Simple Interest Questions and Answers

$text{Principal } = frac{560 times 100}{100 + 4 times 3} = text{Rs.} 500$

Rs. 100 becomes Rs. 300 means interest is Rs. 200 on Rs. 100 in 25 years

$text{Rate} = frac{200 times 100}{100 times 25} = 8$

$=frac{1}{9} text{P} = frac{text{P} times text{R} times text{T} }{100}, text{but } r = t$

Let the annual rent be $x$ so that the investment in $25x$.

$text{ Rate } = frac{x times 100 }{25 times x} = 4$

$text{ A = P } times Big(1 + frac{r}{100} Big)^n$

$= text{P } times Big(1 + frac{5}{100}Big)^3 = text{P } times Big(frac{9361}{8000}Big)$

$therefore$ If P = Rs. 8000, A = Rs. 9261

$therefore$  C.I. =  Rs. 1261

Simple interest on Rs. 8000 for 3 years at 5% = Rs. 1200

Now if C.I. = Rs. 1261, SI = Rs. 1200 and if C.I. = Rs. 1324.50 then

$text{S.I. } = frac{1200}{1261} times 1324.05 = text{ Rs. } 1260$

Let the principal amount P.

$therefore text{ P } + frac{text{ P } times 5 times 4}{100} = 504$
$Rightarrow 120 text{ P } = 50400 Rightarrow text{ P } = 420$
$therefore text{ The amount of Rs. } 420 text{ at }$
$10$
$= 420 + frac{420 times 10 times 5}{100 times 2}$
=Rs. 525

Here, A₁ = Rs. 1586, t₁ = 2 yr, A₂ = 1729 and t₂ = 3 yr

Therefore, Required principle $= dfrac{A_2t_1 - A_1t_2}{t_1 - t_2} = dfrac{1729 times 2 - 1586 times 3}{2-3}$

= Rs. 1300

and required rate R $= bigg[dfrac{(A_2 - A_1) times 100}{A_1t_2 - A_2t_1} bigg]$

$= dfrac{(1729 - 1586) times 100}{1586times 3 - 1729times 2} = dfrac{143 times 100}{1300} = 11$

Let the sum (principal) be Rs. $x$.
Therefore, Simple interest = Rs. $dfrac{x}{2}$

and T = 6 yr, R = 10% per annum

$therefore text{SI } = dfrac{P times R times T}{100}$

$Rightarrow dfrac{x}{2} = dfrac{x times 10 times 6}{100} Rightarrow dfrac{1}{2} = dfrac{6}{10}$

Which is not true, so it is not a possible case.

Simple interest for 4 yr
= Rs. (1360 − 1040) = Rs. 320

Simple interest for 3 yr

= Rs. $bigg(dfrac{320}{4} times3 bigg)$  = Rs. 240

$therefore$  Sum or principle = Rs. (1040 − 240)
= Rs. 800

Total borrowed money = Rs. 40000
and rate of interest = 8%

The interest for 2 yr = $dfrac{4000 times 8 times 2}{100}$
= Rs. 6400

Let he paid Rs. $x$ at the end of second year.
 $therefore$ Interest will be calculated on Rs. $(40000 - x + 6400)$

Interest for 3 yr

$begin{aligned} &= dfrac{(46400 - x) times 3 times 8}{100} &= text{Rs. } dfrac{6}{25}(46400 - x) & therefore dfrac{6}{25} (46400 - x) + 46400 - x = 35960 &Rightarrow x = dfrac{21576 times 25}{31} = text{Rs. }17400 end{aligned}$

Let the amount remaining to pay be Rs. $x$.
 $therefore$ Price of house = Rs. $(x + 8000)$

$begin{aligned} & Rightarrow 9600 - dfrac{x times 4 times 5}{100} =x ~ & Rightarrow 9600 - dfrac{x}{5} =x ~ & Rightarrow 9600 =dfrac{6x}{5} Rightarrow dfrac{9600 times 5}{6} = x ~ & Rightarrow x = 8000 end{aligned}$

 $therefore$ Cash price of the house = Rs. (8000 + 8000) = Rs. 16000

Let the sum be Rs. $x$ and the original rate $r$, then
Simple interest $= dfrac{x times r times 2}{100}$

Now, rate is increased by 3%.

Therefore, New rate $= ( r + 3)$

Because, Simple interest $= dfrac{x times (r+3) times 2}{100}$

$begin{aligned} &therefore dfrac{x times (r + 3) times 2}{100} - dfrac{x times r times 2}{100} = 72 \ &Rightarrow dfrac{(xr + 3x)2}{100} - dfrac{2xr}{100} = 72 \ &Rightarrow dfrac{2xr + 6x -2xr}{100} = 72 \ &therefore x = 1200 end{aligned}$

Here, $I_1 - I_2$ = Rs. 10, $P_1$ = Rs. 1230,
$P_2$ = Rs. 1130 and T = 2 yr

According to formula,

$begin{aligned} &I_2 - I_1 = dfrac{T times R times (P_2 - P1)}{100} \ &-10 = dfrac{R times 2 (1130 - 1230)}{100} \ &-10 = dfrac{-200R}{100} \ &therefore R = 5$