Question

the sum of the digits of a 2- digit number is 13. If the tens digit is 3 more than its ones digit, what is the number

Answer 1

Given

The sum of the digits of a 2- digit number is 13. If the tens digit is 3 more than its ones digit.

To find

• Required number

Solution

Let the tens digit be x and ones digit be y

According 1st condition

• Sum of 2 - digit number = 13
• x + y = 13 -----(i)

According to 2nd condition

• The tens digit is 3 more than its ones digit
• x = 3 + y ----(ii)

Substitute the value of x in eqⁿ (i)

→ x + y = 13

→ 3 + y + y = 13

→ 3 + 2y = 13

→ 2y = 13 - 3

→ 2y = 10

→ y = 10/2 = 5

Putting the value of y in eqⁿ (ii)

→ x = 3 + y

→ x = 3 + 5

→ x = 8

Hence,

• Required numbers = x and y = 8 and 5

Answer 2

Let us take the tens digit to be x.

So, the one's digit becomes y.

∴ Number ⇒ 10x + y

ATQ;

⇒ x + y = 13 → Eq(1)

Also ATQ,

⇒ x + 3 = y → Eq(2)

Adding equations 1 & 2 we get:

⇒ x + y + x - y = 13 + (-3)

⇒ 2x = 10

⇒ x = 10/2

⇒ x = 5

Substitute the value of x in Equation 2.

⇒ x - y = -3

⇒ 5 - y = -3

⇒ 5 + 3 = y

⇒ y = 8

The Number:

⇒ 10x + y

⇒ 10(5) + 8

⇒ 50 + 8

⇒ 58

Final Answer:

58

$\rule{100}{1}$

Detailed Explanation:

Let us take a number 75.

Let the tens digit be 'x', lets the one's digit be 'y'.

Here, the tens digit is 7, and the one's digit is 5. Therefore, the number will be:

⇔ 75 = 10(x) + y

⇔ 75 = 10(7) + 5

⇔ 75 = 70 + 5

⇔ 75 = 75

⇔ LHS = RHS

The same concept is used in the answer.