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Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3.

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MrWhite
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Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3. [#permalink] New post   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) \(3F^2 - 2G^2 = 0\)
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B

Originally posted by MrWhite on 16 Oct 2023, 14:57.
Last edited by Bunuel on 28 Jan 2024, 02:33, edited 2 times in total.
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Re: Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3. [#permalink] New post   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 :)
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Re: Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3. [#permalink] New post   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) \(3F^2 - 2G^2 = 0\)


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
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Re: Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3. [#permalink] New post   25 Oct 2023, 09:05
\(F(0)=-9/2\\
G(0)=-27/4\\
\\
3F=-27/2\\
2G=-27/2. \)

Hence B
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Re: Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3. [#permalink] New post   27 Oct 2023, 04:39
chetan2u JeffTargetTestPrep Could you please provide a solution to this problem?
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Re: Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3. [#permalink] New post   27 Oct 2023, 05:21
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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) \(3F^2 - 2G^2 = 0\)



Two ways I could think of immediately

1. Substitute value for x
Let x=0
F = \(36(2x - \frac{1}{2})^3\) = \(36(- \frac{1}{2})^3=\frac{36}{-8}=\frac{-9}{2}\)
G = \(16(3x - \frac{3}{4})^3\) = \(16(- \frac{3}{4})^3=\frac{16*-27}{64}=\frac{-27}{4}=\frac{3*-9}{2*2}=\frac{3F}{2}\)
Or 2G=3F

2. Solve Algebraically

F = \(36(2x - \frac{1}{2})^3\)
Take out 2 from bracket => \(36*2^3(x-\frac{1}{4})^3 \)
Let \((x-\frac{1}{4})^3=a\), so F=36*8a…..(i)

G = \(16(3x - \frac{3}{4})^3\)
Take out 3 from bracket => \(16*3^3(x- \frac{1}{4})^3=16*27a\)……(ii)

Divide i by ii
\(\frac{F}{G}=\frac{36*8a}{16*27a}=\frac{2}{3}\)
3F=2G


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Re: Let F = 36(2x - 1/2)^3 and G = 16(3x - 3/4)^3. [#permalink]
27 Oct 2023, 05:21
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