Question

The sum of the digits of a two digit number 9.Also nine times this number is twice obtained by reversing the order of the digits. Find the number.

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Answer 1

Step-by-step explanation:

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.

Answer 2

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

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• Sum of 2 digits of 2 digit no. = 9
• nine times the original no. is equal to twice the no. obtain by reversing the digits

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

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• Original no.

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

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Let the original number be 10x + y

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Acc. to the 1st statement :-

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x + y = 9 ---- ( i )

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Interchange no. = 10y + x

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Acc. to the 2nd statement :-

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

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90x + 9y = 20y + 2x

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90x - 2x - 20y + 9y = 0

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88x - 11y = 0

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11(8x - y ) = 0

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8x - y = 0

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• Adding eq ( i ) and ( ii )

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x + y + 8x - y = 9 + 0

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9x = 9

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x = 9 / 9

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x = 1

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• Putting value of x in eq ( i )

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x + y = 9

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1 + y = 9

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y = 9 - 1

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y = 8

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• Finding the original no.

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10x + y

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10 x 1 + 8

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10 + 8

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18

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

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• Original number = 18