Need Help in DS problems

[phoenix9801]

Would you please use Picking Numbers and/or Straightforward Math to solve these questions Please be simple. (not Algebra). Thanks.


1- If M is an integer such that (-2)^2m = 2^9-m, then M= ?

(Please note for this problem I do not understand how (-2)^2 is the same as 2^2m )


4- Is x < y ?

(1) Z < Y

(2) Z < X


5- If n is positive, is square root n > 100 ?

(1) square root n-1 > 99

(2) Square n+1 > 101

[Rich@VeritasPrep]

1- If M is an integer such that (-2)^2m = 2^9-m, then M= ?

(-2)^2m is actually equal to 2^2m. This is because 2 and -2 have the same absolute value, and when you take two numbers that have the same absolute value and you raise them to an even power, you get the same result.

For example:

(-2)^4 = 16 = 2^4

(-3)^2 = 9 = 3^2

So change the equation to:

2^2m = 2^9-m

Since the bases are equal, we can set the exponents equal to each other:

2m = 9 - m

3m = 9

m = 3

[Rich@VeritasPrep]

4- Is x < y ?

(1) Z < Y

Don't know anything about x. INSUFFICIENT

(2) Z < X

Don't know anything about y. INSUFFICIENT

Together, we still don't know anything about the values of x and y. We just know that some value z is smaller than both of them.

You can also show this by picking numbers. Try z=1, x=2, y=3 and z=1, x=3, y=2. INSUFFICIENT

[Rich@VeritasPrep]

5- If n is positive, is square root n > 100 ?

First, simplify the prompt. If sqrt(n) > 100, then n > 10000.

So the prompt really asks if n > 10000.

(1) square root n-1 > 99

Translation: n-1 > (99)^2, which in turn means that n > (99)^2 + 1.

(99)^2 + 1 is less than 10000, so n could be anything between the two values (and thus less than 10000) or anything greater than 10000. INSUFFICIENT.

(2) Square n+1 > 101

Translation: n+1 > (101)^2, which in turn means that n > (101)^2 - 1

(101)^2 - 1 is greater than 10000, so you have this inequality:

n > (101)^2 - 1 > 10000. No matter what, n will always be greater than 10000. SUFFICIENT.