ETS gre guide

[anindita.dc]

Hi have few math question from the book OG 10th ED official guide. I am struggling to solve those problems. need help. My exam date is imminent. please response ASAP.

page 242 #12, page 331 #30, page 362 #13, ,page 366 #29


Thanks in Advance
Anindita

[jen]

Hi Anindita,

On page 242 #12:

An angle bisector cuts an angle in half EXACTLY. Therefore:

The other half of the angle at A is also equal to x.
The other half of the angle at B is also equal to 33.
The other half of the angle at C is also equal to y.

Therefore, the three angles of the big triangle are equal to:
2x
66
2y

2x + 66 + 2y = 180
2x + 2y = 114
x + y = 57

The answer is C.

[jen]

On page 331 #30:

This is a combinations problem I would summarize as "4 pick 2" then "5 pick 4." Also, order does not matter here (picking Motion A and Motion B is the same as picking Motion B and Motion A).

Therefore, the formula can be summed up as:

Everything!
---------------
Picked! Not picked!

In other words:

4!
---
2!2!

and

5!
---
4!1!

The first fraction simplifies to 6. The second simplifies to 5.

Multiply 6 x 5 = 30. (Because for every combination of motions, there are still 5 combos of accessories that could be matched with it.)

The answer is E.

[jen]

On page 362 #13:

We know a and b are negative, but we don't know if they are >-1 or <-1.

Since a negative times a negative is positive, we already know that the values in Column A and Column B will both be positive.

If a and b are < -1 (for instance, -2 and -3), then B will be larger (in this example, Column A = 6 and Column B = 36).

But if a and b are negative FRACTIONS (for instance, -1/2 and -1/3), A will be larger, since proper fractions get smaller when you square them (in this example, Column A = 1/6 and Column B = 1/36).

The answer is D.

When trying numbers, make sure to try the WEIRD ones (whichever weird ones the problem hasn't disallowed):
negatives
fractions
negative fractions
1
0
-1
very extreme numbers (such as 1 million and -1 million)

[jen]

On page 366 #29:

If (x+2) / (y-3) = 0, we know already that y can't be 3, since that would leave the fraction undefined (diving by zero is not allowed).

Now, we can simply multiply both sides by (y-3), which simply gives you x+2 = 0. Therefore, x = -2.

The answer is E.

Another way to look at it is that a fraction is equal to zero when the NUMERATOR is zero and the DENOMINATOR is ANYTHING BUT zero.

Good luck on your GRE!
Jen

[anindita.dc]

Thank you Jen for your help but I like to mention two things.

1) One related to my question . ETS book page 362 #13: I think in actual question a & b less than zero , so they are negative. but u describe as ab > 0: positive . but in my book in qs it mention as: 0>a>b....so how to solve that..

2) In CAT exam 2 ,verbal section #24. In that question I think answer choices are something ambiguous. Because in explanation which is explaining as option (a) :is not mention as in the question.

I am eagerly waiting to hear from you.

Thanks,
Anindita

[jen]

Hi Anindita,

You are correct about p362 #13. It certainly is tricky when they write 0 > a > b instead of what most of us would expect, which is a < b < 0. In any case, I have edited my original response above, so others can benefit as well.

I am going to post my reply to your other question in the verbal forum.

Jen

[midnightviking]

I know it's a bit late. I didn't even know there were forums on this site. But I found it easier to do this on problem #13 on Page 362:

Since 0 > a > b then a < 0 and b < 0. Therefore a*b > 0.

We have
Column A = a*b

and

Column B = (a*b)^2

And since we can divide by always positive number we can divide both sides by a*b. Which leaves us with this:

Column A= 1

Column B= (a*b)^2/(a*b) => (a*b)^(2-1) => a*b

And all we know is a*b > 0 so it can be less than, equal to or greater than 1. Making the choice D.

[tommywallach]

Hey Midnight,

Well done. Spotting those times that you CAN safely divide by variables is tough, but you're absolutely right that we can pull it off here.

-t