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

Answer: Option A

Explanation:

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

By applying above formula we can easily solve this question, as we are already having the simple interest.

\begin{aligned}
=> P = \frac{1230 * 100}{6*5} \\
=> P = 4100
\end{aligned}

Answer: Option A

Explanation:

One thing which is tricky in this question is to calculate the number of days.
Always remember that the day on which money is deposited is not counted while the day on which money is withdrawn is counted.
So lets calculate the number of days now,
Time = (24+31+18) days = 73/365 years = 1/5 years
P = 2000
R = 25/4%

\begin{aligned}
\text{ S.I. = } = \frac{2000 \times 25 }{4 \times 5 \times 100} = 25
\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 C

Explanation:

SI for 3 year = 2600-2240 = 360
SI for 2 year 360/3 * 2 = 240
principal = 2240 - 240 = 2000

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}