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DS question

by sanaa.rizwan » Tue May 14, 2013 12:18 pm
A certain football team played x games last season, of which the team won exactly y games. If tied games were not possible, how many games did the team win last season?

(1) If the team had lost two more of its games last season, it would have won 20 percent of its games for the season.

(2) If the team had won three more of its games last season, it would have lost 30 percent of its games for the season.
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by Brent@GMATPrepNow » Tue May 14, 2013 1:01 pm
sanaa.rizwan wrote:A certain football team played x games last season, of which the team won exactly y games. If tied games were not possible, how many games did the team win last season?

(1) If the team had lost two more of its games last season, it would have won 20 percent of its games for the season.

(2) If the team had won three more of its games last season, it would have lost 30 percent of its games for the season.
Target question: How many games did the team win last season?

Rephrased target question: What is the value of y?

Given: The team played x games and won exactly y games

Statement 1: If the team had lost two more of its games last season, it would have won 20 percent of its games for the season.
So, winning 2 fewer games results in 20% wins
In other words, y - 2 = 20% of x
Simplify to get: y - 2 = 0.2x
Cannot solve this 2-variable linear equation for y, so statement 1 is NOT SUFFICIENT

Statement 2: If the team had won three more of its games last season, it would have lost 30 percent of its games for the season.
Lose 30% = Win 70%
So, winning 3 more games results in 70% wins
In other words, y + 3 = 70% of x
Simplify to get: y + 3 = 0.7x
Cannot solve this 2-variable linear equation for y, so statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined:
We have . . .
y - 2 = 0.2x
y + 3 = 0.7x
We have 2 variables and 2 distinct linear equations.
We can solve this system for y
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

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Brent
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by srcc25anu » Tue May 14, 2013 1:11 pm
Games Played be X
GAmes Won be Y
Games Lost = x-y

St1: If 2 more games were lost, then games lost = X-Y+2; Games WOn = Y-2 and Total games still X
and % wins = 20%
therefore (y-2)/x = 20% or X = 5Y-10
Not sufficient to solve two varibales using one linear eqaution.

St2: if 3 more games were WON, then games won = Y+3; Games Lost = x-y-3 and total games = X
% loses would then be 30%
so (x-y-3) / x = 30% or 7x = 10 y + 30
Again Insufficient to solve for 2 variables using 1 equation.

Together: two equations and two variables to solve. Sufficient
Solving we get x(Total Games) = 10 and y (Games Won) = 4
Ans C
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