Compound Interest Questions and Answers

$text{Amount} = text{ Rs. } 800 times bigg(1 + frac{5}{100} bigg)^2 times bigg(1 + frac{5}{200}bigg)$
= Rs. 904.05

$text{88200} = 80000 Big(1 + frac{x}{100}Big)^2$

$therefore Big(1 + frac{x}{100} Big)^2 = frac{441}{400} = Big( frac{21}{20}Big)^2$

$x = 5$

$text{Principal } = frac{1352}{left( 1 + frac{4}{100}right)^2 }$

$= 1352 times frac{625}{676} = text{Rs. } 1250$

$4x = xBig(1 + frac {r}{100} Big)^2$

$therefore Big(1 + frac{r}{100} Big)^2 = 40 text{ or } 1 + frac{r}{100} = 2$

$text{or } r =100$

$text{Amount} = 800 times Big(1 + frac{5}{100}Big)^3$

$= 800 times frac{9261}{8000} = 926.10$

$therefore text{ Interest} = text{Rs.} 926.10 - text{Rs.} 800 = text{Rs.} 126.10$

$because$ Compounded half yearly

$therefore$ 1 year = 2 half year

and rate 10% annually = $frac{10}{2}$ half yearly

= 5%

$because text{ C.I. } = text{P }Big[big(1+frac{r}{100}big)^n -1 Big] = 10,000$

$= Big[big(1+frac{5}{100}big)^2 -1 Big] = 10,000 times frac{41}{400} = 1,025$

$text{ Required amount } = frac{9261 times 20 times 20 times 20 }{21 times 21 times 21} = 8,000$

$text{P} = frac{1}{Big(1 + frac{tr}{100}Big)} = frac{1000}{Big(1 + frac{10r}{100}Big)}$
$= frac{10000}{10 + r} quad ... (text{I})$

$=\genfrac{}{}{0ex}{}{10000}{10+r}...(\text{I})$

Again

$text{P } = frac{1250}{Big(1 + frac{15r}{100}Big)} = frac{1250 times 20}{20 + 3r} quad ... (text{II})$

By equation I and II

$frac{10000}{10 + r} = frac{1250 times 20}{ 20 + 3r}$

$therefore quad frac{20 + 3r}{10 + r} = frac{1250 times 20 }{10000}$

$quad therefore quad r = 10$

$therefore quad$ Now on putting the value of $r$ in eq. I we get

$text{P } = frac{10000}{10 + 10 } = 500$

Given, principal amount = $P$
Rate of interest, $R = frac{r}{k}$ per annum and
Time, $T = nk$

$begin{aligned} & because A = P bigg(1 + frac{R}{100} bigg)^T hspace{0.5cm} [text{Rule 1. (i)}] \ & therefore A = P bigg(1 + frac{r}{100k} bigg)^{nk} end{aligned}$

Given, $P$ = ₹ 7500, $R$ = 16% and $n$ = 2 yr

$begin{aligned} because A &= P bigg ( 1 + frac{R}{100} bigg) ^n hspace{0.5cm} [text{Rule 1. (i)}] \ & = 7500 bigg ( 1 + frac{16}{100} bigg) ^ 2 \ & = 7500 bigg (frac{116}{100} bigg) ^ 2 \ & = 7500 times frac{29}{25} times frac{29}{25} \ & = 12 times 841 = 10092 end{aligned}$

Let the principal be ₹ $x$.
$therefore$  Rate = ₹ $R$, Amount = ₹ $2x$
and Time = $n$ yr

$begin{aligned} because A &= P bigg ( 1 + frac{R}{100} bigg) ^n hspace{0.2cm} [text{Rule 1. (i)}] &therefore x bigg(1 + frac{R}{100} bigg)^n = 2x &Rightarrow bigg( 1 + frac{R}{100} bigg)^n = 2 hspace{0.4cm} text{...(i)} end{aligned}$

Let it becomes four fold in $N$ yr.
Then, $xbigg(1 + frac{R}{100}bigg)^N = 4x$

$begin{aligned} Rightarrow bigg( 1 + frac{R}{100} bigg)^N = 4 \ Rightarrow 2^2 = bigg( 1 + frac{R}{100} bigg)^N \ Rightarrow bigg( 1 + frac{R}{100} bigg)^{2n} \ = bigg( 1 + frac{R}{100} bigg)^N [text{from Eq. (i)}] \ therefore N = 2n text{ yr} end{aligned}$

Let the principal amount be ₹ $P$.
By given condition, SI = $frac{1}{2}$ CI

$begin{aligned} &Rightarrow frac{P times times 3}{100} = frac{1}{2} & Bigg[400 bigg( 1 + frac{10}{100} bigg) ^2 - 4000 Bigg] [text{Rule 1 (ii)}] end{aligned}$

$begin{aligned} & Rightarrow frac{24P}{100} = frac{1}{2} Bigg[ 4000 times frac{121}{100} - 4000 Bigg] & = frac{1}{2} [4840 - 4000] & Rightarrow frac{24P}{100} = 420 &therefore P = frac{420 times 100}{24} = 1750 end{aligned}$

Given, $P$ = ₹ 1250, $R$ = 8% per annum and $n$ = 2yr

$begin{aligned} because text{Simple interest } = frac{P times R times T}{100} therefore text{Simple interest } = frac{1250 times 8 times 2}{100} end{aligned}$

= ₹ 200

$because$  Compound Interest

$begin{aligned} &= P Bigg[ bigg(1 + frac{R}{100} bigg)^n Bigg] [text{Rule 1. (ii)}] because text{CI } &= 1250 big(1 + frac{8}{100} big)^2 - 1250 ~ &= 1250 times big(frac{108}{100} big)^2 - 1250 ~ &= 1458 - 1250 = 208 end{aligned}$

$therefore$ Difference in SI and CI = ₹ 208 − ₹ 200
= ₹ 8

Given, principal (P) = Rs. 100
Amount (A) received after 2 yr = Rs. 121
Let rate of interest = R% per annum

$begin{aligned} & because A = P bigg(1 + frac{R}{100} bigg)^n hspace{0.2cm} [text {Rule 1. (i)}] & therefore 121 = 100 bigg(1 + frac{R}{100} bigg)^2 ~ & Rightarrow frac{121}{100} = bigg(frac{100+R}{100} bigg)^2 ~ & Rightarrow bigg(frac{11}{10} bigg)^2 = bigg(frac{100+R}{100} bigg)^2 ~ & Rightarrow frac{11}{10} = frac{100+R}{100} ~ & Rightarrow 100+R = frac {11 times 100}{10} & Rightarrow 100+R = 110 & therefore R = 110-100 & therefore text{Rate of interest} = 10$