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

Question

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 https://gmatclub.com/forum/download/file.php?id=81455 GMAT-Club-Forum-yacdafnc.png

MrWhite · Accepted Answer

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

chetan2u · Answer

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

B

gmatophobia · Answer

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

Oppenheimer1945 · Answer

F(0)=-9/2
G(0)=-27/4

3F=-27/2
2G=-27/2.

Hence B

anrocha · Answer

Is is just for me that the display of this question looks wrong?
Looks like a^b^c instead of a(b)^c...

Exam 3 question 16 at Official Exam Mock

Bunuel · Answer

Due to the fractions, (2x - 1/2) and (3x - 3/4) might appear slightly higher than 36 and 16. However, they are not exponents. It looks clearer on our site.

JatinV2 · Answer

Yeah, I got the same question. If you solve inside the bracket for both, (4x-1)^3 can be isolated in both. Solving you get 2F/9=4G/27, 
3F=2G, 
3F-2G=0,
i.e B

KarishmaB · Answer

Look at the options. They are of the form 2F = 3G or 3F = 2G etc. This means that we can equate the expressions and when we do, all terms will match up for the correct option. Let's compare the x^3 terms of F and G since they are the easiest. F = 36 * 2^3*x^3 + ... = 2^5 * 3^2 * x^3 + ... G = 16 * 3^3 *x^3 + ... = 2^4*3^3*x^3+... Comparing the coefficients of their x^3 terms, we see that to equate, we need 3*F = 2*G because then we will get 3F = 3 * 2^5 * 3^2 * x^3 + ... = 2^5*3^3*x^3 +.... and 2G = 2* 2^4*3^3*x^3+...= 2^5*3^3*x^3 +.... which makes the coefficient of x^3 same for both Hence answer (B) Note that if we square F and G, more factors will be required to equate their x^3 terms so 3F^2 = 2G^2 is not possible. GMAT-Club-Forum-sfnuf7mo.png

MarcusGMATBeat · Answer

I think this approach will be much faster instead of plugging number, and easier from my perspective. Just simplify the equation first, and clearly we can tell the options with squaring variable will be eliminated. Then, we test option A & B, can reach for the correct answer with the minimum calculation.

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hemanthPA · Answer

Given X is true for all real numbers, consider x =0 F = 36 x (-1/2)^3 = -9/2 G = 16 x (-3/4)^3 = -27/4 In what cases does this lead to a zero? 3(-9/2) – 2(-27/4) = -27/2 – (-27/2) = 0 So, 3F – 2G = 0 GMAT-Club-Forum-6eeptwgq.png

ameya.satyawadi · Answer

If we go through the two expressions closely they can be deduced through linear equation by taking 8 and 27 common from the brackets respectively
F/G=36(2x−1/2)^3/16(3x-3/4)^3
F/G=36(2(x−1/4))^3/16(3(x-1/4))^3
Since we are taking out 2 and 3 from cubic expression we will multiply it accordingly by taking out from the brackets
=>F/G=36*8(x-1/4)^3/16*27(x-1/4)^3
upon further calculations, we see
=>F/G=2/3
3F-2G=0
Option (B)

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

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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Perfect 805 Score Debrief by Julia

My Rewards