When a particular number is subtracted from each of 7, 9, 11 and 15, the resulting numbers are in proportion. The number to be subtracted is

### Concept & Proportionality Condition Four numbers $a, b, c,$ and $d$ are in proportion if the ratio of the first two equals the ratio of the last two, meaning $\frac{a}{b} = \frac{c}{d}$. We can set up an equation by subtracting an unknown variable from each number. ### Step-by-Step Solution * Let the number to be subtracted be $x$. * The resulting numbers will be $(7 - x)$, $(9 - x)$, $(11 - x)$, and $(15 - x)$. * For these numbers to be in proportion, the following must hold true: $$ \frac{7 - x}{9 - x} = \frac{11 - x}{15 - x} $$ * Cross-multiply to solve for $x$: $$ (7 - x)(15 - x) = (11 - x)(9 - x) $$ * Expand both sides: $$ 105 - 7x - 15x + x^2 = 99 - 11x - 9x + x^2 $$ $$ 105 - 22x + x^2 = 99 - 20x + x^2 $$ * The $x^2$ terms cancel out: $$ 105 - 22x = 99 - 20x $$ * Rearrange to isolate $x$: $$ 105 - 99 = 22x - 20x $$ $$ 6 = 2x $$ $$ x = 3 $$ ### Exam Strategy & Shortcut Instead of solving the quadratic expansion, use the Option Elimination method. Test (a) $1$: $\frac{7-1}{9-1} = \frac{6}{8} = \frac{3}{4}$. And $\frac{11-1}{15-1} = \frac{10}{14} = \frac{5}{7}$. Not equal. Test (b) $2$: $\frac{7-2}{9-2} = \frac{5}{7}$. And $\frac{11-2}{15-2} = \frac{9}{13}$. Not equal. Test (c) $3$: $\frac{7-3}{9-3} = \frac{4}{6} = \frac{2}{3}$. And $\frac{11-3}{15-3} = \frac{8}{12} = \frac{2}{3}$. They match perfectly! ### Common Pitfall A common algebraic error during cross-multiplication is messing up the negative signs when expanding $(a - x)(b - x)$, leading to incorrect linear terms and a wrong answer. ### Final Answer Therefore, the correct answer is **3**.