Question

A number consist of two digit whose sum is 16 if 18 is subtracted from the number. it's digit are reversed.find the number?​

Answer 1

Let ten's digit(first no) be x

Let ten's digit(first no) be xSo ,

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6Therefore the second digit(one's digit)= (6-x)

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6Therefore the second digit(one's digit)= (6-x)Now, original no=10x+(6-x)

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6Therefore the second digit(one's digit)= (6-x)Now, original no=10x+(6-x)According to the question,

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6Therefore the second digit(one's digit)= (6-x)Now, original no=10x+(6-x)According to the question,[10x+(6-x)]-18=[10(6-x)+x]

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6Therefore the second digit(one's digit)= (6-x)Now, original no=10x+(6-x)According to the question,[10x+(6-x)]-18=[10(6-x)+x][10x+6-x]-18=[60-10x+x]

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6Therefore the second digit(one's digit)= (6-x)Now, original no=10x+(6-x)According to the question,[10x+(6-x)]-18=[10(6-x)+x][10x+6-x]-18=[60-10x+x]9x+6-18=60-9x

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6Therefore the second digit(one's digit)= (6-x)Now, original no=10x+(6-x)According to the question,[10x+(6-x)]-18=[10(6-x)+x][10x+6-x]-18=[60-10x+x]9x+6-18=60-9x9x+9x=60+12

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6Therefore the second digit(one's digit)= (6-x)Now, original no=10x+(6-x)According to the question,[10x+(6-x)]-18=[10(6-x)+x][10x+6-x]-18=[60-10x+x]9x+6-18=60-9x9x+9x=60+1218x=72

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6Therefore the second digit(one's digit)= (6-x)Now, original no=10x+(6-x)According to the question,[10x+(6-x)]-18=[10(6-x)+x][10x+6-x]-18=[60-10x+x]9x+6-18=60-9x9x+9x=60+1218x=72x=4

Let ten's digit(first no) be xSo ,Since the sum of two digits is 6Therefore the second digit(one's digit)= (6-x)Now, original no=10x+(6-x)According to the question,[10x+(6-x)]-18=[10(6-x)+x][10x+6-x]-18=[60-10x+x]9x+6-18=60-9x9x+9x=60+1218x=72x=4Hence the number is 10x+(6-x)=10(4)+(6–4) i.e.40+2=42