a two digit number is 3 more than 4 time sum of its .if 18 is added to the number,its digit are reversed .find the number
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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
Hope it helps you
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
Hope it helps you
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