Question

A number constant of two digits who sum is 9 if 27 is subtracted from the number its digits are reversed find the number

Answer 1

Given :

• A number constant of two digits who sum is 9.

• 27 is subtracted from the number its digits are reversed.

To find :

• The number =?

Step-by-step explanation :

Let, the one's digit of two digits number be, x.

Then, the ten's digit of two digits number be, 9 - x.

So,

The number is, 10 (9 - x +x) = 90 - 9x.

After reversing the digit the number will be, 10x + (9 - x) = 9x + 9

According to the question,

➮ 90 - 9x = 9x + 9 + 27

➮ 18x = 90 - 9 - 27

➮ 18x = 54

➮ x = 54/18

➮ x = 3.

Hence,

One's digit, x = 3

Ten's digit, 9 - x = 9 - 3 = 6

Therefore, the number will be, 10 × 6 + 3 = 63.

Answer 2

Answer:

The number is : 63

Given:

➛A number constant of two digits who sum is 9.

➛If 27 is subtracted from the number its digits are reversed.

To Find:

The number

Solution:

We are given,

➛A number constant of two digits who sum is 9.

➛If 27 is subtracted from the number its digits are reversed.

let's the units digit be m and ten's digit be n.

⛬ According to given condition,

so, the number is 10m + n

m + n = 9 .................... ( a )

⛬ According to second condition,

10m + n - 27 = 10n + m

➛ 9m - 9n = 27

➛ m - n = 3...............( b )

Adding ( a ) and ( b )

we get

m + n = 9

+

m - n = 3

---------------------

2m = 12

⛬ m = 12 / 2

⛬ m = 6

Now , putting m = 6 in equation ( a )

we get ,

➛ n = 3

⛬ The number is 63