Question

mason a builds 7 mason b builds 6 mason c builds 5 b works twice as many days and c works half as many as a and b together how many days take them to build 318 meters

Answer 1

Assume a builds 7m per day, b builds 6m per day, c builds 5m per day.

Let A=number of days a builds, B=number of days b builds and C=number of days c builds.

B=2A and C=A/2+B/2=A/2+A=3A/2.

Total number of metres built=318=7A+6B+5C=7A+12A+15A/2=53A/2.

53A/2=318; 53A=636, A=12 days.

Mason a builds 84m, mason b builds 6*24=144 m, mason c builds 5*(12+24)/2=90.

84+144+90=318m.