Question

The sum of two digit number and the number obtained by reversing the order of the digits is 99. Find the number, if the digits differ by 3

Answer 1

Answer:

36 or 63

and the reverse number will be 63 or 63

Step-by-step explanation:

let the first digit of the original no. be x

and the 2nd no. be y

the no. will be 10x+y

the reverse of this will be 10y+x

the of these will be 10x+y+10y+x

                               => 10(x+y) + (x=y)

the sum of these two terms (x and y) can be 9 because if it is larger than suppose 10 thn it will be of two digit and the whole sum will not be 99

so that means

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

                       x-y=3    {given}      -------------2)

                       x=y+3  {transposing y to RHS}   ------------ 3)

             substitute the value of 3)into 1)

         x+y=9

         y+3+y=9

         2y=6

          y=3

   then, x=y+3

             x=6

so the required no. will be 36 or 63

                     

hope it will help you and pls mark me as brainliest

Answer 2

AnswEr:-

Your Answer Is 63 (or 36).

ExplanaTion:-

Given:-

• Sum of two digit number and the number obtained by reversing the order = 99.

• The digits defer by 3.

To Find:-

• The number.

So,

Let the digit at ten's plce be x and the digit at the one's place be y.

• Therefore the number will be 10x+y.

(Because x is at ten's place and y is at one's place)

And if we change the order of digits then y will get the ten's place and x will get the one's place.

• Therefore the number obtained by reversing digits will be 10y+x.

Therefore ATQ,

$\tt\mapsto Difference \:\:Between \:\:The\:\:Digits = 3.\\ \\ \tt\mapsto x - y = 3...eq(1)$

• Also,

$\tt \mapsto(10x + y) + (10y + x) = 99. \\ \\ \tt \mapsto10x + y + 10y + x = 99. \\ \\ \tt \mapsto11x + 11y = 99. \\ \\ \tt \mapsto11(x + y) = 99. \\ \\ \tt \mapsto x + y = \cancel \frac{99}{11} \\ \\ \tt \mapsto x + y = 9...eq(2)$

• By Adding eq(1) and eq(2) we get,

$\tt\mapsto(x - y) + (x + y) = 3 + 9. \\ \\ \tt\mapsto x \: \cancel{ - y} + x \: \cancel{ + y} = 12. \\ \\ \tt\mapsto x + x = 12. \\ \\ \tt\mapsto2x = 12. \\ \\ \tt\mapsto x =\cancel \dfrac{12}{2} . \\ \\ \tt\mapsto x = 6.$

• So by solving we get x=6.

Now by putting the value of x in eq(1) we get,

$\tt\mapsto x - y = 3. \\ \\ \tt\mapsto 6 - y = 3. \\ \\ \tt\mapsto - y = 3 - 6 \\ \\ \tt\mapsto\cancel{ - }y = \cancel{ - }3. \\ \\ \tt\mapsto y = 3.$

• So by solving we get y=3.

$\therefore$ The number is,

$\sf10x + y \\ \\ \sf = 10(6) + 3. \\ \\ \sf = 60 + 3. \\ \\ \sf = 63.$

So the required number is 63 (or it may be 36).

[Because both the numbers satisfy the given conditions].