Question

In a number, these are three digits whose sum is 9. The digit in the uni
place is two times the digit in the hundred place. If the digits e reversed,
the number is increased by 198, find the number.​

Answer 1

Answer:

The original number is 234

Step-by-step explanation:

Let the numbers in original number be =

• Hundreds digit = x
• Units digit = 2x

As the sum of their digits is 9,

$\longrightarrow$ x + 2x + Tens digit = 9

$\longrightarrow$ 3x + Tens digit = 9

$\longrightarrow$ Tens digit = 9 - 3x

Original number:

• Hundreds digit = x
• Tens digit = (9 - 3x)
• Units digit = 2x

$\longrightarrow$ 100(x) + 10(9 - 3x) + (2x)

$\longrightarrow$ 100x + 90 - 30x + 2x

$\longrightarrow$ 72x + 90 ----- (Original number)

Number with reversed digits:

• Hundreds digit = 2x
• Tens digit = (9 - 3x)
• Units digit = x

$\longrightarrow$ 100(2x) + 10(9 - 3x) + x

$\longrightarrow$ 200x + 90 - 30x + x

$\longrightarrow$ 171x + 90 ----- (Number with reversed digits)

According to the Question,

If the digits are reversed, the original number is increased by 198

$\longrightarrow$ (72x + 90) + 198 = 171x + 90

$\longrightarrow$ 72x + 288 = 171x + 90

$\longrightarrow$ 72x - 171x = 90 - 288

$\longrightarrow$ –99x = –198

$\longrightarrow$ x = –198 ÷ –99

$\longrightarrow$ x = 2

Original number =

$\longrightarrow$ 72(2) + 90

$\longrightarrow$ 144 + 90

$\longrightarrow$ 234

Therefore, the original number is 234.