$3625 \times x = 1450$

### Concept & Logic This problem tests your ability to simplify large fractions by identifying **Common Factors**. The equation easily rearranges into a fraction format. The core challenge is breaking down large four-digit numbers into their lowest terms by spotting divisibility rules. ### Step-by-Step Solution **Given:** $$ 3625 \times x = 1450 $$ **Calculation:** Step 1: Isolate $x$. $$ x = \frac{1450}{3625} $$ Step 2: Use divisibility rules to reduce the fraction. Both end in 0 or 5, so they are divisible by 25. Divide the numerator by 25: $1450 \div 25$. (Shortcut: $14 \times 4 + 2 = 58$) Divide the denominator by 25: $3625 \div 25$. (Shortcut: $36 \times 4 + 1 = 145$) $$ x = \frac{58}{145} $$ Step 3: Check for further reduction. Both 58 and 145 are multiples of 29. ($29 \times 2 = 58$) ($29 \times 5 = 145$) $$ x = \frac{2}{5} $$ ### Exam Strategy & Shortcut **Denominator Matching (Reverse Check):** Once you see the options ($\frac{1}{3}, \frac{2}{5}, \frac{1}{5}, \frac{4}{5}, \frac{3}{5}$), four out of five have a denominator of 5. This is a massive hint that 3625 divided by 5 will yield the magic multiplier. $3625 \div 5 = 725$. Now, ask yourself: $725 \times \text{Numerator} = 1450$. Clearly, $725 \times 2 = 1450$. Therefore, the numerator is 2. The fraction is $\frac{2}{5}$. This reverse-engineering approach skips complex prime factorization completely. ### Common Pitfall Students often divide by 5 repeatedly (getting $\frac{290}{725}$, then $\frac{58}{145}$) and stop at $\frac{58}{145}$, assuming it cannot be reduced further because 58 and 145 don't look obviously related. You must memorize prime multiples like 29, 31, and 37 to spot these final reduction steps. ### Final Answer **Therefore, the correct answer is $\frac{2}{5}$.**