\(t>1\)

Quantity A	Quantity B
\(\frac{2t+5}{3}\)	\(3t\)

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

I solved as picking up some values(decimal and whole numbers) for t which was higher than 1. And I got an answer as D. But solution shows something like this: First, multiply both sides by 3 to clear the fraction. Multipying by a positive number does not affect the comparison:
2t + 5 9t
Now, subtract 2t from both columns (again, not affecting the comparison):
5 7t
Because t > 1, 7t is greater than 5. Therefore, Quantity B is greater.

Why is this way is the right thing to do?

t>1
Quantity A is [2t + 5]/[3]
Quantity B is 3t.

Set them equal to each other:
[2t + 5]/[3] =?= 3t
2t + 5 =?= 9t
5 =?= 7t

So when t = 5/7, they're equal, BUT t is > 1, therefore B is greater.

It's the right thing to do to set them "equal" to each other as a hypothetical equality because that's what you have to do to figure out what the solution is for all values of t. But the caveat that t >1 changes the answer and is an important detail to see.

Is it safe bet to choose some values for t and see which one is coming higher for frequent?

Please post a QCQ under the right sub-forum. Ask the GRE expert is for specific questions to ask to the expert such as how can I improve my strategy

Also follow the simple rules for posting and formatting a question

https://gre.myprepclub.com/forum/how-to-po ... 12752.html

The answer id D

If you consider 2 , an integer, Q\ becomes 3 and Q2 is 6

if you take 3/2=1.5

Q1 is 2.6 and Q2 is 4.5

B is the answer

Multiplying both sides by \(3\), we get

\(2t + 5 = 9t\)

Now, we only have to test for \(t=1\) and utmost \(2\), to validate that \(9t\) will always be greater than \(2t + 5\)

Therefore B is the correct answer.

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