If you're enjoying the course, please spread the word on Twitter.

- Video Course
- Video Course Overview
- General GRE Info and Strategies - 7 videos (free)
- Quantitative Comparison - 7 videos (free)
- Arithmetic - 42 videos
- Powers and Roots - 43 videos
- Algebra and Equation Solving - 78 videos
- Word Problems - 54 videos
- Geometry - 48 videos
- Integer Properties - 34 videos
- Statistics - 28 videos
- Counting - 27 videos
- Probability - 25 videos
- Data Interpretation - 24 videos
- Analytical Writing - 9 videos (free)
- Sentence Equivalence - 39 videos (free)
- Text Completion - 51 videos
- Reading Comprehension - 16 videos

- Study Guide
- Philosophy
- Office Hours
- Extras
- Prices

## Comment on

Properties of Fractions - Part II## Dear Brent,

in the fifth problem I am having troubles with the expression's simplification. More specifically, where does that 9 come from? Which property of fractions has been used?

One approach is to simplify the expression.

(27x + 23y)/(3x + 2y) = (27x + 18y + 5y)/(3x + 2y)

= (27x + 18y)/(3x + 2y) + (5y)/(3x + 2y)

= 9 + (5y)/(3x + 2y)

Thank you very much!

## Question link: http:/

Question link: http://gre.myprepclub.com/forum/if-0-y-x-then-which-of-the-following-320...

I'm using the RULE that says: (a + b)/c = a/c + b/c

So, in my solution, I first rewrote 23y as (18y + 5y)

This allowed me to take 27x + 18y + 5y and add brackets as follows to get the equivalent form: (27x + 18y) + 5y

So, we get: (27x + 23y)/(3x + 2y) = [27x + 18y + 5y]/(3x + 2y)

= [(27x + 18y) + 5y]/(3x + 2y)

= (27x + 18y)/(3x + 2y) + 5y/(3x + 2y) {I applied the RULE at the top}

= 9 + [5y/(3x + 2y)]

How did I get 9?

Well 27x + 18y = 9(3x + 2y)

So, (27x + 18y)/(3x + 2y) = (9)(3x + 2y)/(3x + 2y)

= 9 {since (3x + 2y)/(3x + 2y) = 1}

Does that help?

## Thanks for explaining. I too

## Now let's take a closer look

Notice that (5y)/(3y + 2y) = 5y/5y = 1 [since the numerator and denominator are EQUAL]

However, since we're told that y < x, we know that 3y + 2y < 3y + 2x

This means that (5y)/(3x + 2y) < 1, [since the numerator is LESS THAN the denominator]

If (5y)/(3x + 2y) < 1, then we can conclude that 9 + (5y)/(3x + 2y) < 10

I did not understand this part of the solution. Can you please explain?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-0-y-x-then-which-of-the-following-32...

If (5y)/(3x + 2y) < 1, then we can add 9 to both sides of the inequality to get:

(5y)/(3x + 2y) + 9 < 10

Does that help?

Cheers,

Brent

## I just assumed x to be 3 and

## x = 3 and y = 2 is ONE pair

x = 3 and y = 2 is ONE pair of values that satisfy the given restriction that 0 < y < x

However, there are infinitely many possible pairs of values for x and y that satisfy the given restriction.

## On the second reinforcement

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-m-is-the-product-of-all-the-integers...

n = (4)(5).....(10)(11)

m = (2)(3)(4)(5).....(10)(11)

Notice that every value in m is also in n

So, all of m's values will pair with some of n's values to get many 1's

That is: 4/4 = 1, 5/5 = 1, 6/6 = 1, . . . 10/10 = 1 and 11/11 = 1

All that remains, is (2)(3) in the denominator.

Another approach is to first notice that m = (2)(3)(n)

So, n/m = n/(2)(3)(n) = 1/(2)(3) = 1/6

Does that help?

Cheers,

Brent

## Hi there,

I am struggling with the second problem in the practice list.

When using the property (x+y)/z=x/z+y/z

If b/(a+b) = 7/12, then we can also say that (a + b)/b = 12/7

What I don't understand is why we would add 1 next.

the explanation suggest:

Simplify: a/b + 1 = 12/7

Subtract 1 from both sides to get: a/b = 12/7 - 1

Rewrite 1 as follows: a/b = 12/7 - 7/7

Evaluate: a/b = 5/7

Please help break this process down for me! I appreciate your help!

Thank you in advance

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-b-a-b-8385.html

We know that: (a + b)/b = 12/7

Applying the fraction property, we get: a/b + b/b = 12/7

Since b/b = 1, we can write: a/b + 1 = 12/7

Does that help?

Cheers,

Brent

## understood thank you!

## https://gre.myprepclub.com/forum

Hi Brent, I've gone over this problem a few times, but I seem to not understand it. Could you please explain this one again.

Also, for comparison questions, are there certain criteria of a question that would be able to tell me that plugging in numbers could solve the question easily?

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/compare-for-x-2155.html

I posted a 3rd solution here: https://gre.myprepclub.com/forum/compare-for-x-2155.html#p28345

Please let me know if that helps.

Plugging in values has its drawbacks. The biggest problem is that, unless we're able to show conflicting outcomes (meaning the correct answer is D), we can never be 100% certain of the correct answer.

That said, I'd say that plugging in numbers is a great approach when you're not sure how to proceed with a question.

Cheers,

Brent

## Thank you so much. That

## https://gre.myprepclub.com/forum

Can you give me the answer break down for this problem?

## Here's my full solution:

Here's my full solution: https://gre.myprepclub.com/forum/qotd-21-if-bc-0-and-3b-2c-18-then-which...

Cheers,

Brent

## Hey Brent, quick question on

For A of your answer, wouldn't dividing by 2 yield

y>x (which is still a true statement)

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-0-x-y-then-which-of-the-following-mu...

You're absolutely right. Good catch - thanks!

I have edited my solution accordingly.

Cheers,

Brent

## https://gre.myprepclub.com/forum

How come we are able to conclude that 3x+2y > 3y+2y? And therefore, the value of 3x+2y will be less than 1?

Thanks Bretnt for the help!

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/if-0-y-x-then-which-of-the-following-32...

Think of it this way:

We're told that: y < x

Multiply both sides by 3 to get: 3y < 3x

Add 2y to both sides to get: 3y + 2y < 3x + 2y

Does that help?

Cheers,

Brent