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GMATPrep Question

by selfmade » Mon Sep 27, 2010 3:40 pm
If x^4 + y ^4 =100 then greatest possible value of x is in between

1. 0 to 3
2. 3 to 6
3. 6 to 9
4. 9 to 12
5. 12 to 15

I got to answer 3. OA is option 2.[/img]
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by tlt2372 » Mon Sep 27, 2010 4:39 pm
selfmade wrote:If x^4 + y ^4 =100 then greatest possible value of x is in between

1. 0 to 3
2. 3 to 6
3. 6 to 9
4. 9 to 12
5. 12 to 15

I got to answer 3. OA is option 2.[/img]
To get the greatest possible value of x, let y=0. So the greatest x can be is between 3 and 6.
3^4=81.
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by limestone » Mon Sep 27, 2010 6:46 pm
x^4 + y^4 =100
As y^4>=0 then maximum value of x^4 is 100
then x^2 = sqrt(100) = 10 ( why not -10?, as x^2 is always equal to or larger than 0)
Hence x = sqrt(10) (why not -sqrt(10)?, as -sqrt(10) < sqrt(10), and we are asked to pick the maximum x)

For all positive numbers larger than 1: a>b then a^2>b^2 and vice versa
Pick some value from the choices, such as 3
3^2 = 9, while x^2 = 10,as 10>9 then x>3 ( I mean the x that is positive here. There is another x that is negative - the -sqrt(10) "x" that we eliminate above)
6^2 = 36, as 36>10 then x<6
Pick B.
Last edited by limestone on Mon Sep 27, 2010 8:28 pm, edited 1 time in total.
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by Tani » Mon Sep 27, 2010 8:17 pm
3 would be the highest possible integer value of x, but x could be slightly larger than 3 (i.e. up to sqrt of 10 (3.16...)- therefore between 3 and 6, not between 0 and three
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