Question

(ii) The sum of a two-digit number and the number obtained by reversing its digits
is 176. Find the number, if its tens place digit is greater than the units place
digit by 2.​

Answer 1

The sum of a two-digit number and the number obtained by reversing its digits is 176. If its tens place digit is greater than the units place digit by 2 then the number is 97.

Step-by-step explanation:

Let the digits at ones and tens place be x and y respectively

Then, the number obtained = 10y+x

Number obtained by reversing the digits = 10x+y

According to the question

$(10y+x)+(10x+y)=176$

$\implies 11x+11y=176$

$\implies x+y=16$ ...........(1)

Also given that

$y=x+2$

$\implies x-y=-2$ ............(2)

Adding equation (1) and (2)

$2x=14$

$\implies x=7$

Therefore,

$y=16-7=9$

Therefore, the number is

$10\times 9+7$

or, $97$

Hope this answer is helpful.

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Answer 2

Answer:

Step-by-step explanation: