Question

The diait at ones place of a 2-digit number is four times the digit at tens place. The
number obtained by reversing the digits exceeds the given number by 54. Find
the given number.​

Answer 1

Question

The digit at ones place of a 2-digit number is four times the digit at tens place. The number obtained by reversing the digits exceeds the given number by 54. Find the given number.

Answer

Number is 28

$\rule{200}2$

Explanation

Let tens digit be x and ones digit be y.

So, number = 10x + y

The digit at ones place of a 2-digit number is four times the digit at tens place.

As per given condition,

⇒ y = 4x

As per given condition,

⇒ 10x + y + 54 = 10y + x

⇒ 10x - x + y - 10y = - 54

⇒ 9x - 9y = - 54

⇒ 9(x - y) = - 54

⇒ x - y = - 6

⇒ x = y - 6

Substitute value of y = 4x above

⇒ x = 4x - 6

⇒ -3x = - 6

⇒ 3x = 6

⇒ x = 2

Substitute value of x = 2 in y

⇒ y = 4(2)

⇒ y = 8

Therefore,

Number = 10x + y = 10(2) + 8 = 28

Answer 2

||✪✪ GIVEN ✪✪||

• Digit at unit Place = Four Times The Digit at Ten's Place.
• The number obtained by reversing the digits exceeds the given number by 54.

|| ✰✰ ANSWER ✰✰ ||

Let us Assume That, Digit at Unit place is a and Digit at Ten's Place is b .

So,

→ Our Given Number is = (10b + a) .

Now, According To Question :-

→ a = 4b ------------ Equation (1)

Now, when we Reversed The Digits , we get The number is = (10a + b) .

So,

→ (10a + b) - (10b + a) = 54

→ 10a - a + b - 10b = 54

→ 9a - 9b = 54

→ 9(a - b) = 54

→ (a - b) = 6

Putting value of Equation (1) Here, we get,

→ (4b - b) = 6

→ 3b = 6

→ b = 2

Putting This value Again in Equation (1) , we get,

→ a = 4 * 2 = 8

Hence, Our Required Given Number is = (10b + a) = (10*2 + 8) = (20 + 8) = 28 (Ans).