At what rate percent per annum will the simple interest on a sum of money be 2/5 of the amount in 10 years

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 A

Explanation:

Let the sum is 100.

As financier includes interest every six months., then we will calculate SI for 6 months, then again for six months as below:

SI for first Six Months = (100*10*1)/(100*2) = Rs. 5

Important: now sum will become 100+5 = 105

SI for last Six Months = (105*10*1)/(100*2) = Rs. 5.25

So amount at the end of year will be (100+5+5.25)
= 110.25

Effective rate = 110.25 - 100 = 10.25

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 D

Explanation:

We need to calculate the profit of B.
It will be,
SI on the rate B lends - SI on the rate B gets

\begin{aligned}
\text{Gain of B}\\ &= \frac{3500\times11.5\times3}{100} - \frac{3500\times10\times3}{100}\\
= 157.50
\end{aligned}

Answer: Option C

Explanation:

\begin{aligned}
\text{ S.I. = } \frac{P \times R \times T}{100}
\end{aligned}
So, by putting the values in the above formula, our result will be.
\begin{aligned}
\text{ Required result = } \frac{7000 \times 50 \times 9}{3 \times 12 \times 100} = 875
\end{aligned}

[Please note that we have divided by 12 as we converted 9 months in a year format]