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17 Apr 2019, 00:34
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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:
15%
(low)
Question Stats:
84% (01:44) correct
16%
(02:06)
wrong
based on 168
sessions
History
Date
Time
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Not Attempted Yet
Let the function h be defined by \(h(x) = \sqrt{x} + 2\). If \(3h(v) = 18\), then which one of the following is the value of \(h(\frac{v}{4})\) ?
(A) –4
(B) –1
(C) 0
(D) 2
(E) 4
(A) –4
(B) –1
(C) 0
(D) 2
(E) 4
17 Apr 2019, 04:35
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Bunuel
\(h(x) = \sqrt{x} + 2\)
\(h(v) = \sqrt{x} + 2 = 18/3 = 6\)
i.e. \(√x = 4\)
i.e. \(x = 16\)
Now, \(h(\frac{v}{4}) = h(4) = \sqrt{4} + 2 = 4\)
Answer: Option E
17 Apr 2019, 15:03
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Hello GMATinsight!
Why should we have to equal f(x) with f(v)?
I was doing it as follows:
\(h(x) = \sqrt{x} + 2\)
\(h(v) = 18/3 = 6\)
\(h(\frac{6}{4}) = \frac{3}{2}\)
So
\(h(x) = \sqrt{\frac{3}{2}} + 2\)
And then the result would be:
\(\sqrt{6} + 4\)
I am confused =/
Why should we have to equal f(x) with f(v)?
I was doing it as follows:
\(h(x) = \sqrt{x} + 2\)
\(h(v) = 18/3 = 6\)
\(h(\frac{6}{4}) = \frac{3}{2}\)
So
\(h(x) = \sqrt{\frac{3}{2}} + 2\)
And then the result would be:
\(\sqrt{6} + 4\)
I am confused =/
