The sum of the digits of a two-digit number is 11; If the number to the left of the number increases by 2, it becomes 1/8 of the number, what is the number?
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Answer:
Given:
Sum of two digit number = 11
Let unit’s digit be ‘x’
and tens digit be ‘y’,
then x+y=11…(i)
and number = x+10y
By reversing the digits,
Unit digit be ‘y’
and tens digit be ‘x’
and number =y+10x+9
Now by equating both numbers,
y+10x+9=x+10y
10x+y–10y–x=−9
9x–9y=−9
x–y=−1…(ii)
Adding (i) and (ii), we get
2x=10
x=10/2
=5
∴y=1+5=6
By substituting the vales of x and y, we get
Number = x+10y
=5+10×6
=5+60
=65
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