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sqrt of 2014 + 2*(1+2+3+.....
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15 Apr 2021, 02:28
15 Apr 2021, 02:28
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Question Stats:
41% (02:53) correct
58% (02:07) wrong based on 31 sessions
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Quantity A |
Quantity B |
\( \sqrt{2014 + 2 \times (1+2+3+.....+ 2013)}\) |
\((\sqrt{\sqrt{2014} + 1} \times (\sqrt{\sqrt{2014} + \sqrt{2013}} -\sqrt{\sqrt{2014}-\sqrt{2013}}))^2\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
sqrt of 2014 + 2*(1+2+3+.....
[#permalink]
15 Apr 2021, 03:46
15 Apr 2021, 03:46
1
Quote:
Quantity A
\(\sqrt{2014 + 2 * (1+2+3+.....+ 2013)}\)
\(\sqrt{2014 + 2 * (1+2+3+.....+ 2013)}\)
(1+2+3+.....+ 2013) is a sequence with a common difference 1.
In such arithmetic sequences, the sum = average * number of terms.
\(Average = \frac{1+2013 }{ 2} = \frac{2014 }{ 2}\)
Number of terms = 2013, of course.
\(2 * (1+2+3+.....+ 2013) = 2 * \frac{2014 }{ 2} * 2013 = 2014 * 2013\)
\(2014 + 2 * (1+2+3+.....+ 2013) = 2014 + 2014 * 2013 = 2014 ( 1 + 2013) = 2014^2\)
\(\sqrt{2014 + 2 * (1+2+3+.....+ 2013)} = 2014\)
Let's look at another badass expression
Quote:
Quantity B
\(((\sqrt{2014} + 1)^\frac{1}{2} * [(\sqrt{2014} + \sqrt{2013})^\frac{1}{2} - (\sqrt{2014}-\sqrt{2013})^\frac{1}{2}])^2\)
\(((\sqrt{2014} + 1)^\frac{1}{2} * [(\sqrt{2014} + \sqrt{2013})^\frac{1}{2} - (\sqrt{2014}-\sqrt{2013})^\frac{1}{2}])^2\)
Observe that two parts will squared here.
Apply identity expression (a+b)^2 to the following because it will be wholly squared.
\([(\sqrt{2014} + \sqrt{2013})^\frac{1}{2} - (\sqrt{2014}-\sqrt{2013})^\frac{1}{2}]^2\)
\(= \sqrt{2014} + \sqrt{2013} + \sqrt{2014} - \sqrt{2013} - 2
= 2 \sqrt{2014} - 2\)
The remaining part of the expression is also squared.
\(((\sqrt{2014} + 1)^\frac{1}{2})^2 = \sqrt{2014} + 1\)
Now put this all together.
\(((\sqrt{2014} + 1)^\frac{1}{2} * [(\sqrt{2014} + \sqrt{2013})^\frac{1}{2} - (\sqrt{2014}-\sqrt{2013})^\frac{1}{2}])^2\)
\(= (\sqrt{2014} + 1) * (2\sqrt{2014} - 2)\)
\(= (\sqrt{2014} + 1) * 2* (\sqrt{2014} - 1)\)
Apply identity expression \(a^2 - b^2 = (a+b) (a-b)\),
\(2 * ( \sqrt{2014} + 1) * ( \sqrt{2014} - 1) = 2 * (2014 - 1) = 2 * 2013\)
Therefore, quantity A is 2014, while quantity B is 2*2013.
Hence, B is the correct answer.
