Question

A number consists of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.

Answer 1

Answer:

Let the digits in the tens be x and ones place be y.

Hence the number is 10x+y

By reversing 10y+x

Sum of digit =5

⇒x+y=5⟶(i)

Also that when 9 is added to the number the digits get interchanged.

∴(10x+y)+9=(10y+x)

10x+y+9=10y+x=0

9x−9y=−9

x−y=−1⟶(ii)

Adding (i) &(ii) we get ,

⇒x+y=5

⇒x−y=−1

⇒2x=4

⇒x=2

Put x=2 in x+y=5

∴2+y=5

⇒y=3

Hence the no: is 23.

Answer 2

$\huge\mathfrak\red{Answer}$ :-

Let the digits in the tens be x and ones place be y.

Hence the number is 10x+y

By reversing 10y+x

Sum of digit 5

⇒ x+y=5-(i)

Also that when 9 is added to the number the digits get interchanged.

(10x+y)+9=(10y+x)

10x+y+9=10y+x=0

9x-9y=-9

x-y=-1→(ii)

Adding (i) &(ii) we get,

⇒ x+y=5

⇒ x−y=-1

⇒ 2x=4

⇒x=2

Put x=2 in x+y=5

2+y=5

⇒y=3

Hence the no: is 23.

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