Question

The sum of the digits of a two digit number is 10 and the digit at units place is 2/3rd of the
digit at tens place. Find the number.
​

Answer 1

GivEn:

• The sum of the digits of a two digit number is 10.
• The digit at units place is 2/3rd of the digit at tens place.

To find:

• Two digit Number?

Solution:

☯ Let digit at one's place and digit at ten's place be x and y respectively.

Therefore,

• Number = 10y + x

★ According to the Question:

• The sum of the digits of a two digit number is 10.

➯ x + y = 10⠀⠀⠀⠀⠀⠀⠀ ❬ eq (1) ❭

And,

• The digit at units place is 2/3rd of the digit at tens place.

➯ x = 2/3 y⠀⠀⠀⠀⠀⠀ ❬ eq (2) ❭

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

➯ 2/3 y + y = 10

➯ (2y + 3y)/3 = 10

➯ 5y/3 = 10

➯ 5y = 10 × 3

➯ 5y = 30

➯ y = 30/5

➯ y = 6

⠀⠀━━━━━━━━━━━━━━━━

Putting value of y in eq (2),

➯ x = 2/3 × 6

➯ x = 2 × 2

➯ x = 4

Hence, The required two digit number is 64.

Answer 2

Answer:

Given :-

• The sum of the digits of a two digits is 10 and the digits at units place is ⅔ rd of the digits at tens place.

To Find :-

• What is the number.

Solution :-

Let, the one's place be x

And, the ten's place be y

Then, the number will be (10y + x)

According to the question,

⇒ x + y = 10

⇒ x = 10 - y ....... equation no ❶

Again,

⇒ x = $\dfrac{2}{3}y$ ...... equation no ❷

⇒ 3x = 2y

⇒ 3(10 - y) = 2y

⇒ 30 - 3y = 2y

⇒ 30 = 2y + 3y

⇒ 30 = 5y

⇒ - 5y = - 30

⇒ y = $\sf\dfrac{\cancel{- 30}}{\cancel{- 5}}$

⇒ y = 6

Now, by putting y = 6 in the equation no (1) we get,

⇒x = 10 - y

⇒ x = 10 - 6

⇒x = 4

Hence, the required number is,

❐ 10y + x

↦ 10(6) + 4

↦ 60 + 4

➥ 64

$\therefore$ The number is 64 .