Sachin borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends money to Rahul at 25/4% p.a. for 2 years. Find the gain of one year by Sachin.

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:

Two things need to give attention in this question, First we need to calculate gain for 1 year only.
Second, where we take money at some interest and lends at other, then we use to subtract each other to get result in this type of question. Lets solve this Simple Interest question now.

\begin{aligned}
\text{Gain in 2 year = } \\
[(5000 \times \frac{25}{4} \times \frac{2}{100})-(\frac{5000 \times 4 \times 2}{100})] \\
= (625 - 400) = 225 \\
\text{ So gain for 1 year = }\\
\frac{225}{2} = 112.50
\end{aligned}

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 D

Explanation:

We can get SI of 3 years = 12005 - 9800 = 2205

SI for 5 years = (2205/3)*5 = 3675 [so that we can get principal amount after deducting SI]

Principal = 12005 - 3675 = 6125

So Rate = (100*3675)/(6125*5) = 12%

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}