Shortcut Trick
In every set of 10 consecutive numbers (like 1-10, 11-20), there are exactly 2 numbers ending in 1 or 8.
From 1 to 130, there are exactly 130 ÷ 10 = 13 such sets of ten.
Total numbers ending in 1 or 8 = 13 × 2 = 26.
Required percentage = (26 ÷ 130) × 100 = (1 ÷ 5) × 100 = 20%.
∴ The correct answer is 20%.
Alternate Method
Given: Range of numbers from 1 to 130. Required unit digits are 1 or 8.
Formula Used: Percentage = (Count of favorable numbers ÷ Total numbers) × 100
1 130 Total Range (130 numbers) ...1 ...8 x 13 blocks Favorable: 13 × 2 = 26
⇒ Total numbers in the range (1 to 130) = 130
⇒ Numbers ending in 1: 1, 11, 21, 31, 41, 51, 61, 71, 81, 91, 101, 111, 121
⇒ Count of numbers ending in 1 = 13
⇒ Numbers ending in 8: 8, 18, 28, 38, 48, 58, 68, 78, 88, 98, 108, 118, 128
⇒ Count of numbers ending in 8 = 13
⇒ Total favorable numbers = 13 + 13 = 26
⇒ Percentage = (26 ÷ 130) × 100
⇒ Percentage = (1 ÷ 5) × 100 = 20%
∴ The correct answer is 20%.
Additional Information
Percentage Calculation
To find what percent x is of y, use the formula: (x ÷ y) × 100.
Arithmetic Progression (AP) Counting
The number of terms in an AP is n = [(Last Term − First Term) ÷ Common Difference] + 1.
Unit Digit Distribution
In any range of 10n consecutive integers, every digit (0-9) appears exactly n times in the unit's place.