Question

A two digit number is six times the sum of the digits. If the digit in the unit's place is
increased by 2 and the digit in the ten's place is decreased by 2, then the number so formed
is 4 times the sum of its digits. Find the original number. (Ans.: 54)​

Answer 1

Answer:

Thus the number =54

Step-by-step explanation:

Let the  unit digit is y and 10's digit=x

Then number=10x+y

as per given

10x+y=.6*(x+y)=6x+6y

4x-5y=0.........................(1)

If the digit in the unit's place is increased by 2 then digit=y+2

and the digit in the ten's place is decreased by 2, then the 10's gdigit=x-2

So formed number

=10*10's digit + unit digit

=10*(x-2)+y+2=10x+y-18

As per question:

10x+y-18=4(x+y)

10x-4x+y-4y=18

6x-3y=18

2x-y=6

or 4x-2y=12....................(2)

Subtracting(1) from(2) we get

3y=12,y=4

From(2) putting y=4

4x-2*4=12

4x=20

x=5 and y=4

or unit digit=4 and 10's digit=5

Thus the number = 54

Answer 2

SOLUTION :-

Let ,

Digit in one's place = x

Digit in ten's place = y

Two digit number = 10x + y

According to the first condition,

$\longrightarrow\sf 10x+y=6(x+y) \\\\\longrightarrow 10x+y = 6x+6y \\\\\longrightarrow 10x-6x+y-6y=0 \\\\\longrightarrow 4x-5y=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(1)$

According to the second condition,

$\longrightarrow\sf 10(x-2)+(y+2)=4(x+y)\\\\\longrightarrow 10x-20+y+2= 4x+4y \\\\\longrightarrow 10x+y-18 = 4x+4y \\\\\longrightarrow 10x-4x+y-4y = 18 \\\\\longrightarrow 6x-3y=18\\\\\longrightarrow 2x-y =6 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(2)$

Equation (2) × 2,

$\longrightarrow\sf 4x-2y = 12 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ........................(3)$

Equation (3) - Equation (1),

$\longrightarrow\sf 3y= 12 \\\\\longrightarrow \bf y = 4$

Substitute y = 4 in equation (1),

$\longrightarrow\sf 2x-4 = 6 \\\\\longrightarrow 2x= 10 \\\\\longrightarrow \bf x = 5$

Two digit number = 54