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

$\huge\star\underline\mathrm\red{Answer}$

✯ Let the 10s digit of two digit number be x and 1s digit be y

Hence the number is 10x + y

The number of two digits if a two digit number is 10. It means : -

✯ x + y = 10…Eq.1

The digit at the unit place is 2/3rd of the digit at the tens place

✯ y = 2/3x

✯ y = 2x/3.. Eq .2

Now substituting the value of y from Eq.. 2 to Eq..1

☞ x + y = 10

☞ x + 2x/3 = 10

☞ (3x + 2x)/3 = 10

By Cross multipliying : -

☞ 5x = 30

☞ x = 30/5

☞ x = 6

The 10s digit is 6

✯ Substituting value of x in Eq..2 to drive value of y

☞ y = 2x/3

☞ y = 2 × 6/3

☞ y = 2 × 2

☞ y = 4

The 1s digit is $\mathbb{4}$

Therefore, the number is $\mathbb{64}$