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
Given :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.
Solution:
Let the digit in the unit's place be x and the digit at the tens place be y.
Number = 10y + x
The number obtained by reversing the order of the digits is = 10x + y
ATQ :
Condition : 1
x + y = 9……………(1)
Condition : 2
9(10y + x) = 2(10x + y)
90y + 9x = 20x + 2y
20x + 2y – 90y – 9x = 0
11x - 88y = 0
11(x - 8y) = 0
x - 8y = 0 …………(2)
On Subtracting equation (2) from equation (1), we obtain :
x + y = 9
x – 8y = 0
(-) (+) (-)
------------------
9y = 9
y = 9/9
y = 1
On putting y = 1 in eq (2) we obtain :
x - 8y = 0
x - 8 × 1 = 0
x - 8 = 0
x = 8
Now, Number = 10y + x = 10 x 1 + 8 = 10 + 8 = 18
Hence, the number is 18.
Hope this answer will help you…
Some more questions from this chapter :
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Answer:
Step-by-step explanation:
Solution:-
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.