DS Q

[pratyoosh]

Hello,

Can anyone explain the resolution to the below problem:

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.

Regards,
Pratyoosh.

[clock60]

i hope the answer is C
x-total, y- win, (x-y)- lost we need to find y-?
(1) (x-y+2)=0,8(x+2) i see two unknows and one equation insuff
(2) (y+3)=0,7(x+3) again two unknows
both are solvable

i hope that there is no trap as i got fraction as a result, want to think that it is math mistake..

[Night reader]

i hope the answer is C
x-total, y- win, (x-y)- lost we need to find y-?
(1) (x-y+2)=0,8(x+2) i see two unknows and one equation insuff
(2) (y+3)=0,7(x+3) again two unknows
both are solvable

i hope that there is no trap as i got fraction as a result, want to think that it is math mistake..

The answer is C:

Stimuli: 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?
#games=x
#wins=y
#lost games=x-y

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.

[x-y + 2] = x - 0.2*x since no draws were incurred #game=#wins+#lost games
x-y+2-x+0.2*x=0, 2-y+0.2*x=0, y=2+0.2x Insufficient

(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.
x- (y+3)= 0.3*x, 0.7*x = y+3, y=0.7*x-3 Insufficient

Statements 1&2, 0.7*x-3=2+0.2*x, 0.5*x=5, x=10 (#games) (C)

Full solution, y=2+0.2*x, #wins=2+0.2*10= 4, #looses=10-4=6

p.s. no fractions

[pratyoosh]

Thank you both, the resolution was helpful.

i hope the answer is C
x-total, y- win, (x-y)- lost we need to find y-?
(1) (x-y+2)=0,8(x+2) i see two unknows and one equation insuff
(2) (y+3)=0,7(x+3) again two unknows
both are solvable

i hope that there is no trap as i got fraction as a result, want to think that it is math mistake..

The answer is C:

Stimuli: 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?
#games=x
#wins=y
#lost games=x-y

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.

[x-y + 2] = x - 0.2*x since no draws were incurred #game=#wins+#lost games
x-y+2-x+0.2*x=0, 2-y+0.2*x=0, y=2+0.2x Insufficient

(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.
x- (y+3)= 0.3*x, 0.7*x = y+3, y=0.7*x-3 Insufficient

Statements 1&2, 0.7*x-3=2+0.2*x, 0.5*x=5, x=10 (#games) (C)

Full solution, y=2+0.2*x, #wins=2+0.2*10= 4, #looses=10-4=6

p.s. no fractions

[GMATGuruNY]

Hello,

Can anyone explain the resolution to the below problem:

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.

Regards,
Pratyoosh.

Another approach:

Statement 1: Losing 2 more games = winning 20%
y-2 = .2x
2 variables, 1 equation. Insufficient.

Statement 2: Winning 3 more games = losing 30% = winning 70%
y+3 = .7x
2 variables, 1 equation. Insufficient.

Statements 1 and 2 together:
2 variables, 2 equations. Sufficient.

The correct answer is C.

If this were a PS question that we needed to solve:
Dividing the first equation by the second, we get the following ratio:
(y-2)/(y+3) = 2/7
7y-14 = 2y+6
5y = 20
y = 4.