WBahn: I am still catching up on my reading.

I thought this might be relevent. Note that the book dates back to 1934 and the fourth addition is from 1965.
Some things change and some don't. You may not be using a slide rule but many of the hints here still apply.

 

Yeah, but notice that they don't believe in following their own rules. As is often the case with EE and ME texts, they really only give lip service to units and believe that just tacking the units that you want the result to have onto the end is sufficient. Just look at their work on the top of page 32.

I was so very fortunate to have John Trefny (who went on to become the president of the university) for Intermediate Electricity and Magnetism, my first major's tract course. The first day of class he said (as close as I can recall), "Do you want to know what separates you and me from a Nobel Prize winning physicist? It's simple, the people that walk across that stage in Stockholm are religious about two things: they always, always track their units, and they always, always, always ask if the answer makes sense." Clearly a bit of hyperbole, but it set the stage for me to really begin seeing the value of that advice. Before that I had used the "factor label" method for doing conversions, but it had completely escaped me how units are involved in almost every computation you can think of as a fundamental element. And Tref spared no quarter for sloppy units work. The dividends that has paid are unimaginable.

 

@RRITESH KAKKAR

Inasmuch as it is Sunday evening/Monday morning here - I will will not likely have much time today -- however I expect to be 'back' tomorrow at the usual time (i.e. ≈ 8:00 UTC)...

Best regards
HP

 

Hint: in this exercise we're dealing with whole temporal units -- as opposed to fractional units

What is temporal units?

 

Please examine post #937 closely! -- Notice that @WBahn 's approach of tracking/annotating units/entities results in self-documenting work -- thus 'bonding' the 'abstract' to the 'concrete' and, thereby, rendering most common errors obvious! - You will find said practice/habit of significant and increasing value as the complexity of the problems increases

Best regards
HP

How to write that?

 

How to write that?

How to write that?

@RRITESH KAKKAR

It's called 'TEX' formatting --- You may find the below linked thread helpful

http://forum.allaboutcircuits.com/t...f-tex-easy-awesome-looking-posts-howto.21380/

See you tomorrow!

Best regards
HP

 

What is temporal units?

Temporal means of or relating to time -- Hence examples of temporal units include: Seconds, hours, days, years, millennia, etc...

@RRITESH KAKKAR

I apologize for my absence -- Moreover, unfortunately, @Aleph(0) is away and, hence, unable to 'sub' as it were -- I will be available tomorrow however!!!

Best regards
HP

 

A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?

Given that A, B, C represent the task completion times required by the respective workers:
A can do a piece of work in 4 hours;
while A and C together can do it in 2
Note that the above quoted stipulations tell us that A=C

Hence, via:
B and C together can do it in 3 hours
1/(1/4+1/B)=3 --- Solve for B
1/4+1/B=1/3
1/B=1/3-1/4
1/B=0.33-0.25
B=12 HOURS?

 

A does 80% of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?

A=80/100*1/20 OR 1/20*0.80
A+B=1/3 EQ.1
B=? FOR 100% WORK
B=1/3-A
...

 

A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?

Given that A, B, C represent the task completion times required by the respective workers:
A can do a piece of work in 4 hours;
while A and C together can do it in 2
Note that the above quoted stipulations tell us that A=C

Hence, via:
B and C together can do it in 3 hours
1/(1/4+1/B)=3 --- Solve for B
1/4+1/B=1/3
1/B=1/3-1/4
1/B=0.33-0.25
B=12 HOURS?

Correct!

 

how to solve this?
A does 80% of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?

A=80/100*1/20 OR 1/20*0.80
A+B=1/3 EQ.1
B=? FOR 100% WORK
B=1/3-A
...

 

how to solve this?
A does 80% of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?

A=80/100*1/20 OR 1/20*0.80
A+B=1/3 EQ.1
B=? FOR 100% WORK
B=1/3-A
...

Well...

--If 'A' completes 80% of the job in 20 days then clearly 'A' would complete the entire job in 25 days...
--When 'B' joins the task 1/5 of the work remains...

Hence (A's 3_day contribution (i.e. 3/25 of the job) + B's 3_day contribution (i.e. 3/x of the job) = the fraction of the job remaining (i.e. 1/5) = 3/25+3/x = 1/5 (Where x = Days required for 'B' alone to perform the entire task...

Please study the above carefully!

Best regards
HP

PS
@RRITESH KAKKAR
In the interests of precision and/or aesthetics please refrain from expressing non-integral values via decimal expansions -- unless explicitly directed otherwise!

 

Mathematics need clearlty of mind to solve problem...

 

Well...

--If 'A' completes 80% of the job in 20 days then clearly 'A' would complete the entire job in 25 days...
--When 'B' joins the task 1/5 of the work remains...

Hence (A's 3_day contribution (i.e. 3/25 of the job) + B's 3_day contribution (i.e. 3/x of the job) = the fraction of the job remaining (i.e. 1/5) = 3/25+3/x = 1/5 (Where x = Days required for 'B' alone to perform the entire task...

3/25+3/x = 1/5
3/x=1/5-3/25
x=37.5days

 

A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 A.M. while machine P is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished ?


P=1LAKH IN 8HOURS or 1/8 in hours
Q=1/10
R=1/12
All the machines are started at 9 A.M. while machine P is closed at 11 A.M.
so, 14hours for P...............

 

A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 A.M. while machine P is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished ?


P=1LAKH IN 8HOURS or 1/8 in hours
Q=1/10
R=1/12
All the machines are started at 9 A.M. while machine P is closed at 11 A.M.
so, 14hours for P...............

Ask if this makes sense.

You know that the slowest of the machines, working alone, will be done in only 12 hours. So does 14 hours seem reasonable?

How much of the job will remain at 11 AM (i.e., how much of the job can be accomplished in that time by all three machines)?

How much of the job is completed, per hour, when only machines Q and R are running?

 

A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 A.M. while machine P is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished ?

All the machines are started at 9 A.M. while machine P is closed at 11 A.M.
what this line mean?
because P is working super fast than other so, it should closed earlier.

 

How much of the job will remain at 11 AM (i.e., how much of the job can be accomplished in that time by all three machines)?

P+Q+R
1/8+1/10+1/12 = WORK IN 1 HOURS.
=0.30833
or 1/0.30833 total work in 3.24 hours.

=0.30833 *14=4.31hour

 

A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 A.M. while machine P is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished ?

All the machines are started at 9 A.M. while machine P is closed at 11 A.M.
what this line mean?
because P is working super fast than other so, it should closed earlier.

No. It means just what it says.

All three machines are started at 9 A.M.

Then, at 11 A.M., machine P stops working. It doesn't matter why. Maybe it broke down. Maybe the only person authorized to run it got sick and had to leave. Maybe the cleaning lady spilled something on it. What possible difference could it make as far as this problem is concerned? At 11 A.M. it was closed down and the other two machines were left to finish the job.

 

How much of the job will remain at 11 AM (i.e., how much of the job can be accomplished in that time by all three machines)?

P+Q+R
1/8+1/10+1/12 = WORK IN 1 HOURS.
=0.30833
or 1/0.30833 total work in 3.24 hours.

=0.30833 *14=4.31hour

Again, does this make sense?

The slowest machine can do the entire job in 12 hours.

If all three machines ran at the same speed as the slowest machine, how long would it take all three to complete the job?

And, again, where does the 14 come from?

You also are not taking into account the shutdown of machine P at 11 A.M.