Question

A 2-digit number has 5 in the ones place. When 27 is added to the number, the digits are reversed. Find the number.
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Answer 1

Question:-

• A 2-digit number has 5 in the ones place. When 27 is added to the number, the digits are reversed. Find the number.

Answer:-

Required answer = 25

Calculations:-

Let the digits of the number be x and y

Then the number will be 10x + y.

After reversing it will be 10y + x.

According to the question:-

Original number + 27 = reversed number

Now plug the values the given values of original number and the reversed number.

10x + y + 27 = 10y + x

10x - x + y - 10y + 27 = 0

9x - 9y + 27 = 0

x - y + 3 = 0

As given in the question, the ones place of original no. is 5. Then y = 5. Putting the value of y in the equation:

x - 5 + 3 = 0

x - 2 = 0

x = 2

Then the number will be:

10(2) + 5 = 20 + 5 = 25 (Answer)

Answer 2

GiveN:

$\sf{A \:2-digit\: number \:has \:5\: in\: the \:ones\: place.}$

$\sf{When\: 27\: is\: added \:to\: the \:number, the\: digits\: }$$\sf{are \:reversed.}$

To FinD:

$\sf\orange{The\: two-digit\: number?}$

Step-by-Step Explanation:

$\sf\purple{Let\: the\: digits\: of \:the\: number\: be\: x\: and\: y.}$

$\sf\purple{Then \:the \:number \:will\: be \:10x + y.}$

$\sf\purple{After\: reversing \:it \:will\: be\:10y + x.}$

According to question,

$\sf\red{⇒ Original\: number + 27 = Reversed \:number.}$

Now plug the values the given values of original number and the reversed number.

$\sf\blue{⇒ 10x + y + 27 = 10y + x}$

$\sf\blue{⇒ 10x - x + y - 10y + 27 = 0}$

$\sf\blue{⇒ 9x - 9y + 27 = 0}$

$\sf\blue{⇒ x - y + 3 = 0}$

As given in the question, the ones place of original no. is 5. Then, y = 5. Putting the value of y in the equation:

$\sf\pink{⇒ x - 5 + 3 = 0}$

$\sf\pink{⇒ x - 2 = 0}$

$\sf\pink{⇒ x = 2}$

Then the number will be:

$\sf\green{⇒ 10(2) + 5 = 25 (Answer)}$