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645 Q83 V84 DI79
645 Q83 V84 DI79
GMAT 1: 650 Q46 V34
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WE:Engineering (Consulting)
Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3.
[#permalink]
Updated on: 28 Jan 2024, 02:33
Updated on: 28 Jan 2024, 02:33
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Let \(F = 36(2x - \frac{1}{2})^3\) and \(G = 16(3x - \frac{3}{4})^3\). Which of the following is true for all real numbers x?
(A) \(2F - 3G = 0\)
(B) \(3F - 2G = 0\)
(C) \(F^2 - G^2 = 0\)
(D) \(2F^2 - 3G^2 = 0\)
(E) \(3F^2 - 2G^2 = 0\)
(A) \(2F - 3G = 0\)
(B) \(3F - 2G = 0\)
(C) \(F^2 - G^2 = 0\)
(D) \(2F^2 - 3G^2 = 0\)
(E) \(3F^2 - 2G^2 = 0\)
Intern
Joined: 23 Feb 2022
Posts: 36
Given Kudos: 92
Location: New Zealand
Concentration: Strategy, General Management
Schools: Oxford Said'25 (A)
GMAT Focus 1:
645 Q83 V84 DI79
645 Q83 V84 DI79
GMAT 1: 650 Q46 V34
GPA: 8.5/9
WE:Engineering (Consulting)
Re: Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3.
[#permalink]
16 Oct 2023, 15:05
16 Oct 2023, 15:05
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Since x can be any real number, choosing a convenient value for x, let x be 0.
This simplifies F & G as follows:
\(F = 36(-\frac{1}{8}) = -4.5\)
\(G = 16(-\frac{27}{64}) = -6.75\)
Upon cycling through answer choices, \(3F - 2G = 0\) satisfies the solution.
(B)
I was wondering if this is the most optimal solution, or there are other ways to evaluate these types of inequality/algebra questions. Choosing a value for x seems a bit arbitary and prone to error as picking x = 1/4 may have made it trickier. I feel like I got lucky by picking x = 0 and it made things work out. Any help / inputs towards a more robust and efficient method of solving these types of problems will be greatly appreciated!
Update - Ok I just figured an Alternative solution:
Notice \(3x - \frac{3}{4} = \frac{3}{2}(2x - \frac{1}{2})\)
Given the answer choices are after a relationship between F & G, evaluating the ratio \(\frac{F}{G}\)
Let \((2x - \frac{1}{2}) = a\)
\(\frac{F}{G} = \frac{36*a^3 }{ 16*(3a/2)^3} = \frac{2}{3}\)
=> \(3F = 2G \)
=> \(3F - 2G = 0 \)
(B)
I think this might be the appropriate way to solve the problem
This simplifies F & G as follows:
\(F = 36(-\frac{1}{8}) = -4.5\)
\(G = 16(-\frac{27}{64}) = -6.75\)
Upon cycling through answer choices, \(3F - 2G = 0\) satisfies the solution.
(B)
I was wondering if this is the most optimal solution, or there are other ways to evaluate these types of inequality/algebra questions. Choosing a value for x seems a bit arbitary and prone to error as picking x = 1/4 may have made it trickier. I feel like I got lucky by picking x = 0 and it made things work out. Any help / inputs towards a more robust and efficient method of solving these types of problems will be greatly appreciated!
Update - Ok I just figured an Alternative solution:
Notice \(3x - \frac{3}{4} = \frac{3}{2}(2x - \frac{1}{2})\)
Given the answer choices are after a relationship between F & G, evaluating the ratio \(\frac{F}{G}\)
Let \((2x - \frac{1}{2}) = a\)
\(\frac{F}{G} = \frac{36*a^3 }{ 16*(3a/2)^3} = \frac{2}{3}\)
=> \(3F = 2G \)
=> \(3F - 2G = 0 \)
(B)
I think this might be the appropriate way to solve the problem
General Discussion
Quant Chat Moderator
Joined: 22 Dec 2016
Posts: 3130
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Location: India
Concentration: Strategy, Leadership
Re: Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3.
[#permalink]
24 Oct 2023, 22:59
24 Oct 2023, 22:59
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MrWhite wrote:
Let F = \(36(2x - \frac{1}{2})^3\) and G = \(16(3x - \frac{3}{4})^3\). Which of the following is true for all real numbers x?
(A) \(2F - 3G = 0\)
(B) \(3F - 2G = 0\)
(C) \(F^2 - G^2 = 0\)
(D) \(2F^2 - 3G^2 = 0\)
(E) \(3F^2 - 2G^2 = 0\)
(A) \(2F - 3G = 0\)
(B) \(3F - 2G = 0\)
(C) \(F^2 - G^2 = 0\)
(D) \(2F^2 - 3G^2 = 0\)
(E) \(3F^2 - 2G^2 = 0\)
We see that each of the options has a coefficient attached to the terms \(F\) and \(G\) and the difference between the terms is zero. Hence, the question asks us to find the coefficient that must be attached to \(F\) and \(G\) so that both the terms become equal.
\(k_1 * F = k_2 * G\)
\(k_1 * 36(2x - \frac{1}{2})^3 = k_2 * 16(3x - \frac{3}{4})^3\)
We can take a value of x such that one of the fractional parts becomes \(1\)
\(2x - \frac{1}{2} = 1\)
\(x = \frac{3}{4}\)
\(k_1 * 36((2*\frac{3}{4}) - \frac{1}{2})^3 = k_2 * 16((3*\frac{3}{4}) - \frac{3}{4})^3\)
\(k_1 * 36 = k_2 * 16*(\frac{27}{8})\)
\(\frac{k_1}{k_2} =\frac{16*27}{8* 36}\)
\(\frac{k_1}{k_2} =\frac{3}{2}\)
Hence, the ratio of the coefficients = 3:2
Option B
