GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and

Let sqrt(x) = a and sqrt(y) = b
Then a^3 + b^3 = 32 and a^2b + ab^2 = 31 a^2 + b^2 = ?
(a + b)^3 = a^3 + b^3 + 3a^2b + 3ab^2 = 32 + 3*31 = 125 ---> a + b = 5
a^2b + ab^2 = ab(a+b) = ab * 5 =31 ---> ab = 31/5
a^2 + b^2 = (a + b)^2 - 2ab = 25 - 62/5 = 25 - 12.4 = 12.6 D)

I could not solve it. Hence guessed

x√x+y√y=32 and x√y+y√x=31

Subtracting both the equation.

x(√x-√y) +y(√y-√x)=1
(√x-√y)(x-y)=1

so there exist only one condition where, difference of integer and difference between square root of integer yield product 1

ie: x=2 and y=3

so,
x+y=5

consider [square_root]x =a
[square_root]y=b
a^3+b^3=32--1
a^2b+ab^2=3--2
(a+b)^3=125
a+b=5
(a+b)^2=25
(x+y)=25-2ab
=25- 2*31/5 ( Solving from equation 2)
=12.6

x*sqrtx + y*sqrty = 32.....(1) and x*sqrty + y*sqrtx = 32 .....(2)
add equations 1 and 2 and solve
(x+y)(sqrtx + sqrty) = 63 ... (3)

Now, let sqrt x = a and sqrt y = b
then given equations can be re written as
a^3 + b^3 = 32....(4) and a^2b + ab^2 = 31 ......(5)

multiply equation 5 by 3 and add to equation 4
(a+b)^3 = 32 + 3x31 = 125
or a+b = 5
therefore sqrt x + sqrt y = 5

substitute the value in equation 3
x +y = 63/5 = 12.6

Answer:D

Let \(\sqrt{x}\) = m
Let \(\sqrt{y}\) = n

Given

\(m^3 + n^3\) = 32 ---(1)

\(m^{2}*n + n^{2}*m\) = 31 ---(2)

\(mn(m+ n)\) = 31

\((m+n)^3\) = \(m^3 + n^3 + 3mn(m+n)\)

Substituting the values we get -

\((m+n)^3\) = 32 + 3(31)

\((m+n)^3\) = 125

\((m+n)\) = 5

From 2 we know that

\(mn(m+ n)\) = 31

\(mn = 31/5 = approx. 6\)

\((\sqrt{x}+\sqrt{y})\) = 5

Squaring both sides we get -

x + y + 2\(\sqrt{xy}\) = 25

x + y = 25 - 12 = 13

As we have approximated the closest answer is D

Option D

Hello everyone,

The correct answer is D.
I have attached the picture which shows the detailed explanation of the problem.
x+y=12.6

we can write the expression
x√x+y√y=32 and \(x\sqrt{y} + y\sqrt{x} = 31\)
square both sides we get

(x√x+y√y)^2 = 1024 and ( x√y+y√x)^2 = 961
substitute 2xy√xy = 961- x^2y-y^2x in (x√x+y√y)^2 = 1024

we will reduce to
(x-y)^2 ( x+y) = 63
this is possible when both sides are odd
use plug in we see at option B ( 5) this sufficies

OPTION B is correct

x√x + y√y =32 x√y + y√x =31
add two equation together
x√x + x√y +y√y+ y√x =63
take x out and take y out
x(√x +√y)+y(√x +√y)=63
(x+y)(√x +√y)=63

We have on squaring both sides of 1st eq

x^3 + y^3 + 2xy√xy = 32^2

For second we have
y*x^2 + x*y^2 + 2xy√xy = 31^2

Subtracting both we get

(x+y)(x-y)^2 = 63

Now the only valid factor of 63 from the answers could be 12.6

Going with Option D

This is going to be completely unsatisfying for those of you who like elegant MATH, but I'm a huge fan of checking the answer choices and using some logic.
Since swapping the coefficients before the radicals yields 32 in one equation and 31 in the other, x and y can't be THAT far apart from each other.

Look at answer choice B. Let's just try x=3 and y=2. That makes the first equation \(3\sqrt{3}=2\sqrt{2}\) = 5.1+2.8 = 7.9. Woah, that's WAY too small. Now way ar we going to get close with two numbers that sum to 5. B is out. So is A. C is also obviously too small.

We are down to D and E. Let's examine E. If we give most of the 25 to x, \(x\sqrt{x}\) is going to be way bigger than 32 on its own. If we make x and y pretty even, we're still way too big. E is out.

Answer choice D.


If we just want to sanity check D, we can use something like 6 for each. \(\sqrt{6}\) is more than 2 and less than 3. 2.5 should be good enough to know if we are way off the mark.
\(6*2.5 + 6*2.5 = 15+15\). Yeah, I'm cool with that!!

Approached it from a guesstimate point of view. So not sure if this is right:

To find x+y:

Add the two equations in the question and take common, we get:
(x+y) (root(x) + root(y)) = 63

63 is close to 65 which is equal to 13 * 5

so our equation can be written as :
(9+4) (root(9) + root(4) = 65

so (x+y) is close to 13. Of the options given, option D has 12.6 which is the closest. Hence picked that

The answer is 12.6 i.e. D.

Asked: If \(x\sqrt{x} + y\sqrt{y} = 32\) and \(x\sqrt{y} + y\sqrt{x} = 31\), then what is the value of \(x + y\) ?

\(x\sqrt{x} + y\sqrt{y} = 32\)
\(x\sqrt{y} + y\sqrt{x} = 31\)

\(x(\sqrt{x}-\sqrt{y})-y(\sqrt{x}-\sqrt{y}) = 32-31 = 1\)
\((x-y)(\sqrt{x}-\sqrt{y})=1\)
This equation gives an indication that values of x & y are nearly same since differences are very less (1).

\(x(\sqrt{x}+\sqrt{y})+y(\sqrt{x}+\sqrt{y}) = 32+31 = 63\)
\((x+y)(\sqrt{x}+\sqrt{y})= 63\)
\(2x * 2\sqrt{x} = 63\)
\(x\sqrt{x} = \frac{63}{4}\)
x = 6.3
x + y = 2x = 12.6

IMO D

Since x and y are under the square root they are not negative
Since x√y+y√x=31, then x and y cannot be equal to Zero, otherwise, if x=0 then √x=0 then x√y+y√x=0 and not 31,

So x and y are strictly positive,

since x√x+y√y=32 and x√y+y√x=31, we can see that x≈y so from this equation √x+y√y=32 and since x≈y
we can say that 2x√x=32 => x√x=16 =>(x√x)^2=256 => x^3=256 => x≈6,3 =>x+y =12,6=> Answer is D

Let Root(x) = p and Root(y) = q ... (g)

From question,

p^3 + q ^3=32 .... (a)
p^2*q+q^2*p= 31 ... (b)

Now
(p+q)^3=p^3+q^3+3(p^2*q+q^2*p)
=32+3*31
=125
Therefore p+q=5 ... (c)
Root(x) + Root (y) = 5 ... (m)

Squaring both sides,
x+y+2*root(x*y)=25
x+y=25-2*root(x*y) ... (d)

From (b) we have p*q*(p+q)=31
OR [Root(x*y)]*[(Root(x)+Root(y)]=31 .. (n)

From (m), (n) and (d)
x+y=25-2*31/5
OR x+y = 12.6 ANS D