Quantitative Aptitude

Simple and Compound Interest

3,394 Questions
All Simple interest and its inverse forms 768 Compound interest with annual compounding 742 Simple-interest growth and split sums 626 Difference between compound and simple interest 479 Half-yearly and quarterly compounding 252 Changing rates and fractional years 201 Installments (CI based) 190 Ratio of principal to interest 118 Population growth and depreciation 16

Simple and Compound Interest Questions

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

Edwin opened a savings account and deposited Rs. $3000$ as principal. The account earns $5\%$ interest, compounded annually. What is the balance after $10$ years?

  1. $4886$
  2. $4566$
  3. $4658$
  4. $5460$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $P = 3000, r = 5\%, n = 10$ years
We know $A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
Putting all the given values in the formula, we get

$A = 3000\left [\left (1+\dfrac{5}{100}\right)^10\right]$
$A =$ Rs. $4886$

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

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

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.

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 approx interest for the second year.

  1. $2100$
  2. $2200$
  3. $2300$
  4. $2400$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Interst for the first year 

$=Rs \cfrac{15000\times 13\times 1}{100}$
$=Rs1950$
Amount after the first year
$=Rs15000+Rs1950$
$Rs16950$
Interest for the second year
$=Rs\cfrac{16950\times 13\times 1}{100}$
$=Rs2203.5$
$=Rs2200$(approx)

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

Vikram borrowed Rs. $20,000$ for $4\dfrac{1}{2}$ years at $10\%$ per annum, compound annually. How much compound interest would he pay at the end of $4\dfrac{1}{2}$ years?

  1. $30711.22$
  2. $20711.22$
  3. $40711.22$
  4. $10711.22$
Reveal answer Fill a bubble to check yourself
D Correct answer
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. $20000, r = 10\%, n = 1$ and $t =$ $4\dfrac{1}{2}$ years
$A = 20000\left (1+\dfrac{0.1}{1}\right)^{1\times 4.5}$
$A = 20000\times 1.1^{4.5}$
$A = 20000\times 1.535561$
$A =$ Rs. $30711.22$
To find interest we use formula $A = P + I$, since $A = 30711.22$ and $P = 20000$ we have:
$A = P + I$
$30711.22 = 20000 + I$
$I = 30711.22 - 20000 = 10711.22$
Interest, I $=$ Rs. $10711.22$

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

Joshita is having a bank account whose principal is Rs. $12000$ and her bank compounds the interest thrice a year at an interest rate of $15\%$, how much money did she have in her account at the year's end?

  1. $18250.50$
  2. $28250.50$
  3. $38250.50$
  4. $48250.50$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given: $P = 12000, r = 15\%, n = 3$ years
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$A = 12000\left [\left (1+\dfrac{15}{100}\right)^3\right]$
$A =$ Rs. $18250.50$

Multiple choice book keeping and accountancy bill of exchange (trade bill) dishonour of a bill dishonour of bills bills of exchange advantages of bill of exchange

X sold goods to Y on 1st June for 1,50,000. Y immediately accepted a three months bill. On due date Y requested that the bill be renewed for a fresh period of two months. X agrees provided interest at 9% p.a. was paid immediately is cash. What will be the amount of interest in the books of X?

  1. 2,000.

  2. 2,500.

  3. 2,250.

  4. 2,800.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Interest = Principal * Rate * Time. 1,50,000 * 0.09 * (2/12) = 2,250.

Multiple choice business economics and quantitative methods measures of dispersion and skewness shortcut method to find variance and standard deviation variance and standard deviation measures of dispersion

Find the present value of Rs. $10,000$ to be required after $5$ years if the interest rate be $9\%$. Given that $(1.09)^5=1.5386$.

  1. $6,994.42$
  2. $6,949.24$
  3. $6,449.24$
  4. $6,499.42$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, $i=0.09=9\%$
$n=5$
$A _n=10,000$
Required present value $=\displaystyle\frac{A _n}{(1+i)^n}$
$=\displaystyle\frac{10,000}{(1+0.09)^5}$
$=Rs. 6499.42$.

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

Divide Rs. $6500$ in two parts, such that if one part is lent out at $9\%$ per annum and other at $10\%$ per annum, the total yearly income(income from simple interest) is Rs.$605$.

  1. $2000,\,4000$
  2. $2000,\,4500$
  3. $1500,\,4000$
  4. $1500,\,4500$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the on one part $=x$

      the other part $6500-x$
$\Longrightarrow \dfrac{9x}{100}+\dfrac{6500-x}{100}\times 10=605$
$\Longrightarrow 9x+65000-10x=60500$
$\Longrightarrow x=4500$
$6500-x=6500-4500=2000$