Question

the digits at the tens place of a two-digit number is four times that is the unit place.if the digits are reversed,the new number will be 54 less than the original number.find the number
explain me bro ans 82 ​

Answer 1

Given :-

• The digits at the tens place of a two-digit number is four times that is the unit place.if the digits are reversed,the new number will be 54 less than the original number.

To find :-

• Number

Solution :-

Let the ones digit be y then tens digit be x

A two-digit number is four times that is the unit place

→ x = 4y -----(i)

• Original number = 10x + y

• Reversed number = 10y + x

Digits are reversed, the new number will be 54 less than the original number.

→ 10x + y - 54 = 10y + x

→ 10x - x + y - 10y = 54

→ 9x - 9y = 54

→ 9(x - y) = 54

→ (x - y) = 54/9

→ (x - y) = 6

Put the value of x

→ 4y - y = 6 [using (i)]

→ 3y = 6

→ y = 6/3

→ y = 2

Substitute the value of y in equation (i)

→ x = 4y

→ x = 4 × 2

→ x = 8

Hence,

• Tens digit = x = 8
• Ones digit = y = 2

Therefore,

• Original number = 10x + y = 82
• Reversed number = 10y + x = 28

Answer 2

Answer:

The required number is 82.

Step-by-step explanation:

Let the Number at the units place be x.Tenth place = 4x

Then,

➳ Number = 10(x) + 4x

➳ Number = 10x + 4x

➳ Number = 14x

➳ Reversing Number = 10(4x) + x

➳ Reversing Number = 40x + x

➳ Reversing Number = 41x

According to Question now,

➳ 14x - 41x = -54

➳ -27x = -54

➳ x = -54/-27

➳ x = 54/27

➳ x = 2

• Units place = x = 2

• Tenths place = 4x = 4(2) = 8

Therefore,

• Required Number = 10(8) + 2 = 80 + 2 = 82