Question

the sum of the digits of a two digit number is 10 if we reserve the digits the new number will be 54 more than the original number what is the original number

Answer 1

Let the digit at one's place = y
and at ten's place = x
Required no = 10x + y
Now, from the first part of question,
x + y = 10
y = 10 - x ------ (1)
From the second party of question,
After reversing the digits......
Required no = 10y + x

According to question,
10y + x - ( 10x + y ) = 54
=} 10y + x - 10x - y = 54
=} 9y - 9x = 54
=} 9 ( y - x) = 54
=} y - x = 54/9
=} y - x = 6
Therefore, y - x = 6 --------(2)

Now putting the value of (1) and (2),
y - x = 6
=} 10 - x - x = 6
=} 10 - 2x = 6
=} - 2x = 6 - 10
=} - 2x = - 4
=} x = - 4 / - 2
Therefore x = 2

Now ,
y = 10 - x
y = 10 - 2
y = 8

Therefore Required no = 10x + y
= 10 * 2 + 8
= 28


thanks......!!!!!!

Answer 2

Answer:

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