Lesson: Properties of Fractions - Part I

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

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.
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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.
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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

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

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

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

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

Great analogy!!

A few months ago, ETS (the people who make the GRE) made GRE Prep Club remove all official GRE questions from its site.
So, unfortunately there are now many dead links that used to go to official GRE questions.