Question

the sum of the digits of a number consisting of 3 digits is 12 the middle digit is equal to half of the sum of the other two of the order of the digits is reversed the number is diminished by 198 find the number

Answer 1

let the number be xyz where x is in hundred's place, y is in ten's place and z is in unit's place so it can also be expressed as 100x+10y+z . x+y+z=12...(1) y=(x+z)/2 => 2y=x+z ...(2) Substitute (2) in(1), we get 2y+y=12=> y=4...(3) when it is reversed it becomes zxy and can also be expressed as 100z+10y+x . It is given that on reversing, the value gets diminished by 198 so 100x+10y+z-(100z+10y+x) =198=> 99x-99z=198=> x-z=2...(5) When we substitute value of in equation(3) ,we get x+z=8...(4) .Add equations
(4) and (5) , 2x=10=> x=5 and substitute value of x in equation (4), 5-z=2=> z=3. Then the number is 5(100)+4(10)+3= 543. Hope it helps you...

Answer 2

Answer:

Answer is 543

Step-by-step explanation: