GIVEN :

• 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.

To Find :

• The required number = ?

Solution :

Let the ten's digit of two digit number be x and ONE's digit be y.

• The required number = 10x + y

In the question it is given that,The sum of the digits of a two digit number is 10. So, mathematically it can be expressed as :

• x + y = 10 [Equation (i)]

It is also given that,the digit at units place is 2/3rd of the digit at tens place, mathematically it can be expressed as :

• y = ⅔ x [Equation (ii)]

Now,plug in the value of y = ⅔ x from equation (ii) to equation (i) :

→ x + ⅔ x = 10

→ (2x + 3X)/3 = 10

→ 5x = 10 × 3

→ 5x = 30

→ x = 30 ÷ 5

→ x = 6

• Hence,the ten's digit of two digit number is 6.

Now,substitute the value of x = 6 in equation (ii) :

→ y = ⅔ x

→ y = ⅔ × 6

→ y = 2 × 2

→ y = 4

• Hence,the one's digit of two digit number is 4.

★ Finding the required number :

→ Required number = 10x + y

→ Required number = 10 × 6 + 4

→ Required number = 60 + 4

→ Required number = 64

• Hence,the required two digit number is 64.