NIMCET 2014 — Mathematics PYQ

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Question

NIMCET 2014 — Mathematics PYQ

NIMCET | Mathematics | 2014

If $x$ and $y$ are positive real numbers satisfying the system of equations $x^2 + y\sqrt{xy} = 336$ and $y^2 + x\sqrt{xy} = 112$ , then $x + y$ is:

Choose the correct answer:

Explanation

Solution

1. Given Equations:

$x^2 + y\sqrt{xy} = 336$
$y^2 + x\sqrt{xy} = 112$

2. Simplifying the Equations:

Divide the first equation by $\sqrt{x}$ :

x\sqrt{x} + y\sqrt{y} = \frac{336}{\sqrt{x}} \quad \dots(1) \text{}

Divide the second equation by $\sqrt{y}$ :

y\sqrt{y} + x\sqrt{x} = \frac{112}{\sqrt{y}} \quad \dots(2) \text{}

3. Finding the Relationship between $x$ and $y$ :

Since the left-hand sides of equations $(1)$ and $(2)$ are identical ( $x\sqrt{x} + y\sqrt{y}$ ), we can equate the right-hand sides:

\frac{336}{\sqrt{x}} = \frac{112}{\sqrt{y}} \text{}

3\sqrt{y} = \sqrt{x} \quad \dots(3) \text{}

Squaring both sides:

9y = x \quad \dots(4) \text{}

4. Solving for $y$ and $x$ :

Substitute the values from equations $(3)$ and $(4)$ into equation $(1)$ :

9y(3\sqrt{y}) + y\sqrt{y} = \frac{336}{3\sqrt{y}} \text{}

27y\sqrt{y} + y\sqrt{y} = \frac{112}{\sqrt{y}} \text{}

28y\sqrt{y} = \frac{112}{\sqrt{y}} \text{}

28y^2 = 112 \text{}

y^2 = \frac{112}{28} = 4 \text{}

Since $y$ must be a positive real number:

$y = 2$
Using $x = 9y$ , we get $x = 9(2) = 18$

5. Final Calculation:

x + y = 18 + 2 = 20 \text{}

Correct Option: 3 ( $20$ )