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M26-01

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New post 16 Sep 2014, 01:24
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A
B
C
D
E

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  85% (hard)

Question Stats:

53% (02:19) correct 47% (02:13) wrong based on 136 sessions

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New post 16 Sep 2014, 01:24
4
3
Official Solution:

What is the value of \(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}\)?

A. \(2\sqrt{5}\)
B. \(\sqrt{55}\)
C. \(2\sqrt{15}\)
D. 50
E. 60


Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:

Must know for the GMAT: \((x+y)^2=x^2+2xy+y^2\) (while \((x-y)^2=x^2-2xy+y^2\)).

So we get:

\((\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2 =\) \(=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=\)

\(=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(25-10\sqrt{6})\).

Note that sum of the first and the third terms simplifies to \((25+10\sqrt{6})+(25-10\sqrt{6})=50\), so we have

\(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})\) therefore:

\(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=\) \(50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}\).

Also must know for the GMAT: \((x+y)(x-y)=x^2-y^2\), thus

\(50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)} =\) \(= 50+2\sqrt{625-600}=50+2\sqrt{25}=60\).

Recall that we should un-square this value to get the right answer: \(\sqrt{60}=2\sqrt{15}\).


Answer: C
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New post 16 Mar 2016, 06:34
I did not get the part why we need to unsquare the value 60
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New post 16 Mar 2016, 06:38
pkappaga wrote:
I did not get the part why we need to unsquare the value 60


Its done so as to express the final answer in the simplest form, something that GMAT will always ask you to do.
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New post 16 Mar 2016, 07:49
I understand the simplification part but Isn`t the final value 60? I am not sure why you are doing the square root of 60
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New post 16 Mar 2016, 07:53
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pkappaga wrote:
I understand the simplification part but Isn`t the final value 60? I am not sure why you are doing the square root of 60


It is done becaue Bunuel starts with a squared value of the expression that we need to evalute (look below). This is the reason why you then have to take the square root to get the value of the question asked.


You need to find the value of \((\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})\) and NOT \((\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2\)

Bunuel wrote:

Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:

Must know for the GMAT: \((x+y)^2=x^2+2xy+y^2\) (while \((x-y)^2=x^2-2xy+y^2\)).

So we get:

\((\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2 =\) \(=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=\)



Hope this helps.
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New post 13 Sep 2016, 02:36
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Alternative solution: 10*\(\sqrt{6}\) = \(\sqrt{600}\), which is approximately 24.5 as 600 lies between 24*24=576 and 25*25=625
So, \(\sqrt{25+24.5}\) + \(\sqrt{25-24.5}\) = \(\sqrt{49,5}\) + \(\sqrt{0.5}\) ≈ slightly more than 7+ around 0.7*≈ 8

* 0.7*0.7 =0.49 ≈ 0,5

So, we can eliminate A, D and E

B and C
2*\(\sqrt{15}\)≈ 2*4 ≈ 8 - contender

\(\sqrt{55}\) - loser.
First, 55 is closer to 49 than to 64, so \(\sqrt{55}\) is closer to 7.
Second, using guessing technique, we see that 55 has no pair with any other number in answer choices, whilst there is a pair of 60s - 60 and \(\sqrt{60}\).

Hope this helps!
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New post 17 Jan 2017, 02:05
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Alternative solution (much faster in my opinion)

We factor 5 => \(\sqrt{25+10\sqrt{6}\)+\(\sqrt{25-10\sqrt{6}\)=\(\sqrt{5(5+2\sqrt{6})\)+\(\sqrt{5(5-2\sqrt{6})\)

Now we write \(\sqrt{5(5+2\sqrt{6})\) as \(\sqrt{5(3+2+2\sqrt{6})\)=\(\sqrt{5(\sqrt{3}+\sqrt{2})^2\)
and \(\sqrt{5(5-2\sqrt{6})\) as \(\sqrt{5(3+2-2\sqrt{6})\)=\(\sqrt{5(\sqrt{3}-\sqrt{2})^2\)

=> \(\sqrt{5(\sqrt{3}+\sqrt{2})^2\)+\(\sqrt{5(\sqrt{3}-\sqrt{2})^2\) = \(\sqrt{5}(\sqrt{3}+\sqrt{2})\)+\(\sqrt{5}(\sqrt{3}-\sqrt{2})\)=\(\sqrt{5}*\sqrt{3}\) +\(\sqrt{5}*\sqrt{2}\) +\(\sqrt{5}*\sqrt{3}\) -\(\sqrt{5}*\sqrt{2}\) =\(\sqrt{15}+\sqrt{15}\) =\(2\sqrt{15}\) => Answer C
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New post 12 Jun 2017, 09:50
3
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Bunuel wrote:
Official Solution:

What is the value of \(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}\)?

A. \(2\sqrt{5}\)
B. \(\sqrt{55}\)
C. \(2\sqrt{15}\)
D. 50
E. 60


Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:

Must know for the GMAT: \((x+y)^2=x^2+2xy+y^2\) (while \((x-y)^2=x^2-2xy+y^2\)).

So we get:

\((\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2 =\) \(=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=\)

\(=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(25-10\sqrt{6})\).

Note that sum of the first and the third terms simplifies to \((25+10\sqrt{6})+(25-10\sqrt{6})=50\), so we have

\(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})\) therefore:

\(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=\) \(50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}\).

Also must know for the GMAT: \((x+y)(x-y)=x^2-y^2\), thus

\(50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)} =\) \(= 50+2\sqrt{625-600}=50+2\sqrt{25}=60\).

Recall that we should un-square this value to get the right answer: \(\sqrt{60}=2\sqrt{15}\).


Answer: C


Can it be solved the following way?
\(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}=\sqrt{(\sqrt{15}+\sqrt{10})^2}+\sqrt{(\sqrt{15}-\sqrt{10})^2}\)

then it would turn into \(|\sqrt{15}+\sqrt{10}| +|\sqrt{15}-\sqrt{10}|\)
then \(\sqrt{15}+\sqrt{10}+\sqrt{15}-\sqrt{10}=2\sqrt{15}\)
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New post 12 Jun 2017, 13:37
MikeMighty wrote:
Bunuel wrote:
Official Solution:

What is the value of \(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}\)?

A. \(2\sqrt{5}\)
B. \(\sqrt{55}\)
C. \(2\sqrt{15}\)
D. 50
E. 60


Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:

Must know for the GMAT: \((x+y)^2=x^2+2xy+y^2\) (while \((x-y)^2=x^2-2xy+y^2\)).

So we get:

\((\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2 =\) \(=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=\)

\(=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(25-10\sqrt{6})\).

Note that sum of the first and the third terms simplifies to \((25+10\sqrt{6})+(25-10\sqrt{6})=50\), so we have

\(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})\) therefore:

\(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=\) \(50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}\).

Also must know for the GMAT: \((x+y)(x-y)=x^2-y^2\), thus

\(50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)} =\) \(= 50+2\sqrt{625-600}=50+2\sqrt{25}=60\).

Recall that we should un-square this value to get the right answer: \(\sqrt{60}=2\sqrt{15}\).


Answer: C


Can it be solved the following way?
\(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}=\sqrt{(\sqrt{15}+\sqrt{10})^2}+\sqrt{(\sqrt{15}-\sqrt{10})^2}\)

then it would turn into \(|\sqrt{15}+\sqrt{10}| +|\sqrt{15}-\sqrt{10}|\)
then \(\sqrt{15}+\sqrt{10}+\sqrt{15}-\sqrt{10}=2\sqrt{15}\)

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New post 13 Jul 2017, 18:57
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I find the official solution to be very long and it would take me probably 4-5 minutes to finish the answer. By then I would probably pick 60 since i would forget to take the square root.

My approach was using the approximate numbers for each. First I recognized that sqroot of 6 is 2.5 so the second part was equal to 0. The first part was 25+25=50. Square root of 50 is just over 7. Then I looked that answers and found the number closest to the one I got. It will take you probably less than 1:30 to solve using this method.
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New post 11 Oct 2017, 19:50
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Let \(25 + 10\sqrt{6}\ = A\) &
Let \(25 - 10\sqrt{6}\ = B =>\)

\(\sqrt{A}\ + \sqrt{B}\ = x\)
\(A + B + 2\sqrt{AB}\ = x^2 =>\)
\(25 + 10\sqrt{6}\ + 25 - 10\sqrt{6}\ + 2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6})}\ = x^2\)
\(50 + 2\sqrt{25^2 - (10\sqrt{6})^2}\ = x^2 =>\)
\(x^2 = 60\)
\(x = 2\sqrt{15}\)
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New post 19 Jul 2018, 07:19
AlexBunea wrote:
Alternative solution (much faster in my opinion)

We factor 5 => \(\sqrt{25+10\sqrt{6}\)+\(\sqrt{25-10\sqrt{6}\)=\(\sqrt{5(5+2\sqrt{6})\)+\(\sqrt{5(5-2\sqrt{6})\)

Now we write \(\sqrt{5(5+2\sqrt{6})\) as \(\sqrt{5(3+2+2\sqrt{6})\)=\(\sqrt{5(\sqrt{3}+\sqrt{2})^2\)
and \(\sqrt{5(5-2\sqrt{6})\) as \(\sqrt{5(3+2-2\sqrt{6})\)=\(\sqrt{5(\sqrt{3}-\sqrt{2})^2\)

=> \(\sqrt{5(\sqrt{3}+\sqrt{2})^2\)+\(\sqrt{5(\sqrt{3}-\sqrt{2})^2\) = \(\sqrt{5}(\sqrt{3}+\sqrt{2})\)+\(\sqrt{5}(\sqrt{3}-\sqrt{2})\)=\(\sqrt{5}*\sqrt{3}\) +\(\sqrt{5}*\sqrt{2}\) +\(\sqrt{5}*\sqrt{3}\) -\(\sqrt{5}*\sqrt{2}\) =\(\sqrt{15}+\sqrt{15}\) =\(2\sqrt{15}\) => Answer C


Can you please explain this: \(\sqrt{5(3+2+2\sqrt{6})\)=\(\sqrt{5(\sqrt{3}+\sqrt{2})^2\)
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New post 19 Jul 2018, 09:40
illthinker wrote:
Let \(25 + 10\sqrt{6}\ = A\) &
Let \(25 - 10\sqrt{6}\ = B =>\)

\(\sqrt{A}\ + \sqrt{B}\ = x\)
\(A + B + 2\sqrt{AB}\ = x^2 =>\)
\(25 + 10\sqrt{6}\ + 25 - 10\sqrt{6}\ + 2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6})}\ = x^2\)
\(50 + 2\sqrt{25^2 - (10\sqrt{6})^2}\ = x^2 =>\)
\(x^2 = 60\)
\(x = 2\sqrt{15}\)



Same approach as Bunuel, but well laid out. Thank you!
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New post 23 Feb 2019, 13:22
frrcattack wrote:
AlexBunea wrote:
Alternative solution (much faster in my opinion)

We factor 5 => \(\sqrt{25+10\sqrt{6}\)+\(\sqrt{25-10\sqrt{6}\)=\(\sqrt{5(5+2\sqrt{6})\)+\(\sqrt{5(5-2\sqrt{6})\)

Now we write \(\sqrt{5(5+2\sqrt{6})\) as \(\sqrt{5(3+2+2\sqrt{6})\)=\(\sqrt{5(\sqrt{3}+\sqrt{2})^2\)
and \(\sqrt{5(5-2\sqrt{6})\) as \(\sqrt{5(3+2-2\sqrt{6})\)=\(\sqrt{5(\sqrt{3}-\sqrt{2})^2\)

=> \(\sqrt{5(\sqrt{3}+\sqrt{2})^2\)+\(\sqrt{5(\sqrt{3}-\sqrt{2})^2\) = \(\sqrt{5}(\sqrt{3}+\sqrt{2})\)+\(\sqrt{5}(\sqrt{3}-\sqrt{2})\)=\(\sqrt{5}*\sqrt{3}\) +\(\sqrt{5}*\sqrt{2}\) +\(\sqrt{5}*\sqrt{3}\) -\(\sqrt{5}*\sqrt{2}\) =\(\sqrt{15}+\sqrt{15}\) =\(2\sqrt{15}\) => Answer C


Can you please explain this: \(\sqrt{5(3+2+2\sqrt{6})\)=\(\sqrt{5(\sqrt{3}+\sqrt{2})^2\)


I posted a useful heuristic and an explanation for this in M11-18 https://gmatclub.com/forum/m11-183900.html#p2215071

But to summarize, you need to do what it takes algebraically to get a term that looks precisely like \(2\sqrt{some number}\)
i.e. the number "2" multiplied by a square root of some other number

and then you'll be able to recognize a pattern where one number = a+b and the other = a*b. (similar to when you factor a quadratic)

In this case; 5 = 3 + 2
and
6 = 3*2

Look at my explanation in M11-18 for more details
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New post 20 Jul 2019, 11:07
Imo, it's much easier/faster to estimate, especially since D, E are so far apart from A,B,C:
I knew that √5 ≈ 2.25 so let's call √6 ≈ 2.4 (actual is 2.449...)

\(\sqrt{25+10*2.4}+\sqrt{25-10*2.4}\)
\(\sqrt{25+24}+\sqrt{25-24}\)
\(\sqrt{49}+\sqrt{1}\)
\(7 + 1\)
So our answer should be close to 8, D,E are out immediately
C can be rewritten as \(\sqrt{60}\) which is closer than B \(\sqrt{55}\) , so it's C.
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