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sqrt of 6* sqrt of 18/sqrt of 9
[#permalink]
22 Feb 2023, 06:47
22 Feb 2023, 06:47
Expert Reply
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Question Stats:
100% (01:33) correct
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Quantity A |
Quantity B |
\(\frac{\sqrt{6} * \sqrt{18}}{\sqrt{9} }\) |
\(\frac{\sqrt{8} * \sqrt{12}}{\sqrt{6} }\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
sqrt of 6* sqrt of 18/sqrt of 9
[#permalink]
06 Mar 2023, 16:44
06 Mar 2023, 16:44
2
Quantity A
Quantity B
6√∗18−−√9√
6
∗
18
9
8√∗12−−√6√
8
∗
12
6
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Soln:
To solve Quantity A we will simplify the root of numerator, √6∗√18 = √3*2 * √9*2
Next step separate the roots we get √3 * √2* √9 * √2 = 2 * 3√3 = 6√3
Now evaluate it with denominator which is √9 i.e. a perfect square of 3
So,6√3 / 3 = 2√3
For Quantity B we will repeat the procedure √8∗√12 = √4*2 * √4*3
Next step separate the roots we get √4 * √2* √4 * √3 = 4 √2* 3 = 4√6
Now evaluate it with denominator which is √6
So, 4√6 -- √6 = 4
Quantity B > Quantity A
Quantity B
6√∗18−−√9√
6
∗
18
9
8√∗12−−√6√
8
∗
12
6
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Soln:
To solve Quantity A we will simplify the root of numerator, √6∗√18 = √3*2 * √9*2
Next step separate the roots we get √3 * √2* √9 * √2 = 2 * 3√3 = 6√3
Now evaluate it with denominator which is √9 i.e. a perfect square of 3
So,6√3 / 3 = 2√3
For Quantity B we will repeat the procedure √8∗√12 = √4*2 * √4*3
Next step separate the roots we get √4 * √2* √4 * √3 = 4 √2* 3 = 4√6
Now evaluate it with denominator which is √6
So, 4√6 -- √6 = 4
Quantity B > Quantity A
gmatclubot
sqrt of 6* sqrt of 18/sqrt of 9 [#permalink]
06 Mar 2023, 16:44
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