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QUANTITATIVE SECTION 3 -Question 18 of 20|ID: Q01-77
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27 Sep 2017, 05:54
27 Sep 2017, 05:54
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Question Stats:
4% (00:06) correct
95% (01:38) wrong based on 21 sessions
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If a=3, b=5 and c=2, then
A) \(b!>(a+c)!\)
B) \(b!>(ac)!\)
C) \(\frac{b!}{(ac)!}>1\)
D) \(\frac{(bc)!}{a!}<(b + c)!\)
E) \(a! + b! + c!>(abc)!\)
Starting from the fact that option B and C are the same, but it could be a trick, the explanation and the answer to this question are completely messy.
It says "Substitute the values of a, b and c in the expression only \(\frac{b!}{(ac)!}>1\) is true. The correct answer is A".
The fact is that neither \(\frac{b!}{(ac)!}>1\) nor A are right. Without any sign of equality, I would say that no answer is correct. Am I right?
A) \(b!>(a+c)!\)
B) \(b!>(ac)!\)
C) \(\frac{b!}{(ac)!}>1\)
D) \(\frac{(bc)!}{a!}<(b + c)!\)
E) \(a! + b! + c!>(abc)!\)
Starting from the fact that option B and C are the same, but it could be a trick, the explanation and the answer to this question are completely messy.
It says "Substitute the values of a, b and c in the expression only \(\frac{b!}{(ac)!}>1\) is true. The correct answer is A".
The fact is that neither \(\frac{b!}{(ac)!}>1\) nor A are right. Without any sign of equality, I would say that no answer is correct. Am I right?
Re: QUANTITATIVE SECTION 3 -Question 18 of 20|ID: Q01-77
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27 Sep 2017, 07:16
27 Sep 2017, 07:16
Expert Reply
This is a question from our test ?? if yes it should be created by the former mod........and we are having some problem to figure them out and fade out.
Back to the question.
First of all b and c are not the same.
B is \(\frac{b!}{(ac)!}> 0\) if you bring the RHS to the LHS
C is \(\frac{b!}{(ac)!}>\) 1
Now
A) \(\frac{5*4*3*2*1}{5*4*3*2*1} > 0\) that is true 1 > 0
B) \(\frac{5*4*3*2*1}{6*5*4*3*2*1} > 0\) tha is impossible for the test, due to the fact it deals with real numbers and \(\frac{0}{6}\) is infinite. which is not possible in math
All the rest of the answer choices are the same OR impossible OR \(a positive number < 0\) which is impossible as well. Check on your own.
The question is good and sophisticated but we are working to fade out due their complexity and out of scope for the GRE purpose.
Ask anything you need. Even via private message.
regards
Back to the question.
First of all b and c are not the same.
B is \(\frac{b!}{(ac)!}> 0\) if you bring the RHS to the LHS
C is \(\frac{b!}{(ac)!}>\) 1
Now
A) \(\frac{5*4*3*2*1}{5*4*3*2*1} > 0\) that is true 1 > 0
B) \(\frac{5*4*3*2*1}{6*5*4*3*2*1} > 0\) tha is impossible for the test, due to the fact it deals with real numbers and \(\frac{0}{6}\) is infinite. which is not possible in math
All the rest of the answer choices are the same OR impossible OR \(a positive number < 0\) which is impossible as well. Check on your own.
The question is good and sophisticated but we are working to fade out due their complexity and out of scope for the GRE purpose.
Ask anything you need. Even via private message.
regards
Re: QUANTITATIVE SECTION 3 -Question 18 of 20|ID: Q01-77
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28 Sep 2017, 04:42
28 Sep 2017, 04:42
Carcass wrote:
First of all b and c are not the same.
B is \(\frac{b!}{(ac)!}> 0\) if you bring the RHS to the LHS
C is \(\frac{b!}{(ac)!}>\) 1
I am sorry to correct you but B is indeed the same of C because we are dividing both sides of the equation by (ac)! thus we are left with a 1 on the RHS.
Anyway, I'll signal you by pm all the question from the test that seem to be wrong/mispelled
