Question

Two digit number is 6times the sum of its digits .If units digit is increased by 2 and tens digit reduced by 2 the number obtained is 4 times the sum of its digits Find tge original number​

Answer 1

☯ Let the digit at unit place and digit at ten's place be y and x respectively.

Therefore,

• Original number = 10x + y

⠀⠀━━━━━━━━━━━━━━━━━━━━

Now,

★ According to the Question:

• Two digit number is 6 times the sum of its digits.

➯ 10x + y = 6(x + y)

➯ 10x + y = 6x + 6y

➯ 10x - 6x = 6y - y

➯ 4x = 5y

➯ x = 5y/4⠀⠀⠀⠀⠀⠀⠀ ❬ eq (1) ❭

And,

• If units digit is increased by 2 and tens digit reduced by 2 the number obtained is 4 times the sum of its digits.

➯ 10(x - 2) + (y + 2) = 4(x + y)

➯ 10x - 20 + y + 2 = 4x + 4y

➯ 10x + y - 18 = 4x + 4y

➯ 10x - 4x = 4y - y + 18

➯ 6x = 3y + 18

➯ 6x = 3(y + 6)

➯ 2x = y + 6⠀⠀⠀⠀⠀⠀⠀ ❬ eq (2) ❭

Now, Putting eq (1) in eq (2),

➯ 2(5y/4) = y + 6

➯ 10y/4 = y + 6

➯ 10y/4 - y = 6

➯ (10y - 4y)/4 = 6

➯ 6y/4 = 6

➯ 3y/2 = 6

➯ 3y = 12

➯ y = 12/3

➯ y = 4

Now, Putting value of y in eq (1),

➯ x = 5(4)/4

➯ x = 20/4

➯ x = 5

Therefore,

• Digit at unit place, y = 4
• Digit at ten's place, x = 5

∴ Hence, The original number is 54.

Answer 2

Answer:

Let the digit at unit place be y

And digit at tens place be x

Original number = 10x + y

Now,

$\sf 10 x+y = 6(x+y)$

$\sf 10x +y = 6x + 6y$

$\sf 10x - 6x = 6y - y$

$\sf 4x = 5 y$

$\sf x = \dfrac{5y}{4} [EQ 1]$

Now going to second part

$\sf 10(x-2) + (y+2) = 4 x+y$

$\sf 10x - 20 + y + 2 = 4x+y$

$\sf 10x - 4x = 4y - y + 18$

$\sf 6x = 3y + 18$

$\sf 6x = 3(y+6)$

$\sf 2x = y + 6[EQ 2]$

$\sf 2 (\dfrac{5y}{4})= y+6$

$\sf \dfrac{10y}{4} = y + 6$

$\sf \dfrac{10y - 4y}{4} = 6$

$\sf \dfrac{3y}{2} = 6$

$\sf3y = 12$

$\sf y = 4$

$\sf x = \dfrac{5(4)}{4}$

$\sf x = \dfrac {20}{5}$

$\sf x = 4$

Therefore,

Unit place = 4

Tens place = 5

Number = 54

Step-by-step explanation: