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Re: b<a<1 [#permalink]
GreenlightTestPrep wrote:
Carcass wrote:
\(b<a<1\)


Quantity A
Quantity B
\(a^3b^6\)
\(b^6\)



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.



Let's test some possible values of a and b


Great explanation. But what if you take an algebraic approach to this and divide both quantities by b^6
We will be left with a^3 and 1. And since a<1
a^3 will be less than 1
I'd be really grateful if you could clear my confusion out

case i: One possible pair of values is \(a=0.5\) and \(b=0\)
We get:
Quantity A: \(a^3b^6=(0.5^3)(0^6) = 0\)
Quantity B: \(b^6=0^6=0\)
In this case, the two quantities are equal

case ii: Another possible pair of values is \(a=0\) and \(b=-1\)
We get:
Quantity A: \(a^3b^6=(0^3)(-1)^6 = 0\)
Quantity B: \(b^6=(-1)^6=1\)
In this case, Quantity B is greater

Answer: D
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Re: b<a<1 [#permalink]
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I have a hard time to figure out why the students quote a reply of the discussion without any apparent reason.

No writing, no explanation, no doubt

just a quote............ :facepalm_man: :facepalm_man: :facepalm_man:
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Re: b<a<1 [#permalink]
1
My Apologies Sir, I am new to this forum and thus had a hard time figuring it out. My doubt is this

Great explanation. But what if you take an algebraic approach to this and divide both quantities by b^6
We will be left with a^3 and 1. And since a<1
a^3 will be less than 1
I'd be really grateful if you could clear my confusion out
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Re: b<a<1 [#permalink]
1
Hi,

\(b < 1\) so it is possible that \(b = 0\) and we can't divide with \(0\)

Ami2000 wrote:
My Apologies Sir, I am new to this forum and thus had a hard time figuring it out. My doubt is this

Great explanation. But what if you take an algebraic approach to this and divide both quantities by b^6
We will be left with a^3 and 1. And since a<1
a^3 will be less than 1
I'd be really grateful if you could clear my confusion out
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b<a<1 [#permalink]
Expert Reply
Ami2000 wrote:
My Apologies Sir, I am new to this forum and thus had a hard time figuring it out. My doubt is this

Great explanation. But what if you take an algebraic approach to this and divide both quantities by b^6
We will be left with a^3 and 1. And since a<1
a^3 will be less than 1
I'd be really grateful if you could clear my confusion out


no worries

if we put \(a^3b^6=b^6\)

\(a^3=1\)

for being an odd exponent =1 a itself must be 1 so a=1

but we do know that a<1 so your hypothesis is NOt a viable option

The best approach to this kind of question is that a and b could be any possible values , zero included. Therefore, the answer is D
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b<a<1 [#permalink]
Expert Reply
rx10 wrote:
Hi,

\(b < 1\) so it is possible that \(b = 0\) and we can't divide with \(0\)

Ami2000 wrote:
My Apologies Sir, I am new to this forum and thus had a hard time figuring it out. My doubt is this

Great explanation. But what if you take an algebraic approach to this and divide both quantities by b^6
We will be left with a^3 and 1. And since a<1
a^3 will be less than 1
I'd be really grateful if you could clear my confusion out


The new marvel hero is back :tongue_opt3


badass observation BTW
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Re: b<a<1 [#permalink]
1
:D :D :please:

Carcass wrote:
The new marvel hero is back :tongue_opt3


badass observation BTW
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