Question

the sum of digits of two digit number is 11.if the number obtained by reversing the digits is 9 less than the origanal number,find the number

Answer 1

Answer:

5 years or till the child attains age of majority (whichever is earlier). The

Let the 2 digit number as(10a+b)

Number obtain by reversing the digit=(10b+a)

Given,

( a+b=11)~equation-1

also,

=(10a+b)-(10b+a)=9

=10a+b-10b-a=9

=9a-9b=9

=9(a-b)=9

=(a-b)=9–9

=a-b=1

=a+b+a-b=11+1

=2a=12

=a=6

Putting the value of a in equation 1, we get

=6+b=11

=b=11–6

=b=5

Original number=10×6+5=60+5=65

So the require number is 65

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Answer 2

Answer:

65

Step-by-step explanation:

 

Given:

Sum of two digit number = 11

Let unit’s digit be ‘x’

and tens digit be ‘y’,

then x+y=11…(i)

and number = x+10y

By reversing the digits,

Unit digit be ‘y’

and tens digit be ‘x’

and number =y+10x+9

Now by equating both numbers,

y+10x+9=x+10y

10x+y–10y–x=−9

9x–9y=−9

x–y=−1…(ii)

Adding (i) and (ii), we get

2x=10

x=10/2

=5

∴y=1+5=6

By substituting the vales of x and y, we get

Number = x+10y

=5+10×6

=5+60

=65

∴ The number is 65.