Question

- 8 The digit in the ones place of a number is twice the digit in its tens place. If 9 is added to the number, the digits of the sum obtained are reverse of those of the number. Find the number.​

Answer 1

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Let the no be 10x+y

Y is in ones place and x is in tens place According to question the digit in its ones place is twice that of the no which is in tens

place

So x=2y

If 9 is added to the no then the new no is

10x+y +9

And the digits of sum obtained is reversed

So,

10x+y+9 = 10y + x

10x2y+ y + 9 = 10y + 2y

21y+9 = 12y

21y-12y = -9 9y = -9

Y = -1

X= 2y

X= -2

The number is 10x+y

10x-2+-1 = -21

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Hope it's helpful to you

Answer 2

Step-by-step explanation:

Let the no be 10x+y

Y is in ones place and x is in tens place According to question the digit in its ones place is twice that of the no which is in tens

place

So x=2y

If 9 is added to the no then the new no is

10x+y +9

And the digits of sum obtained is reversed

So,

10x+y+9 = 10y + x

10x2y+ y + 9 = 10y + 2y

21y+9 = 12y

21y-12y = -9 9y = -9

Y = -1

X= 2y

X= -2

The number is 10x+y

10x-2+-1 = -21