Answer:

Let, the digit at tens place be x and the digit at units place be y in the original number.

Then the original number =10+y, and the new number obtained by interchanging the digits =10y+x.

By the data, the digit at tens place is double the digit at units place.

∴x=2y

∴x−2y=0       

Again, new number = original number - 27

∴10y+x=10x+y−27

∴10y+x−10x−y=−27

∴−9x+9y=−27            ∴x−y=3   

Subtracting eq. (1) from eq. (2),

x−y=3

x−2y=0

-      +      -

__________

     y=3

Substituting y=3 in x−2y=0

∴x−2(3)=0

∴x−6=0

∴x=6

Then, the original number =10x+y=10(6)+3=63

Hence, the original number is 63.

Answer:

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