Question

Sum of two digit of a number is 11.If the digits are interchanged the number so got is 27 more than it.Find the numbers?​

Answer 1

Answer:

• The original number is 47.

Step-by-step explanation:

Given that:

• Sum of two digit of a number is 11.
• The digits are interchanged the number so got is 27 more than it.

To Find:

• The original number.

Let us assume:

• Tens digit be x.
• Ones digit be y.
• Original number = 10x + y

When digits are interchanged:

• New number = 10y + x

Finding the value of x and y:

Sum of two digit.

⟶ x + y = 11

⟶ x = 11 - y ════(i)

When the digits are interchanged.

⟶ 10x + y + 27 = 10y + x

Substituting the value of x from eqⁿ (i).

⟶ 10(11 - y) + y + 27 = 10y + 11 - y

⟶ 110 - 10y + y + 27 = 9y + 11

⟶ 137 - 9y = 9y + 11

⟶ 9y + 9y = 137 - 11

⟶ 18y = 126

⟶ y = 126/18

⟶ y = 7

In equation (i).

⟶ x = 11 - y

Substituting the value of y.

⟶ x = 11 - 7

⟶ x = 4

Finding the original number:

→ Original number = 10x + y

→ Original number = 10(4) + 7

→ Original number = 40 + 7

→ Original number = 47

Answer 2

Given:-

• Sum of two digit of a number = 11

• The digits are interchanged and the number so got is 27 more than it.

To Find:-

• The Numbers

Solution:-

Let’s represent the tens digit with a and the ones digit with b.

As given,

$⟹\:a + b = 11$

$⟹\:b = 11-a$ ____{1}

The number will be "10a + b".

As given in the question the digits are interchanged. So, According to question

$⟹\:10a + b = 10b + a + 27$

$⟹\:10a + (11 - a) = 10(11 - a) + a + 27$ {from 1}

$⟹\:10a + 11 - a = 110 - 10a + a + 27$

$⟹\:9a + 11 = 110 - 9a + 27$

$⟹\:9a + 11 = 137 - 9a$

$⟹\:18a + 11 = 137$

$⟹\:18a = 126$

$⟹\:a = 126/18$

${\boxed{\red{⟹\:a = 7}}}$

Putting the value of a in {1},

$⟹\:b = 11-a$

$⟹\:b = 11-7$

${\boxed{\red{⟹\:b = 4}}}$

So, the interchanged number is 10a + b = 74.

On interchanging it we get the original number which is 47.

Hence, The original number is 47.

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