sum of digit of a two digit number is 11.The given number is less than the number obtained by reversing the digit by 9.Find the number.
Answers
Answer:
The sum of a 2 digit number is 11. If the number obtained by reversing the digit is 9 less than the original number, what is the number?
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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
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.