Question from Official GMAPTPrep CD

[Venu]

Question 1.
If x is positive, which of the following could be correct ordering of x2,1/x,2x
a) x2 < 2x < 1/x
b) x2 < 1/x < 2x
c) 2x < x2 < 1/x

i) None
ii) a
iii) c
iv) a and b
v) a,b,c

I think that the answer should be (ii) but according to GMAPTPrep CD it is (iv)

Question 2. Data Sufficiency

If z is the median of any 3 positive integers x, y and z then

i) x<y+z
ii) y=z

a) i only is sufficient and ii is not
b) ii only is sufficient and i is not
c) i and ii together are sifficient
d) Both
e) none

Any thoughts ?

[StaceyKoprince]

Hi, Venu

Two things. For your first question, is "x2" supposed to mean x-squared? If so, please write it as x^2 and, if you have to write it with other things (as above), then put it in parentheses: (x^2).

Please also limit yourself to one question per post. I'll answer your first question here after you let me know if that's supposed to read x^2. Please start a separate thread and copy and paste your second question into the new thread. Thanks!

[Venu]

Hi Stacey,

Sorry for the confusion. Yes it is supposed to mean x^2.


If x is positive, which of the following could be correct ordering of x^2 , 1/x , 2x
a) (x^2) < 2x < 1/x
b) (x^2) < 1/x < 2x
c) 2x < (x^2) < 1/x

i) None
ii) a
iii) c
iv) a and b
v) a,b,c

[shaji]

The correct answer is indeed iv.
The ordering is ascending and from a,b,c, & stem it is evident that 0<x<1. c can not be the order since x^2<2x.

Question 1.
If x is positive, which of the following could be correct ordering of x2,1/x,2x
a) x2 < 2x < 1/x
b) x2 < 1/x < 2x
c) 2x < x2 < 1/x

i) None
ii) a
iii) c
iv) a and b
v) a,b,c

I think that the answer should be (ii) but according to GMAPTPrep CD it is (iv)

Question 2. Data Sufficiency

If z is the median of any 3 positive integers x, y and z then

i) x<y+z
ii) y=z

a) i only is sufficient and ii is not
b) ii only is sufficient and i is not
c) i and ii together are sifficient
d) Both
e) none

Any thoughts ?

[StaceyKoprince]

The easiest way to approach this is to try some numbers. We're told x is positive, so I know I can't try negative numbers or zero. I also know (or at least I should :) that I should try both integers and fractions between zero and one. This is simply because these two groups of numbers have some different properties (so know that you should try those).

The most obvious thing is to say that 1/x is the smallest, but none of them start with 1/x... so I'm suspecting a trick. That means I want to try a fraction between zero and one, because I know those are "tricky."

if x = 1/2, then:
x^2 = 1/4
1/x = 2
2x = 1
so the order would be x^2 < 2x < 1/x. So (a) is possible. Eliminate i and iii. (Looks like you got this far)

if x = 3/4, then:
x^2 = 9/16
1/x = 4/3
2x = 3/2
so the order would be x^2 < 1/x < 2x. So (b) is possible. Eliminate ii. (Looks like this is where you had trouble.)

Don't forget to try more than one number - 1/2 is the obvious starting number, but you've got to make sure that other fractions wouldn't do something different...

And (c) isn't going to work b/c the only way to make x^2 smaller than 1/x is for x^2 to be between zero and one. And if that's true, x^2 will be smaller than 2x, because x gets smaller when you square it but larger when you multiply it by 2.

[Venu]

got it Stacey. I has tried only 1/2 ..my bad.


Thanks a lot

[StaceyKoprince]

no problem! :)

[vietst]

Question 1.
If x is positive, which of the following could be correct ordering of x2,1/x,2x
a) x^2 < 2x < 1/x
b) x^2 < 1/x < 2x
c) 2x < x^2 < 1/x

I only help to rule out c.
2x < x^2, then x > 2 or x < 0 (1)
x^2 < 1/x then x^3 < 1 (x>0) then x < 1. together 0<x< 1 (2)
No numbers satisfy both

Question 2. Data Sufficiency

If z is the median of any 3 positive integers x, y and z then

i) x<y+z
ii) y=z

1. x < y + z
x =1, y = 2, z = 3. No
x =1, y=3, z = 2. Yes
Insuff
2. y = z. suff
B

[RonPurewal]

Question 1.
If x is positive, which of the following could be correct ordering of x2,1/x,2x
a) x^2 < 2x < 1/x
b) x^2 < 1/x < 2x
c) 2x < x^2 < 1/x

you can also approach this question by just looking at one of the inequalities at a time (as opposed to both of them in each choice). for instance, given x^2 < 2x < 1/x, you can just think about the x^2 < 2x part, or you can just think about the 2x < 1/x part.

(note: the following solutions do not list numbers greater than 0, because those numbers are irrelevant)
x^2 versus 2x: x^2 is greater if x<0 (irrelevant) or x>2; 2x is greater if 0<x<2
x^2 versus 1/x: x^2 is greater if x>1; 1/x is greater if x<1
2x versus 1/x: 2x is greater if x > root(2)/2; 1/x is greater if x < root(2)/2

now, look at the choices:
(a) left-hand inequality means 0<x<2, right-hand inequality means x < root(2)/2
this is possible, if 0<x<root(2)/2
if you try plugging in x = 1/2, it works (becomes 1/4 < 1 < 2)

(b) left-hand inequality means x<1, right-hand inequality means x > root(2)/2
this is possible, if root(2)/2 < x < 1
try plugging in x = 3/4 --> gives 9/16 < 4/3 < 3/2. this works.

(c) left-hand inequality means x>2, right-hand inequality means x<1
impossible to have both!

so... only (a) and (b) are possible


i'll put the second question into another thread.