Question

Mr john has a son and a daughter,the son being 4years older than the daughter.five years ago,the age of mr john was 23 years more than the sum of the ages of his son and daughter.after 18 years from now,the age of mr john will be seventeen times the difference of ages of his son and daughter.find their present ages

Answer 1

5 years ago:

son's age= x+4
daughter= x
John     =x+x+4+23 (since his age was 23yrs more than the sum of the two)

Present age: (add 5 years to each age)
son =x+4+5
daughter=x+5
John  =2x+27+5=2x+32

Age after 18 years:
son =x+9+18=x+27
daughter=x+5+18=x+23
John        =17{(x+27)-(x+23)} since his age will be 17 time the difference of                                                  the two...therefore his age at this time will be:                                                                  17(x+27-x-23)=17(4)=68
       if his age 18 years later will be 68 the his age now is 68-18=50
        therefore 2x+32=50
                         2x=18  x=9
hence son's age =x+9 =9+9=18
while of the daughter =x+5 =9+5=14

their present ages:
John 50
son 18
daughter 14