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-1<a<0<|a|<b<1

huda

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-1<a<0<|a|<b<1 [#permalink]

Updated on: 05 Jul 2019, 12:06

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\(-1 < a < 0 < |a| < b < 1\)

Quantity A	Quantity B
\(\left[ \begin{array}{cc\|r} \frac{a^2 \sqrt{b}}{ \sqrt{a}} \end{array} \right]^2\)	\(\frac{ab^5}{ ( \sqrt{b} )^4}\)

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.

Originally posted by huda on 05 Jul 2019, 08:28.
Last edited by Carcass on 05 Jul 2019, 12:06, edited 2 times in total.

Edited by Carcass

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GreenlightTestPrep

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Re: -1<a<0<|a|<b<1 [#permalink]

05 Jul 2019, 10:18

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huda wrote:

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.

Concept #1: k² ≥ 0 for all values of k

QUANTITY A
Quantity A = (some value)²
Applying concept #1, we can see that Quantity A is greater than or equal to zero

QUANTITY B
We're told that a is NEGATIVE
We're also told that b is POSITIVE, which means b⁵ is POSITIVE
So, the numerator = (NEGATIVE)(POSITIVE) = NEGATIVE

In the denominator, we have (some value)⁴, which means the denominator must be POSITIVE

So, Quantity B = NEGATIVE/POSITIVE = NEGATIVE
So, Quantity B is some negative number

Answer: A

Cheers,
Brent

D

Carcass

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Re: -1<a<0<|a|<b<1 [#permalink]

05 Jul 2019, 12:07

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Please follow the rules for posting on the board. Images as question are NOT allowed.

https://gre.myprepclub.com/forum/rules-for ... -1083.html

Moreover, format a question in the proper manner.

https://gre.myprepclub.com/forum/how-to-po ... 12752.html

See above my editing.

If you need assistance, send me a private message. I will look to help you.

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AlaminMolla

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Re: -1<a<0<|a|<b<1 [#permalink]

08 Jul 2019, 06:36

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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,

D

Carcass

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Re: -1<a<0<|a|<b<1 [#permalink]

08 Jul 2019, 07:06

Expert Reply

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.

regards

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GreenlightTestPrep

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Re: -1<a<0<|a|<b<1 [#permalink]

08 Jul 2019, 08:34

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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

huda

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Re: -1<a<0<|a|<b<1 [#permalink]

09 Jul 2019, 02:56

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

B

AjanthaJ

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-1<a<0<|a|<b<1 [#permalink]

30 Jan 2021, 01:57

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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.

D

Carcass

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Re: -1<a<0<|a|<b<1 [#permalink]

30 Jan 2021, 02:05

Expert Reply

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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GreenlightTestPrep

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Re: -1<a<0<|a|<b<1 [#permalink]

30 Jan 2021, 05:45

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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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COolguy101

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Re: -1<a<0<|a|<b<1 [#permalink]

22 May 2021, 03:09

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.

D

Carcass

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