Question

a number consists of two digits. the digits in the ten's place is triple the digit in the units place when digits are interchanged the number decreases by 36, then the number is.​

Answer 1

Given :

• A number consists of two digits. the digits in the ten's place is triple the digit in the units place.
• When digits are interchanged the number decreases by 36.

To Find :

• The original 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 ten's digit is triple the digit in the units place.

Equation :

$\sf{x=3y\:\:\:(1)}$

Case 2 :

When the digits are interchanged the new number formed decreases by 36.

Reversed Number = (10y+x)

Equation :

$\implies$ $\sf{10y+x=10x+y-36}$

$\implies$ $\sf{10y-y=10x-x-36}$

$\implies$ $\sf{9y=9x-36}$

$\implies$ $\sf{9x-36=9y}$

$\implies$ $\sf{9x-9y=36}$

$\implies$ $\sf{9(3y)-9y=36}$

$\bold{\big[From\:equation\:(1),\:x\:=\:3y\:\big]}$

$\implies$ $\sf{27y-9y=36}$

$\implies$ $\sf{18y=36}$

$\implies$ $\sf{y=\dfrac{36}{18}}$

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

Substitute, y = 2 in equation (1),

$\implies$ $\sf{x=3y}$

$\implies$ $\sf{x=3\:\times\:2}$

$\implies$ $\sf{x=6}$

$\large{\boxed{\bold{Ten's\:digit\:=\:x\:=\:6}}}$

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

$\large{\boxed{\bold{\red{Original\:Number=\:10x+y=10(6)+2=60+2\:=\:62}}}}$

Answer 2

Given :

• Number consist of 2 digits

• Digit in tens place is triple the digit in units place

• When they are interchanged number decreases by 36

To Find :

• The Number

Solution :

Let the digit at tens place be x and digit at ones place be y

$\sf \star \: Original \: Number = 10x + y$

$\sf \star \: Interchanged\: Number = 10y + x$

It is given that digit in tens place is triple the digit in units place

$\longmapsto \sf \: x = 3y...(i)$

$\sf \star \: Original \: Number = 10(3y)+ y = \boxed{ \sf \: 31y}$

$\sf \star \: Interchanged\: Number = 10y + 3y = \boxed{ \sf \: 13y}$

According to the Question :

$\longrightarrow \sf 13y = 31y - 36$

$\longrightarrow \sf 13y - 31y = - 36$

$\longrightarrow \sf - 18y = - 36$

$\longrightarrow \sf y = \cancel \dfrac{ - 36}{ - 18}$

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

Putting the value of y = 2 in equation i)

$\longmapsto \sf \: x = 3y$

$\longrightarrow \sf \: x = 3(2)$

$\longrightarrow \sf \red{ x =6}$

$\sf \bigstar \large\boxed{ \purple {\sf \: Original \: Number = 31y = 31(2)\: = 62}}$