Question

In a two-digits number, the digit at ten's place

is double the digit at unit's place. If the digitsof the number are interchanged, the new

number so formed is 27 less than the original

number. Find the original number.
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Answer 1

Answer:

$■THE \: ORIGINAL \: NUMBER \: is \: 63■$

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 2

Answer:

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Step-by-step explanation:

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