Free Practice Questions for Simple-and-compound-interest in Maths

Interest for 1 year $=9,720-8,640=$₹ 1080
Interest for 2 year $=1080\times 2=$₹ 2160
Principal $=8,640-2160=$₹ 6480
Rate of interest $=\frac{1080\times 100}{6480\times 1}=16\frac{2}{3}\mathrm{\%}$

The difference between the compound interest and the simple interest for 2 years  
$\begin{array}{rl}& P{\left(\frac{R}{100}\right)}^2=144\\ & P\times \frac{8}{100}\times \frac{8}{100}=144\\ & P=\frac{144\times 100\times 100}{8\times 8}\\ & P=22500\end{array}$

Let $P$ be the principal
According to question
$\begin{array}{rl}& P+\frac{P\times r\times 3}{100}=\frac{P\times r\times 19}{100}\\ & 100P+3Pr=19Pr\\ & 100P=16\mathrm{Pr}\\ & r=\frac{100}{16}=6.25\end{array}$

Let $P_1$ and $P_2$ be the amount given to his daughter and son respectively.

According to question
$P_1\times {(1+\frac{5}{100})}^3=P_2\times {(1+\frac{5}{100})}^2$
$\begin{array}{rl}& P_1\times \left(\frac{105}{100}\right)=P_2\\ & \frac{P_1}{P_2}=\frac{100}{105}=\frac{20}{21}\end{array}$
Given, $(20+21)$ unit $=328000$
41 unit $=328000$
1 unit $=8000$
Share of daughter $=20$ unit $=20\times 8000=160000$

Rate half yearly $=\frac{5}{2}=2.5\mathrm{\%}$
compound interest for 1 year $=3000\times \frac{5}{100}=$₹ 150
Successive interest for 2 years $=2.5\mathrm{\%}+2.5\mathrm{\%}+\frac{2.5\times 2.5}{100}=5.0625\mathrm{\%}$
compound interest for 2 year $=3000\times \frac{5.0625}{100}=$₹ 151.875
Required difference $=$₹ 151.875-₹ 150=1.875= ₹ $1\frac{7}{8}$

Rate $=8\frac{1}{2}=\frac{17}{2}$
Rate for $\frac{1}{4}$ year $=\frac{1}{4}\times \frac{17}{2}=\frac{17}{8}\mathrm{\%}$
Amount $=$ Principal $\times {(1+\frac{\text{ Rate }}{100})}^{\text{Time }}$
Amount $=20000\times (1+\frac{17/2}{100})\times (1+\frac{17/8}{100})$
Amount $=20000\times (1+\frac{17}{200})\times (1+\frac{17}{800})$
Amount $=20000\times \frac{217}{200}\times \frac{817}{800}$
Amount $=\frac{217\times 817}{8}=\frac{177289}{8}=22161.125\approx$₹ $22161$
Compound Interest $=22161-20000=$₹ $2161$

Compound interest at $15\mathrm{\%}$ for $2$ years
$=15+15+\frac{15\times 15}{100}=30+2.25=32.25\mathrm{\%}$
Simple interest at $15\mathrm{\%}$ for $2$ years
$=2\times 15=30\mathrm{\%}$
Difference $=32.25-30=2.25\mathrm{\%}$
$2.25\mathrm{\%}\to$₹ $1944$
$1\mathrm{\%}\to$₹ $864$
Compound interest at $10\mathrm{\%}$ for $2$ years
$=10+10+\frac{10\times 10}{100}=20+1=21\mathrm{\%}$
Hence, $21\mathrm{\%}\to 21\times 864=$₹ $18144$

Interest for 4 years $=16400-13200=3200$
So, sum invested $=13200-3200=10000$
Required $\mathrm{SI}=\frac{10000\times 10\times 16}{100\times 5}$
$=$₹ 3200

Let $P$ be the required sum.
$R=\frac{12}{12}\times 100=10\mathrm{\%}=\frac{1}{10}$
$T=\frac{30}{10}=3$
According to question
$9982.5=\mathrm{P}\times \frac{11}{10}\times \frac{11}{10}\times \frac{11}{10}$
$\begin{array}{rl}& P=\frac{9982.5\times 10\times 10\times 10}{11\times 11\times 11}\\ & P=7500\end{array}$

Let $x$ be the first part.
So, $(2760-x)$ be the second part.
According to question,
$\begin{array}{rl}& x+\frac{x\times 5\times 2}{100}=(2760-\mathrm{x})+\frac{(2760-x)\times 5\times 4}{100}\\ & 110x=276000-100\mathrm{x}+55200-20\mathrm{x}\\ & 230\mathrm{x}=331200\\ & \mathrm{x}=\frac{331200}{230}\\ & \mathrm{x}=1440\end{array}$
Now, second part of investment $=2760-1440=1320$

Let the principal is $\mathrm{P}$ and the interest earned on it in a year is $\mathrm{a}$, then 

Sum after $4$ years
$P+4a$ = ₹ $12740\dots \text{. (i) }$
Sum after $7\frac{1}{2}$ years
$P$ + 7$\frac{1}{2}$ a = ₹ $15925\dots \text{. (ii) }$
By subtracting (i) from (ii)
$\frac{7}{2}a=15925-12740=$₹ $3185$
$a=\frac{6370}{7}=$₹ $910$
So, $P=12740-4\times 910=12740-3640=$₹ $9,100$
Rate $\mathrm{\%}=\frac{910}{9100}\times 100=10\mathrm{\%}$

Rate $=\frac{20\mathrm{\%}}{2}=10\mathrm{\%}$, Time $=3\times 2=6(6$-month cycle $)$
Amount $=3000\times {\left(\frac{110}{100}\right)}^6=3000\times \frac{1331}{1000}\times \frac{1331}{1000}=5314.683$
Compound Interest $=5314.683-3000=2314.683\approx$₹ $2315$

Rate for 2 years $=16\mathrm{\%}$ and rate for $\frac{1}{8}$ years $=16\times \frac{1}{8}=2\mathrm{\%}$
Amount $=8000\times \frac{116}{100}\times \frac{116}{100}\times \frac{102}{100}=10980.096$
Compound Interest $=10980-8000=$₹ $2980$

Given,
Principal $=Rs.10000,\mathrm{R}=14\mathrm{\%}$ p.a. $\mathrm{T}=3$ years
We know that,
Simple interest $=\frac{P\times R\times T}{100}$
$=\frac{10000\times 14\times 3}{100}=$Rs. 4200
So, he has to pay Rs. 14200 to the money lender.

Let the amount be 100 units.
As per the question,
Simple interest $=170-100=70$ units
Simple interest $=\frac{P\times R\times T}{100}$
$70=\frac{100\times R\times 7}{100}\Rightarrow \mathrm{R}=10\mathrm{\%}$
Now compound interest for 3 years at $10\mathrm{\%}$ on 8000 .
$\begin{array}{rl}\text{ Amount }& =\text{ Principal }{(1+\frac{r}{100})}^t\\ & =8000\times \frac{11}{10}\times \frac{11}{10}\times \frac{11}{10}\\ =& 8\times 11\times 11\times 11=10648\end{array}$
Interest $=10648-8000=$₹ 2648

Given,
Compound interest $=1530,\mathrm{r}=4\mathrm{\%}$, time $=2$ years
So,
Total compound interest for 2 years $=4+4+\frac{4\times 4}{100}$
= 8.16 %
8.16 % = 1530
$100\mathrm{\%}=\frac{1530}{8.16}\times 100=$₹ 18750
Principal =₹ 18750
Now Simple interest $=\frac{18750\times 2\times 4}{100}=$₹ 1500

According to question
$\frac{P\times (10-\frac{35}{4})\times 1}{100}$=₹ 84.5

$\frac{5P}{4}=8450$
$P=$ ₹ 6760

Let the rate is $R\mathrm{\%}$,
According to question
$\frac{P\times (R+4)\times 2}{100}-\frac{P\times R\times 2}{100}=480$
$\frac{P\times 2}{100}(R+4-R)=480$
$\frac{P\times 8}{100}=480$
$P=$₹ 6000

According to question
$\begin{array}{rl}& \frac{P\times 7.5\times 8}{100}=\frac{P\times R\times 5}{100}\\ & R=\frac{60}{5}=12\mathrm{\%}\end{array}$

Total interest at $9\mathrm{\%}$ p.a. Compound yearly for 2 years $=9+9+\frac{9\times 9}{100}=18.81\mathrm{\%}$
So $18.81\mathrm{\%}=3762$
Then $100\mathrm{\%}(\mathrm{sum})=\frac{3762}{18.81}\times 100$=₹ 20000