TIME AND WORK

 INTRODUCTION:-

Two quantities are said to vary inversely of the increase or decrease in one quantity causes the decrease or increase in the other quantity. For example:

The number of men doing work is inversely proportional to the time taken to finish a work.

If there are more men at work, less time will be taken to finish the work. If there are  Less men at work,  more time will be taken to finish the work.  Again, to cover a certain distance the speed is inversely proportional to the time taken.

If a vehicle has more speed,  to cover the same distance less time will be required. Less speed, more time taken to  cover the same distance. In solving problems on time and work, we must remember that

• One day’s work = 1/number of days to complete the work.
• Number of days needed to complete the work= 1/ one day’s work
• Time required to do a certain work = work to be done /one day’s work
• Remuneration is in the proportion of work done.

Let’s discuss some problems on time and work.

i)Baban and Raman together erect a shed in 12 days. Baban alone can do this work in 20 days. How much time would Raman take working alone to erect the shed?

Solution:  Baban and Raman together take  = 12 days to erect a shed

So one day’s work of Baban and Raman together = 1/12 part

Now Baban alone can erect the shed in 20 days

Baban’s one day work = 1/20 part

Raman’s one day work = (1/12 – 1/20) = (5-3)/60 = 2/60 = 1/30

Therefore, alone can erect the shed in 30 days.

ii) A can complete 1/5 of a piece of work in 12 hours and B can complete 1/6 of the same work in 15 hours. If A and B both working together in how many hours they can complete the work? (TIME AND WORK MATH)

Solution:  In 12 hours A can complete 1/5 part of work

In 1 hour A can complete 1/5 X 1/12 = 1/60 part of work

In 15 hour B can complete 1/6 part of work

In 1 hour B can complete 1/6 X 1/15 = 1/90 part of work

If they work together then in our hour they will complete (1/60) + (1/90) = (6+4)/360 = 10/360 =1/36

Therefore, A and B can complete the work in 36 hours.

iii)A and B together can dig a pond in 20 days. They worked together for 8 days and B leaves the work.  If A alone can dig the pond in 30 days then how long will A take to finish the work of digging pond? (TIME AND WORK MATH)

Solution: in 20 days A and B can complete digging of pond

In 1 day A and B will do 1/20 part of work

In 8 days A and B will do 8 X 1/20 = 2/5 part of work

After 8 days work left = ( 1-2/5)= (5-2)/5 = 3/5 part

When B has left, A has to complete 3/5 part of work.

In 30 days A can complete 1 work,

In 1 day A can complete 1/30 part of work

A can do 1/30 part in 1 day

A can do 1 part of work in 30 X 1 days

A can do 3/5 part in 30 X (3/5)= 18 days.

Therefore, A can complete the remaining work in 18 days.

iv)P can complete a work in 10 days and Q can do it in 15 days. They worked together 4 days and then   P left the work. In how many days can Q finish the remaining work; if the remuneration for the work is Rs1000, how much amount would each get? ( TIME AND WORK MATH)

Solution:  In 10 days P can complete one work

In 1 day P can do 1/10 part of work

In 15 days Q can complete 1 work

In 1 day Q can do 1/15 part of work

When they work together, their 1 day work = (1/10)+ (1/15)= (3+2)/30= 5/30= 1/6

Their 4 day’s work= (4X 1)/6 = 2/3

After 4 days Q left, so work remained = 1-2/3 = (3-2)/3 = 1/3

Q can do 1/15 part of work in 1 day

Q can do 1 part of work in 15 days

Q can do 1/3 part of work in (15 X 1)/3 = 5 days

Now we have to calculate their remuneration

In 4 days P has done 4 X 1/10 = 2/5 part

So share of P = 2/5 of Rs1000 = Rs400

Share of Q = (1000-400) = Rs 600

Therefore, Q will complete the remaining work in 5 days ; share of P = Rs 400; share of Q= Rs600.

v)If 2 men  or 3 boys take 40 hours to do a certain piece of work, how long will 4 men and 9 boys working together take to complete the work? ( TIME AND WORK MATH)

Solution:- 2 men’s work = 3 boy’s work

1 men’s work = 3/2 boy’s work

4 men’s work= (3 X 4)/2 = 6 boy’s work.

So 4 men + 9 boys = 6 boys + 9 oys = 15 oys

If 3 boys can do one work in 40 hours

1 boy can do one work in 40 X 3  = 120 hours

15 boys can do one work in 120/15 = 8 hours

Therefore, 4 men and 9 boys will complete the work in 8 hours.

vi) A and B together can built a wall in 30 days. If A is twice as good a workman as B, in how many days will A alone finish the work?

Solution: A’s activity is double than B’ activity.

A’s one day work = B’s 2 day’s work

B’s one day work = A’s ½ day work

If A and B work together, they can built a wall in 30 days

So A’s one day work + B’s one day work {A’s ½ day work} = 1/30 part

In (1+1/2 ) 3/2 days A did  1/30 part of work

A’s 1 day work = (1/30) X (2/3) = 1/45 part

Therefore, A alone can complete the work in 45 days.

vii) A,B and C when  working separately can complete  a work in 2,3 and 4 days respectively. If A,B and C work  together and jointly earn Rs3900 for the whole work, what will be the ratio to divide the money? ( TIME AND WORK MATH)

Solution: A’s one day work = ½

B’s one day work = 1/3

C’s one day work 1/4

So, the money will be divided among them,

A,B and C will get money in following ratio = 1/2 : 1/3 : ¼

Or L.C.M. of 2,3 and 4 = 12

So A : B: C :: (1/2 X 12) : (1/3 X 12) : (1/4 X 12) = 6:4:3

Total sum of the ratio = 6+4+3 = 13

Total money = Rs3900

Share of A = {(6 X 13) X 3900} = Rs 1800

Share of B  = {(4/13) X 3900} = Rs 1200

Share of C = {(3/13) X 3900} = Rs 900

Therefore, A will get Rs1800; B will get Rs1200; C will get Rs900.

viii)A and  B together can do a piece of work in 20 days. B and C jointly can complete that work  in 15 days, where  C and A jointly  complete  it in 12 days. How many days are needed to complete the work when both of them are working together? If they are working separately then how much time will be needed to complete the same work? ( TIME AND WORK MATH)

Solution:  A and B can complete a work in 20 days

one day work of A + B = 1/20 part of work

when B and C work  together they   can complete the  work in 15 days

one day work of B + C = 1/15 part of work

C and A can together complete  a work in 12 days

one day work of A + C = 1/12 part of work

Adding up all = A+B +B+C+A+C = 1/20 + 1/15 +1/12

Or,2(A+B+C) = (3+4+5)/60 = 12/60 =1/5

Or, A+B+C =1/10

If A,B and C work together they can finish the work in 10 days.

If they work separately, then (A+B+C) – (A+B) = work of C = (1/10)- (1/20) = (2-1)/20 = 1/20 part in 1 day

(A+B+C) – (B+C) = work of A =( 1/10) –(1/15) = (3-2)/30 = 1/30 part in 1 day,

(A+B+C) – (C+A)= work of B = (1/10)-(1/12) = (6-5)/60= 1/60 part in 1 day.

Therefore, A,B and C can together complete the work  in 10 days.

But, if they work separately then A can finish the work in 30 days,

B can finish the work in 60 days

C can finish the work in 20 days.

ix)Tap A and tap B can fill a tank in 4 hours and 6 hours respectively and tap C present at the bottom which can empty the tank in 12 hours.  When the tank is completely empty  and all the three taps are opened then in how many hours the tank will be full?  

Solution; In one hour, tap A fills 1/4 part of tank,

In one hour, tap B will fill 1/6 part of tank

In one hour tap C will empty 1/12 part of tank

So, in one hour, the portion of tank filled by the taps A,B and C all together =

(1/4 + 1/6) – 1/12 = (3+2-1)/12

=4/12

=1/3

Therefore, when all taps are opened the tank will be full in 3 hours.

x) Pipe A can fill a tank in 10 hours and pipe B can fill the same tank in 20 hours. If both the pipes fill the tank simultaneously. What time will they take to fill the tank? 

Solution: Pipe A can fill tank in 1 hour = 1/10 part

Pipe B can fill tank in 1 hour = 1/20 part

When both the pipes are working  then part of tank filled by pipe A and pipe B

= (1/10)+(1/20)

=(2+1)/30

=3/20

Therefore, time taken by the pipes A and B take to fill the tank is 20/3 hours = 6 hours 40 minutes.