(A + B)'s1day'swork=110C's1day'swork=150(A + B + C)'s1day'swork=(110+150)=650=325........(i)A's1day'swork=(B + C)'s1day'swork........(ii)From(i)and(ii),weget:2×(A's1day'swork)=325⇒A's1day'swork=350∴B's1day'swork(110−350)=250=125So,Balonecoulddotheworkin25days
A's2day'swork=(120×2)=110(A + B + C)'s1day'swork=(120+130+160)=660=110Workdonein3days=(110+110)=15Now,15workisdonein3dys∴Wholeworkwillbedonein(3×5)=15days
His Per day pay = Rs. 26
Total pay employee got = Rs. 829
Total pay he gets if he did not remain idle a single day,
= 26 × 56 = Rs.1456
He Forfeits or fined = 1456 - 829 = Rs. 627
Per day he Forfeits Rs. 7 Means per idle day he loses = 26 + 7 = Rs. 33
So, Total idle days =
62733
= 19 days
Alternatively,
Let he works for X days and remain idle for (56 - X)
Now, According to question,
X × 26 - 7 × (56 - X) = 829
26X + 7X - 392 = 829
33X = 829 + 392 = 1221
X = 37 days
No of days remain idle = 56 -37 =19 days
To do half of the work in one sixth of the time means A is 3 times as efficient as B. To understand this point just think if same work would have been done in one sixth of the time then A would have been six times as efficient as B.
Now they together complete the work in 10 days. B is working with someone who is 3 times as efficient as himself. That means 4 people of B's efficiency finished the work in 10 days. So B alone would have done it in 40 days. A alone would have taken one third of this time.
Alternatively,
Given,
A×16=B×12
So, A = 3B
Now,
Given they together complete the work in 10 days
So, One Day''s work of, (A + B) =
10010
= 10%
(3B + B) = 10%
4B = 10%
So, one day work of B =