Three siblings, Susie, Tommy and Roland are 42 years old together now. When Susie was of Tommy\'s age, Roland was one year older than the triple of Tommy\'s age (at that time). When Tommy will be of Susie\'s age, then Roland will be 10 years younger than the double of Susie\'s age(at that time). How old will Roland be, when Tommy\'s age will be the 3/4 of Susie\'s age (of that time)? Solution let age of susie be s, tommy be t and roland be r. s+t+r = 42...eqn1 susie would be tommy\'s age s-t yrs back age of tommy then would be t - (s-t) = 2t-s yrs rolands age s-t yrs back = r-(s-t) = r+t-s now r+t-s = 1+3(2t-s) =>5t - 2s -r = -1 ....eqn2 tommy will be susie\'s age (s-t) yrs later age of susie would be s+(s-t) = 2s-t age of roland would be r+(s-t) = r+s-t r+s-t = 2(2s-t) -10 =>3s-r -t = 10...eqn3 solving eqn1, 2, 3 we get r = 20 s = 13 t = 9. 3 yrs later tommy\'s age would be 3/4 times that of susie. roland\'s age = 20+3 = 23yrs.