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02 Mar 2022, 07:30
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E
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:
42% (02:04) correct
58%
(01:30)
wrong
based on 182
sessions
History
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What is the value of \(\frac{\sqrt{b}}{a}\) ?
(1) b = a + 2
(2) \(b^2 = a^4\)
Are You Up For the Challenge: 700 Level Questions
(1) b = a + 2
(2) \(b^2 = a^4\)
Are You Up For the Challenge: 700 Level Questions
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
06 Mar 2022, 00:22
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statement 1 is obviously insufficient alone.
From statement 2, we understand that b=+a^2 or b=-a^2
Therefore, statement 2 is also insufficient alone.
On combining the 2 statements, we get that b=a+2 and b=+/-a^2
Case 1 : b=-a^2
b=a+2=-a^2
a^2+a+2=0
This equation has unreal roots
Case 2 : b=+a^2
b=a+2=+a^2
a^2-a-2=0
(a-2)(a+1)=0
a=2 or -1 and b=4 or 1
Therefore, √b/a = +/-1 in both cases.
Therefore, even together the two statements are insufficient. Answer is E
From statement 2, we understand that b=+a^2 or b=-a^2
Therefore, statement 2 is also insufficient alone.
On combining the 2 statements, we get that b=a+2 and b=+/-a^2
Case 1 : b=-a^2
b=a+2=-a^2
a^2+a+2=0
This equation has unreal roots
Case 2 : b=+a^2
b=a+2=+a^2
a^2-a-2=0
(a-2)(a+1)=0
a=2 or -1 and b=4 or 1
Therefore, √b/a = +/-1 in both cases.
Therefore, even together the two statements are insufficient. Answer is E
08 Mar 2022, 15:20
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Bunuel
Statement 1 Alone:
Insufficient.
Statement 2 Alone:
We can move the \(a^4\) over to get \(\frac{b^2}{a^4} = 1\). Then 4th root both sides to get \(|\frac{\sqrt{b}}{a}| = 1\). Note that we must add an absolute value on both sides whenever we do an even root.
Therefore \(\frac{\sqrt{b}}{a}\) could be 1 or -1, so statement 2 alone is insufficient.
Both Statements Combined:
Combined we can throw in \(b = a + 2\) and solve a quadratic equation. We will get \(a^2 - a - 2 = 0\) with \(a = 2\) or \(a = -1\).
Then b = 4 or b = 1 could be true, and if we plug in the pairs respectively to check the value of \(\frac{\sqrt{b}}{a}\), it could be 1 or -1 as discussed earlier, so combined it is still insufficient.
Answer: E
