In a two digit number, digit at the ten's place is twice the digit at unit's place. If the number obtained by interchanging the digit is added to the original number, the sum is 66. Find the number.
Let 10y+x be the digit number.
It is given that the ten's place is twice the digit at unit's
place.
i.e. y=2x ------(1)
Also, (10y+x)+(10x+y)=66
10y+x+10x+y=66
11y+11x=66
⇒x+y=6
⇒x+2x=6 [from (1)]
⇒3x=6
⇒x=2
Form (1), y=2(2)=4
The required number =10(4)+2=42