Sahil took a loan for 6 years at the rate of 5% per annum on Simple Interest, If the total interest paid was Rs. 1230, the principal was

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 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:

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:

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:

Let the principal be P and rate be R

then

\begin{aligned}
\text{ratio = } [\frac{(\frac{P*R*6}{100})}{(\frac{P*R*9}{100})}] \\

= \frac{6PR}{9PR} = 2:3
\end{aligned}