Question 73

If x : y : z = 1 : 2 : 3, then what the value of $$\left(\frac{3x^{2}-2y^{2}+4z^{2}}{x^{2}+2y^{2}+z^{2}}\right)$$

Solution

Given,

x : y : z = 1 : 2 : 3

Let x = a

y = 2a

z = 3a

Putting these value in the equation given in question :

$$\therefore\ \left(\frac{3x^2-2y^2+4z^2}{x^2+2y^2+z^2}\right)$$
$$\therefore\ \frac{3a^2-2\left(2a\right)^2+4\left(3a\right)^2}{a^2+2\left(2a\right)^2+\left(3a\right)^2}$$

$$\therefore\ \frac{3a^2-8a^2+36a^2}{a^2+8a^2+9a^2}$$

$$\therefore\ \frac{31a^2}{18a^2}$$

$$\therefore\ \frac{31}{18}$$

Hence, Option C is correct.

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SSC MTS 2nd Nov 2021 Shift-2

If x : y : z = 1 : 2 : 3, then what the value of $$\left(\frac{3x^{2}-2y^{2}+4z^{2}}{x^{2}+2y^{2}+z^{2}}\right)$$

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