A man took a loan at rate of 12% per annum simple interest. After 3 years he had to pay 5400 interest. The principal amount borrowed by him was.

Answer: Option A

Explanation:

S.I. = 956 - 800 = Rs 156

\begin{aligned}
R = \frac{156*100}{800*3} \\
R = 6\frac{1}{2}\% \\

\text{ New Rate = }6\frac{1}{2}+4 \\
= \frac{21}{2} \% \\

\text{ New S.I. = }800\times\frac{21}{2}\times{3}{100} \\
= 252
\end{aligned}

Now amount will be 800 + 252 = 1052

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Always remember that the day on which money is deposited is not counted while the day on which money is withdrawn is counted.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:

\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 B

Explanation:

Let R% be the rate of simple interest then,

from question we can conclude that

\begin{aligned}
(\frac{5000*R*2}{100}) + (\frac{3000*R*4}{100}) = 2200 \\

<=> 100R + 120R = 2200 \\
<=> R = 10\%


\end{aligned}