Question

nine times a two digit number is the same as twice the number obtained by interchanging the digits of the number if one digit of the number exceeds The Other by 7 find the number​

Answer 1

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Secondary SchoolMath 13 points

Nine time a two digit number is the same as twice the number obtained byinterchanging the digits of the number if one digit of the number exceeds the other number by 7 find the number

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Joshuawoskk

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Let the ten's digit be x and

the unit's digit be y

therefore, the number becomes = 10x+y and

when the number is reversed = 10y+x

Now, when nine times a two-digit number is the same as twice the number obtained by interchanging the digits of the number.

⇒9(10x+y) = 2 (10y+x)

⇒90x+9y = 20y+2x

⇒90x-2x = 20y - 9y

⇒88x=11y

or y = 8x.........(i)

Now, if one digit of the no. exceeds the other by 7, then

y = x+7

⇒8x = x+7 [From (i)]

⇒8x-x=7

⇒7x=7

⇒x=7/7

⇒x=1

From (i) , we get,

y = 8(1)

⇒y = 8

Thus, the number becomes = 10x+y = 10(1)+8=10+8 = 18

Answer 2

Answer:

Step-by-step explanation:

let the units digit is x

and the tens digit is y

the number so obtained is 10y+x

interchanging the digits we get the number as 10x+y

9(10y+x) = 2(10x+y)

90y+9x = 20x+2y

88y-11x = 0     .................1

x=y+7 given

88y - 11(y+7) = 0

88y-11y-77=0

77y=77

y=1

x=8

number is 18