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

Principal:- Rs. 80000

Rate $\quad:10+5+\frac{10 \times 5}{100}$

Time :- 2 years

$\begin{aligned} \text { Interest } &=\text { Rs. } 80000 \times \frac{15.5}{100} \\ &=\text { Rs. } 12,400 \\ \text { Amount } &=80,000+12,400 \\ &=\text { Rs. } 92,400 \end{aligned}$

Let principal be $p$

Rate of interest is $40 \%$ p.a.

And time $=3$ year

$\begin{aligned} \text { After } 3 \text { years total simple interest } &=\frac{p \times 40 \times 3}{100} \\ &=\frac{6 p}{5} \end{aligned}$

$\begin{aligned} \text { Amount after } 3 \text { year } &=\frac{6 p}{5}+\mathrm{p} \\ &=\frac{11 p}{5} \end{aligned}$

Now, according to the given condition,

$\begin{aligned} \frac{11 p}{5} &=9900 \\ \mathrm{p} &=\frac{9900 \times 5}{11} \\ \mathrm{p} &=\text { Rs. } 4500 \end{aligned}$

The principal amount invested by Mohit was Rs. 4500

S.I $=18 \% \times 3$

$=54 \%$

Rs. $8,100=54 \%$

Rs. $37,500=\frac{54 \%}{8,100} \times 37,500=250 \%$

$\mathrm{RT} \%=250 \%$

$\mathrm{R} \times 8=250$

$\mathrm{R}=31.25 \%$

ATQ,

Let salary of Ankit is 1000 units.

Interest of 2 years - 

300               10%   -   30

                                 -  30 $ \quad $3

CI after 2 years = 63 units

63 units = 5670

1 unit = 90

1000 units = 90000

$\begin{aligned} \text { Expenditure } &=90,000 \times \frac{1}{3} \\ &=30,000 \end{aligned}$

2.5 units = 1000

1 unit = 400

Sum = 100 × 400

= 40000

Given
Amount = ₹37,632
Rate = 12%
Time (T) =2 years
$A=P\left(1+\frac{R}{100}\right)^t$
$37632=P\left(1+\frac{12}{100}\right)^2$
$37632=P\left(\frac{112}{100}\right)^2$
$37632=P\left(\frac{28}{25}\right)^2$
$P=\frac{37632 \times 25 \times 25}{28 \times 28}$
P= 30,000
According to question
S.I. $=\frac{P R T}{100}$
S.I. $=\frac{30000 \times 12 \times 4}{100}$
S.I. = 14400

Given Principal amount $\mathrm{P}=$ Rs 500
Time period $\mathrm{T}=4$ years
Rate of interest $\mathrm{R}=8 \%$ p.a.
We know that simple interest $=\frac{(P \times R\times T)}{100}$
On substituting these values in the above equation we get
$
\mathrm{SI}=\frac{(500 \times 8\times 4)}{100}
$
$=\operatorname{Rs} 160$
Amount $=$ Principal amount $+$ Interest
$=\operatorname{Rs} 500+160$
$=\operatorname{Rs} 660$

Interest $=5,310-4500=810$
$810=\frac{P \times R \times T}{100}$
$\Rightarrow 810=\frac{4500 \times R \times 4}{100}$
$\Rightarrow \mathrm{R}=\frac{810}{180}$
$\Rightarrow \mathrm{R}=4.5 \%$

We know SI for the each year is same so we can write it as:

                              SI                 CI

For 1st year  =     6400             6400

For 2nd year =    6400             6400+704

Let x be the required rate.

According to question,

6400 of  x% =704
$6400 \times \frac{x}{100}=704$

$\mathrm{x} \quad=\frac{704}{64}=11 \%$

Rate of SI for two year $=10+10=20 \%$
Rate of CI for two year $=10+10+\frac{10 \times 10}{100}=21 \%$
Given, $(21-20) \%=1200$
$
\begin{aligned}
1 \% &=1200 \\
100 \% &=1,20,000
\end{aligned}
$

Let

 The rate of interest per annum is $r_1$ , $r_2$ , $r_3$ respectively

According to Question

$r_1=5 \%=\frac{1}{20} \quad r_2=6 \%=\frac{3}{50}\quad r_3=8 \%=\frac{2}{25}$

First-year ratio $=20: 21$

Second year ratio $=50: 53$

Third year ratio $=25: 27$

Ratio of principle to Amount $=20 \times 50 \times 25: 21 \times 53 \times 27$ $=25000: 30051$

If the Amount is $=Rs. 30,051$

Then Principle $=Rs. 25,000$

Rate for 8 months $=15 \% \times \frac{8}{12}=10 \%$
Let share of $A$ is $=A$
2 years $=24$ months $=8 \times 3$ months
$n=3$
After 2 years share of $A=A\left(1+\frac{10}{100}\right)^3$
Let share of $B$ is $=B$
$2 \frac{2}{3}$ years $=32$ months $=8 \times 4$ months

$n=3$
After $2 \frac{2}{3}$ years share of $B=B\left(1+\frac{10}{100}\right)^4$
A : B
$\mathrm{A}\left(1+\frac{10}{100}\right)^3: \mathrm{B}\left(1+\frac{10}{100}\right)^4$
$\frac{A}{B}=\left(1+\frac{10}{100}\right)$
$\frac{A}{B}=\frac{11}{10}$
A = $11x$
B = $10x$
$21x$ = ₹ $8,400$
$x$ = ₹ $400$
Share of A = $11x$
$=11 \times $₹ $400$
= ₹ $4,400$

Principal $=$₹ 20,000
Rate $=9 \%$ p.a.
Time $=1$ year and 4 months
Compound interest for 1 year $=9 \%$ of ₹ 20,000=₹ 1800
After 1 year amount $=$₹ 21,800
Compound interest for 4 months
Rate $=9 \% \times \frac{4}{12}=3 \%$
Principal = ₹ 21,800
Compound interest for 4 months = 3% of ₹ 21,800 = ₹ 654
Total compound interest for 1 year and 4 months $=$₹ 1800+₹ 654 = ₹ 2,454

$
\begin{aligned}
&2 \text { year } \mathrm{Cl} \text { and SI difference }=\frac{r^2}{100} \% \\
&=\frac{5^2}{100} \% \\
&=0.25 \% \text { of } 6,000 \\
&\text { = ₹ } 15 \\
&
\end{aligned}
$

R1 = 8%
R2 = 10%
R3 = 12%
Total interest in $\%=\mathrm{R} 1+\mathrm{R} 2+\mathrm{R} 3+\frac{R 1 R 2+R 2 R 3+R 3 R 1}{100}+\frac{R 1 R 2 R 3}{10000}$
$=8 \%+10 \%+12 \%+\frac{8 \%+10 \%+12 \%}{100}+\frac{8 \times 10 \times 12}{10000}$
$=33.056 \%$
Compound interest in Rupees =33.056 % of ₹ 10,000
=₹ 3,305.6

Simple interest at $5 \%=\frac{14000 \times 5 \times 3}{100}=$₹ 2100
Simple interest at $6 \%=\frac{19000 \times 6 \times 3}{100}=$₹ 3420
Total income from the interest in 3 years $=2100+3420=$₹ 5520

Let the principle be x

so, amount be 2x

Interest = 2x-x = x

$\mathrm{X}=\frac{x \times 6 \times r}{100}$
$\mathrm{r}=\frac{50}{3} \%$

When amount is = 5x

Interest = 5x-x = 4x

$4 \mathrm{x}=\frac{x \times t \times 50}{100 \times 3}$
$\mathrm{t}=24$ years

$
\begin{aligned}
&\text { Rate }=\frac{16}{2}=8 \% \\
&\text { Amount }=\mathrm{P}\left(1+\frac{r}{100}\right)^t \\
&5832=5000\left(1+\frac{8}{100}\right)^t \\
&\frac{5832}{5000}=\left(\frac{54}{50}\right)^t \\
&\left(\frac{54}{50}\right)^2=\left(\frac{54}{50}\right)^t \\
&\mathrm{~T}=2
\end{aligned}
$
Interest is compounded half yearly
So required time $=\frac{2}{1}=2$

Difference between CI and SI for 2 year $=\mathrm{P}\left(\frac{r}{100}\right)^2$
$\mathrm{P}\left(\frac{r}{100}\right)^2=152$
$\mathrm{P}\left(\frac{5}{100}\right)^2=152$
$\mathrm{P} \times \frac{1}{20} \times \frac{1}{20}=152$
$\mathrm{P}=152 \times 20 \times 20$
$\mathrm{P}=$₹ 60800