The correct option is (b) - 19.

[as per provisional answerkey]

Solution

To find how many times the digit 5 appears in all two-digit positive integers (from 10 to 99), we must count its occurrences in both the "tens" place and the "units" place.

Step 1: Count occurrences in the Tens place
The digit 5 appears in the tens place in the numbers from 50 to 59.
These numbers are: 50, 51, 52, 53, 54, 55, 56, 57, 58, 59.
Total occurrences in the tens place = 10.

Step 2: Count occurrences in the Units place
The digit 5 appears in the units place in the following two-digit numbers:
15, 25, 35, 45, 55, 65, 75, 85, 95.
Total occurrences in the units place = 9.

Step 3: Calculate the Total
Total occurrences = (Occurrences in Tens place) + (Occurrences in Units place)
Total = $10 + 9 = 19$ .

Note: In the number 55, the digit 5 appears twice. Our method correctly counts one '5' in Step 1 (tens place) and the other '5' in Step 2 (units place).

Why the other options are incorrect

Option (a) - 18: This count is incorrect because it likely misses one occurrence of the digit 5, often by failing to count both 5s in the number 55 or missing one number in the sequence.
Option (c) - 20: This is the count of the digit 5 in the range 1 to 100. In the range 1–100, the digit 5 appears in 5, 15, 25, 35, 45, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 65, 75, 85, 95. This totals 20. However, the question specifically asks for "two-digit" integers, so the single-digit number '5' must be excluded ( $20 - 1 = 19$ ).
Option (d) - 21: This is an overcount that might result from double-counting the number 55 or including numbers outside the specified range (10–99).

Key Concept

Systematic counting by place value (tens and units) ensures that double-digit occurrences like '55' are fully accounted for without over-counting the integers themselves.