Question

If $x={\left(99\right)}^{50}+{\left(100\right)}^{50}$ and $y={\left(101\right)}^{50}$ then which is true:
$\left(A\right)x>y$
$\left(B\right)x<y$

Answer 1

solution :

Given $x=(99{)}^{50}+(100{)}^{50}$

$y=(101{)}^{50}$

Now $x=(99{)}^{50}+(100{)}^{50}={101}^{50}\left[\right(\frac{99}{100}{)}^{50}+\left(\frac{100}{101}{)}^{50}\right]$

$={101}^{50}\left[\right(fractionlessthan1{)}^{50}+(frac.lessthan1{)}^{50}]$

$={101}^{50}\times a$ fraction less than 1

So, clearly, ${101}^{50}$ is larger as ${101}^{50}$ multiplied by a

fraction less than one must be less than the actual

number.