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Root(80)+root(125)

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Bunuel
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Root(80)+root(125) [#permalink] New post   01 Oct 2012, 05:02
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\(\sqrt{80}+\sqrt{125} =\)

(A) \(9\sqrt{5}\)
(B) \(20\sqrt{5}\)
(C) \(41\sqrt{5}\)
(D) \(\sqrt{205}\)
(E) 100
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Bunuel
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Re: Root(80)+root(125) [#permalink] New post   01 Oct 2012, 05:03
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SOLUTION

\(\sqrt{80}+\sqrt{125} =\)

(A) \(9\sqrt{5}\)
(B) \(20\sqrt{5}\)
(C) \(41\sqrt{5}\)
(D) \(\sqrt{205}\)
(E) 100

\(\sqrt{80}+\sqrt{125} =\sqrt{16*5}+\sqrt{25*5}=4\sqrt{5}+5\sqrt{5}=9\sqrt{5}\).

Answer: A.
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Re: Root(80)+root(125) [#permalink] New post   01 Oct 2012, 07:43
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Whe we deal with this problems may be the best thing to do is to break the number into factors

\(\sqrt{80}\)\(=\)\(\sqrt{2*5*2*2*2}\) ---> \(\sqrt{4*4*5}\) ---> \(4\)\(\sqrt{5}\)

\(\sqrt{125}\) \(=\) \(\sqrt{25 *5}\) -----> \(5\) \(\sqrt{5}\)

\(4\)\(\sqrt{5}\) \(+ 5\) \(\sqrt{5}\) \(=\) \(9\)\(\sqrt{5}\)

A should be the answer
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Re: Root(80)+root(125) [#permalink] New post   04 Oct 2012, 00:50
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\(\sqrt{80} = \sqrt{16*5} = 4\sqrt{5}\) ........(1)

\(\sqrt{125} = \sqrt{25*5} = 5\sqrt{5}\) .......(2)

Thus \(4\sqrt{5} + 5\sqrt{5} = 9\sqrt{5}\)

Thus answer is A
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Re: Root(80)+root(125) [#permalink] New post   18 Dec 2013, 14:04
Bunuel wrote:
\(\sqrt{80}+\sqrt{125} =\)

(A) \(9\sqrt{5}\)
(B) \(20\sqrt{5}\)
(C) \(41\sqrt{5}\)
(D) \(\sqrt{205}\)
(E) 100

Practice Questions
Question: 52
Page: 159
Difficulty: 600



The only thing I know is that the answer is just under 24.. And it is definitely not D since D falls between 14 and 15.. And it's clearly not E. And out of the 3 left, B and C are waay too high above 24, and I figured sqrrt of 5 is around "2.something" so A made most sense.

Clearly this approach is very shaky but this time it worked.
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Re: Root(80)+root(125) [#permalink] New post   20 Aug 2014, 23:32
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\(\sqrt{80}\) is approx. 9 and \(\sqrt{125}\) is approx. 11 so the answer should be close to 20. Choice A is the closest.
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Re: Root(80)+root(125) [#permalink] New post   05 Mar 2015, 07:51
I see this question is from a subcategory called "Practice Questions", where can I find that sub?
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Re: Root(80)+root(125) [#permalink] New post   05 Mar 2015, 07:58
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erikvm wrote:
I see this question is from a subcategory called "Practice Questions", where can I find that sub?


Those are practice questions from OG13. You can find all questions HERE.
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Re: Root(80)+root(125) [#permalink] New post   04 May 2016, 10:03
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Bunuel wrote:
\(\sqrt{80}+\sqrt{125} =\)

(A) \(9\sqrt{5}\)
(B) \(20\sqrt{5}\)
(C) \(41\sqrt{5}\)
(D) \(\sqrt{205}\)
(E) 100



Solution:

To solve the problem we must simplify the radicals. Radicals should be simplified whenever possible. Since the square root of a perfect square produces integers, it will often be helpful to locate and simplify perfect squares within a radical expression. Thus, we first locate the perfect squares that divide evenly into 80 and 125, making the simplification of each radical straightforward.

√80 = √16 x √5 = 4√5

√125 = √25 x √5 = 5√5

Now we can add these two results together. Remember to keep the value inside the radical constant and add together the values on the outside.

4√5 + 5√5 = 9√5

The answer is A.
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Re: Root(80)+root(125) [#permalink] New post   04 May 2016, 10:06
A
80=4*4*5
125=5*5*5
Root(80)=4Root(5)
Similarly 125
hence 9Root(5)
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Re: Root(80)+root(125) [#permalink] New post   04 May 2016, 10:19
Bunuel wrote:
\(\sqrt{80}+\sqrt{125} =\)

(A) \(9\sqrt{5}\)
(B) \(20\sqrt{5}\)
(C) \(41\sqrt{5}\)
(D) \(\sqrt{205}\)
(E) 100

Practice Questions
Question: 52
Page: 159
Difficulty: 600


\(\sqrt{80}+\sqrt{125} =\) \(4\sqrt{5}+5\sqrt{5}\) = \(9\sqrt{5}\)

Answer will be (A)
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Re: Root(80)+root(125) [#permalink] New post   22 Mar 2017, 23:29
Prime factorization can really help you and so can converting the roots to fraction exponents.

\(\sqrt { 80 } +\sqrt { 125 } =\\ { ({ 2 }^{ 4 } }\times { 5 }^{ 1 })^{ \frac { 1 }{ 2 } }+{ ({ 5 }^{ 2 }\times { 5 }^{ 1 }) }^{ \frac { 1 }{ 2 } }=\\ ({ 2 }^{ \frac { 4 }{ 2 } }\times { 5 }^{ \frac { 1 }{ 2 } })+({ 5 }^{ \frac { 2 }{ 2 } }\times { 5 }^{ \frac { 1 }{ 2 } })\quad =\\ { 2 }^{ 2 }\sqrt { 5 } +5\sqrt { 5 } =\\ 4\sqrt { 5 } +5\sqrt { 5 } =\\ \Rightarrow 9\sqrt { 5 } \\\)
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Re: Root(80)+root(125) [#permalink] New post   14 Sep 2017, 22:49
√80 ~~ 9
√ 125 ~~ 11
9+11 = 20
√5 ~~2.1
9*2.1~~20
all other answer not near to this
Hence answer must be A
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Re: Root(80)+root(125) [#permalink] New post   03 Jan 2019, 06:29
\(\sqrt{80}\)\(=\)\(\sqrt{16*5}\) ---> \(\sqrt{4*4*5}\)

\(\sqrt{125}\) \(=\) \(\sqrt{25 *5}\) -----> \(5\) \(\sqrt{5}\)

=\(9\)\(\sqrt{5}\)
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Re: Root(80)+root(125) [#permalink] New post   29 Apr 2020, 07:44
Try approximation

√80 ≈ √81 i.e. < 9
√125 ≈ √121 i.e. 11 < value < 12

Therefore sum would be 11.something + (8.9+ something but less than 9)
i.e. ≈ 19

and we know √5 is ≈ 2 as √4 = 2

And 9 * 2 = 18

So (A) is the nearest answer.
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Re: √80+√125= [#permalink] New post   12 Jun 2020, 21:23
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TheUltimateWinner wrote:
\(\sqrt{80} + \sqrt{125} =\)

A. \(9\sqrt{5}\)
B. \(20\sqrt{5}\)
C. \(41\sqrt{5}\)
D. \(\sqrt{205}\)
E. 100


\(\sqrt{80} + \sqrt{125} ≈ 9+11 ≈ 20\)

A. \(9\sqrt{5}\)≈9*2.4≈21 CORRECT
B. \(20\sqrt{5}\)IMPOSSIBLE
C. \(41\sqrt{5}\)IMPOSSIBLE
D. \(\sqrt{205}\)≈14 IMPOSSIBLE
E. 100 IMPOSSIBLE
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Re: Root(80)+root(125) [#permalink] New post   13 Jun 2020, 06:05
√80+√125
=4√5+5√5
=9√5

Answer: A
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Re: Root(80)+root(125) [#permalink] New post   21 Oct 2023, 03:22
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Re: Root(80)+root(125) [#permalink]
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