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

Amount invested by $A=12000$ Rs.
We know that:
Simple interest $=\frac{P\times R\times T}{100}$
$\Rightarrow 3840=\frac{12000\times R\times 4}{100}$
$\Rightarrow R=\frac{3840\times 100}{12000\times 4}$
$\Rightarrow R=8\mathrm{\%}$
If the rate is increased by $2\mathrm{\%}$, then new rate $=$ $8+2=10\mathrm{\%}$
Now,
Amount after 4 years at the rate of $10\mathrm{\%}=$ $12000+\frac{12000\times 10\times 4}{100}$
$=12000+4800$
$=16800\text{ Rs. }$
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 $\mathrm{R}\mathrm{\%}$
We know that:
Simple interest $=\frac{\mathbf{P}\times \mathbf{R}\times \mathbf{T}}{100}$
According to the question,
$\Rightarrow \frac{6400\times R\times 3}{100}+\frac{4000\times R\times 5}{100}=4116$
$\Rightarrow 192R+200R=4116$
$\Rightarrow 392\mathrm{R}=4116$
$\Rightarrow \mathrm{R}=\frac{4116}{392}$
$\Rightarrow \mathrm{R}=10.5\mathrm{\%}$
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,
$\Rightarrow 6900-x=\frac{(x)\times 5\times 3}{100}$
$\Rightarrow \frac{3x}{20}+(x)=6900$
$\Rightarrow \frac{23x}{20}=6900$
$\Rightarrow x=\frac{6900\times 20}{23}$
$\Rightarrow x=6000\mathrm{Rs}$

Amount $=\mathrm{P}{(1+\frac{r}{100})}^t$
According to the question,
$\Rightarrow \frac{8400}{8000}=\frac{P{(1+\frac{r}{100})}^5}{P{(1+\frac{r}{100})}^4}$
$\Rightarrow \frac{21}{20}=(1+\frac{r}{100})$
$\Rightarrow \frac{r}{100}=\frac{21}{20}-1$
$\begin{array}{rl}& \Rightarrow \frac{r}{100}=\frac{21-20}{20}\\ & \Rightarrow \frac{r}{100}=\frac{1}{20}\\ & \Rightarrow r=5\mathrm{\%}\end{array}$
Now,
$P=8400,t=1\text{ year, }r=5\mathrm{\%}$
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 $=\frac{110}{100}\times \frac{110}{100}\times 4000$
$=\frac{121}{100}\times 4000=4840$

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, $3\mathrm{\%}\times 2=762$

$6\mathrm{\%}=762$

$100\mathrm{\%}=\frac{762}{6}\times 100$

$100\mathrm{\%}=12700$

Now,

$x+3x\times r\mathrm{\%}=15748$

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

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

$15\mathrm{\%}=1170$

$100\mathrm{\%}=\frac{1170}{15}\times 100$

$100\mathrm{\%}=7800$

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

We know that:
$\begin{array}{rl}& \text{ Amount }=P{(1+\frac{r}{100})}^t\\ & =132000{(1+\frac{12.5}{100})}^2\\ & =132000{\left(\frac{112.5}{100}\right)}^2\\ & =132000\times \frac{9}{8}\times \frac{9}{8}\end{array}$
$=167062.5$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 $8\mathrm{\%}$ of rate $=x$ Rs.
Money lent at $10\mathrm{\%}$ of rate $=17200-x$ Rs.
According to the question,
Simple interest $=\frac{P\times R\times T}{100}$
$\Rightarrow 3008=\frac{(x)\times 8\times 2}{100}+\frac{(17200-x)\times 10\times 2}{100}$
$\Rightarrow 3008=\frac{4x}{25}+\frac{(17200-x)}{5}$
$\Rightarrow 3008=\frac{4x+86000-5x}{25}$
$\Rightarrow 75200=86000-x$

$\Rightarrow x=86000-75200$
$\Rightarrow x=10800$
Therefore, the money lent at $8\mathrm{\%}$ 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,
$\Rightarrow (x)+\frac{(x)\times r\times 3}{100}=\frac{(x)\times r\times 13}{100}$
$\Rightarrow x=\frac{13xr-3xr}{100}$
$\Rightarrow x=\frac{10xr}{100}$
$\Rightarrow r=10\mathrm{\%}$
Therefore, the value of $\mathrm{r}$ is $10\mathrm{\%}$

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 $=18\times \frac{2}{3}=12\mathrm{\%}$
We know that:
Compound interest $=P{(1+\frac{r}{100})}^t-P$
$\begin{array}{rl}& =12000{(1+\frac{12}{100})}^2-12000\\ & =12000{\left(\frac{112}{100}\right)}^2-12000\\ & =12000\times \frac{112}{100}\times \frac{112}{100}-12000\\ & =15052.8-12000\\ & =3052.80\text{ Rs. }\end{array}$
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 $=\frac{\text{ Principal }\times \text{ Rate }\times \text{ time }}{100}$
$3960=\frac{P\times 8\times 5}{100}$
$P=\frac{396000}{40}=Rs.9900$
Now, Simple interest on the same sum for $6\frac{2}{3}$ years at $12\mathrm{\%}$ 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 =$P{(1+\frac{\text{ Rate }}{100})}^{\text{time }}$
$12777.60=11616(1+\frac{R}{100})$
$1+\frac{R}{100}=\frac{12777.60}{11616}=\frac{11}{10}$
$\frac{R}{100}=\frac{11}{10}-1=\frac{1}{10}$
$R=10\mathrm{\%}$
Hence, Rate of interest = 10%

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

Let sum = Rs. P
$11616=P{(1+\frac{10}{100})}^2$
$11616=P{\left(\frac{11}{10}\right)}^2$
$P=11616\times \frac{10}{11}\times \frac{10}{11}=9600$
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, $R=\frac{9600\times 100}{40000\times 3}=8\mathrm{\%}$

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

$4\mathrm{\%}=\frac{1}{25}$
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.