Hey Brent!

Yes, you are right, I meant to type x-intercepts!

That typo corrected, though, the relationship between the discriminant and the x-intercepts is important: the contemporary GMAT expects you to know the consequences of having b² - 4ac > 0, = 0, and < 0, respectively. There are a few questions in the recent software (Exam Pack 1, Question Pack 1, Quant Focus) that turn on it, as there are on the consequences of a positive/negative a-coefficient, IIRC.

Hi guys:
Could you confirm this problem is wrong? (I know you can solve this with simmetry but if you do the math it does not make sense)
These are my findings:
1. Even if the question stem does not state the problem is about a parabola, the graph loks like it and the OG answer refer to a parabola (OG says in the answer "Second degree eq")
2. The graph implies (0,3) is a point in the parabola. You can notice the y-intercept. This implies that in y = ax^2 + bx + c, c = 3.
3. Because the problem states the graph is simmetric with respect to x=2. That would mean there is minimum point at (2, y). We know the x coordinate of a minimum point of a parabola follows this eq: x= -b/2a. That would mean 4a = -b.
4. We know (1,1) is a point in the parabola. So
y = ax^2 + bx + c
1 = a(1)^2 + b(1) + c
1 = a(1)^2 + b(1) + 3 (because of point 2)
a + b = -2
5. Solving 3. and 4.
a=2/3, b=8/3
6. Parabola formula
y=(2/3)x^2-(8/3)x+3
7. Use parabola formula for point (1/2, 2) and it does not work
2=1/6-4/3+3
2=11/6?????

If you use the graph a little more you can assume the vertex is at (2,0)
Then:
0 = a(2)^2 + b(2) + 3
4a + 2b + 3 = 0
Because x=-b/2a, and (2,0) is the vertex:
4a=-b
Use the 2 eq above:
-b + 2b + 3 = 0
b = -3, a = 3/4 => DOES NOT MAKE SENSE because contradicts what was found in the other scenario.

Could you help me with the math? This is making me crazy.
I'm intending to demonstrate that the answer could also be found using math (symmetry logic is ok for me and for the exam but math should always back up any answer, right?) or that the problem data is incorrect.
I tried graphing the eq I got in 6. in this page:
https://www.mathwarehouse.com/quadratic/ ... rabola.php
I tried MATLAB too and find point (1/2, 2) is not valid.

I guess I can only conclude:
- The graph is not a parabola but some kind of mixed graph for intervals of x.

PS: If you think I'm wrong at using the graph and should only use points (1,1) (1/2, 2) and (2,y) as vertex, try to find the parabola eq. I think you are going to get an invalid result too.

Gustavo

Matt@VeritasPrep wrote:

GMATinsight wrote:Hi Karly,

Please keep in mind that GMAC is NEVER wrong. These question are made scientifically and they don't have any potential to be incorrect.

This is not true. Any standardized test has the potential to err, and most of them make mistakes at some point. The LSAT seems to remove a question from scoring at least once a year due to error or ambiguity, and even a math test as venerable as the AMC has had this issue in the past, so I don't see why the GMAT is exempt.

The difference with the GMAT is that it is very hard to catch their test writers in a mistake, as you'd need to memorize the question, then somehow verify independently after the fact that it was wrong; the question will almost never be released for public scrutiny. (This lack of transparency is a major issue, in my opinion.)

That said, GMAT questions are very seldom wrong or flawed, as they go through many checks and reviews; I'd imagine fewer than 1% of questions on the actual GMAT are flawed or ambiguous. (In other words, if you think there's a flaw, YOU'RE probably wrong ... but not always!)

IF YOU HAVE NO SUBSTANCE/MECHANISM TO PROVE THAT GMAC QUESTIONS ARE INCORRECT THEN I DON'T THINK THAT MISGUIDING THE READER HERE BY INSTILLING THE DOUBT IN THEIR MIND ABOUT GMAC IS A GOOD PRACTICE.

IN YOUR INDIVIDUAL CAPACITY YOU CAN CALL ANYTHING WRONG. I UNDERSTAND WE ARE CITIZENS OF FREE COUNTRIES.

GMATinsight wrote:IF YOU HAVE NO SUBSTANCE/MECHANISM TO PROVE THAT GMAC QUESTIONS ARE INCORRECT THEN I DON'T THINK THAT MISGUIDING THE READER HERE BY INSTILLING THE DOUBT IN THEIR MIND ABOUT GMAC IS A GOOD PRACTICE.

Here we go again.

To weaken a statement like "Please keep in mind that GMAC is NEVER wrong. These question are made scientifically and they don't have any potential to be incorrect," I don't need to show that any questions are wrong, I merely need to show that the potential exists. (You didn't say the test writers had merely a perfect record, you said they also have a perfect future.) Showing that similar mistakes have been made on tests whose contents are actually disclosed seems to meet that burden, and I could pretty easily show that "it's scientific, therefore it's infallible" is a non sequitur.

Beyond that, I think that "we/they never make mistakes" is a flatly authoritarian sentiment that should never be endorsed by anyone, anywhere, period. Human error is everywhere, and the mission of education is to teach students how to investigate any doubts. (As the great Richard Feynman once said, never trust the calculations in any published paper if you haven't derived them yourself.) Even Perelman's proof of the Poincare Conjecture had two (trivial) errors, and that is an epoch-making paper written over the better part of a decade by the foremost mathematician on the planet.

I think this is also a dangerous habit of mind to encourage in future MBA students. I know for a fact that some business schools distribute case assignments with bad math in the spreadsheets/tables/etc. to see if their students will catch the mistakes. Of course if a GMAT question seems flawed, the typical student is likely wrong, but it isn't certain. Even if the student is wrong, (s)he ought to learn how to verify that statement personally, rather than blindly swallowing any assertion.

gpanez wrote:Hi guys: Could you confirm this problem is wrong?

Gustavo, I'm getting the same result.

(.5,2) and (1,1) and seemingly (2,0) are on the parabola, which gives us three equations:

2 = a*(.5)² + b*(.5) + c
1 = a*(1)² + b*(1) + c
0 = a*2² + b*2 + c

This system gives a = 2/3 and b = -3 and c = 10/3, so our parabola is

y = (4/9)x² -3x + (10/3)

Plugging in x = 3 yields y = -5/3.

Adding the seemingly obvious point (0,3) to the mix then breaks the equation. Yuck!

Further, the suggested parabola y = (x-2)² doesn't work, as this equation does not fit the coordinate (.5, 2), which we know is on the graph.

BUT

This graph doesn't have to be a parabola. It could be a 50th degree polynomial for all we know.

Thanks Matt,
I guess the answer explanation confused me when it stated the topic of the problem was "Simmetry, quadratic eqs". I agree we can't assume the graph is a parabola (that was not stated on the question stem)
I think what is a little off is just the reference to quadratic eqs in the answer explanation. Maybe that phrase should be removed.

Regards
Gustavo

gpanez wrote:Thanks Matt,
I guess the answer explanation confused me when it stated the topic of the problem was "Simmetry, quadratic eqs". I agree we can't assume the graph is a parabola (that was not stated on the question stem)
I think what is a little off is just the reference to quadratic eqs in the answer explanation. Maybe that phrase should be removed.

Regards
Gustavo

Yeah, I'm with you! Speaking from experience, there is sometimes a disconnect between the person writing the problem and the person writing the explanation: a classic right hand doesn't know left hand situation.

Karly wrote:

winnerhere wrote:did anyone try to applu

y = ax^2 + bx + c formula

and derive a,b,c for the three equations and find the answer? The answer is different with this method.

I can prove that it is a mistake made by GMAC. I plugged in (1/2,2), (1,1) and (2,0) into the equation:y=ax^2+bx+c and I got when x=3, y should be equal to -5/3.

Did anyone else notice that?

I totally agree. I tried to plug in the general parabolic equation and got a different answer. Moreover i got it as 1/3 for x=3 .