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by dtweah » Thu Feb 11, 2010 6:00 am
mgmt_gmat wrote:70) If x4 + y4 = 100, the greatest possible value of x is between
a. 0 and 3
b. 3 and 6
c. 6 and 9
d. 9 and 12
e. 12 and 15

Please explain.
4(x+Y) =100
We know only one number multiplied by 4 gives positive 100. It is 25. So if both X and Y must sum to 25. If we consider the set of positive integers and zero, then the max each can take is 25. If only positive integers then the max is 24 and min is 1. If we consider all real numbers, then there is NO MAX or MIN since we can stretch x to an infinite number of decimal points and reduce Y by the same amout to infinity. Under this, negative numbers are also possible X could be 50 and y -25 and we still get hundred. Botched question.
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by mohit11 » Thu Feb 11, 2010 7:27 am
I had the following approach to solve this question.

X^4 + Y^4 = 100

and 4^4 = 256

There fore answer has to be below 4, also, since we have to maximize X, we need to minimize Y, and 3^4 = 81 .. however we are not dealing with integers, so we can also consider 3.1 , 3.2 etc.

So the answer is B - between 3 and 6
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by shashank.ism » Thu Feb 11, 2010 7:34 am
mgmt_gmat wrote:70) If x4 + y4 = 100, the greatest possible value of x is between
a. 0 and 3
b. 3 and 6
c. 6 and 9
d. 9 and 12
e. 12 and 15

Please explain.
@mgmt_gmat will u please make it clear whether it is X^4 + y^4 =100 or x4+y4=100
If it is X^4 +Y^4 =100
since X^4 +Y^4 =100
now 1^4 =1
2^4 =16
3^4 = 81
4^4 = 256

Here u can see X is going to lie between 3 and 4 for max value of X.
So B is correct

In case question is X4 +Y4 =100 dtweah approach is very correct.
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by papgust » Thu Feb 11, 2010 8:22 am
To maximize 'x', minimize 'y' as far as possible. Best value of 'y' is 0

x^4 + y^4 = 100

Sub. y=0,

x^4 = 100

3^4 is 81. So, x should be little greater than 3. Best choice is B (3 to 6)
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by komal » Tue Feb 16, 2010 9:24 am
mgmt_gmat wrote:70) If x4 + y4 = 100, the greatest possible value of x is between
a. 0 and 3
b. 3 and 6
c. 6 and 9
d. 9 and 12
e. 12 and 15

Please explain.
Greatest possible value of X when Y^4 is minimum.
The possible minimum value of Y^4 is 0
X^4=100 ---> X^2=10 --> X> 3 AND X<4
B is the answer.
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by LalaB » Tue Dec 13, 2011 9:44 am
to maximize one variable, we need to minimize another one.so, let y=0
then x^4=10^2 x=sqrt10 =3.16
[spoiler]B[/spoiler] is the answ
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