Question

a number consists of two digits whose sum is 9. if 27 is subtracted from the number it digits are reversed. find the number ​

Answer 1

Answer:

Here, we will first consider the two digits of the number as x and y. Then, we will try to form equations using the given conditions to obtain the values of x and y and hence, we can find the number.

We know that if the digits of a two digit number are known then we can easily find the number.

If the first digit of the two digit number is x and the second digit is y, then we can write the given number as 10x+y .

Since, it is given that the sum of the digits of the number is 9. So, we can write the following equation:

x+y=9---------> (1)

It is also given that if we subtract 27 from the number, its digit gets reversed. The reversed number will have now y as its first digit and x as its second digit.

So, the number formed after reversing the digits can be written as 10y+x .

So, on subtracting 27 from the given number and then equating it to its reverse, we get:

10x+y−27=10y+x⇒10x+y−10y−x=27⇒9x−9y=27⇒9(x−y)=27⇒x−y=279=3

Therefore, we have another equation as:

x−y=3---------->(2)

On adding equation (1) and equation (2), we get:

x+y+x−y=9+3⇒2x=12⇒x=122=6

So, the value of x comes out to be = 6.

On substituting x = 6 in equation (1), we get:

6+y=9⇒y=9−6=3

So, the value of y is =3.

Since, the number is of the form of 10x+y, we can write that the number is :

=10×6+3=60+3=63

Hence, the required number is 63.

Hope it helps (◍•ᴗ•◍)❤

Answer 2

$let \: the \: number \: = ab = 10a + b$

$Sum \: of \: Digits \: =9$

$a + b = 9$

$no.27 \: with \: reversed \: digits$

$10a + b = 27 = 10b + a$

$9(a - b) = 27$

$a - b = 3$

$(a + b) + (a - b) = 9 + 3$

$a = 6 \: , \: b = 3$

$Required \: Number \: is \: \: 63$