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Re: -1<a<0<|a|<b<1 [#permalink]
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I am just curious, is that math mathematically valid? I might be terribly wrong. Even that, I wanna share my thought. As we know a must be negative. So in A, how could we validate the square root of something negative?


Thanks,
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Re: -1<a<0<|a|<b<1 [#permalink]
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a is negative because is between -1 and zero.

Must be negative.

However, when it is inside the absolute value, it must be positive. After zero.

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Re: -1<a<0<|a|<b<1 [#permalink]
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AlaminMolla wrote:
I am just curious, is that math mathematically valid? I might be terribly wrong. Even that, I wanna share my thought. As we know a must be negative. So in A, how could we validate the square root of something negative?


Thanks,


You're absolutely correct. It's a bad question.
I didn't look closely inside the bracketed part, since it all gets squared in the end.
If a < 0, then √a is undefined, which renders the question invalid.
The GRE would never have a question like this.

Cheers,
Brent
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Re: -1<a<0<|a|<b<1 [#permalink]
GreenlightTestPrep wrote:
AlaminMolla wrote:
I am just curious, is that math mathematically valid? I might be terribly wrong. Even that, I wanna share my thought. As we know a must be negative. So in A, how could we validate the square root of something negative?


Thanks,


You're absolutely correct. It's a bad question.
I didn't look closely inside the bracketed part, since it all gets squared in the end.
If a < 0, then √a is undefined, which renders the question invalid.
The GRE would never have a question like this.

Cheers,
Brent


:) :) :)
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-1<a<0<|a|<b<1 [#permalink]
1
Carcass GreenlightTestPrep

I got the answer as B in this way. I learnt comparison in this way from Kaplan. Is this valid?

given quantity A :

(a^2 *√b)^2 = (a^4 * b)\ a = a^3 b
___________
(√a)^2


given quantity b :

a * b^5
_______ = ( a * b^5) / b^2 = a * b^3
(√b)^4

comparing qty A and qty B

a^3 * b = a * b^3

simplifying by dividing both sides by ab we get

Qty A = a^2

Qty B = b ^2

we are given a range to pick values for a and b

so when I take a = -0.6 and Mod a = 0.6 which must be less than b. So I pick b = 0.8 and b must be less than 1.

now squaring a and b

I get Qty A = 0.36 and Qty B = 0.64

Now B is the answer isn't it?

Pls correct me if am wrong somewhere.
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Re: -1<a<0<|a|<b<1 [#permalink]
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Hi

the answer is A and I suggest you to read very careful the GreenlightTestPrep explanation.

it is neat and marvelous.

Also read our book for better insights https://gre.myprepclub.com/forum/gre-math- ... -2609.html

Feel free to ask for further assistance

PS: I saw the one you posted method pretty cumbersome
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Re: -1<a<0<|a|<b<1 [#permalink]
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This isn't a good question, and it won't help you prepare for the GRE.
Since \(a\) is negative, \(\sqrt{a}\) is not defined. In other words, it has no real value.

This means evaluating \(\left[ \begin{array}{cc|r} \frac{a^2 \sqrt{b}}{ \sqrt{a}} \end{array} \right]^2\) is similar to evaluating \(\left[ \begin{array}{cc|r} \frac{a^2 \sqrt{b}}{apple} \end{array} \right]^2\)
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Re: -1<a<0<|a|<b<1 [#permalink]
AjanthaJ wrote:
Carcass GreenlightTestPrep

I got the answer as B in this way. I learnt comparison in this way from Kaplan. Is this valid?

given quantity A :

(a^2 *√b)^2 = (a^4 * b)\ a = a^3 b
___________
(√a)^2


given quantity b :

a * b^5

_______ = ( a * b^5) / b^2 = a * b^3
(√b)^4

comparing qty A and qty B

a^3 * b = a * b^3

simplifying by dividing both sides by ab we get

Qty A = a^2

Qty B = b ^2

we are given a range to pick values for a and b

so when I take a = -0.6 and Mod a = 0.6 which must be less than b. So I pick b = 0.8 and b must be less than 1.

now squaring a and b

I get Qty A = 0.36 and Qty B = 0.64

Now B is the answer isn't it?

Pls correct me if am wrong somewhere.

Hi, i got the same as yours, but what i figure out in last is we cannot divide the quantity if it is negative. product of ab is clearly negative thats why we misscalculate the question.
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Re: -1<a<0<|a|<b<1 [#permalink]
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Post locked

The discussion is located here https://gre.myprepclub.com/forum/1-a-0-a-b ... tml#p23077

Regards
Prep Club for GRE Bot
Re: -1<a<0<|a|<b<1 [#permalink]
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