i) The sum of the digits of a two-digit number is 9. Also, nine times this number is
twice the number obtained by reversing the order of the digits. Find the number.
Answers
EXPLANATION.
Let the ten's place be = x
Let the unit place be = y
original number = 10y + x
reversing number = 10x + y
To find the number.
According to the question,
Case = 1.
sum of the digit of two digit number = 9
=> x + y = 9 .......(1)
Case = 2.
Nine times this number is twice the number
obtained by reversing the order of digit.
=> 9 ( 10y + x) = 2 ( 10x + y)
=> 90y + 9x = 20x + 2y
=> 88y = 11x
= 8y = x .....(2)
put the value of x = 8y in equation (1)
we get,
=> 8y + y = 9
=> 9y = 9
=> y = 1
put the Value of y = 1 in equation (2)
we get,
=> x = 8(1)
=> x = 8
Therefore,
original number = 10y + x
=> 10 X 1 + 8
=> 18
original number = 18
Step-by-step explanation:
- The sum of the digits of a two-digit number is 9.
- Nine times this number is twice the number obtained by reversing the order of the digits.
- The original number.
Let the tens digit of the number be x
The ones digit of the number be y
The original number = 10x + y
The reversed number = 10y + x
According to the 1st condition:-
The sum of digits of the number is 9
So:-
→ x + y = 9.........(i)
According to the 2nd condition:-
Nine times this number is twice the number obtained by reversing the order of the digits.
→ 9 × original number = 2 × reversed number
→ 9( 10x + y ) = 2(10y + x)
→ 90x + 9y = 20y + 2x
→ 90x - 2x + 9y - 20y = 0
→ 88x - 11y = 0
Dividing the equation by 11
→ 8x - y = 0........(ii)
Adding equation (i) and (ii)
→ (x + y) + ( 8x - y ) = 9 + 0
→ x + y + 8x - y = 9
→ 9x = 9
→ x = 1
Substituting x = 1 in equation (i)
→ x + y = 9
→ 1 + y = 9
→ y = 9 - 1
→ y = 8
The original number
= 10x + y
= 10( 1) + 8
= 10 + 8
= 18