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26 Jan 2012, 07:35
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
72% (02:15) correct
28%
(02:39)
wrong
based on 414
sessions
History
Date
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Not Attempted Yet
If one integer is greater than another integer by 3, and the difference of their cubes is 117, what could be their sum?
A. 15
B. 7
C. 3
D. 2
E. 1
A. 15
B. 7
C. 3
D. 2
E. 1
26 Jan 2012, 08:55
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Let the two integers be 'a' and 'b'
Then, according to the question
a-b=3 and \(a^3 - b^3\)=117
Now,
\((a-b)^3 = a^3 - b^3 -3ab(a-b)\)
Substituting the values of (a-b) and \(a^3 - b^3\) we get
\(3^3\) = 117 - 3ab*3
27 = 117 - 3ab*3
9ab=90
ab=10
Now to find the value of (a+b) use
\((a+b) = \sqrt{(a-b)^2 +4ab}\)
Again substitute the values
(a+b)=\(\sqrt{9+40}\)
(a+b)=\(\sqrt{49}\)
(a+b)=7
So the answer is B
Then, according to the question
a-b=3 and \(a^3 - b^3\)=117
Now,
\((a-b)^3 = a^3 - b^3 -3ab(a-b)\)
Substituting the values of (a-b) and \(a^3 - b^3\) we get
\(3^3\) = 117 - 3ab*3
27 = 117 - 3ab*3
9ab=90
ab=10
Now to find the value of (a+b) use
\((a+b) = \sqrt{(a-b)^2 +4ab}\)
Again substitute the values
(a+b)=\(\sqrt{9+40}\)
(a+b)=\(\sqrt{49}\)
(a+b)=7
So the answer is B
General Discussion
26 Jan 2012, 09:14
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LM
Actually there are two values of the sum possible.
Let one integer be \(x\) and another \(y=x-3\). Basically we are asked to find the value \(x+y=2x-3\)
Given: \(x^3-y^3=117\) --> \((x-y)(x^2+xy+y^2)=117\) --> substitute \(y\): \((x-x+3)(x^2+x^2-3x+x^2-6x+9)=117\) --> \(x^2-3x-10=0\) --> \(x=-2\) or \(x=5\) --> \(2x-3=-7\) or \(2x-3=7\). Only 7 is among the answer choices.
Answer: B.
