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

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Bunuel
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GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and [#permalink] GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and   11 Jul 2022, 08:00
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Question Stats:

50% (02:27) correct 50% (03:08) wrong based on 189 sessions
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If \(x\sqrt{x} + y\sqrt{y} = 32\) and \(x\sqrt{y} + y\sqrt{x} = 31\), then what is the value of \(x + y\) ?


A. \(\sqrt{5}\)

B. \(5\)

C. \(6.2\)

D. \(12.6\)

E. \(25\)

 


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Bunuel
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Re: GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and [#permalink] GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and   11 Jul 2022, 08:00
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Bunuel
If \(x\sqrt{x} + y\sqrt{y} = 32\) and \(x\sqrt{y} + y\sqrt{x} = 31\), then what is the value of \(x + y\) ?


A. \(\sqrt{5}\)

B. \(5\)

C. \(6.2\)

D. \(12.6\)

E. \(25\)

 


This question was provided by GMAT Club
for the GMAT Club World Cup Competition

Compete, Get Better, Win prizes and more

 


GMAT CLUB Official Explanation:

Step 1:
Let \(\sqrt{x}=m\) and \(\sqrt{y} =n\). Notice that in this case we'd need to find the value of \(x + y=m^2+n^2\).

Then we'd have:
    (i) \(m^3 + n^3 = 32\);
    (ii) \(mn^2 + nm^2 = 31\)

Step 2:
Recall that:
    \((m + n)^3 = m^3 + n^3 + 3mn^2 + 3nm^2\)

Substitute (i) and (ii) in the above:
    \((m + n)^3 = 32 + 3*31\)

Take the cube root:
    \(m + n = 5\)

Step 3:
From (ii) we have that \(mn(n + m) = 31\);

Substitute \(m + n = 5\) in the above:
    \(mn = \frac{31}{5}\)

Step 4:
Recall that \((m + n)^2 = m^2 + n^2 + 2mn\);
Substitute \(m + n = 5\) and \(mn = \frac{31}{5}\) in the above:
    \(5^2= m^2 + n^2 + 2*\frac{31}{5}\);
    \(m^2 + n^2 = 12.6\)

Answer: D.
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Re: GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and [#permalink] GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and   11 Jul 2022, 08:21
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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)
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Re: GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and [#permalink] GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and   11 Jul 2022, 08:47
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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
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Re: GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and [#permalink] GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and   11 Aug 2024, 04:51
can we do it by squaring both the equations and subtracting. we get
\((x+y)(x-y)^2 = 63\)
how can i solve it further?
m getting answers as 3,7, or 9
Bunuel
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Re: GMAT Club World Cup 2022 (DAY 1): If x*x^(1/2) + y*y^(1/2) = 32 and [#permalink]
11 Aug 2024, 04:51
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Bunuel
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