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29 Sep 2021, 01:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
49% (02:01) correct
51%
(01:55)
wrong
based on 267
sessions
History
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Not Attempted Yet
If x and y are positive integers, what is the value of xy?
(1) \(x! = 6y!\)
(2) \(\frac{\frac{x!}{y}}{(x-2)!} = \frac{30}{y}\)
(1) \(x! = 6y!\)
(2) \(\frac{\frac{x!}{y}}{(x-2)!} = \frac{30}{y}\)
29 Sep 2021, 08:06
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Bunuel
Given: x and y are positive integers
Target question: What is the value of xy?
Statement 1: \(x! = 6y!\)
Divide both sides of the equation by \(y!\) to get: \(\frac{x!}{y!} = 6\)
At this point we can see that there are two possible solutions
Solution i: x = 6 and y = 5, in which case xy = (6)(5) = 30
Solution ii: x = 3 and y = 1, in which case xy = (3)(1) = 3
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: \(\frac{\frac{x!}{y}}{(x-2)!} = \frac{30}{y}\)
Multiply both sides of the equation by \(y\) to get: \(\frac{x!}{(x-2)!}=30\)
Some nice cancellations happens here. To see how, notice that we can express the two factorials as follows: \(\frac{(x)(x-1)(x-2)(x-3).....(3)(2)(1)}{(x-2)(x-3).....(3)(2)(1)}=30\)
Simplify: \((x)(x-1)=30\)
Expand left side: \(x^2 - x = 30\)
Set the quadratic equation equal to \(0\) to get: \(x^2 - x - 30 = 0\)
Factor: \((x - 6)(x + 5) = 0\), which means either \(x = 6\) or \(x = -5\)
Since we're told x is positive, it must be the case that \(x = 6\)
HOWEVER, since we don't know the value of \(y\), we can’t answer the target question with certainty
Statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 tells us that EITHER \(x = 6\) and \(y = 5\) OR \(x = 3\) and \(y = 1\)
Statement 2 tells us that \(x = 6\)
So, it must be the case that x = 6 and y = 5, which means xy = (6)(5) = 30
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent
30 Sep 2021, 09:56
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Bunuel
Statement 1:
We can have 3! = 6*1! or 6! = 6*5!. Hence x*y = 3 or 30. Insufficient.
Statement 2:
\(\frac{x!}{(x-2)!} = x*(x - 1)\)
Therefore we have \(\frac{x*(x - 1)}{y} = \frac{30}{y}\),
which is \(x*(x - 1) = 30\). x must be positive, therefore x must be equal to 6. However, we do not know y. Insufficient.
Combined:
We must have x = 6 and y = 5, hence xy = 30. Sufficient.
Ans: C
