A man bought three tables for rs. $2500$, then find the prices of the individual tables?

Question

Problem A man bought three tables for rs. $2500$. He sell the first at $5 $ % loss, second at $5 $ % profit and the third at $10$% profit. Find the cost price of each table if on the whole he neither gains nor losses.

Solution
Let cost price of first table be x

Let cost price of second table be y

then cost price of third table $= 2500-(x+y)$

$0.95x +1.05y +1.10 [2500-(x+y)]=2500$

$3x+y =5000$

What next $?$

How can we find the answer $?$

Answer 1

There's not much you can do, the question does not have a unique solution. You can, however, move forward a bit by considering the fact that all prices must be positive (or at least not negative), so that $2500-(x+y)\geq 0$ which yields $5000=3x+y=2x+(x+y)\leq 2x+2500$, so $x\geq 1250$. In addition since $y\geq 0$ you also have $5000=3x+y\geq 3x$ so $x\leq \frac{5000}{3}$. However, it is still true that for every $x$ in that range you still get a solution.

For example: $x=1250, y=1250, z=0$ is a solution where the loss/profit on the first two tables cancel each other and the third one is meaningless, but $x=1500, y=500, z=500$ is also a valid solution, and so is $x=\frac{5000}{3}, y=0, z=\frac{5000}{6}$.

Answer 2

Let the prices be $x,y,2500-(x+y)=z$ so $x+y+z=2500$ but acc. To condition total loss = total profit so $95x/100=105y/100+110z/100$ . Now just plug in values as per choice as its an open end question one of the answer can be $2000,500,0$ . But that hardly makes sense. Hope it helps you.