Time and Work Questions

2 engines of former type for one hour consumes (2 x 24)/(6 x8) = 1 metric ton 

i.e. 3 engines of latter type consumes 1 ton for one hour

Hence 9 engines consumes 3 tons for one hour

For 15 hours it is 15 x 3 = 45 metric tonnes.

Here given soldier shots 7 shots in 12 min

=> 1 shot doen't take any time in 12 min 

=> Only 6 shots take 12 min 

12 min ------ 6 shots 

90 min ------ ? shots 

=> 90 x (6/12) = 45 

Therefore, total shots fired in 90 minutes = 45 + 1 = 46 shots.

1/15 + 1/12 + 1/R = 1/6, we got R = 60 (it means R will take 60 days to complete the work alone)

so ratio of work done by P:Q:R = 1/15 : 1/12 : 1/60 = 5 : 4 : 1

so R share = (1/10) x 1200 = 120.

Given (3 Men + 4 Women + 6 Children) -----> 9 days

But W = 2M and C = M/2

Now, convert Men and Children into Women by

$$\begin{gathered} 3W2 + 4W + 6W4 \\ => 32 + 4 + 32 = 7 Women \end{gathered}$$ 

Therefore, 7 women alone can complete this work in 9 days.

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.

Since 20% i.e 1/5 typists left the job. So, there can be any value which is multiple of 5 i.e, whose 20% is always an integer. Hence, 5 is the least possible value.

1/3 ---- 8

1 -------?
Hari can do total work in = 24 days

As satya is 60% efficient as Hari, then

Satya = 1/24 x 60/100 = 1/40

=> Satya can do total work in 40 days

1 ----- 40

2/3 ---- ? => 26 2/3 days.

(A+B)'s 15 days work = 120×15=34

Remaining work =  1/4

Now, 1/4 work is done by A in 10 days.

Whole work will be done by A in (10 x 4) = 40 days.