In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘x’ and ‘y’ and mark correct answer. I. \(\frac{{15 \times 4}}{{{x^{4/7}}}} - \frac{{8 \times 3}}{{{x^{4/7}}}} = {x^{10/7}}\) II. y3 + 783 = 999

In the following question, one or two equation(s) is/are given. You have to solve both the equations and find the relation between ‘x’ and ‘y’ and mark correct answer.

I. \(\frac{{15 \times 4}}{{{x^{4/7}}}} - \frac{{8 \times 3}}{{{x^{4/7}}}} = {x^{10/7}}\)

II. y³ + 783 = 999

A x > y

B x ≥ y

C x < y

D x ≤ y

E x = y or the relation cannot be determined

Solution

Correct Answer: Option D

We will solve both the equations separately.

Equation I:

\(\begin{array}{l} \frac{{15 \times 4}}{{{x^{\frac{4}{7}}}}} - \frac{{8 \times 3}}{{{x^{\frac{4}{7}}}}} = {x^{\frac{{10}}{7}}}\\ \Rightarrow \;\frac{{60}}{{{x^{4/7}}}} - \frac{{24}}{{{x^{4/7}}}} = {x^{10/7}}\\ \Rightarrow \;\frac{{36}}{{{x^{4/7}}}} = {x^{10/7}}\\ \Rightarrow \;36 = {x^{\frac{{10}}{7} + \frac{4}{7}}}\\ \Rightarrow \;36 = {x^2} \end{array}\)

⇒ x = ± 6

Equation II:

y³ + 783 = 999

⇒ y³ = 999 – 783

⇒ y³ = 216

⇒ y = 6

Comparing the values of x and y, we get,

x ≤ y

Solution

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