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 $=\frac{1080 \times 100}{6480 \times 1}=16 \frac{2}{3} \%$

The difference between the compound interest and the simple interest for 2 years  
$
\begin{aligned}
&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{aligned}
$

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

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

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

Rate half yearly $=\frac{5}{2}=2.5 \%$
compound interest for 1 year $=3000 \times \frac{5}{100}=$₹ 150
Successive interest for 2 years $=2.5 \%+2.5 \%+\frac{2.5 \times 2.5}{100}=5.0625 \%$
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 $=\frac{1}{4} \times \frac{17}{2}=\frac{17}{8} \%$
Amount $=$ Principal $\times\left(1+\frac{\text { Rate }}{100}\right)^{\text {Time }}$
Amount $=20000 \times\left(1+\frac{17 / 2}{100}\right) \times\left(1+\frac{17 / 8}{100}\right)$
Amount $=20000 \times\left(1+\frac{17}{200}\right) \times\left(1+\frac{17}{800}\right)$
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 $15 \%$ for $2$ years
$
=15+15+\frac{15 \times 15}{100}=30+2.25=32.25 \%
$
Simple interest at $15 \%$ for $2$ years
$
=2 \times 15=30 \%
$
Difference $=32.25-30=2.25 \%$
$2.25 \% \rightarrow $₹ $1944$
$1 \% \rightarrow $₹ $864$
Compound interest at $10 \%$ for $2$ years
$
=10+10+\frac{10 \times 10}{100}=20+1=21 \%
$
Hence, $21 \% \rightarrow 21 \times 864=$₹ $18144$

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 \%=\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{aligned}
&P=\frac{9982.5 \times 10 \times 10 \times 10}{11 \times 11 \times 11} \\
&P=7500
\end{aligned}
$

Let $x$ be the first part.
So, $(2760-x)$ be the second part.
According to question,
$
\begin{aligned}
&x+\frac{x \times 5 \times 2}{100}=(2760-\mathrm{x})+\frac{(2760-x) \times 5 \times 4}{100} \\
&110 x=276000-100 \mathrm{x}+55200-20 \mathrm{x} \\
&230 \mathrm{x}=331200 \\
&\mathrm{x}=\frac{331200}{230} \\
&\mathrm{x}=1440
\end{aligned}
$
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 $ = ₹ $12740 \ldots \text {. (i) }$
Sum after $7 \frac{1}{2}$ years
$P$ + 7$ \frac{1}{2} $ a = ₹ $15925 \ldots \text {. (ii) }$
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 $\%=\frac{910}{9100} \times 100=10 \%$

Rate $=\frac{20 \%}{2}=10 \%$, 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 $=16 \%$ and rate for $\frac{1}{8}$ years $=16 \times \frac{1}{8}=2 \%$
Amount $=8000 \times \frac{116}{100} \times \frac{116}{100} \times \frac{102}{100}=10980.096$
Compound Interest $=10980-8000=$₹ $2980$

Given,
Principal $= Rs.10000, \mathrm{R}=14 \%$ 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 $=\frac{P \times R \times T}{100}$
$70=\frac{100 \times R \times 7}{100} \Rightarrow \mathrm{R}=10 \%$
Now compound interest for 3 years at $10 \%$ on 8000 .
$\begin{aligned} \text { Amount } &=\text { Principal }\left(1+\frac{r}{100}\right)^t \\ &=8000 \times \frac{11}{10} \times \frac{11}{10} \times \frac{11}{10} \\=& 8 \times 11 \times 11 \times 11=10648 \end{aligned}$
Interest $=10648-8000=$₹ 2648

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

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

$\frac{5 P}{4}=8450$
$P=$ ₹ 6760

Let the rate is $R \%$,
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{aligned}
&\frac{P \times 7.5 \times 8}{100}=\frac{P \times R \times 5}{100} \\
&R=\frac{60}{5}=12 \%
\end{aligned}
$

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