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655-705 Level|   Algebra|   Roots|      
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If a < b and 1 + 21 + 123 = a + b, then what is the value of a ? 09 Mar 2020, 03:18
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

61% (02:31) correct 39% (02:44) wrong based on 391 sessions

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If \(a < b\) and \(\sqrt{1 + \sqrt{21 + 12\sqrt{3}}} = \sqrt{a} + \sqrt{b}\), then what is the value of \(a\)?

A. 1
B. 2
C. 3
D. 4
E. 5

Are You Up For the Challenge: 700 Level Questions
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A
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If a < b and 1 + 21 + 123 = a + b, then what is the value of a ? Updated on: 04 Jun 2024, 03:02
Originally posted by GMATWhizTeam on 09 Mar 2020, 05:45.
Last edited by Bunuel on 04 Jun 2024, 03:02, edited 2 times in total.
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Bunuel
If \(a < b\) and \(\sqrt{1 + \sqrt{21 + 12\sqrt{3}}} = \sqrt{a} + \sqrt{b}\), then what is the value of \(a\)?

A. 1
B. 2
C. 3
D. 4
E. 5


Are You Up For the Challenge: 700 Level Questions
Show more

Solution:


    • a < b.
    • \(\sqrt{1+\sqrt{21+12 \sqrt{3}}} = \sqrt{a} + \sqrt{b}\)
Let’s simplify \(\sqrt{1 + \sqrt{21 + 2*6\sqrt{3}}}\) first.
    • \(\sqrt{1 + \sqrt{21 + 2\sqrt{36*3}}}\)
    • \(\sqrt{1 + \sqrt{(\sqrt{12})^2 + (\sqrt{9})^2 + 2\sqrt{12}\sqrt{9}}}\)
    • \(\sqrt{1 + \sqrt{(\sqrt{12} + \sqrt{9})^2}}\)
    • \(\sqrt{1 + \sqrt{12} + \sqrt{9}} \)
    • \(\sqrt{1 + 2\sqrt{3} + 3}\)
    • \(\sqrt{1^2 + 2\sqrt{3} + (\sqrt{3})^2}\)
    • \(\sqrt{(1 + \sqrt{3})^2}\)
    • \(1 + \sqrt{3}\)
Now, \((1 + \sqrt{3}) = \sqrt{a} + \sqrt{b}\)
After comparing both the sides, we get
    • \(a = 1 [a < b]\)
Hence, the correct answer is Option A.­
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If a < b and 1 + 21 + 123 = a + b, then what is the value of a ? 09 Mar 2020, 06:00
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Bunuel
If \(a < b\) and \(\sqrt{1 + \sqrt{21 + 12\sqrt{3}}} = \sqrt{a} + \sqrt{b}\), then what is the value of \(a\)?

A. 1
B. 2
C. 3
D. 4
E. 5


Are You Up For the Challenge: 700 Level Questions

Solution:


    • a < b.
    • \(\sqrt{1+\sqrt{21+12 \sqrt{3}}} = \sqrt{a} + \sqrt{b}\)
Let’s simplify \(\sqrt{1 + \sqrt{21 + 2*6\sqrt{3}}}\) first.
    • \(\sqrt{1 + \sqrt{21 + 2\sqrt{36*3}}}\)
    • \(\sqrt{1 + \sqrt{(\sqrt{12})^2 + (\sqrt{9})^2 + 2\sqrt{12}\sqrt{9}}}\)
    • \(\sqrt{1 + \sqrt{(\sqrt{12} + \sqrt{9})^2}}\)
    • \(\sqrt{1 + \sqrt{12} + \sqrt{9}} \)
    • \(\sqrt{1 + 2\sqrt{3} + 3}\)
    • \(\sqrt{1^2 + 2\sqrt{3} + (\sqrt{3})^2}\)
    • \(\sqrt{(1 + \sqrt{3})^2}\)
    • \(1 + \sqrt{3}\)
Now, \((1 + \sqrt{3})^2 = \sqrt{a} + \sqrt{b}\)
After comparing both the sides, we get
    • \(a = 1 [a < b]\)
Hence, the correct answer is Option A.
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I just want to understand whether there is any possibility of such questions to ever come in GMAT exam? This requires so much of manipulation that it might take more than 5 mins to just understand the manipulation part. And then comes the solution which might require another 5 mins.

Kindly advise.­
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If a < b and 1 + 21 + 123 = a + b, then what is the value of a ? 09 Mar 2020, 06:28
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shameekv1989
GMATWhizard


I just want to understand whether there is any possibility of such questions to ever come in GMAT exam? This requires so much of manipulation that it might take more than 5 mins to just understand the manipulation part. And then comes the solution which might require another 5 mins.

Kindly advise.
Show more


Hey Shameekv,

I can understand that this might seem like a difficult question. However, if you understand the right methodology to solve this question, it won't take you more than 2-2.5 mins.

Also, there is an official question that is based on this kind of question. Here's the link: https://gmatclub.com/forum/the-number-c ... 45521.html

If you'll go to this question, you'll see that this question belongs to Official Quant Review 2019. This is proof that you may come across questions like these.

    Now, as far as solving such questions are concerned, you just need to remember one thing, i.e. you need to use the formula \((a +b)^2 = a^2 + b^2 + 2*a*b\) or \((a - b)^2 = a^2 + b^2 - 2*a*b\) to solve the question.

    The question is basically trying to test, whether the student can simplify the given expression using the above formula or not.

    The trick to solve this question is to focus on the square root part first. Notice how I focused on \(12 \sqrt{3}\) first and separated 2 from it and then tried to bring the rest under a square root and manipulate it to get \(2\sqrt{12}\sqrt{9}\)

I am not saying that doing all this is easy. But with practice and knowing the above logic, you can definitely solve such question. Also, this is a 700-level question, so even if we take about 3 minutes to solve it, it's okay. As long as we are saving time elsewhere. :)
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If a < b and 1 + 21 + 123 = a + b, then what is the value of a ? 30 May 2020, 06:30
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21+12*sqrt(3) = 21+ 2*2*3*sqrt(3)
9+12+ 2*3*(2*sqrt(3))

Expression is Sqrt(1+ 2sqrt(3) + 3)
1+ sqrt(3)
= sqrt(a) + sqrt(b)
as b> a
so a = 1, b = 3

Can also be seen in attachment
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1.PNG [ 18.12 KiB | Viewed 8293 times ]

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If a < b and 1 + 21 + 123 = a + b, then what is the value of a ? 30 May 2020, 07:36
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<quote option="&quot;Bunuel&quot;">If \(a < b\) and \(\sqrt{1 + \sqrt{21 + 12\sqrt{3}}} = \sqrt{a} + \sqrt{b}\), then what is the value of \(a\)?<br />
<br />
A. 1<br />
B. 2<br />
C. 3<br />
D. 4<br />
E. 5<br />
<br />
<br />
<url href="https://gmatclub.com/forum/are-you-up-for-the-challenge-309579.html" option="https://gmatclub.com/forum/are-you-up-for-the-challenge-309579.html"><b><u>Are You Up For the Challenge: 700 Level Questions</u></b></url></quote><br />
<br />
i solved it in one step :lol: and chose A option. Should some expert confirm if my logic is correct i would appreciate :)<br />
<br />
so just square both sides and one can clearly see without hallucination that <b>a = 1</b> :grin:<br />
<br />
\((\sqrt{1 + \sqrt{21 + 12\sqrt{3}}})^2 = (\sqrt{a} + \sqrt{b})^2\)<br />
<br />
= \((1 + \sqrt{21 + 12\sqrt{3}}})= a+2\sqrt{a*b} +b\)­
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If a < b and 1 + 21 + 123 = a + b, then what is the value of a ? 21 Jun 2021, 01:08
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Breakdown the expression-
root(1+ root((21+12root(3)))) = root(1+ root( (2root(3) + 3)^2))
= 1+root(3)
a = 1, b = 3
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If a < b and 1 + 21 + 123 = a + b, then what is the value of a ? 29 Jun 2024, 04:10
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↧↧↧ Detailed Video Solution to the Problem Series ↧↧↧




Given that \(a < b\) and \(\sqrt{1 + \sqrt{21 + 12\sqrt{3}}} = \sqrt{a} + \sqrt{b}\) and we need to find the value of a

Lets start by simplifying \(\sqrt{1 + \sqrt{21 + 12\sqrt{3}}}\)

To simplify this we can start with the inner most square root, and the idea is to convert the terms inside the square root into sqaure of sum of two numbers like \((x + y)^2\)

\(\sqrt{1 + \sqrt{21 + 12\sqrt{3}}}\) = \(\sqrt{1 + \sqrt{9 + 12 + 2 * 3 * 2\sqrt{3}}}\) = \(\sqrt{1 + \sqrt{3^2 + (2\sqrt{3})^2 + 2 * 3 * 2\sqrt{3}}}\)
= \(\sqrt{1 + \sqrt{(3 + 2\sqrt{3})^2}}\) = \(\sqrt{1 + 3 + 2\sqrt{3}}\) = \(\sqrt{4 + 2\sqrt{3}}\)

Following similar logic lets try to write \(4 + 2\sqrt{3}\) like \((x + y)^2\)

=> \(\sqrt{4 + 2\sqrt{3}}\) = \(\sqrt{1 + 3 + 2*1*\sqrt{3}}\) = \(\sqrt{1^2 + (\sqrt{3})^2 + 2*1*\sqrt{3}}\) = \(\sqrt{(1 + (\sqrt{3})^2}\)
= \(1 +\sqrt{3}\) = \(\sqrt{a} + \sqrt{b}\)

Now, a < b
=> \(\sqrt{a}\) = 1 ( as 1 < \(\sqrt{3}\))
=> a = 1

So, Answer will be A
Hope it helps!

Watch the following video to MASTER Roots

­
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If a < b and 1 + 21 + 123 = a + b, then what is the value of a ? 14 Jul 2025, 07:34
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