Question

a two digit number os 3 more than 4 times the sum of digits. if 18 is added to the number, tjr digits are reserved. fimd the number.

Answer 1

Hello .
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
Hope it helps.

Answer 2

Hey dude.....

Very gud mrng....

Solution:

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− 3= 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

Hope this may help u......

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