enigma123 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 ySince we

can answer the

target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,

Brent

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