Question

The digit at the ten’s place of a two digit number is four time that in the unit’s place. If the digits are reversed , the new number will be 54 less than the original numbers. Find the original numbers. Check your solution

Answer 1

Given :

• The digit at the ten’s place of a two digit number is four time that in the unit’s place.
• If the digits are reversed the new number will be 54 less than the original number.

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 ten's place, x is 4 times that of the digit in the units place.

Equation :

$\sf{x=4y\:\:\:(1)}$

Case 2 :

When the digits are reversed, the new number is 54 less than the original number.

Reversed Number = (10y + x)

Equation :

$\longrightarrow$ $\sf{10y+x=10x+y-54}$

$\longrightarrow$ $\sf{10y-y=10x-x-54}$

$\longrightarrow$ $\sf{9y=9x-54}$

$\longrightarrow$ $\sf{9x-54=9y}$

$\longrightarrow$ $\sf{9x-9y=54}$

$\longrightarrow$ $\sf{9(4y)-9y=54}$

$\longrightarrow$ $\sf{36y-9y=54}$

$\longrightarrow$ $\sf{27y=54}$

$\longrightarrow$ $\sf{y=\dfrac{54}{27}}$

$\longrightarrow$ $\sf{y=2}$

Substitute, y = 2 in equation (1),

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

$\longrightarrow$ $\sf{x=4(2)}$

$\longrightarrow$ $\sf{x=8}$

$\large{\boxed{\bold{Ten's\:digit\:=\:x\:=\:8}}}$

$\large{\boxed{\bold{Unit's\:digit\:=\:y\:=\:2}}}$

$\large{\boxed{\bold{Original\:Number\:=\:10x+y=10(8)+2=80+2=82}}}$

$\huge{\underline{\bold{Verification:}}}$

In case 1, the tens digit i.e x is 4 times the digit at the units place, y.

Tens digit = x = 8

Units digit = y = 2.

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

$\longrightarrow$ $\sf{8=4(2)}$

$\longrightarrow$ $\sf{8=8}$

LHS = RHS.

In case 2, the reversed number is 54 less than the original number.

Original Number = 10x + y = 82

Reversed Number = 10y + x = 28

$\longrightarrow$ $\sf{10y+x=10x+y-54}$

$\longrightarrow$ $\sf{28=82-54}$

$\longrightarrow$ $\sf{28=28}$

LHS = RHS.