The given figure shows two triangles ABC and DEF.

What is the value of y?

In ΔABC and ΔEFD:

$$\begin{gathered} \frac{AB}{EF}=\frac{x}{2.4x}=\frac{1}{2.4}=\frac{5}{12} \\ \frac{BC}{FD}=\frac{3.5}{8.4}=\frac{5}{12} \\ \frac{CA}{DE}=\frac{5}{12} \\ \therefore \frac{AB}{EF}=\frac{BC}{FD}=\frac{CA}{DE}=\frac{5}{12} \end{gathered}$$

By the Side-Side-Side similarity postulate, ΔABC is similar to ΔEFD.

Since corresponding angles of similar triangles are equal, ∠C = ∠D

$$\begin{gathered} \therefore y=\frac{y}{3}+40 \\ \Rightarrow \frac{2y}{3}=40 \\ \Rightarrow y=60 \end{gathered}$$