Question

A sum of INR 2800 is to be used to award four prizes. If each prize is INR 200 less than the proceeding prize, find the value of each of these prizes.

Answer 1

Hint: This is a simple arithmetic problem. We will assume the first prize as some variable. Then other three prize values can be considered based on the given constraints in the problem. Further single variable linear equations can be formed and solved accordingly.

Complete step-by-step answer:
Let us suppose that the amount of the first prize be INR x. Since each prize value after the first prize is INR 200 less than the preceding prize. Thus, the amounts of the four prizes are in arithmetic progression means AP.
Thus according to the question,
Amount of the second prize will be x- 200.
Amount of the third prize will be x - 200 - 200 = x - 400.
Amount of the fourth prize will be x- 400 – 200 = x-600.
 Now, as given in the question that the total sum of prizes is INR 2800.
So by the condition given in the question, we have
\[
  x + \left( {x - 200} \right) + \left( {x - 400} \right) + \left( {x - 600} \right) = 2800 \\
   \Rightarrow 4x - 1200 = 2800 \\
   \Rightarrow 4x = 4000 \\
   \Rightarrow x = 1000 \\
 \]
Hence, we got the value of an unknown variable of the above equation, x as 1000.
Amount of the first prize will be, x = INR 1000
Amount of the second prize will be, x- 200 = 1000-200 = INR 800.
Amount of the third prize will be, x – 400 = 1000-400= INR 600.
Amount of the fourth prize will be, x- 600 = 1000-600= INR 400.

Note: Simple algebraic concepts can be utilized for framing the linear equations having one unknown variable. Such equations after solution will give the value of unknown variables. Various mathematical transformations and rules will help to get solutions to such equations.
* Linear equations in one variable are in the form $ax+b=0$.
* Linear equations in two variables are in the form $ax+by+c=0$.