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#A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained? Please solve it in details.

#solution_

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 = 1/6.

Alternative :
(A+B), 1 days work= (1/72+1/48)= 5/144 parr

(A+B), P/2 days work= 5P/288 part

(A+B+C), 1 day work= (1/72+1/48+1/36)= 9/144

(A+B+C), (P+6)/3 days work = 9*(P+6)/432

So,
5P/144+ 9*(P+6)/432= (100-125/3)*1/100= 7/12

=>[ (15P+18*(P+6)]/864= 7/12

=> 12*33P+12*108= 7*864

=> 396P= 4752

=> P= 4752/396= 12 days

Now D, 125/3% of work in= 10 day
100% work in= 10*3*100/125= 10*3*4/5= 8*3= 24

So D need = 24 days
C need= 36 days

12 days work = 12*(1/24+1/36)= 12*(3+2)/72= 5/6

Remain part= 1-5/6= 1/6