Please enter the following details
Free Daily Targets & Mocks
Study Material & Formulas PDFs
1 on 1 Mentorship
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
$$\therefore\ \frac{3a^2-2\left(2a\right)^2+4\left(3a\right)^2}{a^2+2\left(2a\right)^2+\left(3a\right)^2}$$
Given,Â
x : y : z = 1 : 2 : 3Â
Let x = aÂ
y = 2aÂ
z = 3a
Putting these value in the equation given in question :Â
$$\therefore\ \left(\frac{3x^2-2y^2+4z^2}{x^2+2y^2+z^2}\right)$$$$\therefore\ \frac{3a^2-2\left(2a\right)^2+4\left(3a\right)^2}{a^2+2\left(2a\right)^2+\left(3a\right)^2}$$
$$\therefore\ \frac{3a^2-8a^2+36a^2}{a^2+8a^2+9a^2}$$
$$\therefore\ \frac{31a^2}{18a^2}$$
$$\therefore\ \frac{31}{18}$$
Hence, Option C is correct.Â
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
