Math, asked by mistryganesh74, 11 days ago

The digits of a two-digit number differ by 6. If the digits are interchanged and the resulting number is added to the original number, we get 110. What is the original number 2.​

Answers

Answered by velpulaaneesh123
3

Answer:

85 and 58

Step-by-step explanation:

Let the digits of the number be a and b such that the number is (10a+b).

According to the question,

a−b=6 or b−a=6   ____ (1)

10a+b+10b+a=110

a+b=16    _____(2)

Solving both the equations, we have

a=8 and b=5 or a=5 and b=8

Therefore, the required number is 85 or 58.

Answered by varadad25
9

Answer:

The original number is 82.

Step-by-step-explanation:

Let the greater digit at tens place be x.

And the smaller digit at units place be y.

∴ The two digit number = xy

Original number = 10x + y

From the first condition,

x - y = 6

⇒ x = 6 + y

x = y + 6 - - - ( 1 )

Now,

The number obtained by interchanging the digits = yx

New number = 10y + x

From the second condition,

Original number + New number = 110

⇒ 10x + y + 10y + x = 110

⇒ 11x + 11y = 110

⇒ x + y = 10 - - - [ Dividing by 11 ]

⇒ y + 6 + y = 10 - - - [ From ( 1 ) ]

⇒ 2y = 10 - 6

⇒ 2y = 4

⇒ y = 4 ÷ 2

y = 2

By substituting y = 2 in equation ( 1 ), we get,

x = y + 6 - - - ( 1 )

⇒ x = 2 + 6

x = 8

Now,

Original number = 10x + y

⇒ Original number = 10 * 8 + 2

⇒ Original number = 80 + 2

Original number = 82

∴ The original number is 82.

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