Question

15. A two-digit number is 4 more than 6 times the sum of its
digit . If 8 is subtracted from the number the digits are
reversed. Find the number.

Answer 1

i think you've considered 8 instead of 18 by mistake,

Step-by-step explanation:

let the the digit = x

and the unit digit = y

number = 10x+y

when digit are reversed then no. = 10y+x

condition I

x+10y = 6(x+y)+4

x+10y = 6x+6y+4

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

4y = 5x +4 ..................(1)

condition II

x+10y-18 =y+10x-x

9y-18 = 9x

y-2 = x..................(2)

substituting the value of x in eq. 1

4y = 5x + 4

4y = 5(y-2)+4

4y = 5y -10 +4

4y = 5y-6

4y-5y =-6

-y = -6

y= 6

Putting the value of y in eq. 2

y - 2 = x

6 - 2 = x

4 = x

The no. is 46

and reversed no. is 64

I hope it helps you

Answer 2

Answer

Let x= ones digit and y=tens digit

According to the question ,

$∴x+10y=6(x+y)+4 \\ </p><p>⇒10y=6x+6y+4−x \\ </p><p>⇒4y=5x+4</p><p></p><p>$

And , from if 18 is substracted from the number , the digits are reversed ,

[

$∴x+10y−18=y+10x \\ </p><p></p><p>⇒10y−18=y+9x \\ </p><p>⇒9y−18=9x \\ </p><p>⇒y−2=x$

Substitute x value into :

[

$4y=5x+4 \\ </p><p>4y=5(y−2)+4 \\ </p><p>4y=5y−6 \\ </p><p>y=6</p><p></p><p>$

Substitute y into :

[

$]∴y−2=x \\ </p><p>⇒6−2=x \\ </p><p>⇒4=x</p><p>$

∴ The number is 64.

$hope \: its \: help \: u$