solve it please with explanation.
Answers
Let the Work to be Completed be 'W'
Given that Ratio of Efficiency of C and D is 3 : 1
⇒ Work done by C in One day = 3 × Work done by D in One day
Given that If C and D work for alternative days starting from C, they can do the total work in 30 days
⇒ In 30 days : 15 days work is done by C and other 15 days work is done by D
⇒ 15 days Work of C + 15 days Work of D = Total Work
⇒ 15 × C(one day) + 15 × D(one day) = W
⇒ 15 × 3D(one day) + 15 ×D(one day) = W
⇒ 60 × D(one day) = W
⇒ D(one day) = W/60
⇒ C(one day) = 3 × W/60 = W/20
⇒ C alone can Complete the Work in 20 days
⇒ D alone can Complete the Work in 60 days
Given that C worked for 10 days
⇒ 10 days work of C = 10 × W/20 = W/2
⇒ Remaining Work after 10 days working of C : W - W/2 = W/2
Given the Remaining work is completed by M and N in 6 days
⇒ Work of M + Work of N = W/2
⇒ 6 × M(one day) + 6 × N(one day) = W/2
⇒ M(one day) + N(one day) = W/12
Given that Ratio of Efficiency of M and N is 1 : 3
⇒ Work done by N in One day = 3 × Work done by M in One day
⇒ M(one day) + 3M(one day) = W/12
⇒ 4M(one day) = W/12
⇒ M(one day) = W/48
⇒ M alone will complete the Work in 48 days.