the sum of the digits of a two digits number is 9.Also, nine times this number is twice the number obtained by reversing the order of the digits. find the number.
Answers
Answer:
hy
Step-by-step explanation:
Given :-
The sum of digits f a two digit number is 9.
Also nine times this number is twice the number obtained by reversing the order of digit.
To Find :-
The Number
Solution :-
Let the unit digit and tens digits of the number be x and y
Number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
⇒ x + y = 9 ... (i)
⇒ 9(10y + x) = 2(10x + y)
⇒ 88y - 11x = 0
⇒ -x + 8y =0 ... (ii)
Adding equation (i) and (ii), we get
⇒ 9y = 9
⇒ y = 1 ... (iii)
Putting the value in equation (i), we get
⇒ x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18.
Let the two digit number is 10x + y. S
um of the digits of to digit number is
x + y = 9….... (i)
Number obtained by reversing the order of the digits,
we get 10y + x.
Twitching the 10y + x, it is nine times of first number.
∴ 2(10y + x) = 9(10x + 4)
20y + 2x = 90x + 9y
2x – 90x + 20y – 9y = 0
-88x + 11y = 0
88x – 11y = 0
8x – y = 0 ………….. (ii)
By adding eqn. (i) to eqn. (ii),
(in attachment)
Substituting the value of ‘x’ in eqn. (i),
x + y = 9
1 + y = 9
y = 9 – 1
y = 8
∴ x = 1, y = 8
∴ Two digit number = 10x + y
= 10 × 1 + 8
= 10 + 8