Question

the sum of a digit of a two digit number is 11. if the tens place is 5 less than the ones digit, find the number​

Answer 1

Step-by-step explanation:

let the ones digit be x

and the tens digit be y

given,

x+y=11(eq1)

y+5=x (eq 2)

substituting 2 in 1

y+5+y=11

2y=6

y=3

x =y+5

=3+5

=8

number = 10x+y

=10(8)+3

=83

Answer 2

Step-by-step explanation:

Let tens digit be x

Let ones digit be y

Therefore

10x+y

Sum of the digits x+y=11

As they said tens place is 5 less than ones digit

x=y-5

Substitute x value

10(y-5)+y

10y-50+y=11y-50

As x+y=11

1(y-5)+y = y-5+y

=2y-5

2y-5=11

2y=11+5

2y=16

y=8

As y=8

11y-50=11(8)-50=88-50

=38