Question

sum of the digits of a two-digit number is 5 when the digits are reversed the resulting number is greater than the original number by 27 find the number​

Answer 1

Let the number be 10x+y

x+y = 5 …(1)

10x+y-(10y+x) = 27, or

10x-x-10y+y= 27, or

9x-9y = 27, or

x-y = 3 …(2)

Add (1) and (2)

2x= 8, or

x = 4 and y = 1

So the number is 41 and its subsidiary is 14 and the difference between them is 27 and the sum of the digits is 5

Answer 2

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

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• Sum of the digits of 2 digit no. = 5
• The new number formed by reversing the digits is greater than the original by 27

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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 number be 10x + y

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Acc. to the first condition :-

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X + Y = 5 --- ( i )

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Acc. to the second condition :-

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Reversed number = 10y + x

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10x + y + 27 = 10y + x

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10y - y - 10x + x = 27

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9y - 9x = 27

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9 ( y - x ) = 27

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y - x = 27 / 9

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y - x = 3 ---- ( ii )

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Adding eq ( i ) and ( ii )

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x + y + y - x = 3 + 5

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y + y = 3 + 5

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2y = 3 + 5

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2y = 8

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y = 8 / 2

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y = 4

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• putting value of y in eq ( i )

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x + y = 5

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x + 4 = 5

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x = 5 - 4

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x = 1

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• Putting values in the original no.

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Original no. = 10x + y

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Original no. = 10 × 1 + 4

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Original no. = 10 + 4

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Original no. = 14

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

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• Original number is 14