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

A man borrowed money and paid back in two equal annual instalments of Rs. 1,089 , at $10 \%$ compound interest compounded annually for 2 years.
Let Principal $=$ Rs. $P$
Each instalment $=$ Rs. 1089
Principal $=x\left(1+\frac{R}{100}\right)+x\left(1+\frac{R}{100}\right)^{2}=$
$\frac{1089}{\left(1+\frac{10}{100}\right)}+\frac{1089}{\left(1+\frac{10}{100}\right)^{2}}=1089 \times \frac{10}{11}+1089 \times \frac{100}{121}=990+900$
= Rs. 1890
Sum (in Rs.) borrowed = Rs. 1890

$
12 \frac{1}{2} \%=\frac{1}{8}
$
Let Principal $=8^{3}=512$

CI for the 3 years $=64 \times 3+8 \times 3+1=217$.
217 units $=9765$
1 units $=45$
512 units $=512 \times 45=23040$
S.I $=\frac{P \times R \times T}{100}$
$=23040 \times \frac{1}{8} \times \frac{8}{3}$
$=7680$

$\frac{P}{I}=\frac{10}{3}$
$\Rightarrow \frac{I}{P}=\frac{3}{10}$
$\therefore R=\frac{\text { S.I. } \times 100}{P \times T}$
$=\frac{3}{10} \times \frac{100}{15}=2 \%$ per annum

MATHS MIRROR APPROACH -

BASIC APPROACH -

A certain sum doubles itself on simple interest in 10 years.
Let Principal $=$ Rs. $P$
Time $=10$ years
Amount $=$ Rs. $2 \mathrm{P}$
Simple Interest $=$ Rs. $2 P-R s . P=R s . P$
According to question:
$
\begin{aligned}
&\Rightarrow P=\frac{P \times R \times 10}{100} \\
&\Rightarrow R=\frac{100}{10}=10 \%
\end{aligned}
$
Let sum will become 3 times of itself in ' $t$ ' years.
Amount $=$ Rs. 3P
Simple Interest $=$ Rs. $3 P-R s . P=$ Rs. $2 P$
$
\begin{aligned}
&\Rightarrow \quad 2 P=\frac{P \times 10 \times t}{100} \\
&\Rightarrow t=20 \text { years }
\end{aligned}
$
Hence, Sum will become 3 times of itself in 20 years.

MATHS MIRROR APPROACH -

ROI = 8/2 = 4%

Compound interest for a year = 600+600+24 = 1224

BASIC APPROACH -

Principal = Rs. 15000
Interest is compounded semi-annually.
$
\begin{aligned}
&\text { Rate }=\frac{8 \%}{2}=4 \% \\
&\text { Time }=1 \times 2=2 \text { year }
\end{aligned}
$
Compound Interest $=P\left(1+\frac{R}{100}\right)^{t}-P$
Compound Interest $=$$15000\left(1+\frac{4}{100}\right)^{2}-15000$

$=15000 \times \frac{26}{25} \times \frac{26}{25}-15000$

$=$ Rs. $16224-$ Rs. $15000$

$= Rs. 1224$

A certain sum amounts to Rs. $1,543.50$ when invested for 2 years at $5 \%$ per annum compound interest.
Amount $=P\left(1+\frac{\text { Rate }}{100}\right)^{\text {time }}$
$
\begin{aligned}
&\Rightarrow \quad 1543.50=P\left(1+\frac{5}{100}\right)^{2}=P\left(\frac{21}{20}\right)^{2} \\
&\Rightarrow P=1543.50 \times \frac{400}{441}=R s .1400
\end{aligned}
$
Hence, sum = Rs. 1400

A certain sum of money, when invested at a certain rate of simple interest, amounts to Rs. 1,102 in 2 years and Rs. 1,600 in 5 years.
Simple interest for 5 - 2 = 3 years $=$ Rs. $1600 -$ Rs. $1102=$ Rs. 498
Simple interest for 1 years $=\frac{498}{3}=$ Rs. 166
Simple interest for 2 years $=2 \times$ Rs. $166=332$
Principal = Rs. $1102 -$ Rs. $332=$ Rs. 770

A sum of Rs. 16,000 invested at compound interest, amount to Rs. 21,160 in two years.
$
\text { Amount }=P\left(1+\frac{R \text { ate }}{100}\right)^{\text {time }}
$
$\begin{aligned} \Rightarrow & 21160=16000\left(1+\frac{R}{100}\right)^{2} \\ \Rightarrow & \frac{21160}{16000}=\left(1+\frac{R}{100}\right)^{2} \\ \Rightarrow & \frac{2116}{1600}=\left(1+\frac{R}{100}\right)^{2} \\ \Rightarrow &\left(1+\frac{R}{100}\right)=\sqrt{\frac{2116}{1600}}=\frac{46}{40} \\ \Rightarrow &\left(\frac{R}{100}\right)=\frac{6}{40} \end{aligned}$
$
\Rightarrow R=\frac{60}{4}=15 \%
$
Rate per cent per annum $=15 \%$

A man takes a loan of Rs. 6,000 from his friend on $1^{\text {st }}$ January 2019 with the condition that he will repay with accrued simple interest at the rate $6.25 \%$, as and when the interest touches Rs. $75 .$
Simple Interest $=\frac{\text { Principal } \times \text { Rate } \times \text { time }}{100}$
$
\Rightarrow 75=\frac{6000 \times 6.25 \times \text { time }}{100}
$
$\Rightarrow$ Time $=\frac{1200}{6000}=\frac{1}{5}$ years $=\frac{12}{5}$ months $=2$ months $\left(\frac{2}{5} \times 31\right)$ days $=2$ months 13 days
Date on which loan period expire $=1^{\text {st }}$ January $2019+2$ month 13 days $=15^{\text {th }}$ March 2019

We have,Principal (P1) = ₹ 400

Principal (P2) = ₹ 600

Total simple interest = ₹ 90

time = 3 years

For principal (P1) = ₹ 400,

Simple Interest (SI) = (Principal × Time × Rate)/100 = $ \frac{400 × 3 × R}{100}$ = 12 R

For principal (P2) = ₹ 600,

Simple Interest(SI) = (Principal × Time × Rate)/100 = $ \frac{600 × 3 × R}{100}$ = 18 R

According to Question ,

Total simple interest = ₹ 90

∴ 12R + 18R = 90

∴ R = 3%

we have, Principal (P) = Rs.5,000

Time (T) = 2 years

Simple Interest = Rs. 300

rate (R) = ?

Simple interest = $ \frac{P × R × T}{100}$

300 = $ \frac{5000 × R × 2}{100}$

R = $ \frac{300}{100}$

R = $3 \%$

Given:
Principal $=$ Rs. 20000, Time $=1 \frac{1}{2}$ years and Rate of interest $=10 \%$ p.a.
Simple Interest $=\mathrm{P} \times \mathrm{R} \times \mathrm{T}=20000 \times 10 \% \times 1.5=$ Rs. 3000
Compound Interest (6 months compounding $)=P \times(1+R)^{\top}-P$
$\begin{aligned}
&=20000 \times(1+0.05)^{3}-20000 \\
&=20000 \times(1.05)^{3}-20000 \\
&=23152.5-20000 \\
&=\text { Rs. } 3152.5
\end{aligned}
$
Difference between both the interests $=3152.5-3000=$ Rs. $152.5$

Given:
Total interest earned on both the investments for 3 years $=$ Rs. 3,750
$
\begin{aligned}
&\Rightarrow 6500 \times(x) \% \times 3+7500 \times(x-2) \% \times 3=3750 \\
&\Rightarrow 195 x+225 x-450=3750 \\
&\Rightarrow 420 x=3750+450 \\
&\Rightarrow x=4200 / 420 \\
&\Rightarrow x=10
\end{aligned}
$
Therefore, rate of interest on the second investment is $8 \%(10-2)$.

Given:
Interest rate $=5 \%$
Let the sum be Rs. $x$ and no of years be $y$.
Then, interest on that sum $=\frac{3}{8} x$
$
\begin{aligned}
&\Rightarrow \frac{(x) \times 5 \times(y)}{100}=\frac{3 x}{8} \\
&\Rightarrow 40 y=300
\end{aligned}
$
$
\Rightarrow y=300 / 40=7.5 \text { years }
$
i.e. 7 years 6 months

Let the sum of money be P

Total time $=15+15=30$ years.
Therefore, in 30 years it will become four times of itself

Let the principal amount be $x$ Rs.
We know that:
Simple interest $=\frac{P \times R \times T}{100}$
According to the question,
$
\begin{aligned}
&\Rightarrow \frac{(x) \times 12.5 \times 1}{100}-\frac{(x) \times 8 \times 1}{100}=459 \\
&\Rightarrow \frac{12.5 x-8 x}{100}=459 \\
&\Rightarrow \frac{4.5 x}{100}=459 \\
&\Rightarrow x=\frac{459 \times 100}{4.5} \\
&\Rightarrow x=10200 \mathrm{Rs} .
\end{aligned}
$
Therefore, the principal amount is $10200 \mathrm{Rs}$.

MATHS MIRROR SOLUTION

$12 \frac{1}{2} \%= 12.5$

According to the question,

$\Rightarrow12.5\% - 8\%= 459$

$\Rightarrow4.5\%= 459$

$\Rightarrow100\% = \frac{459 \times 100}{4.5}$

$\Rightarrow100\% =10200 $

Let the sum be $y$
According to question
$
\begin{aligned}
&y\left(1+\frac{10.5 \times 5}{100}\right)=27450 \\
&\Rightarrow y(1+0.525)=27450 \\
&\Rightarrow 1.525 y=27450 \\
&\Rightarrow y=18000
\end{aligned}
$

Let the interest rate be $\mathrm{y} \%$ According to the question,
$
12500\left(1+\frac{y}{100}\right)^{2}=14045
$
$\Rightarrow\left(1+\frac{y}{100}\right)^{2}=\frac{14045}{12500}=\frac{2809}{2500}$
$\Rightarrow\left(1+\frac{y}{100}\right)^{2}=\left(\frac{53}{50}\right)^{2}$
$\Rightarrow\left(1+\frac{y}{100}\right)=\frac{53}{50}$
$\Rightarrow \frac{y}{100}=\frac{53}{50}-1$
$\Rightarrow \frac{y}{100}=\frac{3}{50}$
$\Rightarrow y=6$

Let the interest rate be $\mathrm{y} \%$ per annum
According to question
$x\left(1+\frac{y}{100}\right)^{3}=27900$ ......(1)
Also,
$x\left(1+\frac{y}{100}\right)^{6}=41850$ ........(2)
Eq.2 divide by eq.1
$
\frac{x\left(1+\frac{y}{100}\right)^{6}}{x\left(1+\frac{y}{100}\right)^{3}}=\frac{41850}{27900}
$
$
\Rightarrow\left(1+\frac{y}{100}\right)^{3}=1.5
$
Now, put this in Eq. (i)
$
\begin{aligned}
&(x) \times 1.5=27900 \\
&\Rightarrow x=27900 / 1.5=18600
\end{aligned}
$

Amount $=$ Rs. 10000, Time $=2$ years and Interest rate $=5 \%$ p.a. $\mathrm{Cl}$ and $5 \%$ p.a. SI
Compound Interest $=\left[\right.$ Amount $\left.\times(1+\mathrm{R})^{\mathrm{T}}\right]-$ Amount
$
\begin{aligned}
&=\left[10000 \times(1+5 \%)^{2}\right]-10000 \\
&=\left[10000 \times(1.05)^{2}\right]-10000 \\
&=(10000 \times 1.1025)-10000 \\
&=11025-10000=\text { Rs. } 1025 \\
&\text { Simple Interest }=\text { Amount } \times \text { Rate } \times \text { Time } \\
&=10000 \times 5 \% \times 2=\text { Rs. } 1000
\end{aligned}
$
Money, Kanika would have saved if she had taken the loan on simple interest $=\mathrm{Cl}-\mathrm{SI}=1025$
$
-1000=\text { Rs. } 25
$