The distance a spring stretches varies directly with the weight attached to it. A weight of 5 pounds stretches the spring 15 centimeters.…

Direct variation means distance $=k\cdot$ weight. From $15=k(5)$, the constant is $k=3$ cm per pound. For an 8-pound weight the stretch is $3(8)=24$ cm, choice C. Equivalently, the proportion $\frac{15}{5}=\frac{d}{8}$ gives $d=24$. Choice D, $40$, multiplies the 5-pound weight by 8 and ignores the spring's actual behavior. Choices A ($18$) and B ($21$) add to $15$ instead of scaling: direct variation is multiplicative, so adding a fixed amount per extra pound is wrong unless that amount equals $3$ cm per pound, which gives $24$, not $18$ or $21$.