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Re: (x + y)(x − y) = 0 xy ≠ 0

sandy wrote:

\((x + y)(x – y) = 0\)
\(xy \neq 0\)

Quantity A

Quantity B

\(\sqrt[6]{\frac{19}{2x^2}}\)

\(\sqrt{\frac{342}{y^2}}\)

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given.

\((x + y)(x – y) = 0\) means either x=y or x=-y or both
since x and y are in the denominator of the fractions, value of 0 for any of them would make the fraction undefined and therefore it is given \(xy \neq 0\)

I am sure (\(\sqrt[6]{\frac{19}{2x^2}}\)) is meant to be (\(6*\sqrt{\frac{19}{2x^2}}\))

x=y or x=-y or both IMPLIES \(x^2=y^2\)

\(A=\sqrt[6]{\frac{19}{2x^2}}=\sqrt{\frac{36*19}{2x^2}}=\sqrt{\frac{18*19}{y^2}}=\sqrt{\frac{342}{y^2}}=B\)
so equal

C

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Re: If A < 0, 10 < B < 30, and 50 < C < 80, what is the relative

Carcass wrote:

If \(A

A. \(\frac{1}{A} < \frac{1}{B} < \frac{1}{C}\)

B. \(\frac{1}{A} < \frac{1}{C} < \frac{1}{B}\)

C. \(\frac{1}{C} < \frac{1}{A} < \frac{1}{B}\)

D. \(\frac{1}{C} < \frac{1}{B} < \frac{1}{A}\)

E. \(\frac{1}{B} < \frac{1}{C}< \frac{1}{A}\)

remember the rule
1) if A and B are positive and greater than 1..
A>B means \(\frac{1}{A} < \frac{1}{B}\)

So 1/C < 1/B as C>B
Now A is negative and it’s reciprocal will also be negative and thus 1/A will be the least.
B. \(\frac{1}{A} < \frac{1}{C} < \frac{1}{B}\)

B

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Re: The area of △ABD The area of △BCD

Carcass wrote:

Attachment:

GRE – powerprep The area of △ABD The area of △BCD.jpg

Quantity A

Quantity B

The area of \(△ABD\)

The area of \(△BCD\)

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Explanation:

For the \(△ABD\) and \(△BCD\))

AB = BC , i.e. the base of both \(\triangle\) are equal.

Since both \(\triangle\) lies within the \(\triangle ADC,\) therefore the height must be same

Hence the area of \(\triangle ABD\)= area of \(\triangle BCD\)
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Re: If 2x − 3y = 6, then 6y − 4x =

Carcass wrote:

If 2x − 3y = 6, then 6y − 4x =

A. − 12

B. − 6

C. 6

D. 12

E. Cannot be determined

GIVEN: 2x − 3y = 6

We want to determine the value of 6y − 4x

6y − 4x = -4x + 6y
= -2(2x – 3y)
= -2(6)
= -12

Answer: A

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
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Re: Ray is 2 inches taller than Lin, and Ray is 3 inches taller

Carcass wrote:

Ray is 2 inches taller than Lin, and Ray is 3 inches taller than Sam.

Quantity A

Quantity B

The average (arithmetic mean) height of Ray, Lin, and Sam

The median height of Ray, Lin, and Sam

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Let’s write all heights in terms of Ray’s height.

Let R = Ray’s height (in inches)
So, R – 2 = Lin’s height
And R – 3 = Sam’s height

Average height = [R + (R-2) + (R-3)]/3
= [3R – 5]/3
= 3R/3 – 5/3
= R – 5/3

To find the median height, arrange heights in ASCENDING ORDER: R-3, R-2, R
So, the median height = R-2

So, we get:
QUANTITY A: R – 5/3
QUANTITY B: R – 2

Subtract R from both quantities to get:
QUANTITY A: -5/3
QUANTITY B: -2

Since -5/3 is greater than -2, the correct answer is A

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Re: An empty, cube-shaped swimming pool is filled part way with

sandy wrote:

An empty, cube-shaped swimming pool is filled part way with x cubic feet of water. It is then filled the rest of the way with y cubic feet of chlorine. Which of the following, in feet, expresses the depth of the swimming pool?

A. \(x + y\)
B. \(\frac{x + y}{3}\)
C. \(\sqrt{x + y}\)
D. \((x + y)^3\)
E. \(\frac{\sqrt[3]{x+y}}{3}\)

Let the sides and also the height be a each, therefore \(V=a^3\)
Volume is also equal to x+y, therefore \(V=a^3=x+y……a=\sqrt[3]{x+y}\)

C is wrongly written, it should be 3rd root.

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Maths

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Re: c and d are positive integers.

Carcass wrote:

\(c\) and \(d\) are positive integers.

Quantity A

Quantity B

\(\frac{c}{d}\)

\(\frac{c+3}{d+3}\)

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

Explanation

We can solve this question using matching operations
Given:
Quantity A: c/d
Quantity B: (c+3)/(d+3)

Since d is POSITIVE, we can safely multiply both quantities by d to get:
Quantity A: c
Quantity B: d(c+3)/(d+3)

Since d is positive, we know that d+3 is POSITIVE. So, can safely multiply both quantities by d+3 to get:
Quantity A: c(d+3)
Quantity B: d(c+3)

Expand both quantities:
Quantity A: cd + 3c
Quantity B: cd + 3d

Subtract cd from both quantities:
Quantity A: 3c
Quantity B: 3d

Divide both quantities by 3:
Quantity A: c
Quantity B: d

Since we aren’t told anything about the relationship between c and d, the correct answer is D

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c and d are positive integers.

\(c\) and \(d\) are positive integers.

Quantity A

Quantity B

\(\frac{c}{d}\)

\(\frac{c+3}{d+3}\)

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.

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Quant

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Re: X or Y

Explanation

According to the current diagram, y° is larger than x°. But remember, the figures are not always drawn to scale and you are free to make changes provided you keep the information given as a constant. In this case, the constant is that PS = SR.

So keeping line PR the same, where PS = SR, move Q around to see what happens to x and y:

Image

As you can see, it’s impossible to determine which angle is greater without having more information about point Q.

Hence option D is correct.

Attachments

di new.jpg
di new.jpg [ 18.25 KiB | Viewed 7845 times ]

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Re: Calculate cube root

sandy wrote:

\(\sqrt[3]{8 \times 27 \times 64}\)

Useful rule: ∛(xyz) = (∛x)(∛y)(∛z)

So, \(\sqrt[3]{8 \times 27 \times 64}\) = (∛8)(∛27)(∛64)
= (2)(3)(4)
= 24

Answer: 24

Cheers,
Brent
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