A two-digit number is chosen such that it is 18 more when digits are reversed. What is the probability that sum of digits of this number is more than 10 ?
If x and y are digits of a number, then two-digit number=10x+y
Difference between original number & number formed by digits revered= (10y+x) - (10x+y) =18
⇒ y - x = 2
Values of y,x satisfying above condition for a two-digit numbers are (3,1) (4,2) (5,3) (6,4) (7,5) (8,6) (9,7)
Thus there are total 7 numbers 13, 24, 35, 46, 57, 68, 79
Numbers with sum of digits more than 10 are 57, 68, 79 , Total 3
Hence, required probability = 3/7
Correct Option 4)
Which is more difficult ?
solve a mathematics? or Mathematics to solve?
G. C
Pure mathematics is the world's best game. It is more absorbing than chess, more of a gamble than poker, and lasts longer than Monopoly. It's free. It can be played anywhere - Archimedes did it in a bathtub.