Question

the sum of two digit number and the number formed by interchanging its digit is 1 10.in 10 is the subtracted from the first number, the new number is 4 more than 5 times the sum of the digit in the first number full stop find the first number

Answer 1

HEY DEAR ... ✌


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let the original number be (10x+y)

and after interchanging we get (10y+x)

(10x+y) + (10y+x) = 110... ..(1)

(10y+x) - 10 = 4+ 5(x+y)... ...(2)

eqn 1 can be written as 11x +11y = 110

0r x+y = 10....... ..(3)

putting the value of eqn (3) in eqn (2) we get,

(10y+x) - 10 = 4+ 5(10)

or 10y+x = 54+10

or 10y+x = 64.. ...(4)

putting the value of eqn (3) in eqn (4) we get,

10y + (10 - y ) = 64

or 9y = 54

or y = 6.........(5)

putting the value of y in eqn 3 we get

x+6 = 10

or x = 4

THE original number is 46.

HOPE , IT HELPS ... ✌

Answer 2

Here is your solution

Let the digit at ones place = y
Let the digit at tens place = x

original number = 10x + y 

Number formed by reversing the digits =10y + x

On adding original number and Reversed number.

10x+y+10y+x = 110
11x + 11y = 110
11 (x + y ) = 110
x + y = 10
x = 10 - y ................................................. (1)

on subtracting 10 from the original number , we get = 10x + y -10
and

4 more than 5 times the sum of digits = 5(x + y) + 4

A/q
 
10x + y -10 = 5(x + y) + 4
10x + y -10 = 5x +5y + 4
10x -5x + y -5y = 10 +4
5x - 4y = 14.................................................. (2)

putting the value of x in equation (2)

5x - 4y = 14
5×(10-y) - 4y = 14
50-5y-4y =14
-9y = 14 -50
-9y = -36
y = 4

Now

x = 10-y
x = 10 - 4
x = 6

original number or first number = 10x + y  = 10×6+4 = 64

Hope it helps you