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# MGMAT CAT difficult question

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Intern
Joined: 30 Dec 2009
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12 Mar 2010, 09:41
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Question Stats:

100% (02:00) correct 0% (00:00) wrong based on 25 sessions

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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
Senior Manager
Joined: 01 Feb 2010
Posts: 267
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Kudos [?]: 50 [1] , given: 2

Re: MGMAT CAT difficult question [#permalink]

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12 Mar 2010, 09:46
1
KUDOS
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.
Intern
Joined: 30 Dec 2009
Posts: 10
Followers: 0

Kudos [?]: 2 [0], given: 2

Re: MGMAT CAT difficult question [#permalink]

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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!
Manager
Joined: 14 Dec 2009
Posts: 79
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Kudos [?]: 36 [3] , given: 20

Re: MGMAT CAT difficult question [#permalink]

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12 Mar 2010, 10:31
3
KUDOS
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!
Intern
Joined: 30 Dec 2009
Posts: 10
Followers: 0

Kudos [?]: 2 [0], given: 2

Re: MGMAT CAT difficult question [#permalink]

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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!

That helped and I sent you a "kudo".
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Re: MGMAT CAT difficult question [#permalink]

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15 May 2016, 09:35
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Re: MGMAT CAT difficult question   [#permalink] 15 May 2016, 09:35
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