In each game of a certain tournament, a contestant either loses 3 poin

Both statements by themselves are clearly insufficient.

Taking both the statements together:

We have an addition of 4 points. So we need minimum of 2 games which add 2 points each.

Further any 3 point loss making games have to be cancelled out by the 2 point gain games. We need 3 games gaining 2 points each to cancel out 2 games losing 3 points each. 3+2=5 games. We need these 5 games to cancel out.

So we need 5x+2 games to get 104 points, where x>=0 and x represents the number of times the 3 gain making and the 2 loss making games (that cancel out each other) have to be played.

The only number of games satisfying the above equation which is less than 10 games is 2 (when x=0) and 7 (when x=1).

Since we have no single number of games we end up with E.

Answer: E

(1) No information about loosing or gaining of first 100 games,Insufficient

(2) No information about loosing or gaining any game,Insuffifient

(1)+(2) together didn't solve the issue raised above.

Correct Answer E

My approach without solving any equation:

Statement 1: he ends up with 104 points . That means he might have won 2 times so he got +2 and +2 ,or might have lost 2 time -3 and -3 and then won 5 times (gained +2 each time ). One way or another we are not sure about the number of the games he played.---> INSUFF

Statement 2: If Pat played games<10 he might have played 9 , 8 ,7 ,6,5,4 .... we are not sure --> INSUFF

Combining 1 and 2 we know he had 104 points at the end and less than 10 games, AGAIN he might have played a different number of games each time in order to gather 104 points, he might have lost at the beginning and won the rest or vice versa. --> INSUFF

so answer E

Hi All

Thanks for all your very good answers. It took me a while to answer this question. In the end I got it right but I was unsure if I had the right methodology.
Overall I used the same thinking which consists in using test cases. However I'm wondering whether there is a faster solution which would consist in recognizing a pattern:
Can't we just generalize and say that if we have a combination of 1 equation and 1 inequalities with two unknown, therefore it remains insufficient to solve for the 2 unknown ? Using that pattern for other DS questions would help solving other DS questions tremendously faster...

the OG DS problem #282 (OG 2017) used similar reasoning and ended up with E as OA.

Thanks in advance for your feedback on this one.

Target question: How many games did Pat play in the tournament?

Jump straight to...

Statements 1 and 2 combined
There are several CONFLICTING situations that that satisfy BOTH statement 2. Here are two:
Case a: Pat plays 2 games and wins both of them to add 4 points to the 100 points she started with. In this case, Pat plays 2 games
Case b: Pat plays 7 games and wins 5 of them and loses 2 to add 4 points to the 100 points she started with. In this case, Pat plays 7 games
Since we cannot answer the target question with certainty, the combined statements are NOT

Answer:
Cheers,
Brent

YanisBoubenider thefibonacci I still don't get why C is not an option.

if we know that final amount of points is 104, and he played less than 10 games, then the only possible amount of games played is 5. That is, lost 2 (104-6=98) and won 3 (98+6=104). He couldn't lose 3 because it would result in the odd number of loses neither he could lose 4 because he would have to win 6 and that equals 10 which contradicts with the 2nd statement.

I. 104-100 = 4
Min games = 2. But, 4 can also be obtained by scoring 2+2+2+2+2-3-3 = 4. (Games = 7)
II. Games less than 10. 4 or 7 as in 1.
Both Insufficient. Ans=E

Transform the Q: x + y = ?



1) 4 = -3x + 2y
y = 2 + 3x/2

x = 0, y = 2
x = 2, y = 5

Not sufficient

2) x + y < 10
Not sufficient

1+2)

x = 0, y = 2
x = 2, y = 5

Not sufficient

ANSWER: E

Final point = 100+2x -3y ( x = wins, y = losses)
Total Games = x + y

(1) 104 = 100+2x -3y
2x -3y = 4
No separate value for x and y. Insufficient.

(2) x + y <10
Several values for x and y fit this information

Considering both options:
x + y <10, and 2x -3y = 4 , if x = 2 , y = 0 Satisfies the condition, if x = 5, y = 2, still satisfies the condition.

We get two different information for total games. Ans. E.

Hello

Taking : x = no. of games won; y = no. of games lost

Using 1) I created the equation :
2x-3y = 4
2x = 3y + 4
x = (3y + 4)/2. From here we can infer that (3y+4) has to be even so that x is an integer, which means y has to be even.
Taking cases : y=0, x=2: y=2,x=5: y=3,x=8... and so on

Using 2) x+y<10

Combining both 1 and 2:

We get two cases :
Cases 1: y=0,x=2
Case 2: y=2, x=5

Since we dont get a unique case, both statements together are insufficient.

Please verify if my approach is correct.

Thanks,
Akshit

Really interesting question. Here's my take on this.

In each game of a certain tournament, a contestant either loses 3 points or gains 2 points. If Pat had 100 points at the beginning of the tournament, how many games did Pat play in the tournament?

(1) At the end of the tournament, Pat had 104 points
(2) Pat played fewer than 10 games

Let's work with the question stem first.

Equation can be: 100+2w-3l
To find: w+l.

Statement 1:

100+2w-3l= 104
we can make this: 2w+3l=4
2w=4+3l
Since we know the value has to be integral. We can take values. But we can keep the even and odd rule in mind.

We know 4+3l has to be even to be divisible by 2w.
Since, even + even=even. We can infer that 3l is also even. Making l even.
So let's only see values of l which are even: 0,2,4,6,8,10......till infinity.
For l=0, w=2
For l=2, w= 5
For l=4, w= 8

.... There are multiple values resulting in 104 points. Hence, insufficient.

We can remove A and D.

Statement 2: He played less than 10 games.
This is clearly insufficient.

We can remove B

Combining both statements:

As we saw there are multiple versions for his final score, and on adding this filter of less than 2 games, he still would have 2 versions remaining. Making this insufficient.

For l=0, w=2.......He played 2 games
For l=2, w= 5........He played 7 games.
Both leading to a 104 result.

E is the answer.

 
Answer: Option E

Video solution by GMATinsight

GMAT insight -

That trick where the first variable changes by the coefficient of the second variable is gold. Thanks for that...

According to me, the answer should be (c)- Both statements are together sufficient. Why are we trying to solve it just by Algaebric equations? We know that we had 100 points and we want to reach 104. If we know that number of games are < 10, then there is only 1 possibility by Hit and Trial method that this is possible- he wins 2 games. 102+2+2= 104.

­First of all, this is an official question, and the OA is given as E, not C. This means that the OA is indeed E. Next, it seems that you missed the discussion above altogether because many posts provide two possible combinations: 2 wins and 0 losses, and 5 wins and 2 losses.

You are absolutely right! Thanks!
­

Why cant we substitute the value of one variable and get the answer? For example -
2x - 3y = 4 - from statement 1
x + y <10 - from statement 2

2x = 4+3y
this deduces to y <1.5 when input in statement 2 i.e., y can be 0 or 1
when we input both values in 100+2x-3y = 4, we get x =2 for y =0, and x = 3.5 for y = 1 (invalid)

So, C, both statements are sufficient together. Please correct if I have made a silly mistake or understood it wrongly.