Quicker Maths

## You can make a difference

Posted on October 11, 2010

I remember back to early 2005 (before going for my post graduation), when I’d just started preparing for some competitive examination, that even though I had great ambitions, my knowledge, expertise, and confidence as a student was sorely lacking. My love for internet (as a source of information) made me stumble through various places on the web to know maths short cuts and quick maths tricks, but without much to gain from it.

Now, after having being there and facing it all as a student first and then as a facilitator (though I take good number of Quants classes for CAT aspirants, I would not liked to be called a tutor or a teacher. These titles should be reserved for people of great wisdom and learning) for over 4-5 years I felt I had a really good grip on things;  and even then, there were many things I am still struggling with.

Now with the help of QuickerMaths.com, I want to create a platform, easily accessible to the student fraternity where they can get what they are looking for; at least things related to Quantitative Aptitude, Logical Reasoning, Analytical Reasoning and Critical Reasoning - a platform where you can ask questions and where you can give answers.

## Let The Process Begin

I’m starting the process by posting below an interesting formula of the topic- time and work . The aim is to initiate the process of collaborative learning. So anyone who wants to make a difference can post a comment with any explanation, formula, shortcut, trick, solved examples, etc. on the same topic (Time and Work for this post). You can also help by sharing the information of this platform with your friends or anyone who wants to help others and learn from others in the process.

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## Time and Work Formula

I think the single most useful formula for the topic Time and Work is

N1H1D1E1W2 = N2H2D2E2W1

Where:

N1 and N2 = number of person

H1 and H2 = Hours worked by per person per day (assumed constant)

D1 and D2 = days

E1 and E2 = Efficiency

W1 and W2= Amount of work done

Consider this example to understand the applicability of this formula –

A piece of work can be done by 16 men in 8 days working 12 hours a day.How many men are needed to complete another work, which is three times the first one,in 24 days working 8 hours a day. The efficiency of the second group is half that of the first group?

Solution –

N1H1D1E1W2 = N2H2D2E2W1

16*12*8*1*3 = N2*8*24*0.5*1

N2 = (16*12*8*1*3)/ (8*24*0.5*1) = 48

So number of men required is 48.

Note – you can remove anything from formula is not given in the question. For example if the question would have been –

“A piece of work can be done by 16 men in 8 days working 12 hours a day.How many men are needed to complete another work, which is three times the first one,in 24 days working 8 hours a day.”

The applicable formula would have been –

N1H1D1W2 = N2H2D2W1

Since nothing is mentioned about efficiency, we remove it from both sides.

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#### Posted by Vineet Patawari

1. 1 man, 3 women and 4 boys can do a piece of work in 96 hours. 2 men and 8 boys can do it in 80 hours. 2 men and 3 women can do it in 120 hours. 5 men and 12 boys can do it in how many hours?

• m+3w+4b=96———————(1)
2m+8b=80————————–(2)
or m+4b=160———————–(3)
Since, the no of persons are decreasing, so no of hours will increase.
2m+3w=120————————(4)
From (1) and (3),
3w=(96*160)/(160-96)=240 hrs——–(5)
From (4) & (5)
2m=(240*120)/(240-120)=240 hrs——–(6)
1m can do in 240*2=480 hrs—————-(7)
From (2) &(6)
8b=(240*80)/(240-80)=120 hrs
So, 1b=120*8=960 hrs—————(8)
If one man can do in 480 hrs, then 5m can do in 480/5=96 hrs—-(9)
If one boy can do in 960 hrs, then 12b can do in 960/12=80 hrs—(10)
Together (5m+12b) can do in

2. how they calculated workdone is 3

3. in that example 16*12*8*1*3 = N2*8*24*0.5*1 how 1*3 came

4. P, when working alone, can complete a work in 30 days. He can complete the same work in 20 days when working with Q. Working with Q and R, P can complete the same work in 10 days. In how many days can Q and R together do the same work
?

• Q & R together will complete the work in 15 days.

• how? Can u pls explain abhishek

• I did it lik this –
3 workers (p,q & r) = 10 days
2 workers (Q & R) = ?

3*10/2*X = 15 days

Is this the correct method?

• @monica: its correct.

• @Monica: Sorry but….it should be……like this:

P can do in 30 days

P&Q can do in 20 days

P,Q&R can do in 10 days
———————————————–
find LCM of 30,20,10= 60 = total work
————————————————

P can do in 30 days => per unit work= 60/30=2 work per day by P……………….eq(1)

P&Q can do in 20 days => per unit work= 60/20=3 work per day by P&Q => Q’s per unit work= 3-2=1 per unit work (from eq.1.)

P,Q&R can do in 10 days => per unit work= 60/10=6 work per day
=> R’s per unit work= 6-2-1= 3 per unit work.

**Hence, Q&R will complete the work (which we have calculated by taking LCM ,i.e, 60) in : total work/ per unit work by Q&R = 60/(1+3)= 60/4= 15 days…..ans!!

Follow the same concept….as by your method may pose some probs…..Your method is suitable for Chain rule kind of time & work problem.

• tks a lot abhishek. Ur way of teaching is easy to understand:)

5. If 12 men or 18 women can reap a field in 14 days , then the same number of days that 8 men and 16 women will take to reap the same field is:

6. 4 men & 10 boys dig 60 meters in 8 days, 10 men &16 boys dig 90 meters in 6 days,then how many boys with 10 men could dig 180 meters in 8 days ?

• [(4M+10B)*8]/[(10M+16B)*6]=60/90 => M=2B

let, number of boys be ‘x’.

now, [(4M+10B)*8]/[(10M+x)*6]=60/180
=> [18B*8]/[(20B+x)*8]=60/180 ………….since, M=20B
=> x=34B

Hence, 10 men and “34 boys” will be required to dig 180mts in 8 days!!

• Plese explain this line.. How you got this??

[(4M+10B)*8]/[(10M+x)*6]=60/180

• @Gayatri: Plese explain this line.. How you got this??

[(4M+10B)*8]/[(10M+x)*6]=60/180
———————————————————-
Just take any one of the conditions given,i.e, (4M+10B)*8=60
or (10M+16B)*6=90…….and equate with (10M+x)*6=180….where ‘x’ is required number of boys.

According to : N1H1D1E1W2 = N2H2D2E2W1 (explained above)

you can equate any two given conditions and equate given values and solve the equation..
Now, firstly I equated the towo given conditions to find relation between boys and men….which came out to be: M=2B

So, in next step I equated with the required equations.

• I understood but how come u have written it as (10M+x)*6).. AS mentioned in the question it is asked for 8 days.. Is it your typing mistake???

• @Gayatri: I understood but how come u have written it as (10M+x)*6).. AS mentioned in the question it is asked for 8 days.. Is it your typing mistake???
——————————————-

Ohh…..pardon me. Its a typing mistake of mine -It should be

[(4M+10B)*8]/[(10M+x)*8]=60/180

By mistake I took the no. of days val in denominator as 6…it should be 8 days as per the question!!

Thanks for pointing that out….but the no. of boys required=x=34.

7. A can do a piece of work in 14 days while B can do it in 21 days.They began together and worked at it for 6 days.Then A fell ill and B had to complete the remaing work alone.In how many days was the work completed?

• concept: (no. of days needed when working alone)*(work done per unit time) = Total work done

A B (A+B) B
————————————————-
total days when working alone: 14 21 6 6

No. of work done/day : 3 2 (3+2=5) 2
————————————————–
Total work : 42 42 30 (42-30=12)
————————————————–

Hence, B alone worked for 6 more days after A left the job. Thus, the work (here, LCM of 21 & 14) will be completed in 12 days (since, A&B worked for 6 days and completed 30 work. So, remaining work,i.e., 12 will be done by B alone, which will be in 12/2=6 days)….thus, in all 12 days.