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06 Apr 2020, 04:22
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D
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55% (02:36) correct
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The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is
A. 95
B. 85
C. 58
D. 50
E. 40
Are You Up For the Challenge: 700 Level Questions
A. 95
B. 85
C. 58
D. 50
E. 40
Are You Up For the Challenge: 700 Level Questions
06 Apr 2020, 05:15
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Bunuel
The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is
A. 95
B. 85
C. 58
D. 50
E. 40
[textarea]
A. 95
B. 85
C. 58
D. 50
E. 40
[textarea]
a*b = 616
616 = 1*616
2*308
4*154
7*88
8*77
11*56
14*44
22*28
\(\frac{a^{3} - b^{3} }{ (a-b)^{3}} = \frac{157}{3}\)
It seems possible only with 22*28 because only these are so close that their cubes will be close to cube of their difference
Also, the difference of these numbers is a multiple of 3 which is necessary to meet the condition of denominator
11304/216 = 157/3
i.e. a and b are 22 and 28
i.e. Sum = 22+28 = 50
Answer: Option D
06 Apr 2020, 05:27
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(a^3-b^3)/(a-b)^3=157/3
(a-b)(a^2+ab+b^2)/(a-b)^3=157/3
(a^2+ab+b^2)/(a-b)^2=157/3
3(a^2+b^2)+3ab=157(a^2+b^2) -314ab
317ab =154(a^2+b^2)
Now ab =616
Therefore
a^2 +b^2 =1268
Therefore
a=28 b=22
Hence a+b=50
Option D is the answer
Posted from my mobile device
(a-b)(a^2+ab+b^2)/(a-b)^3=157/3
(a^2+ab+b^2)/(a-b)^2=157/3
3(a^2+b^2)+3ab=157(a^2+b^2) -314ab
317ab =154(a^2+b^2)
Now ab =616
Therefore
a^2 +b^2 =1268
Therefore
a=28 b=22
Hence a+b=50
Option D is the answer
Posted from my mobile device
