तीन क्रमागत विषम संख्याओं का औसत, पहली संख्या के एक-तिहाई से 16 अधिक है । सबसे बड़ी संख्या होगी :

The correct answer is (B) 25.

• Given Data and Mathematical Principle: Let the three consecutive odd numbers be represented as $x$, $x+2$, and $x+4$. The average of these three numbers is given by $\frac{x+(x+2)+(x+4)}{3}$$\frac{x + (x+2) + (x+4)}{3}$. According to the problem statement, this average is 16 more than one-third of the first number ($x$), which can be expressed algebraically as $\frac{x+x+2+x+4}{3}=\frac{x}{3}+16$$\frac{x + x + 2 + x + 4}{3}$ = $\frac{x}{3}$ + 16.
• Step-by-Step Algebraic Derivation: Simplify the left side of the equation: $\frac{3x+6}{3}=x+2$$\frac{3x + 6}{3}$ = x + 2. Now equate it to the right side: $x + 2 =$$\frac{x}{3}$ + 16. Subtract $\frac{x}{3}$$\frac{x}{3}$ from both sides: $x -$$\frac{x}{3}$ = 16 - 2, which gives $\frac{2x}{3}=14$$\frac{2x}{3}$ = 14. Multiplying both sides by 3 yields $2x = 42$, so $x = 21$. The first number is 21, the second is 23, and the largest number is $x+4 = 21+4 = 25$.
• Verification and Option Analysis: The three consecutive odd numbers are 21, 23, and 25. Their average is $\frac{21+23+25}{3}=\frac{69}{3}=23$$\frac{21+23+25}{3}$ = $\frac{69}{3}$ = 23. One-third of the first number (21) is $\frac{21}{3}=7$$\frac{21}{3}$ = 7. The difference between the average (23) and 7 is $23 - 7 = 16$, which precisely matches the condition given in the question. Hence, options (A), (C), and (D) are incorrect.