Question

10. A two-digit number is 4 more than 6 times the sum of its digits. If 18 is subtracted from the
number, the digits are reversed. Find the number.
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Answer 1

Step-by-step explanation:

Let x= ones digit and y=tens digit

According to the question ,

∴x+10y=6(x+y)+4

⇒10y=6x+6y+4−x

⇒4y=5x+4

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

∴x+10y−18=y+10x

⇒10y−18=y+9x

⇒9y−18=9x

⇒y−2=x

Substitute x value into :

4y=5x+4

4y=5(y−2)+4

4y=5y−6

y=6

Substitute y into :

∴y−2=x

⇒6−2=x

⇒4=x

∴ The number is 64

Answer 2

Step-by-step explanation:

Let

the ones digit=y

tens digit=10x

${:}\longrightarrow$the number is=10x+y

According to question

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

${:}\longrightarrow$6x+6y+4=10x+y

${:}\longrightarrow$$10x-6x+y-6y+4=0$

${:}\longrightarrow$4x-5y+4=0 ${\dots}(1)$

Let's go to second case

10x+y-18=10y+x

${:}\longrightarrow$$10x-x+y-10y-18=0$

${:}\longrightarrow$$9x-9y-18=0 {\dots}(2)$

Now

from eq (1)

4x-5y+4=0

${:}\longrightarrow$$4x+4=5y$

${:}\longrightarrow$$y={\frac {4x+4}{5}}{\dots}(3)$

Substitute the value of y in eq (2)

${:}\longrightarrow$$9x- {\frac {9 (4x+4)}{5}})-18$

${:}\longrightarrow$$9x-{\frac {36x+36}{5}}=18$

${:}\longrightarrow$$9x-36x+36=90$

${:}\longrightarrow$$-27x=90-36$

${:}\longrightarrow$$-27x=54$

${:}\longrightarrow$$x={\frac {54}{-27}}$

${:}\longrightarrow$${\underline{\boxed{\bf{x=(-2)}}}}$

Substitute the value of x in eq(3)

$y={\frac {4 (-2)+4 }{5}}$

${:}\longrightarrow$$y={\frac {-8+4}{5}}$

${:}\longrightarrow$${\underline{\boxed{\bf {y={\frac {-4}{5}}}}}}$

Hence

the number is

10x+y=

${:}\longrightarrow$$10 (-2)+{\frac {-4}{5}}$

${:}\longrightarrow$$20+{\frac {-4}{5}}$

${:}\longrightarrow$${\frac {101+4}{5}}$

${:}\longrightarrow$${\frac {105}{5}}$

${:}\longrightarrow$$21$

$\therefore$The number is 21