Question

Two digit number thistime by multiplying the sum of digits by eight and then subtracting five or so it is obtained by multiplying the difference of the digits by 16 and adding three find the number

Answer 1

Answer:

Step-by-step explanation:

Let the two digit number be 10x + y where x is the tens digit and y is the ones digit.

Now, according to the question.

10x + y = 8(x + y) - 5

10x + y = 8x + 8y - 5

10x - 8x + y - 8y = - 5

2x - 7y = - 5  .................(1)

And,

10x + y = 16(x - y) + 3

10x + y = 16x - 16y + 3

10x - 16x + y + 16y = 3

- 6x + 17y = 3  ................(2)

Now, multiplying the equation (1) by 17 and (2) by 7, we get

34x - 119y = - 85 ...............(3)

- 42x + 119y = 21 ..............(4)

Now, adding (3) and (4), we get

  34x - 119y = - 85

- 42x + 119y =   21

_________________

 - 8x             = - 64

_________________

⇒ 8x = 64

x = 64/8

x = 8 

So, tens digit is 8.  

Substituting the value of x = 8 in (1), we get

2x - 7y = - 5

2*8 - 7y = - 5

16 - 7y = - 5

- 7y = - 5 - 16

- 7y = - 21

7y = 21

y = 21/7

y = 3

Ones digit is 3.

So, the required number is 83.

Answer 2

Solution :

Let the ten's place digit be r & unit's place digit be m.

$\boxed{\bf{Original\:number=10r+m}}}}$

A/q

$\underbrace{\bf{1_{st}\:Case\::}}}}$

$\longrightarrow\sf{8(r+m)-5=10r+m}\\\\\longrightarrow\sf{8r+8m-5=10r+m}\\\\\longrightarrow\sf{8r-10r+8m-m=5}\\\\\longrightarrow\sf{-2r+7m=5....................(1)}$

$\underbrace{\bf{2_{nd}\:Case\::}}}}$

$\longrightarrow\sf{16(r-m)+3=10r+m}\\\\\longrightarrow\sf{16r-16m+3=10r+m}\\\\\longrightarrow\sf{16r-10r-16m-m=-3}\\\\\longrightarrow\sf{6r-17m=-3..........................(2)}$

$\underline{\boldsymbol{Using\:Substitution\:method\::}}}$

From equation (2),we get;

$\longrightarrow\sf{6r-17m=-3}\\\\\longrightarrow\sf{6r=-3+17m}\\\\\longrightarrow\sf{r=-3+17m/6......................(3)}$

∴ Putting the value of r in equation (1),we get;

$\longrightarrow\sf{-2\bigg(\dfrac{-3+17m}{6} \bigg)+7m=5}\\\\\\\longrightarrow\sf{\dfrac{6-34m}{6} +7m=5}\\\\\\\longrightarrow\sf{6-34m+42m=30}\\\\\longrightarrow\sf{6+8m=30}\\\\\longrightarrow\sf{8m=30-6}\\\\\longrightarrow\sf{8m=24}\\\\\longrightarrow\sf{m=\cancel{24/8}}\\\\\longrightarrow\bf{m=3}$

∴ Putting the value of m in equation (3),we get;

$\longrightarrow\sf{r=\dfrac{-3+17(3)}{6} }\\\\\\\longrightarrow\sf{r=\dfrac{-3+51}{6} }\\\\\\\longrightarrow\sf{r=\cancel{\dfrac{48}{6} }}\\\\\longrightarrow\bf{r=8}$

Thus;

$\boxed{\sf{The\:number=10r+m=[10(8)+3]=[80+3]=\boxed{\bf{83}}}}}$