If a and b are constants, what is the value of a ?
(1) a < b
(2) (t − a)( t − b) = t² + t − 12, for all values of t.
Answer: C
Source: Official guide
If a and b are constants, what is the value of a ?
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Target question: What is the value of a?BTGModeratorVI wrote: ↑Thu Jun 18, 2020 5:54 amIf a and b are constants, what is the value of a ?
(1) a < b
(2) (t − a)( t − b) = t² + t − 12, for all values of t.
Answer: C
Source: Official guide
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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Solution:BTGModeratorVI wrote: ↑Thu Jun 18, 2020 5:54 amIf a and b are constants, what is the value of a ?
(1) a < b
(2) (t − a)( t − b) = t² + t − 12, for all values of t.
Answer: C
Source: Official guide
We need to determine the value of a.
Statement One Alone:
Knowing a < b does not allow us to determine the value of a. Statement one alone is not sufficient.
Statement Two Alone:
Expanding the left side of the equation, we have:
t^2  (a + b)t + ab = t^2 + t  12
Equating like terms from both sides, we have:
(a + b) = 1 and ab = 12
a + b = 1 and ab = 12
The two numbers that add up to 1 and multiply to 12 are 4 and 3. However, it could be the case that a = 4 and b = 3 or the case that a = 3 and b = 4. Statement two alone is not sufficient.
Statements One and Two Together:
With the two statements, we see that a must be 4 and b must be 3 since a < b. Both statements together are sufficient.
Answer: C
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