I got C but the answer is E.
A is insufficient because we don't know whether x and y are fractions or integers.
B is insufficient for the same reason.
so, I chose C..
Experts pls help..
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GMAT Prep DS - Inequalities
Source: Beat The GMAT — Data Sufficiency |
Let a= x/z and b = y/z
Is a^4 + b^4 > 1
1. a^2 + b^2 >1
a^4 + b^4 + 2a^2b^2 > 1
a^4 + b^4 -2a^2b^2 > 0
a^4 + b^4 > 1/2 (insufficient)
2. a + b > 1 when z is +ve
a^2+b^2 > 1/2
a^4 + b^4 > 2a^2b^2 = 2(1/4)(1/4) = 1/8
a^4 + b^4 > 1/8 Insufficient.
Combined together, a^4 + b^4 > 1/2
Insufficient.
Forget abt z being negative.
Is a^4 + b^4 > 1
1. a^2 + b^2 >1
a^4 + b^4 + 2a^2b^2 > 1
a^4 + b^4 -2a^2b^2 > 0
a^4 + b^4 > 1/2 (insufficient)
2. a + b > 1 when z is +ve
a^2+b^2 > 1/2
a^4 + b^4 > 2a^2b^2 = 2(1/4)(1/4) = 1/8
a^4 + b^4 > 1/8 Insufficient.
Combined together, a^4 + b^4 > 1/2
Insufficient.
Forget abt z being negative.
Hi palvarez,palvarez wrote:Let a= x/z and b = y/z
Is a^4 + b^4 > 1
1. a^2 + b^2 >1
a^4 + b^4 + 2a^2b^2 > 1
a^4 + b^4 -2a^2b^2 > 0
a^4 + b^4 > 1/2 (insufficient)
2. a + b > 1 when z is +ve
a^2+b^2 > 1/2
a^4 + b^4 > 2a^2b^2 = 2(1/4)(1/4) = 1/8
a^4 + b^4 > 1/8 Insufficient.
Combined together, a^4 + b^4 > 1/2
Insufficient.
Forget abt z being negative.
I can't get how did you come up with a^4 + b^4 + 2a^2b^2 > 1
If you don't mind pls explain it in more detail
Abdulla
a^2 + b^2 > 1
square on both sides.
a^4 + b^4 + 2.a^2. b^2 > 1
another obvious fact: (a2 - b^2)^2 > 0
a+b >= k, what's the min value of a^2+b^2? twice the geometric mean. Compute the minimum of the geometric mean when a = b = k/2.
square on both sides.
a^4 + b^4 + 2.a^2. b^2 > 1
another obvious fact: (a2 - b^2)^2 > 0
a+b >= k, what's the min value of a^2+b^2? twice the geometric mean. Compute the minimum of the geometric mean when a = b = k/2.
Hi Palvarez,
Thanks for the response. However, how do u get this statement: -
a^4 + b^4 -2a^2b^2 > 0
Pls clarify
Thanks
LA
Thanks for the response. However, how do u get this statement: -
a^4 + b^4 -2a^2b^2 > 0
Pls clarify
Thanks
LA
lance770 wrote:Hi Palvarez,
Thanks for the response. However, how do u get this statement: -
a^4 + b^4 -2a^2b^2 > 0
Pls clarify
Thanks
LA
(a^2-b^2)^2 >= 0 (a square is +ve)
a^4 +b^4 - 2a^2b^2 >= 0
a^4 + b^4 >= 2a^2b^2
Or average (arithmetic mean) >= geometric mean
I don't understand the explanation given by Palvarez... No offence.
Look at this thread for explanations given by Ron Purewal and Ian Stewart.
https://www.beatthegmat.com/gmatprep-is- ... 23339.html
Look at this thread for explanations given by Ron Purewal and Ian Stewart.
https://www.beatthegmat.com/gmatprep-is- ... 23339.html
Just adding more steps for others to understand
\Assume a = x/z, b = y/z
Is a^4 + b^4 > 1
1. a^2 + b^2 >1
(a^2+b^2)^2 > 1, squaring a positive doesn't change the inequality.
a^4 + b^4 + 2a^2b^2 > 1 ---(1)
(a^2-b^2) >= 0 (any square is +ve; this is the same used in deriving AM >= GM inqualitty, gm = geometric mean and am = arithmetic mean)
a^4 + b^4 -2a^2b^2 >= 0 -- (2)
Add (1) and (2): a^4 + b^4 >= 1/2 (insufficient)
2. a + b > 1 when z is +ve
a^2+b^2 + 2ab > 1 (square the above) --(3)
(a-b)^ 2>= 0
a^2 + b^2 -2ab >= 0 --- (4)
Add (3) and (4): a^2+b^2 > 1/2
Now square it again and use (a^2 - b^2)^2 and sum them up, you will get a^4+b^4 >= 1/8 (insufficient)
------------------------------------
There is a faster way to do all these things
a^2+b^2 >= 2ab (arithmetic mean >= geometric mean)
Now we need to get the minimum value of 2ab when a + b > 1: how to go about?
just assume a = b and solve a+b =1, ya end up with a = b = 1/2
Substitute these in the above inequality.
2ab = 2(1/2)(1/2) = 1/2
a^2 + b^2 > =1/2
a^4+b^4 >= 2a^2b^2 (arith mean >= geometric mean)
What is a^2 when a^2 = b^2 and a^2 + b^2 = 1/2
a^2 = b^2 = 1/4
therfore a^4 + b^4 >= 2(1/4(1/4)
a^4+b^4 >= 1/8
\Assume a = x/z, b = y/z
Is a^4 + b^4 > 1
1. a^2 + b^2 >1
(a^2+b^2)^2 > 1, squaring a positive doesn't change the inequality.
a^4 + b^4 + 2a^2b^2 > 1 ---(1)
(a^2-b^2) >= 0 (any square is +ve; this is the same used in deriving AM >= GM inqualitty, gm = geometric mean and am = arithmetic mean)
a^4 + b^4 -2a^2b^2 >= 0 -- (2)
Add (1) and (2): a^4 + b^4 >= 1/2 (insufficient)
2. a + b > 1 when z is +ve
a^2+b^2 + 2ab > 1 (square the above) --(3)
(a-b)^ 2>= 0
a^2 + b^2 -2ab >= 0 --- (4)
Add (3) and (4): a^2+b^2 > 1/2
Now square it again and use (a^2 - b^2)^2 and sum them up, you will get a^4+b^4 >= 1/8 (insufficient)
------------------------------------
There is a faster way to do all these things
a^2+b^2 >= 2ab (arithmetic mean >= geometric mean)
Now we need to get the minimum value of 2ab when a + b > 1: how to go about?
just assume a = b and solve a+b =1, ya end up with a = b = 1/2
Substitute these in the above inequality.
2ab = 2(1/2)(1/2) = 1/2
a^2 + b^2 > =1/2
a^4+b^4 >= 2a^2b^2 (arith mean >= geometric mean)
What is a^2 when a^2 = b^2 and a^2 + b^2 = 1/2
a^2 = b^2 = 1/4
therfore a^4 + b^4 >= 2(1/4(1/4)
a^4+b^4 >= 1/8
1. X^2 + y^2 > z^2 => x^4 + y^4+ 2x^2y^2 > z^4
2. x+y > z => x^2 + y^2 + 2xy > z^2
using 1 and 2 : z^2 + 2xy > z^2 => 2xy>0 we can only infer x and y are of same sign
Answer is E.
2. x+y > z => x^2 + y^2 + 2xy > z^2
using 1 and 2 : z^2 + 2xy > z^2 => 2xy>0 we can only infer x and y are of same sign
Answer is E.
GmatVerbal,
I love how you can solve this in 3 lines where others take half a page. But your concise solution fails to explains your rationale. You may understand it, but I sure don't. So, if your intent is for others to wonder in amazement about how it is that you can solve difficult problems in 3 lines -
Oooh, aaaaah...
I love how you can solve this in 3 lines where others take half a page. But your concise solution fails to explains your rationale. You may understand it, but I sure don't. So, if your intent is for others to wonder in amazement about how it is that you can solve difficult problems in 3 lines -
Oooh, aaaaah...
