays back Rs. 50. So the principal for the second year is Rs. 160.
$\therefore$ Amount of Second year

$\displaystyle=160\left[1+\frac{5}{100}\right]$

$\displaystyle=160\times\frac{21}{20}=168$

$\therefore$ At the end of year he pays back Rs. 50. 
So rest amount = 168 -50 = Rs. 118.
This amount is principal amount for third year.
$\therefore$ Amount of third year

$\displaystyle=118\left[1+\frac{5}{100}\right]$

$\displaystyle=118\times\frac{21}{20}=123.90$

At the end of years he pays back Rs. 50.
So rest amount
$=123.90-50=Rs. 73.90$
This amount is principal for fourth year.
$\therefore$ Amount of fourth year

$\displaystyle=73.90\left[1+\frac{5}{100}\right]$

$\displaystyle=73.90\times\frac{21}{20}=77.59$

At the end of year he pays back Rs. 50.
So rest amount
$=77.595-50=Rs. 27.59$
$\therefore$ At the end of fourth year he owes Rs. 27.59.

Lakshman borrowed Rs. $20$ lakhs as housing loan from ICICI at $10\%$ p.a to be repaid in $10$ years. if the EMI is Rs. $2500$ per lakh, find how much he pays as interest in the first month. Find also he principal repaid then.

  1. Rs. $33333.34$

  2. Rs. $33344.64$

  3. Rs. $36543.45$

  4. Rs. $54600$


Correct Option: A
Explanation:

Interest at $10\%$ for the first month for Rs. $20$ lakhs $=$ $2000000 \times \dfrac{1}{12}\times \dfrac{10}{100}=16666.66$
EMI for one month, for $20$ lakhs $= 2500 \times  20 =$ Rs. $50000$
Hence principal repaid $=$ Rs. $50000 - 16666.66 =$ Rs. $33333.34$

A sum of Rs $550$ was taken as a loan. This is to be paid back in two equal instalments. If the rate of interest be $20\%$ compounded annually, then the amount of each instalment will be

  1. Rs $360$

  2. Rs $350$

  3. Rs $340$

  4. Rs $300$


Correct Option: B
Explanation:
Let $x$ be each installment
After paying the first installement $x$
the remaininng principle is $ 550\times1.2-x$
This then compounded yearly should be equal to the second installment
$\left(550\times1.2-x\right)\times 1.2= x$
$ 550\times 1.2^2-x\times 1.2 = x$
$792=2.2\times x$
$x=360$
Thus eachn installemnt should be $Rs.360$

A television set is sold for Rs. $10000$ cash on Rs. $2000$ cash down followed by six equal instalments of Rs. $1600$ each. What is the rate of interest?

  1. $50\%$

  2. $60\%$

  3. $70\%$

  4. $80\%$


Correct Option: B
Explanation:

Given, $n = 6, I =$ Rs. $1600$, 

$E = 2000 + 6 \times  1600 - 10000 = $ Rs. $1600$
We know $R = \dfrac{2400E}{n(n+1)I-2E}$
$\Rightarrow R = \dfrac{2400\times 1600}{6(6+1)1600-2\times 1600}$
$\Rightarrow R = 60\%$
Thus, the rate of interest is $60\%$.

Raghav buys a shop for $Rs. 1,20,000$. He pays half of the amount in cash and agrees to pay the balance in $12$ annual installments of $Rs. 5000$ each. If the rate of interest is $12\%$ and he pays with the installment the interest due on the unpaid amount find the total cost of the shop.

  1. $Rs. 1,60,800$

  2. $Rs. 1,66,800$

  3. $Rs. 1,68,800$

  4. $Rs. 1,60,000$


Correct Option: B
Explanation:
Given that: 
Raghav buys a shop for $Rs.1,20,000.$
He pays half of the amount in cash $= \dfrac{120000}{2}\Rightarrow Rs.60,000$

Balance amount to be paid $= 120000 - 60000 \Rightarrow Rs. 60000.$

Given that amount of each installment $=Rs. 5000.$

He agrees to pay the balance in $12$ annual installments with interest of $12\%.$

 Amount of the $1^{st}$ installment 
$\Rightarrow 5000 + \dfrac{12}{100}\times   60000$

$\Rightarrow 5000 + 600 \times 12$

$\Rightarrow 5000 + 7200$

$\Rightarrow Rs. 12,200.$


 Amount of the $2^{nd}$ installment
$ \Rightarrow 5000 + \dfrac{12}{100} \times (60000 - 5000)$
$\Rightarrow 5000 + \dfrac{12}{100}\times  55000$
$\Rightarrow 5000 + 550 \times 12$
$\Rightarrow 5000 + 6600$
$\Rightarrow Rs. 11,600.$

As the amount paid for installment is $12200,11600,....... $ so It forms an $AP.$

The first term $a = 12,200$
Common Difference $d =  11600 - 12200\Rightarrow-600$
Total number of terms $n = 12.$

We know that sum of $n$ terms in $AP$
$\Rightarrow \dfrac{n}{2}[2a + (n-1) d]$

 Therefore the total cost of the shop
 $\Rightarrow 60000 +\dfrac{ 12}{2}[2(12200) + (12-1) \times (-600)]$

$\Rightarrow 60000 + 6(24400 - 6600)$
$\Rightarrow 60000 + 6 \times 17800$
$\Rightarrow 60000 + 106800$
$=Rs. 1,66,800.$

Hence, the total cost of the shop $= Rs.1,66,800.$

What sun will become Rs 9826 in 18 months if the rate of interest is $\displaystyle 1\frac{1}{2}$% per annum and the interest is compounded half-yearly?

  1. Rs 9466.54

  2. Rs 9646.54

  3. Rs 9566.54

  4. Rs 9456.54


Correct Option: A
Explanation:

r $ = 2\dfrac {1}{2} $ % $ = \dfrac {5}{2} $ % n $ = 18 $ months $ = \dfrac {3}{2} $ years

When the interest is compounded half yearly,

$ A=P\left( 1+\dfrac { r }{ 2\times 100 }  \right) ^{ n\times 2 } $

$ => 9826 = P\left( 1+\dfrac { \dfrac {5}{2} }{ 2\times 100 }  \right) ^{ \dfrac {3}{2}\times 2 } $
$ => 9826 = P( \dfrac {81}{80}) ^{ 3 } $
$ => P =Rs 9466.54 $

The NAV of a unit in mutual fund scheme is Rs $10.65$, then find the amount required to buy $500$ such units.

  1. Rs.$5325$

  2. Rs.$5235$

  3. Rs.$53250$

  4. Rs.$5350$


Correct Option: A
Explanation:

Amount required to buy $500$ such units

$=10.62\times500$
 $=Rs.5325$

What annual instalment will discharge a debt of Rs. $2710$ due in $4$ years at $7\%$ simple interest?

  1. Rs. $1000$

  2. Rs. $613$

  3. Rs. $225$

  4. Rs. $150$


Correct Option: B
Explanation:

Let the annual installment be $x$

Now, $ x$ accumulates an interest of $7.1$ every year
$\implies$
$x$ deposited in first year becomes $1.21x$ in next $3$ years
$x$ deposited in second year becomes $1.11x$ in next $2$ years
$x$ deposited in first year becomes $1.07x$ in next $4$ years
$\implies1.21x+1.41x+1.07x+x=2710\ \implies 4.92x=2710\ \implies x=Rs.613$