MGMAT Challenge Problem 9/2/02

[cdemisch]

In this data sufficiency question, the two statements seem to contradict each other; although it doesnt effect the answer, it was confusing to me when solving:

In the game Cako, a player is awarded one tick for every third Alb captured, and one click for every fourth Berk captured. The total score is equal to the product of clicks and ticks. If a player has a score of 77, how many Albs did he capture?

(1) The difference between Albs captured and Berks captured is 7.

(2) The number of Albs captured is divisible by 4.

Statement 1 means that the player must have captured 35 Albs; 35 is obviously not divisible by 4 as statement 2 says.
Looking at Statement 2 they player captured either 4 or 232 Albs..

Are there ever questions where the statements contradict, or is this just an error in the construction of the question?

[StaceyKoprince]

Haven't forgotten about you! The two statements are not supposed to contradict each other. Someone's checking into this one and will get back to you ASAP.

[StaceyKoprince]

cdemisch, you have indeed found an error! The two statements do not contradict and, in this problem, they do. We've fixed the problem in our database, so all future students will benefit from your sharp eye.

Also, you've earned yourself something free from the ManhattanGMAT stores for finding the error. Please email danielle@manhattangmat.com (I've told her about you) to make arrangements for us to send it to you. Please reference this problem and give her the username "cdemisch" and the correct password for that account in order to verify your identity.

Thanks!!

[unique]

cdemisch,

can you solve this please.

[StaceyKoprince]

Nope, s/he can't actually - there's an error with the problem. :)

[Jeff]

Actually, this problem is testing knowledge of modular arithmetic - that is arithmetic with remainders.

Consider how many ticks you get for each Alb captured

Alb Tick
0 0
1 0
2 0
3 1
4 1
5 1
6 2

The point here is that you do not get fractional ticks (T) for each Alb captured - rather you get one for every third Alb (A) and one click (C) for every fourth Berk (B).

So from the question stem, we know the T + C = 77

Now consider 1). We are told that A - B = 7. It is tempting to see this as sufficient. Turn Alb and Berk into ticks and clicks and you might get A/3 + B/4 = 77. You have two equations in two variables. If you solve, you get A = 135 and B = 128. This gives you 45 ticks and 32 clicks. Now when you look at 2) and see that Alb is divisible by four, there appears to be a conflict between the two statements.

The problem with the above is that the equation you should have sent up above is Q (A/3) + Q (B/4) =77 where Q(x/y) is the integer quotient, excluding the remainder. So eg Q(10/5) =2 and Q(13/5) =2. Now, (1) does not yield a unique solution. A =135, B=128 and A =136, B=129 and A=137 and B=130 all yield T =45 C = 32. Looking just at the Alb for this example, see 135/3 = 45 r0, 136/3 = 45 r1, 137/3 = 45 r2.

Now consider 2. It's clearly insufficient by itself. Consider 1 and 2 together. Of the three possible solutions we found in 1), only A=136, B=129 has A divisible by 4. So 1) and 2) are sufficient together and the answer is C.


I hope you add this problem back into the database - after a couple of problems like this, the real exam will seem easy. ;)

[unique]

Actually, this problem is testing knowledge of modular arithmetic - that is arithmetic with remainders.

Consider how many ticks you get for each Alb captured

Alb Tick
0 0
1 0
2 0
3 1
4 1
5 1
6 2

The point here is that you do not get fractional ticks (T) for each Alb captured - rather you get one for every third Alb (A) and one click (C) for every fourth Berk (B).

So from the question stem, we know the T + C = 77

Now consider 1). We are told that A - B = 7. It is tempting to see this as sufficient. Turn Alb and Berk into ticks and clicks and you might get A/3 + B/4 = 77. You have two equations in two variables. If you solve, you get A = 135 and B = 128. This gives you 45 ticks and 32 clicks. Now when you look at 2) and see that Alb is divisible by four, there appears to be a conflict between the two statements.

The problem with the above is that the equation you should have sent up above is Q (A/3) + Q (B/4) =77 where Q(x/y) is the integer quotient, excluding the remainder. So eg Q(10/5) =2 and Q(13/5) =2. Now, (1) does not yield a unique solution. A =135, B=128 and A =136, B=129 and A=137 and B=130 all yield T =45 C = 32. Looking just at the Alb for this example, see 135/3 = 45 r0, 136/3 = 45 r1, 137/3 = 45 r2.

Now consider 2. It's clearly insufficient by itself. Consider 1 and 2 together. Of the three possible solutions we found in 1), only A=136, B=129 has A divisible by 4. So 1) and 2) are sufficient together and the answer is C.


I hope you add this problem back into the database - after a couple of problems like this, the real exam will seem easy. ;)

Jeff,


Q1.Was the discrepancy in the original problem the 'product' of T x C, because you have done T + C = 77 . I never saw the revised question.
Q2. Statement A says difference between Albs and Berks is 7. Does that mean a - b = 7 or it could also mean b-a=7

[StaceyKoprince]

The only discrepancy in the original question was that the two statements contradicted each other.

The revised question is:

In the game Cako, a player is awarded one tick for every third Alb captured, and one click for every fourth Berk captured. The total score is equal to the product of clicks and ticks. If a player has a score of 77, how many Albs did he capture?

(1) The difference between Albs captured and Berks captured is 7.

(2) The number of Albs captured is divisible by 5.