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If a^6 + b^6 = 144 then the greatest possible value for b is between

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If a^6 + b^6 = 144 then the greatest possible value for b is between  [#permalink]

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New post 10 Sep 2017, 05:30
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A
B
C
D
E

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67% (01:27) correct 33% (02:01) wrong based on 169 sessions

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Re: If a^6 + b^6 = 144 then the greatest possible value for b is between  [#permalink]

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New post Updated on: 10 Sep 2017, 08:20
1
greatest possible value of b is when a is zero.

so b^6 = 144
b^3 = 12 b = 2.xxx
b is between 3 and 4 so

Ans D

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Originally posted by Ejiroghene on 10 Sep 2017, 06:58.
Last edited by Ejiroghene on 10 Sep 2017, 08:20, edited 1 time in total.
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Re: If a^6 + b^6 = 144 then the greatest possible value for b is between  [#permalink]

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New post 10 Sep 2017, 07:14
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2
It's a D. Between 2 and 4.

B is maximum when value of A is zero.

And since no condition has been given for A and B to be integers, so B can take decimal values also.

Again b^6=144
b^3=12.
So B can also be 2.1 or 2.2.
But it cannot be 3 or greater than 3.
So it has to be somewhere between 2 and 3.
So option D holds good in this case.



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If a^6 + b^6 = 144 then the greatest possible value for b is between  [#permalink]

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New post 10 Sep 2017, 07:20
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Bunuel wrote:
If a^6 + b^6 = 144 then the greatest possible value for b is between

A. 8 and 10
B. 6 and 8
C. 4 and 6
D. 2 and 4
E. 2 and 0


The question states that \(b^6 = 144-a^6\). Hence for \(b\) to be maximum \(a=0\)
so \(b^6=144\) taking square root of both the sides, we get
\(b^3 = 12\). Now,
\(8<12<27\) or \(8<b^3<27\). Taking cube root of the inequality we get
\(2<b<3\). Hence \(b\) is definitely greater than \(2\) but less than \(3\) (to be precise; \(b^3 = 12\), or \(b= 2.289428\))

Ideally the greatest possible value of \(b\) is between \(2\) & \(3\), but we don't have any option stating that.
Hi Bunuel can you confirm whether Option D is correct and there are no typo errors
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If a^6 + b^6 = 144 then the greatest possible value for b is between  [#permalink]

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New post Updated on: 17 Oct 2017, 00:56
niks18 wrote:
Bunuel wrote:
If a^6 + b^6 = 144 then the greatest possible value for b is between

A. 8 and 10
B. 6 and 8
C. 4 and 6
D. 2 and 4
E. 2 and 0


The question states that \(b^6 = 144-a^6\). Hence for \(b\) to be maximum \(a=0\)
so \(b^6=144\) taking square root of both the sides, we get
\(b^3 = 12\). Now,
\(8<12<27\) or \(8<b^3<27\). Taking cube root of the inequality we get
\(2<b<3\). Hence \(b\) is definitely greater than \(2\) but less than \(3\) (to be precise; \(b^3 = 12\), or \(b= 2.289428\))

Ideally the greatest possible value of \(b\) is between \(2\) & \(3\), but we don't have any option stating that.
Hi Bunuel can you confirm whether Option D is correct and there are no typo errors


niks18
hi

I agree with you completely
though I approached the problem similar way, I don't understand how everybody is claiming "b" to be between 2 and 4

certainly, we need some Bunuel here ..

Originally posted by gmatcracker2018 on 16 Oct 2017, 22:47.
Last edited by gmatcracker2018 on 17 Oct 2017, 00:56, edited 2 times in total.
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Re: If a^6 + b^6 = 144 then the greatest possible value for b is between  [#permalink]

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New post 16 Oct 2017, 23:22
gmatcracker2017 wrote:
niks18 wrote:
Bunuel wrote:
If a^6 + b^6 = 144 then the greatest possible value for b is between

A. 8 and 10
B. 6 and 8
C. 4 and 6
D. 2 and 4
E. 2 and 0


The question states that \(b^6 = 144-a^6\). Hence for \(b\) to be maximum \(a=0\)
so \(b^6=144\) taking square root of both the sides, we get
\(b^3 = 12\). Now,
\(8<12<27\) or \(8<b^3<27\). Taking cube root of the inequality we get
\(2<b<3\). Hence \(b\) is definitely greater than \(2\) but less than \(3\) (to be precise; \(b^3 = 12\), or \(b= 2.289428\))

Ideally the greatest possible value of \(b\) is between \(2\) & \(3\), but we don't have any option stating that.
Hi Bunuel can you confirm whether Option D is correct and there are no typo errors


niks18
hi

I agree with you completely
though I approached the problem similar way, I don't understand how everybody is claiming "b" to be between 2 and 4

certainly, we need some Bunuel here ..


The greatest possible value of b turns out to be ~2.3, which is indeed between 2 and 3, but it will also be true to say that it's between -1,000 and 10,000 isn't it? Between x and y, here means in the range from x to y. So, we can say that 2.3 is between 2 and 4.
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Re: If a^6 + b^6 = 144 then the greatest possible value for b is between  [#permalink]

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New post 17 Oct 2017, 00:57
gmatcracker2017 wrote:
niks18 wrote:
Bunuel wrote:
If a^6 + b^6 = 144 then the greatest possible value for b is between

A. 8 and 10
B. 6 and 8
C. 4 and 6
D. 2 and 4
E. 2 and 0


The question states that \(b^6 = 144-a^6\). Hence for \(b\) to be maximum \(a=0\)
so \(b^6=144\) taking square root of both the sides, we get
\(b^3 = 12\). Now,
\(8<12<27\) or \(8<b^3<27\). Taking cube root of the inequality we get
\(2<b<3\). Hence \(b\) is definitely greater than \(2\) but less than \(3\) (to be precise; \(b^3 = 12\), or \(b= 2.289428\))

Ideally the greatest possible value of \(b\) is between \(2\) & \(3\), but we don't have any option stating that.
Hi Bunuel can you confirm whether Option D is correct and there are no typo errors


niks18
hi

I agree with you completely
though I approached the problem similar way, I don't understand how everybody is claiming "b" to be between 2 and 4

certainly, we need some Bunuel here ..


thanks a lot to you Bunuel

you are super helpful
thanks 8-)
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Re: If a^6 + b^6 = 144 then the greatest possible value for b is between  [#permalink]

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New post 19 Oct 2017, 10:21
Bunuel wrote:
If a^6 + b^6 = 144 then the greatest possible value for b is between

A. 8 and 10
B. 6 and 8
C. 4 and 6
D. 2 and 4
E. 2 and 0


Since we want to maximize the value of b, we want to minimize the value of a^6. Since a has an even power, the minimum value of a^6 is 0, which occurs when a = 0 (note: if a ≠ 0, then a^6 > 0). Thus, when a^6 = 0, we have:

b^6 = 144

Since 2^6 = 64 and 3^6 = 729, b is greater than 2 but less than 3.

Answer: D
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Re: If a^6 + b^6 = 144 then the greatest possible value for b is between &nbs [#permalink] 19 Oct 2017, 10:21
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