the sum of the digits of a two digit number is ll. when the digit are interchanged the new number so formed becomes 45 more then the ogiginal number. find the number.
Answers
ᴀɴsᴡᴇʀ
- let the unit's digit be x
- then, ten's digit = 11-x
number formed by these digits:
= 10 × ten's digit + unit's digit
= 10(11-x) + x
= 110 - 10x + x
= 110 - 9x .........(i)
when the digits are interchange,
- unit's digit become = 11-x
- and ten's digit become = x
then, number formed on interchange the digits;
= 10 × ten's digit + unit's digit
= 10x + 11 - x
= 9x + 11 ........(ii)
and it is given that the new number is 45 more then the original number:
➪ new number = original number + 45
➪ new number - original number = 45
now put the value of new number and original number from eq.(i) and (ii)
➪ 9x + 11 - (110 - 9x) = 45
➪ 9x + 11 - 110 + 9x = 45
➪ 9x + 9x + 11 - 110 = 45
➪ 18x - 99 = 45
➪ 9(2x - 11) = 45
➪ 2x - 11 = 45/9
➪ 2x - 11 = 5
➪ 2x = 5 + 11
➪ 2x = 16
➪ x = 16/2
➪ x = 8
hence, unit's digit is 8
now, put the value of x in eq.(i)
➪ ten's digit = 11 - x
➪ ten's digit = 11 - 8
➪ ten's digit = 3
hence, ten's digit is 3
therefore, the original number is 38
_____________________
for verification →
from equation (i)
orginal number = 110 - 9x
➪ orginal number = 110 - 9 × 8
➪ orginal number = 110 - 72
➪ orginal number = 38
hence, varified!