Given:
In a two-digit number, the tens digit is 5 less than the units digit.
The number itself is five more than three times the sum of its digits.
To find:
The number.
Solution:
Let the two digit number is ab. So,
[tex]ab=10a+b[/tex]
Tens digit is 5 less than the units digit.
[tex]a=b-5[/tex] ...(i)
The number itself is five more than three times the sum of its digits.
[tex]10a+b=3(a+b)+5[/tex]
[tex]10a+b=3a+3b+5[/tex]
[tex]10a+b-3a-3b=5[/tex]
[tex]7a-2b=5[/tex] ...(ii)
Using (i) and (ii), we get
[tex]7(b-5)-2b=5[/tex]
[tex]7b-35-2b=5[/tex]
[tex]5b=5+35[/tex]
[tex]5b=40[/tex]
[tex]b=8[/tex]
Putting b=8 in (i), we get
[tex]a=8-5[/tex]
[tex]a=3[/tex]
Therefore, the required number is 38.
