If $x : y = 3 : 7$, then $\frac{6x^2 – 2xy + y^2}{xy}$ is equal to
[amp_mcq option1=”$\frac{62}{21}$” option2=”$\frac{61}{21}$” option3=”$\frac{63}{21}$” option4=”$\frac{64}{21}$” correct=”option2″]
This question was previously asked in
UPSC CISF-AC-EXE – 2024
We can express x and y in terms of a common variable. Let $x = 3k$ and $y = 7k$ for some non-zero constant k.
We need to evaluate the expression $\frac{6x^2 – 2xy + y^2}{xy}$.
Substitute $x = 3k$ and $y = 7k$ into the expression:
Numerator:
$6x^2 – 2xy + y^2 = 6(3k)^2 – 2(3k)(7k) + (7k)^2$
$= 6(9k^2) – 2(21k^2) + 49k^2$
$= 54k^2 – 42k^2 + 49k^2$
Combine the terms with $k^2$:
$= (54 – 42 + 49)k^2$
$= (12 + 49)k^2$
$= 61k^2$.
Denominator:
$xy = (3k)(7k) = 21k^2$.
Now, form the ratio:
$\frac{6x^2 – 2xy + y^2}{xy} = \frac{61k^2}{21k^2}$.
Since k is a non-zero constant, $k^2$ is also non-zero, so we can cancel $k^2$ from the numerator and the denominator.
The expression simplifies to $\frac{61}{21}$.
$\frac{6x^2 – 2xy + y^2}{xy} = \frac{\frac{6x^2}{y^2} – \frac{2xy}{y^2} + \frac{y^2}{y^2}}{\frac{xy}{y^2}} = \frac{6(\frac{x}{y})^2 – 2(\frac{x}{y}) + 1}{(\frac{x}{y})}$.
Substitute the value of $\frac{x}{y} = \frac{3}{7}$:
$= \frac{6(\frac{3}{7})^2 – 2(\frac{3}{7}) + 1}{(\frac{3}{7})} = \frac{6(\frac{9}{49}) – \frac{6}{7} + 1}{\frac{3}{7}}$
$= \frac{\frac{54}{49} – \frac{42}{49} + \frac{49}{49}}{\frac{3}{7}} = \frac{\frac{54 – 42 + 49}{49}}{\frac{3}{7}} = \frac{\frac{61}{49}}{\frac{3}{7}}$
$= \frac{61}{49} \times \frac{7}{3} = \frac{61}{7 \times 7} \times \frac{7}{3} = \frac{61}{7 \times 3} = \frac{61}{21}$.
Both methods yield the same result.