Given:
The tens digit of a two digit number is 5 greater the units digit.
If you subtract double the reversed number from it, the result is a fourth of the original number.
To find:
The original number.
Solution:
Let n be the two digit number and x be the unit digit. Then tens digit is (x+5) and the original number is:
[tex]n=(x+5)\times 10+x\times 1[/tex]
[tex]n=10x+50+x[/tex]
[tex]n=11x+50[/tex]
Reversed number is:
[tex]x\times 10+(x+5)\times 1=10x+x+5[/tex]
[tex]x\times 10+(x+5)\times 1=11x+5[/tex]
If you subtract double the reversed number from it, the result is a fourth of the original number.
[tex]11x+50-2(11x+5)=\dfrac{1}{4}(11x+50)[/tex]
[tex]11x+50-22x-10=\dfrac{1}{4}(11x+50)[/tex]
[tex]40-11x=\dfrac{1}{4}(11x+50)[/tex]
Multiply both sides by 4.
[tex]160-44x=11x+50[/tex]
[tex]160-50=11x+44x[/tex]
[tex]110=55x[/tex]
Divide both sides by 55.
[tex]\dfrac{110}{55}=x[/tex]
[tex]2=x[/tex]
The unit digit is 2. So, the tens digit is [tex]2+5=7[/tex].
Therefore, the original number is 72.
