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 compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

Elsa puts Rs. $1000$ into an account to use for school expenses. The account earns $10\%$ interest, compounded annually. How much will be in the account after $3$ years?

  1. $1210$
  2. $2210$
  3. $3210$
  4. $4210$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $P = 1000, r = 10\%, n = 2$ years
We know the formula $A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
Putting all the given values in the formula, we get

$A = 1000\left [\left (1+\dfrac{10}{100}\right)^2\right]$
$A =$ Rs. $1210$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

Calculate the amount and the compound interest on a sum of Rs. $20000$ at the end of $3$ years at the rate of $10\%$ p.a. compounded annually.

  1. $6620$
  2. $5620$
  3. $3410$
  4. $2386$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $P = 20000, r = 10\%, n = 3$ years
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$A = 20000\left [\left (1+\dfrac{10}{100}\right)^3\right]$
$A =$ Rs. $26620$
Compound interest $=$ Amount $-$ Principal
CI $= 26620 - 20000$
CI $=$ Rs. $6620$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

A man invests Rs. $3000$ for four years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs. $4500$. Calculate the rate of interest per annum.

  1. $10\%$
  2. $20\%$
  3. $50\%$
  4. $30\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $P =$ Rs. $3000$

At the end of $1$ year, the investment amounts to Rs. $4500$.
Thus, $A = $ Rs. $4500$
Interest $= A - P$
Therefore, $I = 4500 - 3000 = 1500$, $T = 1 $ year.
$R = \dfrac{1500\times 100}{3000\times 1}$
$R = 50\%$
Therefore, the rate of interest per annum is $50\%$.

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

A sum of Rs. $1728$ becomes Rs. $3375$ in $3$ years at compound interest, compound annually. Find the rate of interest.

  1. $10\%$
  2. $15\%$
  3. $20\%$
  4. $25\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given: $P = 1728, A = 3375, n = 23$ years

We need to find rate of interest i.e. $r\%$
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$\Rightarrow 3375 = 1728\left [\left (1+\dfrac{r}{100}\right)^3\right]$
$\Rightarrow \dfrac{3375}{1728}=\left (1+\dfrac{r}{100}\right)^3$
$\Rightarrow \left (\dfrac{12}{15}\right)^3 = \left (1 +\dfrac{r}{100}\right)^3$
Cubing on both the sides, we get
$\dfrac{12}{15}=1+\dfrac{r}{100}$
Thus $r = 20\%$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

Raghu borrowed Rs. $25,000$ at $20\%$ p.a. compounded half-yearly. What amount of money will discharge his debt after $1 \displaystyle \frac{1}{2}$ years ?

  1. Rs. $28,275$
  2. Rs. $30,275$
  3. Rs. $33,275$
  4. Rs. $35,275$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have, $P=$ Rs. $25,000,\,R=20\%\,p.a.=\dfrac{20}{2}\%\ \text{,half-year}$ and $T=1\dfrac{1}{2}$ years $=3$ half-years.

$\Rightarrow$  $A=P\left (1+\dfrac{R}{100}\right)^T$.
$\Rightarrow$  $A=25,000\times \left (1+\dfrac{20}{2\times 100}\right)^3$
$\Rightarrow$  $A=25,000\times \left (\dfrac{11}{10}\right)^3$
$\Rightarrow$  $A=25,000\times \dfrac {1331}{1000}$
$\Rightarrow A=$ Rs. $33,275.$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

What is the difference between the compound interests on Rs. $10,000$ for $\displaystyle1\frac{1}{2}$ years at $4\%$ per annum compounded yearly and half-yearly?

  1. Rs. $4.04$
  2. Rs. $4.08$
  3. Rs. $4.12$
  4. Rs. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Rightarrow$  C.I. when interest compounded yearly = $[10000\times (1+\dfrac{4}{100})^1\times (1+\dfrac{\dfrac{1}{2}\times 4}{100})]-10000$


$\Rightarrow$  C.I. when interest compounded yearly = $10000\times \dfrac{26}{25}\times \dfrac{51}{50}-10000$

$\therefore$   C.I. when interest compounded yearly = $10608-10000=Rs.608$
$\Rightarrow$ C.I. when interest is compounded half-yearly = $[10000\times (1+\dfrac{4}{2\times 100})^3]-10000$

$\Rightarrow$   C.I. when interest is compounded half-yearly = $[10000\times (\dfrac{51}{50})^2]-10000$

$\Rightarrow$   C.I. when interest is compounded half-yearly = $10612-10000=$ Rs.$612.08$
$\therefore$    Difference between C.I = Rs $612.08-$Rs.$608=$Rs.$4.08$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

A sum of Rs. $15,000$ is invested for $3$ years at $10 \%$ per annum compound interest. Calculate the interest for the second year.

  1. Rs. $1,680$
  2. Rs. $1,650$
  3. Rs. $1,710$
  4. Rs. $1,640$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Here $P=$ Rs. $15,000$ and $R=10\%$.

C.I. for second year $=$ $P\times \dfrac{R}{100}\times \left (1+\dfrac{R}{100}\right)$
C.I. for second year $=$ $15000\times \dfrac{10}{100}\times \left (1+\dfrac{10}{100}\right)$
$=$ $1500\times \dfrac{11}{10}$
$=$ Rs. $1650.$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

Aman borrowed Rs.$1,20,000$ for $2$ years at $8$ % per year compound interest. Calculate the final amount at the end of the second year.

  1. $1,39,968$
  2. $1,38,968$
  3. $1,39,743$
  4. $1,39,928$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Rightarrow$  Here, $P=$Rs.$1,20,000,\,T=2\,$years and $R=8\%$

$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$

$\Rightarrow$  $A=120000\times (1+\dfrac{8}{100})^2$

$\Rightarrow$  $A=120000\times (\dfrac{27}{25})^2$

$\Rightarrow$  $A=$Rs.$1,39,968$.

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

Find the compound interest on Rs. $2,000$ for $2$ years, compounded annually at $10\%$ per annum.

  1. $400$
  2. $410$
  3. $420$
  4. $430$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$


$\Rightarrow$  $A=2000\times (1+\dfrac{10}{100})^2$

$\Rightarrow$   $A=2000\times \dfrac{121}{100}$

$\Rightarrow$  $A=Rs.2420.$

$\therefore$    $C.I.=A-P=Rs.2420-Rs.2000=Rs.420.$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

Find the compound interest on Rs. $20,000$ for $2$ years, compounded annually at $10\%$ per annum.

  1. $4200$
  2. $4300$
  3. $4400$
  4. $4500$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$


$\Rightarrow$  $A=20000\times (1+\dfrac{10}{100})^2$

$\Rightarrow$  $A=20000\times \dfrac{121}{100}$

$\Rightarrow$  $A=Rs.24,200.$

$\therefore$  $C.I.=A-P=Rs.24,200-Rs.20,000=Rs.4200.$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

Find the compound interest on Rs. $100,000$ for $2$ years, compounded annually at $10\%$ per annum.

  1. $20000$
  2. $21000$
  3. $22000$
  4. $23000$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Principal $=100,000$
$t=2 $ years
$r=10 $% per annum


$A=P(1+\dfrac{R}{100})^T$

$\Rightarrow$  $A=100000\times (1+\dfrac{10}{100})^2$

$\Rightarrow$  $A=100000\times \dfrac{121}{100}$

$\Rightarrow$  $A=Rs.1,21,000$

$\Rightarrow$  $C.I.=A-P=Rs.1,21,000-Rs.1,00,000=Rs.21,000$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

The Amount on Rs. $20,000$ at $\displaystyle6\frac{1}{4}\%$ per annum compunded annually for $2$ years $73$ days, is:

  1. $57205$
  2. $56405$
  3. $22860$
  4. $50000$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $P=$ Rs. $20,000,\,T=2$ years $73$ days $=2\dfrac{1}{5}$ years and $R=6\dfrac{1}{4}\%=\dfrac{25}{4}\%$

$\Rightarrow$ $A=20000\times \left (1+\frac{\frac{25}{4}}{100}\right)^2\times (1+\frac{\frac{1}{5}\times \frac{25}{4}}{100})$
$\Rightarrow$ $A=20000\times \left (1+\dfrac{1}{16}\right)^2\times \left (1+\dfrac{1}{80}\right)$
$\Rightarrow$ $A=20000\times \left (\dfrac{17}{16}\right)^2\times \left (\dfrac{81}{80}\right)$
$\Rightarrow$ $A=20000\times \dfrac{289}{256}\times \dfrac{81}{80}$
Therefore, $A=$ Rs. $22860$

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

What sum of money will amount to Rs. $18738$ in four years at $17$% per annum compounded yearly?

  1. $10,000$ Rs.
  2. $12,000$ Rs.
  3. $13,000$ Rs.
  4. $15,000$ Rs.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The genetic formula used in calculating compound interest is $A=P(1+r/n)^{(nt)}$
A = final amount (P+interest)
P = the principal amount
r = the annual interest rate
n = the number of times that interest is compounded per year
t = the number of years
Given : $A= Rs \,18738 ; R=17\%; t = 4\,year $
So,
$18738 = P(1+0.17)^{4}$
$18738=P(1.17)^{4}$
$18783=P(1.8738)$
$P= Rs 10,000$
Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

A sum of Rs.$15,000$ is invested for $3$ years at $13$ % per annum compound interest. Calculate the compound interest.

  1. $6500.435$
  2. $6689.245$
  3. $6643.455$
  4. $6276.585$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have,

$P=Rs.\ 15000$
$T=3\ years$
$R=13\%$
$C.I=?$

We know that
$A=P\left(1+\dfrac{R}{100}\right)^T$

$A=15000\left(1+\dfrac{13}{100}\right)^3$

$A=15000\left(\dfrac{113}{100}\right)^3$

$A=15000\left(\dfrac{113\times 113\times 113}{100\times 100\times 100}\right)$

$A=15\left(\dfrac{113\times 113\times 113}{1000}\right)$

$A=Rs.\ 21,643.455$

So, the compound interest
$=21.643.455-15000=Rs.\ 6643.455$

Hence, this is the answer.

Multiple choice mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

Raman borrowed Rs.$1,20,000$ for $4$ years at $8$ % per year compound interest. Calculate the final amount at the end of four years.

  1. $1,63,250$
  2. $1,53,250$
  3. $1,63,700$
  4. $1,66,250$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have,

$P=Rs.\ 120, 000$
$T=4\ years$
$R=8\%$
$A=?$

We know that
$A=P\left(1+\dfrac{R}{100}\right)^T$

So,
$A=120000\left(1+\dfrac{8}{100}\right)^4$

$A=120000\left(\dfrac{108}{100}\right)^4$

$A=120000\left(\dfrac{108\times 108\times 108\times 108}{100\times 100\times 100\times 100}\right)$

$A=12\left(\dfrac{11664\times 11664}{10000}\right)$

$A=Rs.\ 163258.675\approx Rs.\ 163250$

Hence, this is the answer.