Let the one's digit be x and ten's digit be y.
∴ Original number = x + 10y
If the digits are interchanged,
∴ Interchanged number = y + 10x
Now according to 1st condition,
⇒ x + y = 11
⇒ x = 11 - y -(1)
According to 2nd condition,
⇒ y + 10x = x + 10y + 27
⇒ y + 10x - x - 10y = 27
⇒ 9x - 9y = 27
⇒ 9(x - y) = 27
⇒ x - y = 27/9
⇒ x - y = 3
⇒ 11 - y - y = 3 [Substituting value from (1)]
⇒ 11 - 2y = 3
⇒ -2y = 3 - 11
⇒ -2y = -8
⇒ y = -8/-2
∴ y = 4
So, x = 11 - 4
∴ x = 7
∴ Original number = 7 + 10(4)
⇒ Original number = 7 + 40
∴ Original number = 47
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Solution :
Let the one's digit be x and ten's digit be y.
∴ Original number = x + 10y
If the digits are interchanged,
∴ Interchanged number = y + 10x
Now according to 1st condition,
⇒ x + y = 11
⇒ x = 11 - y -(1)
According to 2nd condition,
⇒ y + 10x = x + 10y + 27
⇒ y + 10x - x - 10y = 27
⇒ 9x - 9y = 27
⇒ 9(x - y) = 27
⇒ x - y = 27/9
⇒ x - y = 3
⇒ 11 - y - y = 3 [Substituting value from (1)]
⇒ 11 - 2y = 3
⇒ -2y = 3 - 11
⇒ -2y = -8
⇒ y = -8/-2
∴ y = 4
So, x = 11 - 4
∴ x = 7
∴ Original number = 7 + 10(4)
⇒ Original number = 7 + 40
∴ Original number = 47
∴ The required number = 47.