Question : Simplify: $\frac{256 x^4-16 y^4}{\left(80 x^2-20 y^2\right)\left(16 x^2+4 y^2\right)}$
Option 1: $5$
Option 2: $\frac{1}{20}$
Option 3: $\frac{1}{5}$
Option 4: $\frac{2}{5}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $\frac{1}{5}$
Solution : To simplify the given expression, let's factorize the numerator and the denominator first: Numerator $=256x^4-16y^4$ $=16(16x^4-y^4)$ $=16(4x^2+y^2)(4x^2-y^2)$ Denominator $=\left(80x^2-20y^2\right)\left(16x^2+4 y^2\right)$ $=20(4x^2-y^2)×4(4x^2+y^2)$ $=80(4x^2+y^2)(4x^2–y^2)$ So, now $\frac{256x^4–16y^4}{\left(80x^2–20y^2\right)\left(16x^2+4 y^2\right)}$ $=\frac{16(4x^2+y^2)(4x^2-y^2)}{80(4x^2+y^2)(4x^2–y^2)}$ $=\frac{1}{5}$ Hence, the correct answer is $\frac{1}{5}$.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : The simplified value of $\small \left (1-\frac{2xy}{x^{2}+y^{2}}\right )\div\left (\frac{x^{3}-y^{3}}{x-y}-3xy\right)$ is:
Question : If $x^4+y^4+x^2 y^2=17 \frac{1}{16}$ and $x^2-x y+y^2=5 \frac{1}{4}$, then one of the values of $(x-y)$ is:
Question : Simplify the following equation. What is the difference between the two values of $x$? $7 x+4\left\{x^2 \div(5 x \div 10)\right\}-3\left\{5 \frac{1}{3}-x^3 \div\left(3 x^2 \div x\right)\right\}=0$
Question : Simplify the following: $\frac{\cos x-\sqrt{3} \sin x}{2}$
Question : What is the LCM of $\left(8 x^3+80 x^2+200 x\right)$ and $\left(4 x^4+16 x^3-20 x^2\right)$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile