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M11-18

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M11-18  [#permalink]

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New post 15 Sep 2014, 23:44
2
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

55% (00:55) correct 45% (01:07) wrong based on 139 sessions

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Re M11-18  [#permalink]

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New post 15 Sep 2014, 23:44
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Official Solution:

\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

A. 1.0
B. 1.7
C. 2.0
D. 2.4
E. 3.0

The expression \(7 + \sqrt{48}\) can be rewritten as \(4 + 4\sqrt{3} + 3\) which is the same as \((2 + \sqrt{3})^2\). Therefore,\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = 2 + \sqrt{3} - \sqrt{3} = 2\).

Answer: C
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Re: M11-18  [#permalink]

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New post 22 Nov 2014, 18:10
Bunuel wrote:
Official Solution:

The expression \(7 + \sqrt{48}\) can be rewritten as \(4 + 4\sqrt{3} + 3\) which is the same as \((2 + \sqrt{3})^2\). Therefore,\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = 2 + \sqrt{3} - \sqrt{3} = 2\).

Answer: C


Is there anything specific here that we should pick up on, that would tell us that we should break up the problem this way?

the √48 can be broken down. I get that, and it seems intuitive enough. But is there a way to realize that we need to break the 7 into 4+3, and then use foil?

I guess Im just trying to figure out what the "real time" thought process should be when one sees this.

Thanks
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Re: M11-18  [#permalink]

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New post 23 Nov 2014, 06:07
JackSparr0w wrote:
Bunuel wrote:
Official Solution:

The expression \(7 + \sqrt{48}\) can be rewritten as \(4 + 4\sqrt{3} + 3\) which is the same as \((2 + \sqrt{3})^2\). Therefore,\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = 2 + \sqrt{3} - \sqrt{3} = 2\).

Answer: C


Is there anything specific here that we should pick up on, that would tell us that we should break up the problem this way?

the √48 can be broken down. I get that, and it seems intuitive enough. But is there a way to realize that we need to break the 7 into 4+3, and then use foil?

I guess Im just trying to figure out what the "real time" thought process should be when one sees this.

Thanks


I guess such kind of tricks should come with practice.
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M11-18  [#permalink]

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New post 05 Apr 2015, 12:23
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Bunuel wrote:
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

A. 1.0
B. 1.7
C. 2.0
D. 2.4
E. 3.0


This question can a be a little harder if answers will not be so diverse.
Because now we can calculate roughly
\(\sqrt{48}= 6.9\)

\(\sqrt{7 + 6.9}= 3.8\)

\(3.8-\sqrt{3}=2.1\)

So nearest answer is C.
IMHO if there will be answer with 2.1 and 1.9 than rough calculation will be imposible and task will be more useful.
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M11-18  [#permalink]

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New post 03 May 2015, 01:40
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Another smart way of doing this is like this:
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = X\)
\(\sqrt{7 + \sqrt{48}} = X + \sqrt{3}\)
\(7 + \sqrt{48} = 7 + 4*\sqrt{3} = (X^2+3) + 2*X*\sqrt{3}\)
Its not hard to figure out that X = 2
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Re: M11-18  [#permalink]

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New post 27 Apr 2016, 09:21
Harley1980 wrote:
\(\sqrt{7 + 6.9}= 3.8\)


I have approximate for root 3 memorized but should I know roughly square root of higher numbers like 14 as well?
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Re: M11-18  [#permalink]

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New post 27 Apr 2016, 09:56
redfield wrote:
Harley1980 wrote:
\(\sqrt{7 + 6.9}= 3.8\)


I have approximate for root 3 memorized but should I know roughly square root of higher numbers like 14 as well?


Hi,
No you don't have to, whenever you encounter something of this sort, there would be better ways to do it, as GMAT is not likely to test you \(\sqrt{14}\) etc..

But say you get this kind of problems, you can always approximate..
\(\sqrt{14}=\sqrt{3}*\sqrt{5}\) = 1.7*2.2 = 3.8
so you can work with lower numbers too to get the answer
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Re: M11-18  [#permalink]

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New post 27 Apr 2016, 10:38
1
For square roots:

1. Find the nearest perfect square

\(\sqrt{14}\) would be \(\sqrt{16}=4\).


2. Divide the original number by the answer

\(14/4=3.5\)


3. Find average of square root of perfect square and your answer in part 2 above

\((3.5+4)/2=3.75\)

This approximation is accurate enough for any GMAT level questions I have come across.

For the example above. Simplify to

\(\sqrt{14} - \sqrt{3} = 3.75-1.7\)

Round down as the original question should have been slightly lower than \(\sqrt{14}\).

Answer = 2.0 (c)
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Re M11-18  [#permalink]

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New post 08 Jun 2016, 21:09
I think this is a high-quality question and I agree with explanation.
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Re: M11-18  [#permalink]

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New post 02 Jul 2016, 07:08
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Great question. Could not figure out that the way Bunuel has solved it.
I guessed the answer in following way though -
48 is close to 49 which is square of 7
Now sqrt of 48 will be around 7
Now it becomes sqrt of (7+7) which srt of 14. It is close to sqrt of 16 which is equal to 4.
Now 4 - sqrt of 3 = 4-1.7 = 2.3

Now answer will be slightly smaller than 2.3 as we considered sqrt of 16 rather than original sqrt of 14.
So answer will be smaller than 2.3 and within given answers, it will be equal to 2.

The way Bunual has solved it is very good. New learning. Thanks.
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Re: M11-18  [#permalink]

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New post 04 Jul 2016, 20:25
Form an equation. Square it both the side. You will get answer as 2.

Or, Use ball park figure.. it will give u approx. 1.9-2.2 range. answer 2 satisfies this. sqrt(3)=1.73
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Re M11-18  [#permalink]

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New post 08 Nov 2016, 01:47
I think this is a high-quality question and I agree with explanation. Pretty innovative question, and very close options!
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Re: M11-18  [#permalink]

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New post 26 Jan 2019, 14:57
Bunuel wrote:
JackSparr0w wrote:
Bunuel wrote:
Official Solution:

The expression \(7 + \sqrt{48}\) can be rewritten as \(4 + 4\sqrt{3} + 3\) which is the same as \((2 + \sqrt{3})^2\). Therefore,\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = 2 + \sqrt{3} - \sqrt{3} = 2\).

Answer: C


Is there anything specific here that we should pick up on, that would tell us that we should break up the problem this way?

the √48 can be broken down. I get that, and it seems intuitive enough. But is there a way to realize that we need to break the 7 into 4+3, and then use foil?

I guess Im just trying to figure out what the "real time" thought process should be when one sees this.

Thanks


I guess such kind of tricks should come with practice.



I came up with a generalized approach.

Visually superimpose the following form onto your expression (or sub-expression):
\(a±b\sqrt{c}\)

Then test to see if
\(\sqrt{a-c} = \frac{b}{2}\)

If so then you can rearrange your initial expression as
\((a-c) ± b\sqrt{c} + c\)

Try this with the problem example, where a = 7; b = 4; and c = 3

Also see with another example here:

\(14 + 6\sqrt{5}\)

a = 14
b = 6
c = 5

Test it:
\(\sqrt{14-5} = \frac{6}{2}\)

3 = 3

It passes the test, so we can rewrite the original expression as:

\(9 + 6\sqrt{5} + 5\)

Which can then be factored into
\((3+\sqrt{5})^2\)


I'm not sure if there are enough occurrences of this to justify memorizing this trick, but it's a tool you can use nonetheless.

I personally like Magsy's method.
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Re: M11-18   [#permalink] 26 Jan 2019, 14:57
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