Question

The middle digit of a number between 100 and 1000 is zero and the sum of the other digit is 13.If the digits are reversed,the number so formed exceeds the original number by 495.find the number.

Answer 1

Hlw mate


Solution:

Let the number in 100s digit be x,

Let the number in 1s digit be y,
 
Then,the number would be x0y : 100x+y

Condition 1: Sum of the digits : x+y=11.....(i)

If it is reversed ,it would be y0x : 100y+x,

The difference between them:    100x+y -(100y+x) = -495
                                                               100x+y-100y-x =-495
                                                                           99x-99y  = -495  (divided by 99)
                                                                                      x-y=-5.....(ii)              

Adding (i) and (ii), we get,

                                 (x+y)+(x-y)=11+(-5)
                                      x+y+x-y=11-5
                                                2x=6
                                               
                                              ∴ x=3
Substituting value of x in (i),
                                           x+y=11
                                           3+y=11
                                               y=11-3

                                            ∴ y=8

                  ∴The number is 308

Answer 2

Answer:

the answer above answered is wrong..

Step-by-step explanation:

the no. is 409