A snapshot $1\frac{7}{8}" \times 2\frac{1}{2}"$ is to be enlarged so that the longer dimension is $4"$. What will be the dimension of the shorter side? (DMRC, 2003)

### Concept & Proportional Scaling When an image is enlarged, the ratio of its dimensions remains constant. Therefore, the ratio of the new sides must equal the ratio of the original sides. $$ \frac{\text{New Shorter Side}}{\text{Original Shorter Side}} = \frac{\text{New Longer Side}}{\text{Original Longer Side}} $$ ### Step-by-Step Solution * Identify original dimensions and convert them to improper fractions: * Original shorter side = $1\frac{7}{8} = \frac{15}{8}"$. * Original longer side = $2\frac{1}{2} = \frac{5}{2}"$. * New longer side = $4"$. Let the new shorter side be $x$. * Set up the proportion: $\frac{x}{15/8} = \frac{4}{5/2}$. * Simplify the right side: $\frac{4}{5/2} = 4 \times \frac{2}{5} = \frac{8}{5}$. * Solve for $x$: $x = \frac{8}{5} \times \frac{15}{8}$. * The 8s cancel out, and $15 \div 5 = 3$. * $x = 3"$. ### Exam Strategy & Shortcut Find the enlargement multiplier directly: $\text{Multiplier} = \frac{\text{New Longer}}{\text{Old Longer}} = \frac{4}{5/2} = \frac{8}{5}$. Multiply the old shorter side by this multiplier: $\frac{15}{8} \times \frac{8}{5} = 3$. ### Common Pitfall Incorrectly identifying which side is the "longer" or "shorter" side initially, or making arithmetic errors when dividing by a fraction. ### Final Answer Therefore, the correct answer is **$3"$**.