Question

A two-digit number is 7 times the sum of its digits. The number formed by reversing its

digits is 18 jess than the original number. What is the number?
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Answer 1

Explanation:

this is Answer of your question

Answer 2

Given :

• A two-digit number is 7 times the sum of its digits.
• Number formed by reversing its digits is 18 jess than the original number.

To Find :

• The 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 two digit number i.e (10x + y) is 7 times the sum of the digits i.e (x+y)

Equation :

$\longrightarrow$ $\sf{10x+y=7(x+y)}$

$\longrightarrow$ $\sf{10x+y=7x+7y}$

$\longrightarrow$ $\sf{10x-7x=7y-y}$

$\longrightarrow$ $\sf{3x=6y}$

$\sf{x=\dfrac{6y}{3}\:\:\:(1)}$

Case 2 :

The number formed when the digits of the two digit number is reversed is 18 less than the original two digit number.

Reveresed Number = (10y + x)

Equation :

$\longrightarrow$ $\sf{10y+x=10x+y-18}$

$\longrightarrow$ $\sf{10y-y=10x-x-18}$

$\longrightarrow$ $\sf{9y=9x-18}$

$\longrightarrow$ $\sf{9x-18=9y}$

$\longrightarrow$ $\sf{9x-9y=18}$

Divide throughout by 9,

$\sf{\cancel{\dfrac{9}{9}}x\:-\:\cancel{\dfrac{9}{9}}y\:=\:\cancel\dfrac{18}{9}}$

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

From equation (1), x = 6y/3,

$\longrightarrow$ $\sf{\dfrac{6y}{3}\:-\:y=2}$

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

$\longrightarrow$ $\sf{\dfrac{\cancel{3}y}{\cancel{3}}\:\:=2}$

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

Substitute, y = 2 in equation (1),

$\longrightarrow$ $\sf{x=\dfrac{6y}{3}}$

$\longrightarrow$ $\sf{x=\dfrac{6\:\times\:2}{3}}$

$\longrightarrow$ $\sf{x=\cancel\dfrac{12}{3}}$

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

$\large{\boxed{\sf{\red{Ten's\:digit\:=\:x\:=\:2}}}}$

$\large{\boxed{\sf{\red{Unit's\:digit\:=\:y\:=\:4}}}}$

$\large{\boxed{\sf{\purple{Original\:Number\:=\:(10x+y)=10(2)+4=20+4=24}}}}$