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28 Jan 2022, 08:15
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
62% (02:55) correct
38%
(02:43)
wrong
based on 130
sessions
History
Date
Time
Result
Not Attempted Yet
Given, \(\frac{x^3 + 12x}{6x^2 + 8} = \frac{y^3 + 27y}{9y^2 + 27}\). What is the value of x/y?
(A) \(\frac{3}{4}\)
(B) \(\frac{6}{5}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{3}{2}\)
(E) \(\frac{5}{6}\)
Are You Up For the Challenge: 700 Level Questions
(A) \(\frac{3}{4}\)
(B) \(\frac{6}{5}\)
(C) \(\frac{2}{3}\)
(D) \(\frac{3}{2}\)
(E) \(\frac{5}{6}\)
Are You Up For the Challenge: 700 Level Questions
28 Jan 2022, 11:03
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Bunuel
What is componendo-dividendo?
If a/b = c/d, then, we have: (a+b)/(a-b) = (c+d)/(c-d)
This is the idea of componendo-dividendo (it is easy to prove how).
To validate: 6/4 = 3/2 => (6+4)/(6-4) = (3+2)/(3-2)
Applying componendo-dividendo in this question:
\(\frac{x^3 + 12x}{6x^2 + 8} = \frac{y^3 + 27y}{9y^2 + 27}\)
=> \(\frac{x^3 + 12x + 6x^2 + 8}{x^3 + 12x - 6x^2 - 8} = \frac{y^3 + 27y + 9y^2 + 27}{y^3 + 27y - 9y^2 - 27}\)
=> \(\frac{x^3 + 6x^2 + 12x + 8}{x^3 - 6x^2 + 12x - 8} = \frac{y^3 + 9y^2 + 27y + 27}{y^3 - 9y^2 + 27y - 27}\)
=> \(\frac{(x + 2)^3}{(x - 2)^3} = \frac{(y + 3)^3}{(y - 3)^3}\)
Taking cube-root:
=> \(\frac{(x + 2)}{(x - 2)} = \frac{(y + 3)}{(y - 3)}\)
Applying componendo-dividendo again:
\(\frac{(x + 2 + x - 2)}{(x + 2 - x + 2)} = \frac{(y + 3 + y - 3)}{(y + 3 - y + 3)}\)
=> \(\frac{2x}{4} = \frac{2y}{6}\)
=> \(\frac{x}{2} = \frac{y}{3}\)
=> \(\frac{x}{y} = \frac{2}{3}\)
Answer C
The better and FASTER alternative:
Use the options:
Option C: x/y = 2/3 => Let x = 2kand y = 3k
Substitute in the given equation and verify if it holds true.
Similarly, try the other options.
Option C will hold true, the others won't.
General Discussion
28 Jan 2022, 08:47
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Instead of solving the eqution I tried putting values from options.
For option c if we put x=2 and y=3, the two sides of the eqation will be equal.
So, IMO option C is correct ans.
Posted from my mobile device
For option c if we put x=2 and y=3, the two sides of the eqation will be equal.
So, IMO option C is correct ans.
Posted from my mobile device
