Question

In two digits number sum of digits 8, reverse number 18 less than the original number, find the original number.

Answer 1

Answer:

The original number is 53.

Step-by-step explanation:

Let,

Units digit = x

Tens digit = 8 - x

★ No. :

⇒ 10 (8 - x) + x

⇒ 80 - 10x + x

⇒ 80 - 9x

★ Reverse number :

⇒ 10x +(8 - x)

⇒ 9x + 8

Reverse number 18 less than the original number

★ According to the Question :

⇒ 80 - 9x = 9x + 8 + 18

⇒ 80 - 9x = 9x + 26

⇒ 80 - 26 = 9x + 9x

⇒ 54 = 18x

⇒ x = 54/18

⇒ x = 3

Units digit = 3

• Tens digit = 8 - x

⇒ 8 - 3

⇒ 5

Tens digit = 5

Number = 53

Therefore, the original number is 53.

Answer 2

Answer:

Given :-

• In two digits number sum of digits is 8, reverse number is 18 less than the original number.

To Find :-

• What is the original number.

Solution :-

Let,

$\mapsto \bf Digit\: at\: unit's\: place =\: a$

$\mapsto \bf Digit\: at\: ten's\: place =\: 8 - a$

❒ Original Number :

$\leadsto \sf Original\: Number =\: 10(8 - a) + a$

$\leadsto \sf Original\: Number =\: 10(8) - 10(a) + a$

$\leadsto \sf Original\: Number =\: 80 - 10a + a$

$\leadsto \sf\bold{\pink{Original\: Number =\: 80 - 9a}}$

❒ When number are reversed :

$\leadsto \sf Reversed\: Number =\: 10a + 8 - a$

$\leadsto \sf Reversed\: Number =\: 10a - a + 8$

$\leadsto \sf\bold{\pink{Reversed\: Number =\: 9a + 8}}$

According to the question,

$\implies \sf 80 - 9a =\: 9a + 8 + 18$

$\implies \sf 80 - 9a =\: 9a + 26$

$\implies \sf - 9a - 9a =\: 26 - 80$

$\implies \sf {\cancel{-}} 18a =\: {\cancel{-}} 54$

$\implies \sf 18a =\: 54$

$\implies \sf a =\: \dfrac{\cancel{54}}{\cancel{18}}$

$\implies \sf a =\: \dfrac{3}{1}$

$\implies \sf\bold{\purple{a =\: 3}}$

Hence, the required original number is :

$\longrightarrow \sf Original\: Number =\: 80 - 9a$

$\longrightarrow \sf Original\: Number =\: 80 - 9(3)$

$\longrightarrow \sf Original\: Number =\: 80 - 27$

$\longrightarrow \sf\bold{\red{Original\: Number =\: 53}}$

${\small{\bold{\underline{\therefore\: The\: original\: number\: is\: 53\: .}}}}$