Quantitative Aptitude · Commerce Accountancy

Interest and Annuities

638 Questions

Interest and annuities represent a critical quantitative aptitude section focusing on the mathematical calculation of simple interest, compound interest, and future values of investments. Questions challenge candidates to determine maturity values, compute recurring deposit returns, and calculate prevailing interest rates. Mastery of this topic is essential for scoring high in banking and SSC examinations.

Simple and compound interestFuture value of annuitiesRecurring deposit calculationsInterest rate determinationPresent value formulas

Interest and Annuities Questions

Multiple choice maturity value of recurring deposits banking maths

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

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

Multiple choice maturity value of recurring deposits banking maths

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.

  1. $2$ years
  2. $3$ years
  3. $4$ years

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice maturity value of recurring deposits banking maths

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?

  1. Rs 6000

  2. Rs 9000

  3. Rs 8000

  4. Rs 7000
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice maturity value of recurring deposits banking maths

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.

  1. Rs.$8412$
  2. Rs.$8421$
  3. Rs.$2481$
  4. Rs.$1234$
Reveal answer Fill a bubble to check yourself
B Correct answer
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. $8421$

Multiple choice investement and financial planning banking compound interest comparing quantity maths

You invest Rs. $3,000$ in a two year investment that pays you $12\%$ p.a. Calculate the future value of the investment.

  1. Rs. $3,367.20$
  2. Rs. $3,673.20$
  3. Rs. $3,763.20$
  4. Rs. $3,736.20$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$F=C.F. (1+i)^n$
Where, $F=$ Future value
$C.F. =$ Cash flow $=$Rs. $3,000$
$i=$ rate of interest $=0.12$
$n=$ time period $=2$
$F=$Rs. $3,000(1+0.12)^2$
$=$Rs. $3,000\times 1.2544$
$=$Rs. $3,763.20$

Multiple choice investement and financial planning banking compound interest comparing quantity maths

A man borrows $Rs. 6000$ at $5\% $ $C.I.$ per annum$.$ if the repays $Rs.1200$ at the end of the each year$,$ find the amount of the loan outstanding at the beginning of the third year$.$ 

  1. Rs 4155.0

  2. Rs 5555.5

  3. Rs 5452.0

  4. Rs 4452.5

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given Principle amount $= Rs . 6000$
And$,$
Rate of interest $= r = 5\% $ compounded annually

So$,$
Interest after $1 year = 6000 × \dfrac{5}{100} × 1 = 60× 5 = Rs . 300$
Total money owed after $1 year = 6000 + 300 = 6300$

And$,$
$Rs. 1200$ paid $,$ So
Total money starting of second year $= 6300 - 1200 = Rs.5100$

And$,$
Interest after $2 year = 5100 × 5 × 1100 = 51× 5 = Rs . 255$
money owned after $2 year = 5100 + 255 = 5355$

And$,$
$Rs. 1200$ paid $,$ So
Total money outstanding starting of Third year $= 5355 - 1200$ $= Rs.4155$

Hence,
option $(A)$ is correct answer.
Multiple choice investement and financial planning banking compound interest comparing quantity maths

A man borrowed Rs.4000 at 10% per annum compound interest.At the end of each year he has repaid Rs.1000.The amount of money he still incurs after the third year is -------- .

  1. Rs.2740

  2. Rs.2104

  3. Rs.2014

  4. Rs.3400
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Year 1: 4000 * 1.1 = 4400; 4400 - 1000 = 3400. Year 2: 3400 * 1.1 = 3740; 3740 - 1000 = 2740. Year 3: 2740 * 1.1 = 3014; 3014 - 1000 = 2014.

Multiple choice investement and financial planning banking compound interest comparing quantity maths

Mr. Dua invested money in two schemes P and Q offering compound interest @ 8 p.c.p.a. and 9 p.c.p.a respectively. if the total amount of interest accrued two schemes together in two years was Rs 4818.30 and the total amount invested was Rs 27, 000, what was the amount invested in Scheme P?

  1. Rs 12, 000

  2. Rs 13,500

  3. Rs 15, 000

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let p invested Rs x Then q invested Rs $(27000-x)$
$\therefore x(1+\dfrac{8}{100})^{2}-1+(27000-x)(1+\dfrac{9}{100})^{2}-1=4818.30$
$\Rightarrow (x\times \dfrac{104}{625})+\dfrac{1881(27000-x)}{10000}= \dfrac{481830}{100}$
$\Rightarrow 1664x+1881(27000-x)=48183000$
$\Rightarrow (1881x-1664x)=50787000-48183000$
 Or $217x=2604000$
  Or $x=12000 Rs$

Multiple choice investement and financial planning banking compound interest comparing quantity maths

A family made a down payment of \$75 and borrowed a set of encyclopedias that cost \$400. The balance with interest was paid in 23 monthly payments of \$16 each and a final payment of \$9. What was the per cent of interest to the borrowed sum?

  1. 12 %

  2. 14 %

  3. 16 %

  4. 18 %
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total cost =\$ 400
Down payment=\$ 75
Remaining amount=$400-75=$ 325\$
Balanced paid in 24 months=\$23 \times 16+9=$ 377$
Difference=$377-325=$52$
The per cent of interest to the borrowed sum=$\frac{52}{325}\times 100=16$


Multiple choice investement and financial planning banking compound interest comparing quantity maths

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 $
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $=$ $400+440+484+532.40=1856.40$

Multiple choice investement and financial planning banking compound interest comparing quantity maths

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. $28.14$
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice investement and financial planning banking compound interest comparing quantity maths

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. $54600$
Reveal answer Fill a bubble to check yourself
A Correct answer
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. $33333.34$

Multiple choice investement and financial planning banking compound interest comparing quantity maths

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 $300$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $Rs.360$
Multiple choice investement and financial planning banking compound interest comparing quantity maths

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. $Rs. 1,60,000$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $= Rs.1,66,800.$
Multiple choice investement and financial planning banking compound interest comparing quantity maths

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $