Time and Work Questions

Let work done by 1 man in i day be m

and Let work done by 1 boy in 1 day be b

From the given data,

4(5m + 3b) = 23

20m + 12b = 23....(1)

2(3m + 2b) = 7

6m + 4b = 7 ....(2)

By solving (1) & (2), we get

m = 1, b = 1/4

Let the number of required boys = n

6(7 1 + n x 1/4) = 45

=> n = 2.

Ratio of efficiency of A to B is 3 : 2

Let, In 8 days they complete = (3 + 2) × 8 = 40 units

So, total work = 40 × 12/5 units

Time taken by B alone to complete whole work = 40 ×  12/5 ×2 = 48 days 

Raghu can complete the work in (12 x 9)hrs = 108 hrs.

 Arun can complete the work in (8 x 11)hrs = 88 hrs.

Raghu's 1 hrs work = 1/108 and Arun's 1 hrs work = 1/88

(Raghu + Arun)'s 1 hrs work = 1108+188=492376

 
So, both Raghu and Arun will finish the work in 237649 hrs

Number of days of 12 hours each=237649×112 = 19849=4349 days

Work done by P and Q in the first two hours, working alternately

= First hour P + Second hour Q

⇒14+112=13

work is completed in 2 hours

Then, the total time required to complete the work by P and Q working alternately=2 x 3= 6hours

Thus, work will be completed at 3pm.

For 30 days he receives

30 x 25 = 750
               

Actually he gets 425

Therefore, Fine is 750 - 425 =  325

=> No of days absent = 325/(25+7.50)= 10

Suppose A, B and C take x, x/2 and x/3 respectively to finish the work. 

Then, (1/x + 2/x + 3/x) = 1/2 

=> 6/x = 1/2      => x = 12 

So, B takes 6 hours to finish the work.

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.

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.