Question

the sum of the digits of a two digit number is 9.the no. obtained by reversing the order of the digits of the no. exceeds the given the given no. by 27.find the given no.



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Answer 1

Answer:

Original number=36

Reversed number=63

Given:

Sum of two digit=9

After reversing it exceeds 27

To find:

The given numbers

Solution:-

Let the ten's digit be x and unit digit be y

Original number=10x+y

After reversing

number=10y+x

Given,

x+y=9

y=9-x .........i)

Now A/Q

10x+y=10y+x+27

10x-x+y-10y=27

9x-9y=27

9(x-y)=27

x-y=3 ........ii)

Solving i) and ii)

y-x+x+y=3+9

2y=12

y=6

•x=9-6

=3

______________

Original number=10x+y

=30+6

=36

Reversed number=10y+x

=60+3

=63

Answer 2

Answer:

Let the unit digit be x and ten's digit be y.

Then, Number = 10y + x

According to the question,

Sum of digits, x + y = 9 --> ( i )

Also,

10y + x - ( 10x +y) = 27

9y - 9x = 27

y - x = 3 --> ( ii )

Adding (i) and (ii),

2y = 12

y = 6

x + y = 9

x = 9 - 6 = 3

x = 3

Number = 10y +x = 10×6 + 3 = 63

Number = 63 or 36 ( After reversing)