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

Calculate the amount and the compound interest on:
Rs. 8,000 for $1\dfrac{1}{2}$ years at 10% per annum compounded yearly.

  1. Rs. 9,230 and Rs. 1,230

  2. Rs. 10,000 and Rs. 2,000

  3. Rs. 9,735 and Rs. 1,735

  4. Rs. 8,442 and Rs. 442
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Amount $A=P(1+r)^{t}$


Here, $P=8000, r=\dfrac{10}{100}=0.1, t=1\dfrac12=\dfrac32=1.5$

$\therefore A=8000(1+0.1)^{1.5}$
         $=8000(1.1)^{1.5}$
         $=8000(1.1537)$
         $=9230$

Compound interest$=9230-8000=1230$

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

Saurabh invests Rs. 48,000 for 7 year at 10% per annum compound interest. Calculate: The interest for the first year.

  1. Rs. 4,800

  2. Rs. 4,900

  3. Rs. 5,000

  4. Rs. 5,100
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$P=Rs.48000$


$Rate=10%$


$Time =1 year$

C.I for 1 year=S.I for 1 year$=\dfrac{PRT}{100}$

$\Rightarrow \dfrac{48000\times 10\times 1}{100}=Rs.4800$

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

Ramesh invests Rs. 12,800 for three years at the rate of 10% per annum compound interest. Find the sum due to Ramesh at the end of the first year

  1. Rs. $14,080$
  2. Rs. $15,691$
  3. Rs. $17,776$
  4. Rs. $148,342$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Principal$=Rs.12800$


Rate$=10$%

Time$=1$ year

C.I for 1 year $=$ S.I for 1 year$=\dfrac{PRT}{100}$

$\Rightarrow \dfrac{12800\times 10\times 1}{100}=Rs.1280$

$Amount=P+S.I=12800+1280=Rs.14080$

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

A sum of Rs. $13,500$ is invested at $16\%$ per annum compound interset for $5$ years. Calculate the interest for the first year

  1. Rs. $2,690$
  2. Rs. $2,460$
  3. Rs. $2,230$
  4. Rs. $2,160$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$P=13500$

$R=16$%
$T=1 year$

$\Rightarrow$ C.I for 1 year $=$ S.I of 1 year $=\dfrac{13500\times 16\times 1}{100}=Rs.2160$

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

Saurabh invests Rs. 48,000 for 7 year at 10% per annum compound interest. Calculate: The interest for the third year 

  1. Rs. 6,411

  2. Rs. 5,808

  3. Rs. 5,269

  4. Rs. 4,922
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Sum=rs.48000$


$Rate=10%$

$Time =2 \ year$

Amount after 2 year
$Amount=P\left(1+\dfrac{R}{100}\right)^t$

$\Rightarrow 48000\left(1+\dfrac{10}{100}\right)^2$

$\Rightarrow 48000\times \dfrac{110}{100}\times \dfrac{110}{100}=Rs.58080$

For third  year

$Sum=58080$

$Time =1 \ year$

Then amount after 3rd year$=58080\left(1+\dfrac{10}{100}\right)$

$\Rightarrow 58080\times \dfrac{110}{100}=Rs.63888$

$\therefore $Interest for third year$=63888-58080=Rs.5808$

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

A sum of Rs. $13,500$ is invested at $16\%$ per annum compound interest for $5$ years. Calculate the interest for the second year, correct to the nearest rupee.

  1. Rs. 2,106

  2. Rs. 2,279

  3. Rs. 2,506

  4. Rs. 2,792
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Sum=Rs.13500$

$Rate=16%$
Amount at the end of first year$=P\left(1+\frac{R}{100}\right)^T$

$\Rightarrow 13500\left(1+\dfrac{16}{100}\right)$

$\Rightarrow 13500\times \dfrac{116}{100}=Rs.15660$

For the second year

$Sum=Rs.15660$

Then amount after second year$=15660\left(1+\dfrac{16}{100}\right)$

$\Rightarrow 15600\times \dfrac{116}{100}=Rs.18165.6$

$\therefore$ Interest for second year$=18165.6-15660=Rs.2505.6=Rs.2506$

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

A sum of money placed out at compound interest amounts to Rs. 20,160 in 3 years and to Rs. 24,192 in 4 years. Calculate amount in 2 years.

  1. Rs. 12,500

  2. Rs. 16,800

  3. Rs. 19,600

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

Amount in three year $=Rs. 20160$


Amount in four year $=Rs.24192$

Interest in 1 year $=24192-20160=Rs.4032$

Let the rate of interest $=R$%

C.I fir 1 year=S.I for 1 year$=\frac{PRT}{100}$

$\Rightarrow 4032=\dfrac{20160\times R\times 1}{100}$

$\Rightarrow R=\dfrac{4032\times 100}{20160\times 1}=20$%

Amount in 3 years $=$ Rs. $20160$

Let the sum be x
$\therefore 20160=x \left(1+\dfrac{20}{100} \right)$

$\Rightarrow 20160=x\times \dfrac{120}{100}$

$\Rightarrow x=\dfrac{20160\times 100}{120}=Rs.16800$

Hence the amount in $2$ year is Rs. $16800.$


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

The compound interest, calculated by yearly, on a certain sum of money for the second year is Rs. 864 and the third year is Rs. 933.12. calculate the rate of interest and the compouned interest on the same sum and at the same rate, for the fourth year.

  1. 0
  2. $Rs. 1007.77$
  3. $Rs. 1000$
  4. $Rs. 1100$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rate of interest=$\frac{Difference  in  interest  of  two  consecutive periods\times 100}{C.I  of  preceding year\times Time}\times 100$

$\Rightarrow \frac{933.12-864}{864\times 1}\times 100$
$\Rightarrow \frac{69.12}{864}\times 100=8$%

C.I for the forth year=$933.12+8\% of 933.12$
$\Rightarrow 933.12+\frac{8}{100}\times 933.12$
$\Rightarrow 933.12+74.65=Rs.1007.77$


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

Find the amount of Rs. $  8000$ for $3$ years, compounded annually at $10 \%$ per annum. Also, find the compound interest.

  1. $2648$
  2. $2640$
  3. $2348$
  4. $2216$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Here, $P=$ Rs. $8000,R=10\%$ per annum and $n=3$ years

Using the formula, $A=P\left (1+\dfrac {R}{100}\right)^n$
Amount after $3$ years $=8000\times \left (1+\dfrac {10}{100}\right)^3$
$=8000\times \dfrac {11}{10}\times \dfrac {11}{10}\times \dfrac {11}{10}$
$=10648$
Thus, amount after $3$ years $=$ Rs. $1648$.
And compound interest $=$ Rs. $(10648-8000)=2648$.

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

If the present value of my investment is Rs. $2,000$ and the rate of interest is $5\%$ compounded annually, what will the value be after $5$ years?

  1. $2341.12$
  2. $1348.90$
  3. $2552.56$
  4. $3129.10$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, Principal $P=2,000, r=5\%, n=5$ years

We know the formula of compound interest,
$A = P\left (1+\dfrac {r}{100}\right)^n$
$ = 2000\left (1+\dfrac{5}{100}\right)^5$
$ = 2000\left (1+\dfrac{5}{100}\right)^5$
$ = 2000 \times 1.05^5$
$ =$ Rs. $2552.56$

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

The principal that amounts to Rs. $4913$ in $4\dfrac{1}{2}$ years at $6\%$ per annum compound interest, compounded annually, is

  1. $3179.81$
  2. $3579.81$
  3. $3779.81$
  4. $3979.81$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given, $A =$ Rs. $4913$, $ r = 6\%$, $n = 4$ and $t =$ $4\dfrac{1}{2}$ years
Therefore, $4913 = P\left (1+\dfrac{0.06}{1}\right)^{1\times 4.5}$
$\Rightarrow 4913 = P\times 1.06^{4.5}$
$\Rightarrow 4913 = P\times 1.2998$
$\Rightarrow P = \dfrac{4913}{1.2998}$
$\Rightarrow P =$ Rs. $3779.81$

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

On what sum will the compound interest for $4\dfrac{1}{2}$ years at $10\%$ amount to Rs. $4620$?

  1. $2008.67$
  2. $3008.67$
  3. $4008.67$
  4. $5008.67$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given, $A =$ Rs. $4620$, $r = 10\%$, $n = 1$ and $t =$ $4\dfrac{1}{2}$ years
Therefore, $4620 = P\left (1+\dfrac{0.1}{1}\right)^{1\times 4.5}$
$\Rightarrow 4620 = P\times 1.1^{4.5}$
$\Rightarrow 4620 = P\times 1.535561$
$\Rightarrow P = \dfrac{4620}{1.535561}$
$\Rightarrow P =$ Rs. $3008.67$

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

The compound interest on Rs. $4000$ at $3\%$ per annum for $3\dfrac{1}{2}$ years, compounded annually, is

  1. $135.99$
  2. $235.99$
  3. $335.99$
  4. $435.99$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given:
$P =$ Rs. $4000, r = 3\%, n = 1$ and $t = 3.5$ years
$\Rightarrow A = 4000\left (1+\dfrac{0.03}{1}\right)^{1\times 3.5}$
$\Rightarrow A = 4000\times 1.03^{3.5}$
$\Rightarrow A = 4000\times 1.108997$
$\Rightarrow A =$ Rs. $4435.99$
To find interest, we use formula 

$A = P + I$
Since $A = 4435.99$ and $P = 4000$, 
we have $A = P + I$
$\Rightarrow 4435.99 = 4000 + I$
$\Rightarrow I = 4435.99 - 4000 = 435.99$
Interest, $I =$ Rs. $435.99$

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

Lauren puts Rs. $600.00$ into an account to use for school expenses. The account earns $3\%$ interest, compounded annually. How much will be in the account after $6\dfrac{1}{2}$ years?

  1. $527.1$
  2. $627.1$
  3. $727.1$
  4. $827.1$
Reveal answer Fill a bubble to check yourself
C 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. $600, r = 3\%, n = 1 $ and $t =$ $6\dfrac{1}{2}$ years
$A = 600\left (1+\dfrac{0.03}{1}\right)^{1\times 6.5}$
$A = 600\times 1.03^{6.5}$
$A = 600\times 1.211831$
$A =$ Rs. $727.1$

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

Babu opened a savings account and deposited Rs. $1200.00$ as principal. The account earns $10\%$ interest, compounded annually. What is the balance after $2\dfrac{1}{2}$ years?

  1. Rs. $1522.87$
  2. Rs. $1400.67$
  3. Rs. $1456.37$

  4. None of these
Reveal answer Fill a bubble to check yourself
A 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. $1200, r = 10\%, n = 1$ and $t =$ $2\dfrac{1}{2}$ years
$A = 1200\left (1+\dfrac{0.1}{1}\right)^{1\times 2.5}$
$A = 1200\times 1.1^{2.5}$
$A = 1200\times 1.269059$
$A =$ Rs. $1522.87$