Need Help in DS problems -3

[phoenix9801]

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

6- Is x Between 0 and 1 ?

(1) x^2 is less then X

(2) X^3 is positive


7- Is p^2 an odd integer ?

(1) P is an odd integer

(2) square root P an odd integer

[Rich@VeritasPrep]

Anytime you run into a problem that involves possible fractions and/or possible negative numbers, you want to consider four key ranges:

1. x < -1
2. -1 < x < 0
3. 0 < x < 1
4. x > 1

Let's see how that works for this question:

Is x Between 0 and 1 ?

(1) x^2 is less then X

x^2 is always either zero or positive. It can't be zero, because then x^2 wouldn't be less than X. So we know x^2 is positive.

If 0 < x < 1 (i.e. if x is a proper fraction), then x^2 is less than x. For example (1/2)^2 = 1/4 < 1/2.

If x > 1, then x^2 is greater than x. For example, 2^2 = 4 > 2.

So x MUST be between 0 and 1. SUFFICIENT

(2) X^3 is positive

If a number is cubed, and the result is positive, the number itself must be positive. So if x^3 is positive, then x must be positive.

You can show this by picking numbers in either of our positive ranges:

For 0 < x < 1, pick x = 1/2. (1/2)^3 = 1/8, which is positive
For x > 1, pick x = 2. 2^3 = 8, which is positive.

Because both of these ranges work, you can't say for sure whether x is between 0 and 1. INSUFFICIENT

[Rich@VeritasPrep]

7- Is p^2 an odd integer ?

(1) P is an odd integer

Pick any even number, and you'll find that you always get an even number if you square it. For example, 2^2 = 4, 4^2 = 16, etc etc. The only way p^2 can be odd is if p itself is odd. For example, 3^2 = 9, 5^2 = 25, etc etc.

SUFFICIENT

(2) square root P an odd integer

Same thing. If the square root of P is an odd integer, then P itself must be an odd integer. For example, sqrt(9)=3, sqrt(25)=5, etc etc.

And if p is an odd integer, then p^2 must be odd as well. SUFFICIENT.