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 mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

Gopal has a cumulative deposit account and deposits Rs. $900 $per month for a period of $4$ years. If he gets Rs.$ 52,020$ at the time of maturity, find the rate of interest.

  1. $5\%$
  2. $2\%$
  3. $10\%$
  4. $12\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Installment per month $\left( P \right) = Rs. 900$
No. of months $\left( n \right) = 4 \text{ years} = 12 \times 4 = 48 \text{ months}$
Let rate of interest be $r \%$ per annum
$t = \cfrac{n \left( n + 1 \right)}{2\times 12} = \cfrac{48 \times 49}{24} = 98$
$\therefore \; S.I. = P \times \cfrac{n \left( n + 1 \right)}{2\times 12} \times \cfrac{r}{100}$
$\Rightarrow \; S.I. = 900 \times \cfrac{48 \left( 48 + 1 \right)}{2\times 12} \times \cfrac{r}{100} = Rs. 882 r$
Maturity value $= Rs. \left(900 \times 48 + 882 r \right) = Rs \left( 43200 + 882 r \right)$
maturity value $= Rs. 52020$
$\therefore \; 43200 + 882 r = 52020$
$\Rightarrow \; 882 r = 52020 - 43200$
$\Rightarrow \; r = \cfrac{8820}{882} = 10 \%$
Hence, rate of interest $10 \%$.
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

If the compound interest on an amount of $29000$ in two years is $9352.5$, what is the rate of interest?

  1. $11\%$
  2. $9\%$
  3. $15\%$
  4. $18\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that 


$\Rightarrow Total\space amount=P(1+\dfrac{R}{100})^n$

Here $P=29000; \space n=2;\space interest=9352.5$

$\Rightarrow 29000+9352.5=(29000)(1+\dfrac{R}{100})^2$

$\Rightarrow 38352.5=(29000)(1+\dfrac{R}{100})^2$

$\Rightarrow 1.3225=(1+\dfrac{R}{100})^2$

$\Rightarrow 1+\dfrac{R}{100}=1.15$

$\Rightarrow \dfrac{R}{100}=0.15$

$\Rightarrow R=15$

Therefore, Rate of interest is $15\%$

Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

The difference between simple and compound interest on sum of $10000$ is $64$ for $2$ years. Find the rate of interest.  

  1. $8$
  2. $64$
  3. $4$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Simple Interest $=\dfrac{PNR}{100}$

Compound Interest $=P\left(1+\dfrac{R}{100}\right)^N-P$
Now,
$P\left(1+\dfrac{R}{100}\right)^N-P$ $-\dfrac{PNR}{100}=64$

$\left[10000\times \left(1+\dfrac{R}{100}\right)^2-10000\right]-\left(\dfrac{10000\times R\times 2}{100}\right)=64$

$\Rightarrow$  $10000\left[\left(1+\dfrac{R}{100}\right)^2-1-\dfrac{2R}{100}\right]=64$

$\Rightarrow$  $10000\left[\dfrac{(100+R)^2}{10000}-1-\dfrac{2R}{100}\right]=64$

$\Rightarrow$  $10000\left[\dfrac{10000+200R+R^2-10000-200R}{10000}\right]=64$

$\Rightarrow$  $R^2=64$

$\Rightarrow$  $R=8$

$\therefore$  $Rate=8\%$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

A certain amount of money deposited for compound interest, becomes 3 times in 3 years. In how many years will that amount be 27 times the deposited amount if it is given for the same rate of interest?

  1. 9

  2. 6

  3. 12

  4. 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$A=3P$

For $t=3$
So, $3P=P(1+\cfrac{R}{100})^3\implies R=(3^{2/3}-1)100$
Now, new amount $=27P$
So, $27P=P(1+\cfrac{R}{100})^t$
So, $\implies 27P=P(1+\cfrac{(3^{2/3}-1)100}{100})^t$
$\implies t=9$ years

Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

A sum of money doubled in $10$ years. The rate of interest per annum is?

  1. $20\%$
  2. $15\%$
  3. $18\%$
  4. $10\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the principal be P. Since the sum doubles in 10 years, the simple interest equals the principal (SI = P). Using the simple interest formula SI = (P * R * T) / 100, we substitute P = (P * R * 10) / 100, which simplifies to 100 = 10 * R, yielding R = 10%.

Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

If simple interest on a sum of money for $3$ years is Rs. $240$ and compound interest on the sum at same rate for $2$ years is Rs. $170$, then the rate $\%$ p.a. is 

  1. $16\%$
  2. $8\%$
  3. ${ 12 }\dfrac12\%$
  4. ${ 1 }\dfrac18\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
simple interest for one year$=\dfrac{240}{3} =Rs. 80$

simple Interest for two year$=80×2= Rs.160$

Compound interest for two years$=Rs. 170$

Difference for $2$ year$=170−160=Rs.10$

Hence

$Rate( \%)=\dfrac{10}{80}×100$

$=12\dfrac{1}{2}\%$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

Manish invested a sum of money at CI. It amounted to Rs 2420 in 2 years and Rs 2662 in 3 years. Find the rate percent per annum.

  1. 5%

  2. 10%

  3. 20%

  4. 15%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Last year interest $= 2662 - 2420 =\ Rs. 242$

Difference between SI and CI for $2$ years, Difference $= P{\left[\dfrac{R}{100}\right]}^{2}$

$\therefore\,$Rate $\%=\dfrac{242\times 100}{2420\times 1}=10\%$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

The CI on a sum of Rs 625 in 2 years is Rs 51. Find the rate of interest.

  1. 4%

  2. 3%

  3. 2%

  4. 1%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know that $A = C.I + P$
$A=625+51=676$
Using the formula $A=P{\left[1+\dfrac{R}{100}\right]}^{n}$
$676=625{\left[1+\dfrac{R}{100}\right]}^{2}$
$\Rightarrow\,\dfrac{676}{625}={\left[1+\dfrac{R}{100}\right]}^{2}$
$\Rightarrow\,{\left[1+\dfrac{R}{100}\right]}^{2}=\dfrac{676}{625}$
$\Rightarrow\,1+\dfrac{R}{100}=\sqrt{\dfrac{676}{625}}$
$\Rightarrow\,1+\dfrac{R}{100}=\dfrac{26}{25}$
$\Rightarrow\,\dfrac{R}{100}=\dfrac{26}{25}-1=\dfrac{26-25}{25}=\dfrac{1}{25}$
$\Rightarrow\,R=\dfrac{100}{25}=4$
$\therefore\,$Rate of interest $R=4\%$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

At what rate per cent of simple interest will the interest on Rs.3,750 be one-fifth of itself in 4 years? To what will it amount in 15 years?

  1. 6 % and Rs.6,562.50

  2. 8 % and Rs.6,562.50

  3. 5 % and Rs.6,562.50

  4. 4 % and Rs.6,562.50
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Simple Interest $ I = \cfrac {PNR}{100} $
Given,
$ P = Rs  3,750 $
$ I = \cfrac {1}{5}  \times P =  Rs  750 $
$ N = 4  years $
$ R = ? $
So, $ => I = \cfrac {PNR}{100} $
$ => 750  = \cfrac {3,750 \times 4 \times R }{100} $
$ =>R = 5 $ %
And for $ 15 $ years, interest
$ I =  \cfrac {PNR}{100} $
$ => I  = \cfrac {3,750 \times 15 \times 5 }{100} = Rs. 2,812.5 $
And Amount $ = I + P = Rs.  3,750  + Rs.  2,812.5  = Rs.  6,562.5  $

Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

In a simple interest. at what rate percent per annum will a sum of money double in 8 years?

  1. $12.5 \%$
  2. $10.5 \%$
  3. $12.0 \%$
  4. $15.5 \%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the initial sum of money be $P$.

Let time in years be $t$ and rate be $r$.

$Final sum = 2\times P$
$Simple interest = \dfrac {P\times t\times r}{100}$
$Total sum = P+\dfrac {P\times t\times r}{100}$
$2P=P+\dfrac{P\times t\times r}{100}$
$P = \dfrac{P\times t\times r}{100}$
given time = 8 years
$r=100/8=12.5\% $

Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

A certain sum of money amounts to $Rs.\,756$ in $2$ years and to $Rs.\,873$ in $3\displaystyle\frac{1}{2}$ years at a certain rate of simple interest. What is the rate of interest per annum?

  1. $\;11\%\,p.a.$
  2. $\;12\%\,p.a.$
  3. $\;13\%\,p.a.$
  4. $\;14\%\,p.a.$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Amount in $2$ years $=$ Rs. $756$

Amount is $3\dfrac{1}{2}$ years $=$ Rs. $873$
$\therefore$ Interest for $1\dfrac{1}{2}$ years $=$ Rs. $873-$ Rs. $756=$ Rs. $117$
Interest for $2$ years $=\dfrac{117}{\frac{3}{2}}\times 2=\dfrac{117\times 2\times 2}{3}=$ Rs. $156$
Interest for $1$ year $=\displaystyle\frac{117\times2}{3}=$ Rs. $78$
$\therefore$ Principal $=\text{Amount}-\text{Interest}$ ....(for $2$ years) $=$ Rs. $756-$ Rs. $156=$ Rs. $600$
$\therefore$ Rate of interest $=\displaystyle\frac{78\times100}{600\times1}=\,13\%$ p.a.

Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

A man invested Rs. $1000$ on simple interest at a certain rate and Rs. $1500$ at $2\%$ higher rate. The total interest in three years is Rs. $390$. What is the rate of interest for Rs. $1000$?

  1. $4\%$.
  2. $5\%$.
  3. $6\%$.
  4. $7\%$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the interest rate at which $Rs.\,1000$ is invested is $r\%$ 
Then Rs. $1500$ is invested at $(r+2)\%$

Then according to the question, we have
$\displaystyle\frac{1000\times\,r\times\,3}{100}+\displaystyle\frac{1500\times(r+2)\times3}{100}=390$
$\Rightarrow\;30r+45r+90=390$
$\Rightarrow\;75r=300$
$\Rightarrow\;r=4\%$

Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

A person lends $40\%$ of his sum of money at $15\%\,p.a.$, $50\%$ of rest at $10\%\,p.a.$ and the rest at $18\%\,p.a.$ rate of interest. What would be the annual rate of interest, if the interest is calculated on the whole sum?

  1. $\;13.4\%$
  2. $\;14.33\%$
  3. $\;14.4\%$
  4. $\;13.33\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the whole sum be Rs. $100$. 

Then, sum at $15\%$ p.a. $=$ Rs. $40$
 Remaining sum $=$ Rs. $60$
$\therefore$ Sum at $10\%$ p.a. $=50\%$ of Rs. $60=$ Rs. $30$ and sum at  $18\%$ p.a. $=$ Rs. $30$

$\therefore \text {S.I.} $ on Rs. $100$ for $1$  year $=\begin{pmatrix}40\times\displaystyle\frac{15}{100}\times1\end{pmatrix}+\begin{pmatrix}30\times\displaystyle\frac{10}{100}\times1\end{pmatrix}+\begin{pmatrix}30\times\displaystyle\frac{18}{100}\times1\end{pmatrix}$
$ =$ Rs. $(6+3+5.4)=$ Rs. $14.4$
Hence, required rate $=14.4\%$.

Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

The compound interest on a sum for two years is Rs. $832$ and the simple interest on the same sum at the same rate for the same period is Rs. $800$. What is the rate of interest ?

  1. $6\%$
  2. $8\%$
  3. $10\%$
  4. $12\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the sum be Rs. $P$ and rate of interest per annum be $R\%$
Then $\displaystyle P\left [ \left ( 1+\frac{R}{100} \right )^{2}-1 \right ]-\frac{2PR}{100}=$ Rs. $832-$ Rs. $800=$ Rs. $32$
$\displaystyle \Rightarrow P\left [ 1+\frac{2R}{100}+\frac{R^{2}}{10000}-1 \right ]-\frac{2PR}{100}=32$
$\displaystyle \Rightarrow \frac{PR^{2}}{10000}=32$

$\Rightarrow PR\times R=320000$ ..........(i)
Also $\displaystyle \frac{2PR}{100}=800$ (S.I)
$\Rightarrow PR=40000$ .........(ii)
$\displaystyle \therefore$ From (i) and (ii), we have
$ 40000 \times  R = 320000$  

$\displaystyle \Rightarrow$ $R=8\%$ p.a.