Practicing Success

Target Exam

CUET

Subject

General Test

Chapter

Quantitative Reasoning

Topic

Algebra

Question:

If x = 7 and y = 2 then, what is the value of $\frac{x^3-y^3}{x^2-y^2}-\frac{3xy}{x+y}$ ?

Options:

28/9

21/10

25/9

26/9

Correct Answer:

25/9

Explanation:

x = 7 and y = 2

$\frac{x^3-y^3}{x^2-y^2}-\frac{3xy}{x+y}$

= $\frac{7^3-2^3}{7^2-2^2}-\frac{3 × 7 × 2}{7 + 2}$

= \(\frac{343}{45}\) - \(\frac{42}{9}\)

= \(\frac{125}{45}\) = \(\frac{25}{9}\)