Question

in a two digit number the sum of two digits is 8 if 18 is added to the number its digits are reversed find the number​

Answer 1

Answer:

Let one digit of the number be x , then its ten digits digits will be (8−x) .

So , given number =10(8−x)+x

=80−10x+x

=80−9x

Now if digits are reversed , then Number obtained

=10x+(8−x)

=10x+8−x

=9x+8

According to question

(80−9x)+18=9x+8

98−9x=9x+8

90=18x

x=5

⇒ given number =80−9x

=80−9×5

=80−45

=35

∴ the number is 35

Step-by-step explanation:

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Answer 2

• The number is 35

Step-by-step explanation:

Let,

• The tense digit be 'x'
• The Ones digit be 'y'
• The number be '10x + y' ------- (1)

According to the question,

x + y = 8

y = 8 - x -------- (2)

Solving,

⟼ 10x + y = 10x + (8 - x)

⟼ (10x - x) + 8

⟼ 9x + 8

Reversed number,

⟼ 10y + x = 10 (8 - x) + x

⟼ (80 - 8x) + x

⟼ 80 - 9x

When 18 is added to the number, its digits are interchanged,

⟼ (9x + 8) + 18 = 80 - 9x

⟼ 9x + 9x = 80 - 18 - 8

⟼ 18x = 54

⟼ x = 54 / 18

⟼ x = 3

Substituting 'x' in (2),

⟼ y = 8 - x

⟼ y = 8 - 3

⟼ y = 5

Substituting 'x' & 'y' in (1),

⟼ 10x + y

⟼ 10 (3) + 5

⟼ 30 + 5

⟼ 35

• The number is 35