Question

the digit at the tens place of a two digit number is 4 times the digit at ones place if the sum of this number and the number formed by reversing the digit is 55 find the number​

Answer 1

Given :

• The digit at the tens place of a two digit number is 4 times the digit at ones place.
• The sum of this number and the number formed by reversing the digit is 55.

To Find :

• The original two digit number.

Solution :

Let the digit at the tens place be x.

Let the digit at the units place be y.

Original Number = (10x + y)

Case 1 :

The digit at the tens place is 4 times the digit at the ones place.

Equation :

$\sf{\longrightarrow{x=4y}}$

$\sf{x=4y}$

Case 2 :

The sum of original number and the number formed by reversing the digit is 55.

Reversed Number = (10y + x)

Equation :

$\sf{\longrightarrow{10x+y+10y+x=55}}$

$\sf{\longrightarrow{10x+x+10y+y=55}}$

$\sf{\longrightarrow{11x+11y=55}}$

$\sf{\longrightarrow{11(x+y)=55}}$

$\sf{\longrightarrow{x+y=\dfrac{55}{11}}}$

$\sf{\longrightarrow{x+y=5}}$

$\sf{\longrightarrow{4y+y=5}}$

$\bold{\big[From\:equation\:(1)\:x\:=\:4y\:\big]}$

$\sf{\longrightarrow{5y=5}}$

$\sf{\longrightarrow{y=\dfrac{5}{5}}}$

$\sf{\longrightarrow{y=1}}$

Substitute, y = 1 in equation (1),

$\sf{\longrightarrow{x=4y}}$

$\sf{\longrightarrow{x=4(1)}}$

$\sf{\longrightarrow{x=4}}$

$\large{\boxed{\tt{Tens\:digit\:=\:x\:=\:4}}}$

$\large{\boxed{\tt{Units\:digit\:=\:y\:=\:1}}}$

$\large{\boxed{\tt{Original\:Number=\:10(x)+y=10(4)+1=40+1=41}}}$