A sum of money at simple interest amounts to Rs. 2240 in 2 years and to Rs. 2600 in 5 years. What is the principal amount

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 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:

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}

Answer: Option C

Explanation:

SI for 3 year = 2600-2240 = 360
SI for 2 year 360/3 * 2 = 240
principal = 2240 - 240 = 2000

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}