Question

1) 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.
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Answer 1

#Brainly.in

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.

Answer 2

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

11x + 11y = 176

x + y = 16 ...........(1)

Also given that

y = x + 2

x - y= -2 ............(2)

Adding equation (1) and (2)

2x = 14

x = 7

Therefore,

y = 16-7 = 9

Therefore, the number is

10 × 9 + 7

= 97