Question

the sum of the digits of a two-digit is 10. if the number formed by reversing the digit is greater than the original number by 36. find the original number. check your solution

Answer 1

Let the required two-digit number be 10x+y where x and y denote the digits of the number.

Sum of digits is given as 10. Therefore,

x+y=10⋯(1)

The reverse of the required number is 10y+x. [The digits are reversed]

Let us assume for convenience, the required number is greater than its reverse.

From the problem, the difference between the required number and it's reverse is 36. Therefore,

(10x+y)−(10y+x)=36

10x+y−10y−x=36

9x−9y=36

(÷9)⟹x−y=4⋯(2)

Solve equations (1) and (2).

x+y=10

x−y=4

⟹2x=14 [by eliminating 'y' from both equations].

⟹x=142=7

Substitute 'x' value in equation (1).

7+y=10⟹y=10−7=3

x=7,y=3

Now substitute 'x' and 'y' values to get the required number.

10x+y=10(7)+3=70+3=73

Now let's check whether the found number meets the required conditions:

Sum of digits = 7+3 = 10

Difference between number and it's reverse = 73–37 = 36.

Therefore the number is correct.

Answer 2

Let the two digit number be 10x + y.

y + x= 10 → (1)

The number formed by reversing the digits will be 10y + x.

(10y + x) - (10x + y) = 36

= 10y + x - 10x - y = 36

= 9y - 9x = 36

= 9(y - x) = 36

y - x = 36 ÷ 9 = 4 → (2)

(1) + (2)

= y + x + y - x = 10 + 4

= 2y = 14

y = 14 ÷ 2 = 7

(1) - (2)

= (y + x) - (y - x) = 10 - 4

= y + x - y + x = 6

= 2x = 6

x = 6 ÷ 2 = 3

10x + y = 10 × 3 + 7 = 30 + 7 = 37

Let's check.

3 + 7 = 10

The number got by interchanging the digits = 10y + x = 73

73 - 37 = 36

∴ 37 is the answer.

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