Target question: What is the value of a?

Statement 1:a < b
Definitely NOT SUFFICIENT

Statement 2: (t − a)( t − b) = t² + t − 12
Factor: t² + t − 12 = (t + 4)(t - 3)
Rewrite in terms of (t - a) and (t - b) to get: t² + t − 12 = (t - -4)(t - 3)
There are two possible cases:
Case a: a = -4 and b = 3, in which case a = -4
Case b: a = 3 and b = -4, in which case a = 3
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 2 tells us that EITHER a = -4 and b = 3 OR a = 3 and b = -4
Statement 2 tells us that a < b, which means it MUST be the case that a = -4 and b = 3
So, a = -4
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C
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Ans C:
St. 2: Solving Quadratic, we get
(t-a)(t-b) = (t+4)(t-3)
St. 1: a<b
Therefore, a=-4 and b= +3

Statement 1: a<b - Clearly not sufficient

Statement 2: Ignore the LHS. \(t ^2 + t − 12\) The factors of this equation are +4 and -3. Still not sufficient.

Combining 1 and 2:
a=-4 and b=3 since a<b.

Hence C
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GMATPrepNow Could you help me with a small query?

For statement 2



Could we have done this as below and would it be fruitful?

(t − a)( t − b) = t² + t − 12
(t - a) = t² + t − 12
t - a = t² + t − 12
12 = t² + a

and ( t − b) = t² + t − 12
t - b = t² + t − 12
12 = t² + b Therefore since, t² + b = t² + a we get a = b ? (Is this right?)

Thank you
Dablu

I think you're confusing (t − a)( t − b) = t² + t − 12 with (t − a)( t − b) = 0
If (t − a)( t − b) = 0, then we know that either (t − a) = 0 or ( t − b) = 0
We're using the principle that says: If AB = 0, then either A = 0, or B = 0

The same strategy doesn't apply when the quadratic expression does not equal 0.
For example, if AB = 6, we can't conclude that A = 6 or B = 6

So, if (t − a)( t − b) = t² + t − 12, we can't then conclude that either (t - a) = t² + t − 12 or (t - b) = t² + t − 12

Cheers,
Brent
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Hi GMATPrepNow


Thank you for your reply. Definitely cleared my confusion!

Hey Brent

Since it says in statement B "for all values of t", I just plugged in 0 to get a*b=12 and even combining that with stmt. 1, I figured I can´t get a unique value. Where did I go wrong?

Thanks,

Philipp

Yes, why can we not just plug in 0 for t ?

philippi Moritz279

I will just try to answer what you're asking.

(2) (t−a)(t−b)=t^2+t−12 for all values of t.
If you say t=0
then equation becomes (0-a)(0-b)=0+0-12 which is (-a)(-b)=-12 which is ab=-12. We know that a and b are constants, so values must be 3/4 with one of the two being negative.
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TheNightKing

Yes but another solutions could be -2 and 6 for ab=-12.

Moritz279

No. \(t^2+t-12\) still has to hold true. +6-2 gives you 4. You need 1 which is only possible by +4-3.
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Asked: If a and b are constants, what is the value of a?

(1) \(a < b\)
Since a and b are unknown
NOT SUFFICIENT

(2) \((t − a)( t − b) = t ^2 + t − 12\), for all values of t.
t=-4 or t=3
a=-4 or 3
NOT SUFFICIENT

(1)+(2)
(1) \(a < b\)
(2) \((t − a)( t − b) = t ^2 + t − 12\), for all values of t.
a=-4 and b=3 since a<b
SUFFICIENT

IMO C

Posted from my mobile device
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Thanks, I get it now :D

Posted from my mobile device

Posted from my mobile device

PL correct me...statement 2 is true for all value of t. If we take t=0 we get ab=-12
From statement 1, a can be -2/-3/-4 and subsequently b can be 2/4/3
Hence we can not get exact value of a. So answer can be E

Posted from my mobile device

Even though I chose C, the way I attempted the question was very time consuming. In the second statement, I started equating (t−a)(t−b) = t^2+t−12. When I saw the answer behind the book, it solved t^2+t−12 first. My question is, is there any way of knowing which side of equation should be solved in order to solve the question within time limit?

For Statement 2 -

Why can't we take t = 0. In that case ab = -12 . Not sufficient.

Combining 1 and 2,

a,b and ab = -12
If a=-4, b=3 OR If a=-6, b=2. Not suff.

Can you please tell at which step I'm going wrong? Rather why can't we take t=0?