Question

The ones digit of a 2-digit number is twice the tens digit. When the number formed by reversing the digits is added to the original number, the sum is 99. Find the original number.

Don't spam ❌❌
useless answers will be reported.
​

Answer 1

Step-by-step explanation:

Let the two digit number is xy

y is digit at unit's place

x is digit at ten's place

⇒ Number can be written as = 10x+y

According to the conditions : y=2x ___ (1)

and

(10x+y)+(10y+x)=99

⇒11x+11y=99

⇒x+y=9 (dividing by 11)

Now x+2x=9 (from (1))

⇒3x=9

⇒x=9/3=3

y=2x=6

⇒ Number is 36

Answer 2

GivEn:

• The ones digit of a 2-digit number is twice the tens digit.
• When the number formed by reversing the digits is added to the original number, the sum is 99.

To find:

• The Original number?

Solution:

☯ Let x be the ten's place and y be the one's place of a 2 - digit number.

Therefore,

• Original number = 10x + y
• Number after reversing = 10y + x

Now,

★ According to the Question:

• The ones digit of a 2-digit number is twice the tens digit.

➯ y = 2x⠀⠀⠀⠀⠀⠀⠀❬ eq (❶) ❭

And,

• When the number formed by reversing the digits is added to the original number, the sum is 99.

★ (original number) + (number formed after reversing) = 99

➯ 10x + y + 10y + x = 99

➯ 11x + 11y = 99

➯ 11(x + y) = 99

➯ x + y = 99/11

➯ x + y = 9⠀⠀⠀⠀⠀⠀⠀❬ eq (❷) ❭

⠀━━━━━━━━━━━━━━━━━━━━━

Substituting eq (1) in eq (2),

➯ x + 2x = 9

➯ 3x = 9

➯ x = 9/3

➯ x = 3

Now, Substitute value of x in eq (1),

➯ y = 2 × 3

➯ y = 6

∴ The original number is 36.