A and B together can finish a job in 20 days. B and C together can finish the same job in 30 days. If A and C together can finish it in 24 days, in how many days can A alone finish the job?
– A and B together finish the job in 20 days: a + b = 1/20 (Work done in one day)
– B and C together finish the job in 30 days: b + c = 1/30
– A and C together finish the job in 24 days: a + c = 1/24
– Add the three equations: (a + b) + (b + c) + (a + c) = 1/20 + 1/30 + 1/24
– 2a + 2b + 2c = (6 + 4 + 5) / 120 (LCM of 20, 30, 24 is 120)
– 2(a + b + c) = 15 / 120 = 1/8
– a + b + c = 1/16 (Combined work rate of A, B, and C in one day)
– To find the work rate of A (a), subtract the work rate of B and C together (b + c) from the combined work rate:
– a = (a + b + c) – (b + c)
– a = 1/16 – 1/30
– Find a common denominator (LCM of 16 and 30 is 240):
– a = (15/240) – (8/240) = (15 – 8) / 240 = 7/240
– A’s work rate is 7/240 of the job per day.
– The time taken by A alone to finish the job is the reciprocal of A’s work rate:
– Time for A = 1 / (7/240) = 240 / 7 days.
– Convert the improper fraction to a mixed number: 240 ÷ 7 = 34 with a remainder of 2.
– So, 240/7 days = 34 and 2/7 days.
b = (a+b+c) – (a+c) = 1/16 – 1/24 = (3 – 2)/48 = 1/48. Time for B = 48 days.
c = (a+b+c) – (a+b) = 1/16 – 1/20 = (5 – 4)/80 = 1/80. Time for C = 80 days.