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MGMAT CAT difficult question

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MGMAT CAT difficult question [#permalink]

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New post 12 Mar 2010, 09:41
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This is from an MGMAT CAT exam. Any help in how to arrive at the correct solution is greatly appreciated. I am uncertain as to the steps!

If the sum of the cubes of a and b is 8 and a6 – b6 = 14, what is the value of a3?

1. 7/4
2. 23/6
3. 39/8
4. 6
5. 22/3
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Re: MGMAT CAT difficult question [#permalink]

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New post 12 Mar 2010, 09:46
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l02g060 wrote:
This is from an MGMAT CAT exam. Any help in how to arrive at the correct solution is greatly appreciated. I am uncertain as to the steps!

If the sum of the cubes of a and b is 8 and a6 – b6 = 14, what is the value of a3?

1. 7/4
2. 23/6
3. 39/8
4. 6
5. 22/3


a^3 + b^3 = 8
a^6 - b^6 = (a^3 + b^3)(a^3 - b^3) = 14
a^3 - b^3 = 7/4
2a^3 = 39/4
a^3 = 39/8 hence 3 or C.

Whats the OA.
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Re: MGMAT CAT difficult question [#permalink]

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New post 12 Mar 2010, 10:22
bangalorian2000 wrote:
l02g060 wrote:
This is from an MGMAT CAT exam. Any help in how to arrive at the correct solution is greatly appreciated. I am uncertain as to the steps!

If the sum of the cubes of a and b is 8 and a6 – b6 = 14, what is the value of a3?

1. 7/4
2. 23/6
3. 39/8
4. 6
5. 22/3


a^3 + b^3 = 8
a^6 - b^6 = (a^3 + b^3)(a^3 - b^3) = 14
a^3 - b^3 = 7/4
2a^3 = 39/4
a^3 = 39/8 hence 3 or C.

Whats the OA.


Thanks....my computer crashed and I can't retrieve OA, but will let you know when I can. How did you arrive at 2a^3= 39/4...I am temporarily brain dead.

Thanks again!
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Re: MGMAT CAT difficult question [#permalink]

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New post 12 Mar 2010, 10:31
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I used the same approach:
1. \(a^3+b^3=8\)
2. \(a^6-b^6=14\) from which \((a^3-b^3)(a^3+b^3)=14\) \(=>\) \(a^3+b^3= \frac{14}{a^3-b^3}=8\)(from the first equation) \(=>\) \(14=(a^3-b^3)*8\) \(=>\) \(a^3-b^3=\frac{14}{8}=\frac{7}{4}\)
Now combine the two equations:
\(a^3+b^3=8\)
\(a^3-b^3=\frac{7}{4}\)

The result is:
\(2a^3=8+\frac{7}{4}=\frac{39}{4}\)

So, \(a^3=\frac{\frac{39}{4}}{2}=\frac{39}{8}\)

Hope that helps! :-D
Intern
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Joined: 30 Dec 2009
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Re: MGMAT CAT difficult question [#permalink]

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New post 12 Mar 2010, 14:22
Igor010 wrote:
I used the same approach:
1. \(a^3+b^3=8\)
2. \(a^6-b^6=14\) from which \((a^3-b^3)(a^3+b^3)=14\) \(=>\) \(a^3+b^3= \frac{14}{a^3-b^3}=8\)(from the first equation) \(=>\) \(14=(a^3-b^3)*8\) \(=>\) \(a^3-b^3=\frac{14}{8}=\frac{7}{4}\)
Now combine the two equations:
\(a^3+b^3=8\)
\(a^3-b^3=\frac{7}{4}\)

The result is:
\(2a^3=8+\frac{7}{4}=\frac{39}{4}\)

So, \(a^3=\frac{\frac{39}{4}}{2}=\frac{39}{8}\)

Hope that helps! :-D


That helped and I sent you a "kudo".
Re: MGMAT CAT difficult question   [#permalink] 12 Mar 2010, 14:22
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