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Math Expert V
Joined: 02 Sep 2009
Posts: 56302

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2
7 00:00

Difficulty:   65% (hard)

Question Stats: 56% (01:43) correct 44% (01:59) wrong based on 86 sessions

### HideShow timer Statistics $$\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?$$

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

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Math Expert V
Joined: 02 Sep 2009
Posts: 56302

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3
4
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,$$\sqrt{7 + \sqrt{48}} - \sqrt{3} = 2 + \sqrt{3} - \sqrt{3} = 2$$.

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Manager  Joined: 08 Feb 2014
Posts: 204
Location: United States
Concentration: Finance
GMAT 1: 650 Q39 V41 WE: Analyst (Commercial Banking)

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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,$$\sqrt{7 + \sqrt{48}} - \sqrt{3} = 2 + \sqrt{3} - \sqrt{3} = 2$$.

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
Math Expert V
Joined: 02 Sep 2009
Posts: 56302

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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,$$\sqrt{7 + \sqrt{48}} - \sqrt{3} = 2 + \sqrt{3} - \sqrt{3} = 2$$.

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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Retired Moderator Joined: 06 Jul 2014
Posts: 1224
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40 ### Show Tags

1
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
$$\sqrt{48}= 6.9$$

$$\sqrt{7 + 6.9}= 3.8$$

$$3.8-\sqrt{3}=2.1$$

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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Manager  Joined: 17 Mar 2015
Posts: 116

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2
3
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
Current Student S
Joined: 18 Aug 2014
Posts: 324

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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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Math Expert V
Joined: 02 Aug 2009
Posts: 7764

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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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Intern  Joined: 13 Feb 2016
Posts: 6
Location: United States
GMAT 1: 710 Q48 V41 GMAT 2: 760 Q49 V45 ### Show Tags

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 $$\sqrt{14}$$.

Intern  Joined: 01 Apr 2014
Posts: 43
Schools: ISB '17
GMAT 1: 530 Q35 V28 GPA: 2.5

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I think this is a high-quality question and I agree with explanation.
Intern  Joined: 14 Mar 2015
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Location: India
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1
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.
Current Student B
Joined: 13 Feb 2015
Posts: 8
Location: India
GPA: 3.68
WE: Information Technology (Consulting)

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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
Intern  Joined: 02 Jan 2015
Posts: 5

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I think this is a high-quality question and I agree with explanation. Pretty innovative question, and very close options!
Intern  B
Joined: 14 Oct 2018
Posts: 6
Location: United States
Schools: Stanford '21 (S)

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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,$$\sqrt{7 + \sqrt{48}} - \sqrt{3} = 2 + \sqrt{3} - \sqrt{3} = 2$$.

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.

***UPDATE*** (scroll down past the following until the next "***UPDATE***" tag)

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.

***UPDATE***
There is a similar opportunity to apply this in M26-01... and in light of the example there, I am restating a method that works better, which effectively boils down to just pulling out a $$2$$ from the square root (but leave everything else in there!) .....

M26-01 has part where you get to an expression of $$5 + 2\sqrt{6}$$

Here 5-6 = -1... so my first thought was to adjust my formula to be absolute value (|a-c|), but what seems to be the best heuristic is to manipulate a number to get a 2*square root or just look for a 2 * square root... (a more visual solution)

e.g. let's go back to the original example

$$14 + 6\sqrt{5}$$
Re-write as

$$14 + 2\sqrt{5}(3)$$

Get the $$3$$ inside the square root:

$$14 + 2\sqrt{5*9}$$

Well, now you can recognize your 2 * square root expression to be reminding you of

$$2\sqrt{ab}$$

from

$$a +$$ $$2\sqrt{ab}$$ + $$b$$

so split up your 14 into: 5 + 9

And your overall expression is now:

$$5+ 2\sqrt{5*9} + 9$$

which factors to:

$$(\sqrt{5} + 3)^2$$

e.g. #2; now visually, the M26-01 example should be come clear

$$5 + 2\sqrt{6}$$

$$(3+2) + 2\sqrt{(3*2)}$$

$$(3+2) + 2\sqrt{(3*2)}$$

$$(\sqrt{3} + \sqrt{2})^2$$

....

Finally, e.g. 3; the example from this problem... perhaps even more useful than how Bunuel broke it down, let's do this:

$$7 + \sqrt{48}$$ = $$7 + 2\sqrt{12}$$

And intuitively, we should immediately be able to see the numbers 3 and 4 (3+4 = 7; and 3*4 = 12)

so $$\sqrt{3}$$ and $$\sqrt{4}$$ (or 2) become our numbers...

$$(\sqrt{3} + 2)^2$$

Note that the trick here is that for $$\sqrt{48}$$ we are ONLY trying to pull out a "2" on the outside rather than trying to maximally simplify the square root ( $$2\sqrt{12}$$ instead of. $$4\sqrt{3}$$ )

This may be counter-intuitive, but if you simply remember that "I need to pull out a 2 that touches the outside of the square root sign directly" that would be a very useful real-time way of visually thinking about it.
Intern  B
Joined: 08 Feb 2019
Posts: 2

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I tried it using this way

√48 < 7
=> < 14
=> √‾(7+√48)‾‾ < 4 (since 4*4 = 16)
=> (√‾(7+√48)‾ - √3‾) < (4 - √3‾)
=> (√‾(7+√48)‾ - √3‾) < 2.3

so C

√3‾ = 1.732 Re: M11-18   [#permalink] 14 Apr 2019, 02:57
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# M11-18

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