If x and y are positive, which is greater between A and B?

Question

(1) A =  (x^3+y^3)^\frac{1}{3}

(2) B =  (x^2+y^2)^\frac{1}{2}

Are You Up For the Challenge: 700 Level Questions

chetan2u · Answer

We have to look at the two statements combined..

take the sixth power..

(1)  A^6=((x^3+y^3)^\frac{1}{3})^6=(x^3+y^3)^2=x^6+y^6+2x^3y^3

(2)  B^6=((x^2+y^2)^\frac{1}{2})^6=(x^2+y^2)^3=x^6+y^6+3x^2y^4+3x^4y^2

when we check the two, we have to compare  2x^3y^3  and  3x^2y^4+3x^4y^2 .....Now at least one of the term  3x^2y^4  or  3x^4y^2  will surely be greater than  2x^3y^3 .
Thus B>A

C

chetan2u · Answer

No, they are different 
A=(1^3+1^3)^(1/3)=2^(1/3)
B=(1^2+1^2)^(1/2)=2^(1/2)
B>A

ShankSouljaBoi · Answer

Hi chetan2u , i have realized my mistake. Thank you

Regards

veron90 · Answer

Hi  chetan2u  
The question doesn't specify that x and y are integers.

B<A in this case.

Will "E" not be the ans?

chetan2u · Answer

No, still the above would stand till the time x and y are positive

DixitNiket · Answer

Yes, made sense. Taking first the 6th power and then comparing between the uncommon terms. Got it

felicity615 · Answer

Hi  chetan2u  , Just wondering how to get from  (x^2 + y^2)^3  to  x^6+ y^6 + 3x^2y^4 + 3x^4y^2  ? Hope you can advise. Thanks!

chetan2u · Answer

GMAT may not require you to know but direct formula is 
 (a+b)^3=a^3+b^3+3ab(a+b)……. (x^2 + y^2)^3=(x^2)^3+(y^2)^3+3x^2y^2(x^2+y^2)

If you do not the formula, the way would be. 
  (x^2 + y^2)^3=(x^2+y^2)* (x^2 + y^2)^2= (x^2 + y^2)((x^2)^2+(y^2)^2+2x^2y^2) 
Open the bracket further to get the answer

Fdambro294 · Answer

We need to know the value of A and B to answer the question stem. S1 alone and S2 alone are not sufficient.

(S1 & S2)

The inequality according to the question stem:

Is: (x^3 + y^3)^(1/3) < (x^2 + y^2)^(1/2) ?

As X and Y are both positive values, the cube and square of each value will be positive. Therefore, both sides of the inequality are positive values. We can raise each side to the 6th power to obtain more workable expressions.

(x^3 + y^3)^(2) < (x^2 + y^2)^(3) ?

-next, we can use the Square of Sum and Cube of a Sum "templates" to expand each expression:

(x)^6 + (2)(x^3)(y^3) + (y)^6 < (x)^6 + 3(x)^4 * (y)^2 + 3(x)^2 * (y)^4 + (y)^6 ?

-subtract (x)^6 and (y)^6 from each side, leaving us with:

(2)(x^3)(y^3) < 3(x)^4 * (y)^2 + 3(x)^2 * (y)^4 ?

-again, since X and Y are positive values, the square of each results in a non-zero, positive value We can Divide each side of the inequality by (x^2)(y^2)

2(x)(y) < 3(x)^2 + 3(y)^2 ?

case 1: x > y

2(x) (y) will be less than < 3(x)(x) alone, so surely if we add another positive value to it 3(y)^2, the right side will contain a larger value.
YES

case 2: x = y

2(x)(y) < either of the two terms on the right hand side: Let x = y = k

2(k)(k) < 3(k)(k) ---- as a matter of numerical principle: 2 times a positive value is always less than 3 times the same positive value
YES

case 3: x < y

2(x)(y) will be less than < 3(y)(y) alone, so surely if we add another positive value to it 3(x)^2, the right side will contain a larger value.
YES

under every possible scenario, A < B
*C* Sufficient Together

If x and y are positive, which is greater between A and B?

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

If x and y are positive, which is greater between A and B?

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards