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98^(1/2) + 72^(1/2) =
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05 Feb 2021, 09:13
05 Feb 2021, 09:13
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91% (01:08) correct
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\(\sqrt{98}+ \sqrt{72} =\)
A. \(\sqrt{170}\)
B. \(\sqrt{232}\)
C. \(\sqrt{286}\)
D. \(\sqrt{338}\)
E. \(\sqrt{420}\)
A. \(\sqrt{170}\)
B. \(\sqrt{232}\)
C. \(\sqrt{286}\)
D. \(\sqrt{338}\)
E. \(\sqrt{420}\)
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Re: 98^(1/2) + 72^(1/2) =
[#permalink]
05 Feb 2021, 09:35
05 Feb 2021, 09:35
Carcass wrote:
\(\sqrt{98}+ \sqrt{72} =\)
A. \(\sqrt{170}\)
B. \(\sqrt{232}\)
C. \(\sqrt{286}\)
D. \(\sqrt{338}\)
E. \(\sqrt{420}\)
A. \(\sqrt{170}\)
B. \(\sqrt{232}\)
C. \(\sqrt{286}\)
D. \(\sqrt{338}\)
E. \(\sqrt{420}\)
The fastest approach here is to call up the onscreen GRE calculator to evaluate \(\sqrt{98}+ \sqrt{72} =\) and then calculate each answer choice until you get a match.
That said, here's another approach:
\(\sqrt{98}+ \sqrt{72} = \sqrt{49 \times 2}+ \sqrt{36 \times 2} \)
\(=7\sqrt{2}+ 6\sqrt{2} \)
\(=13\sqrt{2}\)
\(=\sqrt{169} \times \sqrt{2}\)
\(=\sqrt{169 \times 2}\)
\(=\sqrt{338}\)
Answer: D
gmatclubot
Re: 98^(1/2) + 72^(1/2) = [#permalink]
05 Feb 2021, 09:35
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