Question

In a Three digit number, when the tens and hundreds digit are interchanged the new number is 54 more than three times the original number. If 198 is added to the number, the digits are reversed. The tens digit exceeds the hundreds digit by twice as that of the tens digit exceeds the unit digit. Find the original number

Answer 1

Answer:

Step-by-step explanation:

Answer 2

• The three digit numbers is 153.

Step-by-step explanation:

Let,

• The Hundred's digit be 'x'
• Tense digit be 'y'
• Ones digit be 'z

The three digit original number will be,

• 100x + 10y + z

When tense and hundreds are interchanged, the value is 54 more than 3 times the original number,

• 100y + 10x + z = 54 + [3(100x + 10y + z)] ---------- (1)

If 198 is added to the number, the digits are reversed,

• 100x + 10y + z + 198 = 100z + 10y + x ---------- (2)

The tens digit exceeds the hundreds digit by twice as that of the tens digit exceeds the unit digit,

• (y - x) = 2(y - z) --------- (3)

Solving (1),

⟼ 100y + 10x + z = 54 + [3 (100x + 10y + z)]

⟼ 100y + 10x + z = 300x + 30y + 3z + 54

⟼ 100y + 10x + z - 300x - 30y - 3z = -54

⟼ (300x - 10x) - (100y - 30y) + (3z - z) = -54

⟼ 290x - 70y + 2z = -54 ------- (4)

Solving (2),

⟼ 100x + 10y + z + 198 = 100z + 10y + x

⟼ 100x + 10y + z - 100z - 10y - x = -198

⟼ (100x - x) + (10y - 10y) - (100z - z) = -198

⟼ 99x + 0 - 99z = -198

⟼ 99x - 99z = -198 (÷99)

⟼ x - z = -2

⟼ x = z - 2 -------- (5)

Solving (3),

⟼ (y - x) = 2(y - z)

⟼ (y - x) = 2y - 2z

⟼ y - x - 2y - 2z = 0

⟼ x + (2y - y) = 2z

⟼ x + y = 2z

⟼ y = 2z - z + 2

⟼ y = z + 2 -------- (6)

Substituting 'x', 'y' in (4),

⟼ 290x - 70y + 2z = -54

⟼ 290 (z - 2) - 70 (z + 2) + 2z = -54

⟼ 290z - 580 - 70z - 140 + 2z = -54

⟼ 222z = 666

⟼ z = 666 / 222

⟼ z = 3

Substituting 'z' in (5),

⟼ x = z - 2

⟼ x = 3 - 2

⟼ x = 1

Substituting 'z' in (6),

⟼ y = z + 2

⟼ y = 3 + 2

⟼ y = 5

• ∴ The three digit numbers is 153.