Math, asked by kaustubhdesale, 6 months ago

the sum of the digits of two-digit numbers is 9. also,nine times this numbers is twice the number obtained by reversing the order of the digits find the numbers.​

Answers

Answered by ap5495989
4

Answer:

ANSWER

Let the ten's digit no. be x and one's digit no. be y.

So the no. will be = 10x+y.

Given : x+y=9-----(I)

9(10x+y)=2(10y+x) ⇒88x−11y=0 -----(II)

On solving I and II simultaneously you will get x=1 and y=8.

Therefore your desired no. is 18.

Answered by deepalunthi878
1

Step-by-step explanation:

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.

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