Let the sum of the odd positive integers from $1$ to $100$ be $’x’,$ and the sum of the even positive integers from $150$ to $200$ be $’ y’.$
Now, $x = 1 + 3 + 7 + \ldots + 99$
Here, we have $a = 1,d = 2,l = 99$
$T_{n} = l = a+(n-1)d$
$\implies 99 = 1 + (n-1)2$
$\implies 2n – 2 = 98$
$\implies n = 50$
$\therefore x = \frac{50}{2} (1 + 99) = 2500 \quad [\because S_{n} = \frac{n}{2}(a + l)]$
And, $y = 150 + 152 + 154 + \ldots + 200$
Here, we have $a = 150,d = 2,l = 200$
$T_{n} = l = a+(n-1)d$
$200 = 150 + (n – 1)2$
$\implies 2n - 2 = 50$
$\implies n = 26$
$\therefore y = \frac{26}{2} (150 + 200) = 13 \cdot 350 = 4550$
Now, the required ratio $ = \dfrac{x}{y} = \dfrac{2500}{4550} = \dfrac{50}{91}.$
So, the correct answer is $(C).$