<p>A five-digit number is written down at random. The probability that the number written is such that when the digits are put in the reversed order the new number is also a five-digit number equal to the original number, is upto two places of decimal</p>
Step-by-Step Solution
Key Concept: A five-digit number that reads the same forwards and backwards must be a palindrome of the form abcba, where the first digit a ≠ 0 (for both original and reversed number to be five-digit). Count favorable outcomes (choices for a, b, c) and divide by total five-digit numbers.
<p><strong>Step 1:</strong> Identify the structure of five-digit palindromes. A five-digit palindrome has the form abcba where digits read the same forwards and backwards.</p><p><strong>Step 2:</strong> Determine constraints. For a five-digit number: first digit a ∈ {1,2,...,9} (9 choices, a ≠ 0). For the reversed number to also be five-digit: last digit a ≠ 0 (already satisfied). Middle digits: b ∈ {0,1,...,9} (10 choices) and c ∈ {0,1,...,9} (10 choices).</p><p><strong>Step 3:</strong> Count favorable outcomes = 9 × 10 × 10 = 900 (palindromes where both original and reverse are five-digit).</p><p><strong>Step 4:</strong> Count total five-digit numbers. First digit has 9 choices (1-9), remaining four digits have 10 choices each = 9 × 10⁴ = 90,000.</p><p><strong>Step 5:</strong> Calculate probability = 900/90,000 = 1/100 = 0.01.</p><p>∴ Answer: A (0.01)</p>
Correct Answer: A