Two typists can together type a report in 7.5 minutes. If the 1st typist types alone for 3 minutes and then the 2nd typist types alone for 10 minutes, they complete 3/4 of the total work. How many minutes would the slower typist take to complete the work alone?

ATQ, Let the 1st typist take x minutes and the 2nd typist take y minutes alone. Together, 1/x + 1/y = 1/(7.5) = 2/15 ---- (i) Work done in 3 minutes by 1st and 10 minutes by 2nd is 3/4, so 3/x + 10/y = 3/4 ---- (ii) Let 1/x = u and 1/y = v. From (i): u + v = 2/15 ---- (A) From (ii): 3u + 10v = 3/4 ---- (B) From (A), u = 2/15 − v. Substitute in (B): 3(2/15 − v) + 10v = 3/4 6/15 − 3v + 10v = 3/4 2/5 + 7v = 3/4 7v = 3/4 − 2/5 = (15 − 8)/20 = 7/20 v = 1/20 ⇒ y = 20 u = 2/15 − 1/20 = (8 − 3)/60 = 5/60 = 1/12 ⇒ x = 12 Slower typist takes = 20 minutes.