2 DS problems that kick my butt

gmatleyFool · Posted: Tue Aug 14, 2007 2:17 pm Post subject: 2 DS problems that kick my butt

Problem #1

Is x^4 + y^4 > z^4?

(1) x^2 + y^2 > z^2
(2) x + y > z

Problem #2

If ax + b = 0, is x > 0?

(1) a + b > 0
(2) a - b > 0

Can anybody offer insight on how to tackle this type of problem??

beny · Posted: Tue Aug 14, 2007 5:49 pm Post subject:

1. a+b > 0

a = -1, b = 2, x = 2
a = 2, b = 1, x = -.5
NOT SUFFICIENT

2. a-b > 0

a = 1, b = -2, x = 2
a = -1, b = -2, x = -2
NOT SUFFICIENT

Together:
a = 6, b = -3, x = .5
a = 6, b = 3, x = -.5
NOT SUFFICIENT.

Ergo, E.

beny · Posted: Tue Aug 14, 2007 5:54 pm Post subject:

I believe the answer to the other question is A.

I didn't work out any math, but it seems pretty intuitive.

ratindasgupta · Posted: Wed Aug 15, 2007 5:34 am Post subject:

givemeanid · Posted: Wed Aug 15, 2007 7:16 am Post subject:

Is x^4 + y^4 > z^4?

(1) x^2 + y^2 > z^2
(2) x + y > z

1. x^2 + y^2 > z^2
Let x^2 = a, y^2 = b and z^2 = c
Since they are squares, a, b and c are all > 0.
So, we are given a + b > c and the question is whether a^2 + b^2 > c^2

If a = 3, b = 4, c = 6; a+b = 3+4 = 7 > c; a^2+b^2 = 9+16 < 36 (c^2)
If a = 1/4, b = 1/4, c = 1/3; a+b = 1/4+1/4 = 1/2 > 1/3 (c); a^2+b^2 = 1/16+1/16 = 1/8 > 1/9
INSUFFICIENT.

2. x+y > z
Applying same argument as above (although, in this case, x,y and z can be -ve), this is also INSUFFICIENT.

Combining both:
If x = 3, y = 4, z = 4; x+y = 3+4 = 7 > z; x^2+y^2 = 9+16 > 16 (z^2); x^4+y^4 = 81+256 > 256 (z^4)
If x = 1/2, y = 3/4, z = 1; x+y = 1/2+3/4 = 5/4 > 1 (z); x^2+y^2 = 1/4+9/16 = 13/16 < 1; x^4 + y^4 = 1/16+81/256 = 97/256 < 1 (z^4).
INSUFFICIENT.

Just looking at the question, I knew the answer had to be D or E because essentially, you are given a+b > c and being asked whether a^2+b^2 > c^2.
In case of statement 1, a = x^2, b = y^2, c = z^2. In case of statement 2, a = x, b = y, c = z. So, if one was sufficient, then the other one should be sufficient as well. And if one wasn't, then other won't either.
In our case, one isn't. So, the other one isn't either.

ratindasgupta · Posted: Wed Aug 15, 2007 8:56 am Post subject:

Combining both:
If x = 3, y = 4, z = 4; x+y = 3+4 = 7 > z; x^2+y^2 = 9+16 > 16 (z^2); x^4+y^4 = 81+256 > 256 (z^4)
If x = 1/2, y = 3/4, z = 1; x+y = 1/2+3/4 = 5/4 > 1 (z); x^2+y^2 = 1/4+9/16 = 13/16 < 1; x^4 + y^4 = 1/16+81/256 = 97/256 < 1 (z^4).
INSUFFICIENT.

givemeanid,
In the 2nd assumption where x = 1/2, y = 3/4, z = 1;
the data given is that x^2 + y^2 > z^2. But your assumption doesnt match that data. x^2+y^2 = 1/4+9/16 = 13/16 < 1; Isnt that a wrong assumption then?

L'il confused here...

givemeanid · Posted: Wed Aug 15, 2007 9:42 am Post subject:

Hit me. Twice. Doing these at work is ALWAYS a bad idea.

However, the lines along which I was thinking:

If x, y and z all > 1, then obviously, 1 and 2 are sufficient.
Also, making x or y or z < -1 would not help much since powers of 2 and 4 would be positive and keep increasing anyway.

So, 0 to 1 interval sounds like the one we need to check.
Now, if we have 0<x<1, y = let's say 2 and z just slightly larger than 2 so as to satisfy statement 1, can we have 4th power increase enough to compensate.

x=1/2, y=2, z=2.01 (or in other words 2+1/100)
x+y = 2 + 1/2 = 2.5 > z
Statement 2 is satisfied.

x^2 + y^2 = 4 + 1/16
z^2 = (2+1/100)^2 = 4 + 2*2*1/100 + 0.01*0.01 = 4 + 0.04 + 0.0001
Do the comparison
4 + 1/16 ? 4 + 0.0401
1/16 ? 0.0401
0.0625 > 0.401
So, statement 1 is satisfied.

Now, x^4 + y^4 = 1/16 + 16
z^4 = (z^2)^2 = (4 + 0.0401)^2 = 16 + 8*0.0401 + 0.0016...
Do the comparison
1/16 + 16 ? 16 + 8*0.0401 + 0.0016
0.0625 ? 0.32 + 0.0016
0.0625 < 0.32 + 0.0016
So, x^4 + y^4 < z^4

Now, these calculations seem lengthy and tedious but write it out on a piece of paper and they seem to fly. Still, it took me little over 2 min. to do it on paper. (The earlier post was done all on notepad Smile

) but I figured this one deserved a notebook and pencil which I try to avoid at work for obvious reasons!)

The gist is, to get y (or x) and z real close to each other and buffer out x (has to < 1) such that x^2 compensates the difference between z^2 - y^2 but x^4 does not compensate the diff between z^4 - y^4.

If I came across this during real test, I would've guessed D or E and moved on depending on how much time I had left compared to how many questions left and how much progress I was able to make under real time test conditions.

And if there is some other mistake here, I will refrain from doing these at work in future.

ratindasgupta · Posted: Wed Aug 15, 2007 9:59 am Post subject:

man ur good Shocked