a two digit no. is 3 more than 4 times the sum of its digits. If 18 is added to the no., its digits are reversed. find the no.
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Let the digit in ones place be x and tens place be y
Original number = 10y + x
Number formed by reversing the digits = 10x + y
Given 10y + x = 4(x + y) + 3
⇒ 10y + x − 4x − 4y = 3
⇒ 6y − 3x = 3
⇒ 2y − x = 1 → (1)
Also given that when 18 is added to the number the digits gets interchanged.
Hence (10y + x) + 18 = 10x + y
⇒ 9x − 9y = 18
⇒ x − y = 2 → (2)
Add (1) and (2), we get
2y − x = 1
x − y = 2
---------------
y = 3
Put y=3 in x − y = 2, we get
x − 3 = 2
∴ x = 5
Hence (10y + x) = 10(3) + 5 = 35
Original number = 10y + x
Number formed by reversing the digits = 10x + y
Given 10y + x = 4(x + y) + 3
⇒ 10y + x − 4x − 4y = 3
⇒ 6y − 3x = 3
⇒ 2y − x = 1 → (1)
Also given that when 18 is added to the number the digits gets interchanged.
Hence (10y + x) + 18 = 10x + y
⇒ 9x − 9y = 18
⇒ x − y = 2 → (2)
Add (1) and (2), we get
2y − x = 1
x − y = 2
---------------
y = 3
Put y=3 in x − y = 2, we get
x − 3 = 2
∴ x = 5
Hence (10y + x) = 10(3) + 5 = 35
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