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

Amount invested by $A=12000$ Rs.
We know that:
Simple interest $=\frac{P \times R \times T}{100}$
$
\Rightarrow 3840=\frac{12000 \times R \times 4}{100}
$
$
\Rightarrow R=\frac{3840 \times 100}{12000 \times 4}
$
$
\Rightarrow R=8 \%
$
If the rate is increased by $2 \%$, then new rate $=$ $8+2=10 \%$
Now,
Amount after 4 years at the rate of $10 \%=$ $12000+\frac{12000 \times 10 \times 4}{100}$
$
=12000+4800
$
$
=16800 \text { Rs. }
$
Therefore, the amount received by A after 4 years is $16800 \mathrm{Rs}$.

Let the certain sum be P Rs.

Therefore, it will become $6 \frac{1}{4}$ times in 10 years

Let the rate of interest be $\mathrm{R} \%$
We know that:
Simple interest $=\frac{\mathbf{P} \times \mathbf{R} \times \mathbf{T}}{100}$
According to the question,
$
\Rightarrow \frac{6400 \times R \times 3}{100}+\frac{4000 \times R \times 5}{100}=4116
$
$
\Rightarrow 192 R+200 R=4116
$
$
\Rightarrow 392 \mathrm{R}=4116
$
$
\Rightarrow \mathrm{R}=\frac{4116}{392}
$
$
\Rightarrow \mathrm{R}=10.5 \%
$
Therefore, interest paid by Ganesh $=\frac{4000 \times 10.5 \times 5}{100}=Rs. 2100$ 

Principal = 9000 Rs.
Rate = 4%
Simple interest = 10080 – 9000 = 1080 Rs.
We know that:
Simple interest $=\frac{P \times R \times T}{100}$
$
\Rightarrow 1080=\frac{9000 \times 4 \times T}{100}
$
$
\Rightarrow \mathrm{T}=\frac{1080 \times 100}{9000 \times 4}
$
$
\Rightarrow \mathrm{T}=3 \text { years }
$
Now,
$
\Rightarrow 6900-x=\frac{(x) \times 5 \times 3}{100}
$
$
\Rightarrow \frac{3 x}{20}+(x)=6900
$
$
\Rightarrow \frac{23 x}{20}=6900
$
$
\Rightarrow x=\frac{6900 \times 20}{23}
$
$
\Rightarrow x=6000 \mathrm{Rs}
$

Amount $=\mathrm{P}\left(1+\frac{r}{100}\right)^{t}$
According to the question,
$
\Rightarrow \frac{8400}{8000}=\frac{P\left(1+\frac{r}{100}\right)^{5}}{P\left(1+\frac{r}{100}\right)^{4}}
$
$
\Rightarrow \frac{21}{20}=\left(1+\frac{r}{100}\right)
$
$
\Rightarrow \frac{r}{100}=\frac{21}{20}-1
$
$
\begin{aligned}
&\Rightarrow \frac{r}{100}=\frac{21-20}{20} \\
&\Rightarrow \frac{r}{100}=\frac{1}{20} \\
&\Rightarrow r=5 \%
\end{aligned}
$
Now,
$
P=8400, t=1 \text { year, } r=5 \%
$
Simple interest $=\frac{P \times R \times T}{100}$
$
=\frac{8400 \times 5 \times 1}{100}
$
$
=\mathrm{Rs} \text {. }420
$
Therefore, the amount at the end of $6^{\text {th }}$ year $=8400+420=Rs.8820$ 

Remaining amount at the end of $1^{\text {st }}$ year $=8000\times\frac{110}{100}-2800$
$
=8800-2800=6000
$
Remaining amount at the end of $2^{\text {nd }}$ year $=6000\times\frac{110}{100} -2600$
$
=6600-2600=4000
$
So, amount after 4 year $=\frac{110}{100} \times\frac{110}{100} \times4000$
$
=\frac{121}{100} \times 4000=4840
$

Let the sum be Rs. x.

Then, as per question,

[(x) + (x) × 3 × (r + 2)%] – [(x) + (x) × 3 × (r)%] = 16510 – 15748

⇒ 0.03xr + 0.06x – 0.03xr = 762

⇒ 0.06x = 762

⇒ x = 762/0.06 = 12700

Now,

(x) + (x) × 3 × (r)% = 15748

⇒ 12700 + 12700 × 3 × r% = 15748

⇒ 381r = 15748 – 12700

⇒ r = 3048/381 = 8%

MATHS MIRROR SOLUTION

Difference between amount $= 16510 – 15748= 762$

As we know, increasing the interest rate by 2% gives Rs 762 more in $3$ year

So, $3\%\times 2  =  762$

$6\%=762$

$100\%= \frac{762}{6}\times100$

$100\%=12700$

Now,

$x + 3x \times r\% = 15748$

⇒ 12700 + 12700 × 3 × r% = 15748

⇒ 381r = 15748 – 12700

⇒ r = 3048/381 = 8%

Let the sum be Rs. A.
Then, as per question,
Simple interest at $x \%$ p.a. for $2 \frac{1}{2}$ years $+585=$ Simple interest at $(x+3) \%$ p.a. for $2 \frac{1}{2}$ years
$
\Rightarrow A \times(x) \% \times 2.5+585=A \times(x+3) \% \times 2.5
$
$
\Rightarrow 0.025 \mathrm{Ax}+585=0.025 \mathrm{Ax}+0.075 \mathrm{~A}
$
$
\Rightarrow 0.075 A=585
$
$
\Rightarrow A=585 / 0.075=7800
$
Therefore, the sum is Rs. 7800 .
Now, simple interest on Rs. 7800 for $4 \frac{2}{3}$ years at $14 \%$ p.a.
$
=7800 \times 4 \frac{2}{3} \times 14 \%
$
$=7800 \times \frac{14}{3} \times 14 \%$
= Rs. 5096

MATHS MIRROR SOLUTION

As we know, increasing the interest rate by 3% gives Rs 585 more in $2 \frac{1}{2}$ year

So, $3\%\times 2 \frac{1}{2} = 585$

$3\%\times\frac{5}{2}=585$

$15\%= 1170$

$100\%= \frac{1170}{15}\times100$

$100\%=7800$

Now, simple interest on Rs. 7800 for $4 \frac{2}{3}$ years at $14 \%$ p.a.
$
=7800 \times 4 \frac{2}{3} \times 14 \%
$
$=7800 \times \frac{14}{3} \times 14 \%$
= Rs. 5096

We know that:
$
\begin{aligned}
&\text { Amount }=P\left(1+\frac{r}{100}\right)^{t} \\
&=132000\left(1+\frac{12.5}{100}\right)^{2} \\
&=132000\left(\frac{112.5}{100}\right)^{2} \\
&=132000 \times \frac{9}{8} \times \frac{9}{8}
\end{aligned}
$
$= 167062.5 $Rs.

Therefore, the price of the scooter = 167062.5 – 107062.5 = 60000 Rs.

Let the rate of interest be R%

We know that:

Amount = 

 5547 = 4800 

  = 

 R = 

 R = 7.5%

Therefore, the rate of interest is 7.5%

Hence, option A is correct.

Simple Interest = Rs. 81

 Principal = Rs. 900

 Rate of interest = 4.5%

 ⇒

⇒

 Hence, Time taken = 2 years

A sum of₹7,500 amounts to₹8,748 after 2 years at a certain rate per cent per annum compounded annually.

Amount = 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

Now, Principal = Rs. 7500

Time = 3 years

Rate =

Simple Interest =

Let the sum = P Rs.

We know that:

Therefore, it becomes 9 times in 8 years

Let the money lent at $8 \%$ of rate $=x$ Rs.
Money lent at $10 \%$ of rate $=17200-x$ Rs.
According to the question,
Simple interest $=\frac{P \times R \times T}{100}$
$\Rightarrow 3008=\frac{(x) \times 8 \times 2}{100}+\frac{(17200-x) \times 10 \times 2}{100}$
$\Rightarrow 3008=\frac{4 x}{25}+\frac{(17200-x)}{5}$
$\Rightarrow 3008=\frac{4 x+86000-5 x}{25}$
$\Rightarrow 75200=86000-x$

$
\Rightarrow x=86000-75200
$
$
\Rightarrow x=10800
$
Therefore, the money lent at $8 \%$ rate $=10800$ Rs.
Hence, option C is correct.

Let the sum be x Rs.
We know that:
Simple interest $=\frac{P \times R \times T}{100}$
According to the question,
$\Rightarrow(x)+\frac{(x) \times r \times 3}{100}=\frac{(x) \times r \times 13}{100}$
$\Rightarrow x=\frac{13 x r-3 x r}{100}$
$\Rightarrow x=\frac{10 x r}{100}$
$\Rightarrow r=10 \%$
Therefore, the value of $\mathrm{r}$ is $10 \%$

Given:
Principal $=12000$ Rs.
If the interest is compounded 8 monthly, then
Time $=1 \frac{1}{3}$ years $=\frac{4}{3} \times \frac{3}{2}=2$ years
Rate $=18 \times \frac{2}{3}=12 \%$
We know that:
Compound interest $=P\left(1+\frac{r}{100}\right)^{t}-P$
$
\begin{aligned}
&=12000\left(1+\frac{12}{100}\right)^{2}-12000 \\
&=12000\left(\frac{112}{100}\right)^{2}-12000 \\
&=12000 \times \frac{112}{100} \times \frac{112}{100}-12000 \\
&=15052.8-12000 \\
&=3052.80 \text { Rs. }
\end{aligned}
$
Therefore, the compound interest $=3052.80$ Rs.

The simple interest on a sum for 5 years at 8% p.a. is Rs. 3,960.
Let sum = Rs. P
Simple Interest $=\frac{\text { Principal } \times \text { Rate } \times \text { time }}{100}$
$
3960=\frac{P \times 8 \times 5}{100}
$
$
P=\frac{396000}{40}=R s .9900
$
Now, Simple interest on the same sum for $6 \frac{2}{3}$ years at $12 \%$ p.a $=\frac{9900 \times 12 \times \frac{20}{3}}{100}$= Rs. 7920

A sum amount of Rs. 11,616 in 2 years and to Rs. 12,777.60 in 3 years, when the interest is compounded annually.
Principal for 3rd year =Rs. 11616

Amount after 3rd year = Rs. 12177.60

Amount =$P\left(1+\frac{\text { Rate }}{100}\right)^{\text {time }}$
$12777.60=11616\left(1+\frac{R}{100}\right)$
$1+\frac{R}{100}=\frac{12777.60}{11616}=\frac{11}{10}$
$\frac{R}{100}=\frac{11}{10}-1=\frac{1}{10}$
$R=10 \%$
Hence, Rate of interest = 10%

Sum amounts of Rs. 11,616 in 2 years.

Let sum = Rs. P
$11616=P\left(1+\frac{10}{100}\right)^{2}$
$11616=P\left(\frac{11}{10}\right)^{2}$
$P=11616 \times \frac{10}{11} \times \frac{10}{11}=9600$
Hence, Sum = Rs. 9600

Simple interest received each year
$=\frac{(56000-49600)}{(5-3)}$
$=\frac{6400}{2}=3200$ rupees
So, Principal $=49600-3 \times 3200$
$=40000$
Now $, \mathrm{P}=40000, \mathrm{SI}=9600, \mathrm{~T}=3$ years
So, $R=\frac{9600 \times 100}{40000 \times 3}=8 \%$

According to question,
New rate of interest $=8+2=10 \%$
Interest received in 10 years $=100 \%$ -------------------(sum becomes double)
So, required Time $=\frac{100}{10}=10$ years

$4 \%=\frac{1}{25}$
So,
$\begin{array}{cc}\mathrm{P} & \mathrm{I} \\ 25 \times 26 & 26 \times 26 \\ (25)^{2} & (26)^{2}\end{array}$
Now, $(26)^{2}=980$
So, $P=25 \times 26+(25)^{2}=1275$
$=\frac{980}{26 \times 26} \times 1275$
$\approx 1850$ rupees.