Question

One of the two digits of a two digit is two times the other digit. If we interchange the digits of this two-digit number and add the resulting number to the original number, we get 99. What is the original number?

Answer 1

Given :

• One of the two digits of a two digit is two times the other digit.

• We interchange the digits of this two-digit number and add the resulting number to the original number, we get 99.

To find :

• The original number =?

Step-by-step explanation :

Let one digit be x.

Then, the other digit is 2x.

Then, the original number will be, 10(x + 2x) .

On simplifying 10(x + 2x) . We get,

= 10x + 3x

= 12x.

Now,

After interchanging the digits, we get,

= 10(2x + x)

= 10 × 2x + x

= 20x + x

= 21x.

Now,

After adding both numbers, we get,

➮ 12x + 21x = 99

➮ 33x = 99

➮ x = 99/33

➮ x = 3.

Therefore, We got the value of, x = 3.

Hence,

The original number,

= 10(x + 2x)

On, putting the value of x = 3 we get,

= 10(3 + 2 × 3)

= 30 + 6

= 36.

Therefore, the original number = 36.

Answer 2

$\bf\large{\underline{Question:-}}$

One of the two digits of a two digit is two times the other digit. If we interchange the digits of this two-digit number and add the resulting number to the original number, we get 99. What is the original number?

$\bf\large{\underline{Solution:-}}$

• Let one digit (once place) is = x
• Another digit ( tens place) is = 2x

→tens place + once place

→10 (2x) + 1(x)

Then,

→The number is = 20x+x = 21 x

On interchanging

→Tens place + once place

→10(x). + 1×(2x)

★Number is 10x+2x = 12x

Now,

Now, when we add the number ,we get = 99

Now, when we add the number ,we get = 99So,

$\tt → 21x+12x=99\\\tt→ 33x=99 \\\tt→ x=\frac{99}{33} \\\tt→ x=3$

Now,

Now,putting x=3 in 21x

Hence,

the two digit number is 21×3 = 63

$\rule{230}2$

★ Extra knowledge

It is not necessary that the original number is = 63

if we consider once place is = 2x

and tens place is = x

then the number will be = 36