Question

The sum of the digits of a two digit number is 13. If 27 is added to the number, the place of the digits are reversed. Then the required number is??
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Answer 1

Answer:-

Let the digit at ten's place be x and digit at one's place be y.

⟶ The number = 10x + y.

Given:

Sum of the digits = 13

⟹ x + y = 13

⟹ x = 13 - y -- equation (1)

Also given that,

If 27 is added to the number, the digits are reversed.

• Number formed by reversing the digits = 10y + x.

According to the above condition,

⟹ 10x + y + 27 = 10y + x

⟹ 27 = 10y + x - 10x - y

⟹ 27 = 9y - 9x

Substitute the value of x from equation (1).

⟹ 27 = 9y - 9(13 - y)

⟹ 27 = 9y - 117 + 9y

⟹ 27 + 117 = 18y

⟹ 144 = 18y

⟹ 144/18 = y

⟹ 8 = y

Substitute the value of y in equation (1).

⟹ x = 13 - 8

⟹ x = 5

Hence,

• The number = 10(5) + 8 = 50 + 8 = 58.

∴ The required two digit number is 58.

Answer 2

$\boxed{\mathfrak{ let \: the \: no \: be \: \:}} = { \mathfrak{X \: Y}}$

$\boxed{ \mathfrak{X + Y = 13}} \: \: \: \: \: \: \: \: (1)$

$\boxed{ \mathfrak{XY + 27 = XY}} \: \: \: \: \: (2)$

$\boxed{ \mathfrak{10X + Y + 27 = 10Y + X}}$

$\boxed{ \mathfrak{9X + 27 = 9Y}}$

$\boxed{ \mathfrak{X + 3 = Y}} \: \: \: \: \: \: \: \: (3)$

$\boxed{ \mathfrak{adding \: (1) \: and \: (2)}}$

$\boxed{ \mathfrak{2X = 10}}$

$\boxed{ \mathfrak{Y = 13 - 5 =q 8}}$

$\boxed{ \mathfrak{10×5+8}}$

$\huge \boxed{ \mathfrak{the \: number \: is \: 58}}$