Question

If the Josephites soccer team won a total of c games this season and last season, and the team won b fewer games this season than last season, how many games did they win last season?
(A). $\dfrac{{b - c}}{2}$
(B). $\dfrac{{c - b}}{2}$
(C). $\dfrac{{b + c}}{2}$
(D). $b + \dfrac{c}{2}$
(E). $c + \dfrac{b}{2}$

Answer 1

Hint: This particular question belongs to algebraic equations. First, we’ll assume some initial variables and then we’ll analyse the question. According to the question we’ll try to make algebraic equations and then solve them to get an answer.

Complete step-by-step answer:
In the question, it says a team won c games in two seasons. That is the current season and previous season. Also, the same team won b fewer games. It means, if it would have won 10 games in the previous season then this season it won only 10-b games. (That’s how algebraic questions work).
Here, we need to find the total matches won in last year’s season.
To calculate this, we’ll first let the total number of matches won last season is y and matches won this season is x.
Now according to the given condition, c is the sum of total matches won in both seasons so,
$x + y = c - - - - - (1)$
On using the condition, team won b fewer matches then previous season, we get,
$x = y - b - - - - - - (2)$
In order to solve this question, we’ll simply solve equation (1) and (2).
On putting the value of x in equation (1) from equation (2).
$
  y - b + y = c \\
   \Rightarrow 2y = c + b \\
   \Rightarrow y = \dfrac{{c + b}}{2} \\
 $
We had to find the number of matches won last year’s season and we assumed it as y. Now we have got its value as $\dfrac{{c + b}}{2}$.
Hence, Option C is the correct option.

Note: In topics like algebraic equations, we use to assume the required value as some variable. Later as per the given conditions we try to find its value. In higher studies, whenever we assume such variables, we need to keep the domain in mind. Topics like linear algebra domains of a variable play a crucial role to get the problem solved.