Question

sum of digits of two digit number is 9.When the digits are interchanged the new number is greater than the original number by 9.Find the original number.​

Answer 1

Sᴏʟᴜᴛɪᴏɴ:

$\underline{\boxed{\red{\bf Original \: Number = 45}}}$

Step-by-step explanation:

Let the digits at ten's place be x. And, the digit at unit's place be y.

$★ \boxed{\bf Original \: Number = 10x + y}$

Also, it is given that the sum of the digits is 9.

⇒ x + y = 9____(i)

According to the question, If the digits are interchanged, the new number is greater than the original number by 9.

Adding equation (i) and (ii),

$\implies \sf x + y + x - y = 9 - 1\\\\\implies \sf 2x = 8\\\\\implies \sf x =\frac{8}{2}\\\\\boxed{\bf \therefore x = 4}$

Substituting the value of x in (i),

$\implies \sf x + y = 9 \\\\\implies \sf 4 + y = 9\\\\\implies \sf y = 9-4\\\\\boxed{\therefore \bf y = 5}$

Now, Original number = 10x + y

                                     = 10 * 4 + 5

                                     = 40 + 5

                                     = 45

Hence, the required number is 45.

Answer 2

Answer:

Original number is 45

Step-by-step explanation:

For original number,

Let the unit digit be 'x'.

Then the tens digit will be '9-x'.

Therefore, Original Number = 10(9-x)+x

= 90-10+x

= 90-9x

For new number,

The unit digit will be '9-x'.

Then the tens digit will be 'x'.

Therefore, New Number = 10x+(9-x)

= 10x+9-x

= 9x+9

A/q,

$(9x + 9) - (90 - 9x) = 9 \\ = >9x + 9 - 90 + 9x \: = 9 \\ = > 18x - 81 = 9 \\ = > 18x = 90 \\ = > x = 90 \div 18 \\ = > x = 5$

Therefore, Original number is 45

hope it helps!