Question

A two digit no is 3 more than 4 times the sum of its digits if 8 is added to a number , its digits are reversed find the number

Answer 1

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

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Answer 2

solutions :-

Let ones place digit be x .

Tens place digit be y .

Original number = 10y + x

Number formed by reversing the digits = 10x + y

A two digit number is 3more than 4times the sum of its digits .

A/q

=>10y + x = 4(x + y) + 3 
=>10y + x − 4x − 4y = 3
 => 6y − 3x = 3 
=>2y − x = 1..................(1) 

Again when 18 is added to the number then digits get interchanged. 

(10y + x) + 18 = 10x + y 

=>9x − 9y = 18
=> x − y = 2...................(2) 

On adding equation (1) and (2):-

 2y − x +(x-y)= 1+2
2y-x+x-y = 3
Y = 3

Putting the value of y im equation (2)

=> x − y = 2
=>x- 3 =2
=>x= 2+3
=>x =5

Hence

Number = 10y +x= 10×3+5=35

Thanks