A hydrocarbon of formula $\mathrm{C}{}_7\mathrm{H}{}_{10}$ is completely hydrogenated over Pt. One mole of it takes up exactly 2 mol of $\mathrm{H}{}_2$ and gives a single saturated-type product of formula $\mathrm{C}{}_7\mathrm{H}{}_{14}$. How many rings does the hydrocarbon contain?

DU of C7H10 = (14 + 2 - 10)/2 = 3, so rings + pi bonds = 3. Hydrogenation of pi bonds consumes H2 (a C=C takes 1 mol, a C-triple-bond-C takes 2 mol), so 2 mol H2 removes 2 pi-bond equivalents, leaving 3 - 2 = 1 ring. The product C7H14 has DU = 1, which is that remaining ring.