Question

A number consist of two digits whose sum is 5. When the digits are reversed, the number becomes greater by 9. Find the number?​

Answer 1

Answer:

32 or 23

Step-by-step explanation:

Let number be 10x+y

x+y=5.........i

10y+x+9=10x+y

9x-9y=9

x-y=1........ii

By elimination method

2y=4

y=2

Put it in i

x=3

So no. could be 32 or 23(by reversing the order of digits)

Answer 2

Given:

• Sum of two digits of a number is 5.
• When the digits are reversed, the number becomes greater by 9.

To find:

• Original number?

Solution:

☯ Let the digits in the tens and ones place be x and y respectively.

Therefore,

• Orginal Number = 10x + y

★ According to the Question:

• Sum of two digits of a number is 5.

➯ x + y = 5

➯ y = 5 - x⠀⠀ ⠀⠀❬ eq (1) ❭

⠀⠀━━━━━━━━━━━━━━━━━━━━━

Also,

• When the digits are reversed, the number becomes greater by 9.

Number after reversing the digits = 10y + x

➯ (10x + y) + 9 = 10y + x

➯ 10x - x + 9 = 10y - y

➯ 9x + 9 = 9y

➯ 9x - 9y = -9

➯ 9(x - y) = -9

➯ x - y = -9/9

➯ x - y = - 1⠀⠀⠀⠀ ❬ eq (2) ❭

⠀⠀━━━━━━━━━━━━━━━━━━━━━

Now, Putting value of eq (1) in eq (2),

➯ x - (5 - x) = - 1

➯ x - 5 + x = - 1

➯ 2x - 5 = - 1

➯ 2x = - 1 + 5

➯ 2x = 4

➯ x = 4/2

➯ x = 2

Putting value of x in eq (1),

➯ y = 5 - 2

➯ y = 3

Therefore,

• Tens digit of the number, x = 2
• Ones digit of the number, y = 3

∴ Hence, The original number is 23.