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605-655 (Medium)|   Algebra|   Roots|      
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Bunuel
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If a, b, and c are positive integers, is b = ac^(-1/2) 10 Oct 2017, 00:44
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61% (01:16) correct 39% (01:18) wrong based on 474 sessions

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If a, b, and c are positive integers, is \(b = \frac{a}{\sqrt{c}}\)?


(1) \(c = \frac{a^2}{b^2}\)

(2) \(\sqrt{c} = \frac{b}{a}\)
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A
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If a, b, and c are positive integers, is b = ac^(-1/2) Updated on: 10 Oct 2017, 06:18
Originally posted by GMATinsight on 10 Oct 2017, 04:43.
Last edited by GMATinsight on 10 Oct 2017, 06:18, edited 2 times in total.
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Bunuel
If a, b, and c are positive integers, is \(b = \frac{a}{\sqrt{c}}\)?


(1) \(c = \frac{a^2}{b^2}\)

(2) \(\sqrt{c} = \frac{b}{a}\)
Show more

Question: is \(b = \frac{a}{\sqrt{c}}\)?

Statement 1: \(c = \frac{a^2}{b^2}\)
Taking square root both sides
√c = a/b
SUFFICIENT


Statement 2: \(\sqrt{c} = \frac{b}{a}\)
@a = b = c = 1 Answer to the question is YES
@a = 1, b=2, c = 4 Answer to the question is NO
NOT SUFFICIENT


Answer: Option A
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If a, b, and c are positive integers, is b = ac^(-1/2) 10 Oct 2017, 04:46
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If a, b, and c are positive integers, is \(b = \frac{a}{c}\)?


(1) \(c = \frac{a^2}{b^2}\)

(2) \(\sqrt{c} = \frac{b}{a}\)

Question: is \(b = \frac{a}{c}\)?

Statement 1: \(c = \frac{a^2}{b^2}\)
If a = b = c = 1 then answer to the question is YES
if a = b = 2 then c = 1 then answer to the question is NO

NOT SUFFICIENT


Statement 2: \(\sqrt{c} = \frac{b}{a}\)

If a = b = c = 1 then answer to the question is YES
if a = b = 2 then c = 1 then answer to the question is NO

NOT SUFFICIENT


After combining the two statements

If a = b = c = 1 then answer to the question is YES
if a = b = 2 then c = 1 then answer to the question is NO

NOT SUFFICIENT

Answer: Option E
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Sorry GMATinsight, the question is "If a, b, and c are positive integers, is \(b = \frac{a}{\sqrt{c}}\)?" NOT "If a, b, and c are positive integers, is \(b = \frac{a}{c}\)?". Edited.
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If a, b, and c are positive integers, is b = ac^(-1/2) 10 Oct 2017, 05:00
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Bunuel
If a, b, and c are positive integers, is \(b = \frac{a}{\sqrt{c}}\)?


(1) \(c = \frac{a^2}{b^2}\)

(2) \(\sqrt{c} = \frac{b}{a}\)
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(1) \(c = \frac{a^2}{b^2}\)
=> taking square root
=> \(\sqrt{c} = \frac{a}{b}\)
=> \(b = \frac{a}{\sqrt{c}}\)
Sufficient

(2) \(\sqrt{c} = \frac{b}{a}\)
=> \(b =a\sqrt{c}\)
In this case :
if c = 1 => \(b =a\sqrt{c}\) => \(b = \frac{a}{\sqrt{c}}\)
if c = 4 => \(b =a\sqrt{c}\) is not same as \(b = \frac{a}{\sqrt{c}}\)
Insufficient

Answer: A
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If a, b, and c are positive integers, is b = ac^(-1/2) 07 Jun 2018, 09:19
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Bunuel
If a, b, and c are positive integers, is \(b = \frac{a}{\sqrt{c}}\)?


(1) \(c = \frac{a^2}{b^2}\)

(2) \(\sqrt{c} = \frac{b}{a}\)
Show more

Given: a,b,c are +'ve
To Calculate:\(b=\frac{a}{\sqrt{c}}\)

Statement 1:
\(c = \frac{a^2}{b^2}\)
Since, \(\sqrt{x^2}=|x|\)
and a,b,c are all +'ve [Already stated in question]
Therefore, \(\sqrt{x^2}=x\)
Using the above, we can root both sides(Since, all the variables are Positive, there will be no loss of solutions)
\(\sqrt{c}=\frac{a}{b}\)
Rearranging the terms, we get
\(b = \frac{a}{\sqrt{c}}\)
Hence Sufficient.

Statement 2:
\(\sqrt{c} = \frac{b}{a}\)
Rearranging the terms, we get:
\(b=a\sqrt{c}\)
Which is not the term we desire.
Hence Insufficient.

Answer: A

However, I do not see the need of Value Substitution while evaluating Statement 2.
Can someone throw some light on why did we do that ?
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If a, b, and c are positive integers, is b = ac^(-1/2) 07 Jun 2018, 09:36
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[quote="Rumanshu1990"][quote="Bunuel"]If a, b, and c are positive integers, is [m]b = \frac{a}{[square_root]c[/square_root]}[/m]?


(1) [m]c = [fraction]a^2/b^2[/fraction][/m]

(2) [m][square_root]c[/square_root] = [fraction]b/a[/fraction][/m][/quote]

[b]Given:[/b] a,b,c are +'ve
[b]To Calculate[/b]:[m]b=[fraction]a/[square_root]c[/square_root][/fraction][/m]

[b]Statement 1:[/b]
[m]c = [fraction]a^2/b^2[/fraction][/m]
Since, [m][square_root]x^2[/square_root]=|x|[/m]
and a,b,c are all +'ve [Already stated in question]
Therefore, [m][square_root]x^2[/square_root]=x[/m]
Using the above, we can root both sides(Since, all the variables are Positive, there will be no loss of solutions)
[m][square_root]c[/square_root]=[fraction]a/b[/fraction][/m]
Rearranging the terms, we get
[m]b = \frac{a}{[square_root]c[/square_root]}[/m]
[b]Hence Sufficient.[/b]

[b]Statement 2:[/b]
[m][square_root]c[/square_root] = [fraction]b/a[/fraction][/m]
Rearranging the terms, we get:
[m]b=a[square_root]c[/square_root][/m]
Which is not the term we desire.
[b]Hence Insufficient.[/b]

[b]Answer: A[/b]

However, I do not see the need of Value Substitution while evaluating Statement 2.
Can someone throw some light on why did we do that ?[/quote]

Hello

Value substitution is just one of the techniques which often comes in very handy while solving DS questions, especially for someone who is not so good in algebra and advanced formulas, or if one is not so good in mathematics.

Even for people who are good in formulas/algebra etc, sometimes value substitution saves time. That's why you will often see experts solving questions using this technique.

Although I agree this questions second statement can probably be very easily resolved using what you suggested.

[size=80][b][i]Posted from my mobile device[/i][/b][/size]
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If a, b, and c are positive integers, is b = ac^(-1/2) 26 Jun 2019, 23:11
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Bunuel
If a, b, and c are positive integers, is \(b = \frac{a}{\sqrt{c}}\)?


(1) \(c = \frac{a^2}{b^2}\)

(2) \(\sqrt{c} = \frac{b}{a}\)
Show more

In GMAT if x^2 = 25, the x can be +/-5
But root(25) will have only one answer= 5.

1. Square root both sides:

\(b = \frac{a}{\sqrt{c}}\) Sufficient

2. b= a* root(c)
Yes for all a=b=c=1
No for a!=b!=c!=1. Insufficient.
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If a, b, and c are positive integers, is b = ac^(-1/2) 23 Jun 2023, 23:31
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