Reema took a loan of Rs 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest.

Answer: Option D

Explanation:

We need to calculate the total amount to be paid by him after 4 years, So it will be Principal + simple interest.

So,
\begin{aligned}
=> 500 + \frac{500*5*4}{100}
=> Rs. 600
\end{aligned}

Answer: Option D

Explanation:

Here firstly we need to calculate the principal amount, then we can calculate the new rate.

\begin{aligned}
P = \frac{S.I. * 100}{R*T} \\
P = \frac{840 * 100}{5*8} \\
P = 2100 \\

\text{Required Rate = } \frac{840 * 100}{5*2100} \\
R = 8\%\\

\end{aligned}

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}

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 B

Explanation:

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

\end{aligned}