Problem: A train starts from station A with some passengers, At station B, 10% of the passengers get down and 100 passengers get in . At station C, 50% get down and 25 get in. A station D, 50%  get down and 50 get in making the total number of passengers 200. How many passengers did board the train at station A?

Correct Answer: 500

Explanation:

Let, x passengers were boarded at station A.

After getting down 10% passengers, the number of passengers = x-10% of x = x-$\frac{10x}{100}$=x-$\frac{x}{10}$ =$\frac{9x}{10}$

After getting in 100 passengers, the number of passengers at station B = $\frac{9x}{10}$+100= $\frac{9x+1000}{10}$

After getting down 50% passengers, the number of passengers = $\frac{9x+1000}{10}$-50% of ($\frac{9x+1000}{10}$)

= $\frac{9x+1000}{10}-\frac{50}{100}(\frac{9x+1000}{10})$

= $\frac{9x+1000}{10}-\frac{1}{2}(\frac{9x+1000}{10})=\frac{1}{2}(\frac{9x+100}{10})$

Afer getting in 25 passenger, the number of passengers at station c

= $\frac{1}{2}(\frac{9x+1000}{10})+25=\frac{9x+1000}{20}+25=\frac{9x+1500}{20}$

After getting down 50% passengers, the number of passengers = $\frac{9x+1500}{20}-50% of \frac{9×1500}{20}$

= $\frac{9x+1500}{20}-\frac{50}{100}(\frac{9x+1500}{20})$-$\frac{1}{2}(\frac{9x+1500}{20})$

= $\frac{1}{2}(\frac{9x+1500}{20})=\frac{9x+1500}{40}$

= The number of passengers at station D will be, $\frac{9x+1500}{40}+50=200$

=> $\frac{9x+1500}{40}+50=200-50$

=> $\frac{9x+1500}{40}$ =150

=> 9x+1,500 =6,000

=>9x=6,000-1,500 = 4,500

so, x=$\frac{4500}{9}$ = 500

500 passengers were boarded at station A.