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 $10\mathrm{\%}$ compound interest compounded annually for 2 years.
Let Principal $=$ Rs. $P$
Each instalment $=$ Rs. 1089
Principal $=x(1+\frac{R}{100})+x{(1+\frac{R}{100})}^2=$
$\frac{1089}{(1+\frac{10}{100})}+\frac{1089}{{(1+\frac{10}{100})}^2}=1089\times \frac{10}{11}+1089\times \frac{100}{121}=990+900$
= Rs. 1890
Sum (in Rs.) borrowed = Rs. 1890

$12\frac{1}{2}\mathrm{\%}=\frac{1}{8}$
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 $=\frac{P\times R\times T}{100}$
$=23040\times \frac{1}{8}\times \frac{8}{3}$
$=7680$

$\frac{P}{I}=\frac{10}{3}$
$\Rightarrow \frac{I}{P}=\frac{3}{10}$
$\therefore R=\frac{\text{ S.I. }\times 100}{P\times T}$
$=\frac{3}{10}\times \frac{100}{15}=2\mathrm{\%}$ 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. $2P-Rs.P=Rs.P$
According to question:
$\begin{array}{rl}& \Rightarrow P=\frac{P\times R\times 10}{100}\\ & \Rightarrow R=\frac{100}{10}=10\mathrm{\%}\end{array}$
Let sum will become 3 times of itself in ' $t$ ' years.
Amount $=$ Rs. 3P
Simple Interest $=$ Rs. $3P-Rs.P=$ Rs. $2P$
$\begin{array}{rl}& \Rightarrow 2P=\frac{P\times 10\times t}{100}\\ & \Rightarrow t=20\text{ years }\end{array}$
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.
$\begin{array}{rl}& \text{ Rate }=\frac{8\mathrm{\%}}{2}=4\mathrm{\%}\\ & \text{ Time }=1\times 2=2\text{ year }\end{array}$
Compound Interest $=P{(1+\frac{R}{100})}^t-P$
Compound Interest $=$$15000{(1+\frac{4}{100})}^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 $5\mathrm{\%}$ per annum compound interest.
Amount $=P{(1+\frac{\text{ Rate }}{100})}^{\text{time }}$
$\begin{array}{rl}& \Rightarrow 1543.50=P{(1+\frac{5}{100})}^2=P{\left(\frac{21}{20}\right)}^2\\ & \Rightarrow P=1543.50\times \frac{400}{441}=Rs.1400\end{array}$
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.
$\text{ Amount }=P{(1+\frac{R\text{ ate }}{100})}^{\text{time }}$
$\begin{array}{rl}\Rightarrow & 21160=16000{(1+\frac{R}{100})}^2\\ \Rightarrow & \frac{21160}{16000}={(1+\frac{R}{100})}^2\\ \Rightarrow & \frac{2116}{1600}={(1+\frac{R}{100})}^2\\ \Rightarrow & (1+\frac{R}{100})=\sqrt{\frac{2116}{1600}}=\frac{46}{40}\\ \Rightarrow & \left(\frac{R}{100}\right)=\frac{6}{40}\end{array}$
$\Rightarrow R=\frac{60}{4}=15\mathrm{\%}$
Rate per cent per annum $=15\mathrm{\%}$

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 $6.25\mathrm{\%}$, 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 $(\frac{2}{5}\times 31)$ 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 = $\frac{400\times 3\times R}{100}$ = 12 R

For principal (P2) = ₹ 600,

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

300 = $\frac{5000\times R\times 2}{100}$

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

R = $3\mathrm{\%}$

Given:
Principal $=$ Rs. 20000, Time $=1\frac{1}{2}$ years and Rate of interest $=10\mathrm{\%}$ p.a.
Simple Interest $=\mathrm{P}\times \mathrm{R}\times \mathrm{T}=20000\times 10\mathrm{\%}\times 1.5=$ Rs. 3000
Compound Interest (6 months compounding $)=P\times (1+R{)}^{\mathrm{\top }}-P$
$\begin{array}{rl}& =20000\times (1+0.05{)}^3-20000\\ & =20000\times (1.05{)}^3-20000\\ & =23152.5-20000\\ & =\text{ Rs. }3152.5\end{array}$
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
$\begin{array}{rl}& \Rightarrow 6500\times (x)\mathrm{\%}\times 3+7500\times (x-2)\mathrm{\%}\times 3=3750\\ & \Rightarrow 195x+225x-450=3750\\ & \Rightarrow 420x=3750+450\\ & \Rightarrow x=4200/420\\ & \Rightarrow x=10\end{array}$
Therefore, rate of interest on the second investment is $8\mathrm{\%}(10-2)$.

Given:
Interest rate $=5\mathrm{\%}$
Let the sum be Rs. $x$ and no of years be $y$.
Then, interest on that sum $=\frac{3}{8}x$
$\begin{array}{rl}& \Rightarrow \frac{(x)\times 5\times (y)}{100}=\frac{3x}{8}\\ & \Rightarrow 40y=300\end{array}$
$\Rightarrow y=300/40=7.5\text{ years }$
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,
$\begin{array}{rl}& \Rightarrow \frac{(x)\times 12.5\times 1}{100}-\frac{(x)\times 8\times 1}{100}=459\\ & \Rightarrow \frac{12.5x-8x}{100}=459\\ & \Rightarrow \frac{4.5x}{100}=459\\ & \Rightarrow x=\frac{459\times 100}{4.5}\\ & \Rightarrow x=10200\mathrm{Rs}.\end{array}$
Therefore, the principal amount is $10200\mathrm{Rs}$.

MATHS MIRROR SOLUTION

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

According to the question,

$\Rightarrow 12.5\mathrm{\%}-8\mathrm{\%}=459$

$\Rightarrow 4.5\mathrm{\%}=459$

$\Rightarrow 100\mathrm{\%}=\frac{459\times 100}{4.5}$

$\Rightarrow 100\mathrm{\%}=10200$

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

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

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

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