Question

One of the digits of a two digit no. is twice the other digit. The sum of the original number and the number formed by reversing the digits is 132. Find the number.​

Answer 1

Answer:

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

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One of the digits of a two digit no. is twice the other digit .

The sum of the original number and the number formed by reversing the digits is 132.

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

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The original number = ??

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

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Let the original no. be 10x + y

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No. formed by reversing the digits = 10y + x

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Acc. to the first statement :-

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x = 2y --- ( i )

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Acc. to the second statement :-

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10x + y + 10y + x = 132

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11x + 11y = 132

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11( x + y ) = 132

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x + y = 132 / 11

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x + y = 12 --- ( ii )

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Putting value of x in eq ( ii ) from eq ( i )

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2y + y = 12

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3y = 12

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y = 12 / 3

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y = 4

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Putting value of y in eq ( i )

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x = 2y

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x = 2 × 4

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x = 8

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Finding the original no.

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Original no. = 10x + y

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Original no. = 10 × 8 + 4

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Original no. = 80 + 4

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Original no. = 84

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

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Required number = 84

Answer 2

Answer:

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