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## Comment on

Properties of Fractions - Part I## In the 2nd link below the

## It comes down to how we

It comes down to how we divide fractions:

(a/b) ÷ (c/d) = (a/b) x (d/c)

So, 250/(1/3) = 250 ÷ (1/3) = 250 x (3/1) = 250 x 3

Likewise, 250/(1/2) = 250 ÷ (1/2) = 250 x (2/1) = 250 x 2

## Hello Brent, I have a doubt

https://www.youtube.com/watch?time_continue=3&v=pzoiS2v-L8E&ab_channel=MagooshGRE

I got this question in my powerprep practice test. I used one of the property suggested here which is : Increase the numerator and denominator with the same amount and the fraction approaches one.

Using that property I thought answer would be B which is B > A, because fraction in qty B is increased by same amount in the numerator and denominator and hence it is approaching one.

Why is this understanding incorrect? Because the answer here is D, as per video (by plugging in values to fraction). Please explain.

Also i'm giving my GRE tomorrow so could you please clarify this doubt today at any time please? I get a score of 150 in Verbal and 150 in maths. I could have done better in Maths, but I'm losing points because of my silly mistakes. Like I'm reading 2 dozens as one dozen sometimes which is very silly I know. Some 4 digit numbers addition mistakes. Such kind of silly mistakes I have been committing always. Any final tips please?

## In the video above, we state

In the video above, we state that, if we increase the numerator and denominator by the same amount then the fraction approaches 1. This is 100% true.

For example, if we take 1/2 and add 100 to top and bottom, we get 101/102, which is a lot closer to 1 than 1/2 is. So, 101/102 > 1/2

So, when we start with a fraction that's LESS THAN 1 and add the same value to the top and bottom, the resulting fraction is greater than the original fraction.

-------------------------------------

IMPORTANT: What if we start with a fraction that's GREATER THAN 1?

For example, if we take 3/2 and add 100 to top and bottom, we get 103/102. Here, 103/102 is a lot closer to 1 than 3/2 is. In this case, 103/102 < 3/2

So, when we start with a fraction that's GREATER THAN 1 and add the same value to the top and bottom, the resulting fraction is less than the original fraction.

-------------------------------------

IMPORTANT: What if we start with a fraction that's EQUAL to 1?

For example, if we take 1/1 and add 100 to top and bottom, we get 101/101. Here, both fractions equal 1.

As you can see, the rule holds true. HOWEVER, we need to pay close attention to whether the original fraction is less than, greater than or equal to 1.

Does that help?

Cheers,

Brent

## Thank you Brent.

## A googol is the number that

I do not understand this question. It says 1 followed by 100 zeroes which means 10^99 right? Sorry I get a bit confused here with the statement 1 followed by 100 zeroes.

## Let's look at some examples:

Let's look at some examples:

10^1 = 10 (1 followed by 1 zero)

10^2 = 100 (1 followed by 2 zeros)

10^3 = 1,000 (1 followed by 3 zeros)

10^4 = 10,000 (1 followed by 4 zeros)

10^5 = 100,000 (1 followed by 5 zeros)

So, 10^99 = 1 followed by 99 zeros

And 10^100 = 1 followed by 100 zeros

Does that help?

Cheers,

Brent

## Hello Brent,

What is the solution for final practice question?

Pls explain

## Question link: http://www

Question link: http://www.urch.com/forums/gre-math/157976-logical-reasoning.html

The correct answer for that question is A.

You can see my solution on the same thread.

Cheers,

Brent

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

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/gre-math-challenge-104-x-710.html

It's a common belief that, in order to compare two quantities, we must know the SPECIFIC values of both quantities, otherwise the answer must be D. This, however, is not true. All we need to do is determine whether one quantity is greater than the other quantity.

Consider this rudimentary example:

x is a POSITIVE integer, and y is a NEGATIVE integer.

QUANTITY A: x

QUANTITY B: y

Even though SPECIFIC values of x and y, we can be certain that Quantity A is greater than Quantity B.

The same concept applies to the linked question. Even though we don't know the specific value of x, we can still be certain that Quantity B is greater than Quantity A.

This means that, for ALL values of x that satisfy the condition that x > 0, Quantity B will be greater than Quantity A.

Does that help?

Cheers,

Brent

## Hello Brent!

Please can you explain a little more why you add 1075 in the following way :

= 1 + 0 + 7 + 5 + a bunch of zeros

Thank you

## I'm happy to help!

I'm happy to help!

ASIDE: In the future, please provide a link to the question you're referring to.

Question link: https://gre.myprepclub.com/forum/a-googol-is-the-number-that-is-written-...

In my solution, I found that G/8 + G/5 + G/4 + G/2 = (1075)(10^97)

10^97 = 10,000,000,000....etc until we have 97 0's in total.

So, (1075)(10^97) = 1075000000000000000000000....

We're asked to find the sum of all digits in the number 1075000000000000000000000....

So, the SUM = 1 + 0 + 7 + 5 + a bunch of zeros

= 13

Does that help?

Cheers,

Brent

## Thank you Brent! :)

## Not sure whether anyone will

## Great analogy!!

Great analogy!!