If a sum of money doubles itself in 8 years at simple interest, the ratepercent per annum is

Answer: Option B

Explanation:

\begin{aligned}
P = \frac{S.I. * 100}{R*T}
\end{aligned}

So by putting values from our question we can get the answer

\begin{aligned}
P = \frac{4016.25 * 100}{9*5} \\
= 8925
\end{aligned}

Answer: Option B

Explanation:

Let rate = R% then Time = R years.
\begin{aligned}
=> \frac{1200*R*R}{100}=432 \\
=> R^2 = 36 \\
=> R = 6\%

\end{aligned}

Answer: Option B

Explanation:

\begin{aligned}
\text{S.I.} = \frac{P*R*T}{100} \\
=> P = \frac{S.I. * 100}{R*T} \\
=> P = \frac{5400 * 100}{12*3} = Rs 15000
\end{aligned}

Answer: Option B

Explanation:

Lets assume that three parts are x, y and z.
Simple Interest, R = 5%

From question we can conclude that, x + interest (on x) for 2 years = y + interest (on y) for 3 years = z + interest (on y) for 4 years

\begin{aligned}
\left( x + \frac{x*5*2}{100} \right) = \left( y + \frac{y*5*3}{100} \right) = \left( z + \frac{z*5*4}{100} \right)\\
\left( x + \frac{x}{10} \right) = \left( y + \frac{3y}{20} \right) = \left( z + \frac{z}{5} \right) \\
=> \frac{11x}{10} = \frac{23y}{20} = \frac{6z}{5} \\

\text{lets assume k = }\frac{11x}{10} = \frac{23y}{20} = \frac{6z}{5} \\
\text{then }x = \frac{10k}{11} \\
y = \frac{20k}{23}\\
z = \frac{5k}{6}\\
\text{we know x+y+z = 2379}\\
=> \frac{10k}{11} + \frac{20k}{23} + \frac{5k}{6} = 2379\\
\text{10k*23*6+20k*11*6+5k*11*23=2379*11*23*6}\\
\text{1380k+1320k+1265k=2379*11*23*6}\\
\text{3965k=2379*11*23*6}\\
k = \frac{2379*11*23*6}{3965}\\
\text{by putting value of k we can get x} \\
x = \frac{10k}{11} \\
=>x = \frac{10}{11}*\frac{2379*11*23*6}{3965}\\
=>x = \frac{10*2379*23*6}{3965}\\
= \frac{2*2379*23*6}{793}\\
= 2 * 3 * 23 * 6 = 828
\end{aligned}

Answer: Option A

Explanation:

\begin{aligned}
\text{S.I.} = \frac{P*R*T}{100} \\
=> R = \frac{S.I. * 100}{P*T}
\end{aligned}

So, S.I = 15500 - 12500 = 3000.

\begin{aligned}
=> R = \frac{3000 * 100}{12500*4} = 6\%
\end{aligned}