Solution

The correct answer is option 3.

Key Points

By using inclusion exclusion principle, |AUB|= |A|+|B|-|A∩B|

Number of odd numbers in the range of (1-100)=100 ÷ 2=50 (1,3,5,7....97,99)=|A|

Number of squares=10 (1,4,9,16,25,36,49,64,81,100)=|B|

Number of odd numbers and squares=5 (1,9,25,49,81)=|A∩B|

Number of positive integers not exceeding 100 that are either odd or the square=|AUB|= 50+10-5=55

∴ Hence the correct answer is 55.