Question

5..The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the numbers get reversed. Find the number

Answer 1

Answer:

☯ One's place = 3

☯ Ten's place = 6

☯ Original number = 36

Step-by-step explanation:

→ Let the unit's digit to be x

→ Now, the ten's digit will be (9 - x)

→ Original number = 10(9 - x) + x = 90 - 10x + x = 90 - 9x

★ On interchanging the digits :

→ The new number = 10x + (9 - x) = 10x + 9 - x = 9x + 9

★ According to the question :

→ New number = Original number + 27

→ 9x + 9 = 90 - 9x + 27

= 9x + 9 = 117 - 9x

= 9x + 9x = 117 - 9

= 18x = 108

= x = 108/18

= x = 6

★ Now, the original number will be :

→ 90 - (9 × 6)

= 90 - 54

= 36

__________________________

Answer 2

Given,

• The sum of the digits of a two digit number is 9.
• If 27 is added to it then the digit of the numbers get reversed .

To Find,

• Two Digit Number  

Solution,

⇒Suppose the Ten's Digit number be a

And, Suppose the one's Digit number be b  

Therefore,

• Two Digit Number = 10a + b  
• Interchange Number = 10b + a  

According to the First Condition :-

• The sum of the digits of a two digit number is 9.

$\implies \sf{a + b = 9}$

$\implies \boxed{\sf{a = 9 - b }}$

According to the Second Condition :-

• If 27 is added to it then the digit of the numbers get reversed .

$\implies \sf{10a + b + 27 = 10b + a}$

$\implies \sf{10a - a + 27 = 10b - b}$

$\implies \sf{9a + 27 = 9b}$

$\implies \sf{9(9 - b) + 27 = 9b}$

(Put the Value of a From the First Condition)

$\implies \sf{81 - 9b + 27 = 9b}$

$\implies \sf{108 = 9b + 9b}$

$\implies \sf{18b = 108}$

$\implies \sf{b = \dfrac{108}{18}}$

$\implies \boxed{\sf{b = 6}}$

Now Put the Value of b in First Condition :-

$\implies \sf{a = 9 - b}$

$\implies \sf{a = 9 - 6}$

$\implies \boxed{\sf{a = 3}}$

Therefore,

$\boxed{\large{\bold{\red{Two \ Digit \ Number = 10a + b = 10(3) + 6 = 36}}}}$

$\rule{200}2$