OFFICIAL EXPLANATION


As Distance $\(=\)$ Speed $\(\times\)$ Time, time taken to travel $\(X\)$ miles being $\(t\)$ seconds imply Speed $\(=\frac{X}{t}\)$ miles $\(/\)$ sec ond

So, the time taken to travel 900 miles would be $\(\frac{ Distance}{ Speed}=\frac{900}{\frac{X}{t}}\)$ seconds $\(=\frac{900 t}{X} \sec\)$ seconds which is clearly greater than $\(\frac{15t}{X}\)$

Hence column A has higher quantity, so the answer is (A).