6M+8B=1/10
26M+48B=1/2

we will solve for m and b??

Solve for whichever variable you wish -- then substitute the solution for said variable in the other equation

 

60M+80B=1
52M+96B=1

 

60M+80B=1 *6
52M+96B=1 *5

360m+480b=6
260m+480b=5
m=1/100
b=1/200

what is different between men and boy?
they both should be same

 

what is different between men and boy?
they both should be same

As you have shown the men work twice as effectively as the boys...

m=1/100
b=1/200

 

If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

Yes.
M=1/100 or 100days
M=1/200 or 200days
the time taken by 15 men and 20 boys
M*15=15/100
B*20=20/200*20
15/100+20/200=4days

In my opinion

 

The images are of NOS line output transformers -- I use them in power supplies for radiography equipment...

what is this NOS Transformer ?

 

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?

A=1/4 in work done in 1 hr
B+C=3 hrs
1/B+1/C=1/3 in work done in 1 hr
A+C=2
1/A+1/C=1/2 in work done in 1 hr
b=?

is this right?

 

Yes.
M=1/100 or 100days
M=1/200 or 200days
the time taken by 15 men and 20 boys
M*15=15/100
B*20=20/200*20
15/100+20/200=4days

In my opinion

We don't do 'opinions' - only facts -- That said - you're correct!

1/(15/100 + 20/200)=4

 

what is this NOS Transformer ?

NOS = New Old Stock --- IOW it is unused but has probably been resting on a 'shelf' for ~40 Years...

 

how the nos used?

 

A=1/4 in work done in 1 hr
B+C=3 hrs
1/B+1/C=1/3 in work done in 1 hr
A+C=2
1/A+1/C=1/2 in work done in 1 hr
b=?

is this right?

anyway conc on this problem only.

 

how the nos used?

Again, 'NOS' is descriptive of condition -- not function

 

anyway conc on this problem only.

Please state the exercise such that I may distinguish the original question from your work...

 

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?


A=1/4 in work done in 1 hr
B+C=3 hrs
1/B+1/C=1/3 in work done in 1 hr
A+C=2
1/A+1/C=1/2 in work done in 1 hr
b=?

 

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

 

Inasmuch as you seem to have retired for the nonce -- I'll leave you with this:

Ordinarily I'd advise writing three equations (i.e. solutions of each variable in terms of the other two) for simultaneous solution...

Once again, however, it seems we've been offered a 'trick question' and, hence, a 'short cut'...

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

@RRITESH KAKKAR

While I'm not in the habit of conducting 'walkthroughs' -- I felt it appropriate to assist you in 'spotting' trick questions and otherwise making labor-saving observations...

Best regards
HP

 

Yes.
M=1/100 or 100days
M=1/200 or 200days
the time taken by 15 men and 20 boys
M*15=15/100
B*20=20/200*20
15/100+20/200=4days

In my opinion

So if someone asked you what was

15/100 + 20/200

you would say that it is 4?

Your sloppiness with units is going to keep getting you into trouble.

\(
\( 15 men \) \(\frac{1}{100} \frac{\(\frac{job}{day}\)}{man} \) \; + \; \( 20 boys \) \(\frac{1}{200} \frac{\(\frac{job}{day}\)}{boy} \) \; = \; \frac{15}{100} \frac{job}{day} \; + \; \frac{20}{200} \frac{job}{day} \; = \; \frac{1}{4} \frac{job}{day}
\)

or 1 job per 4 days.

 

So if someone asked you what was

15/100 + 20/200

you would say that it is 4?

Your sloppiness with units is going to keep getting you into trouble.

\(
\( 15 men \) \(\frac{1}{100} \frac{\(\frac{job}{day}\)}{man} \) \; + \; \( 20 boys \) \(\frac{1}{200} \frac{\(\frac{job}{day}\)}{boy} \) \; = \; \frac{15}{100} \frac{job}{day} \; + \; \frac{20}{200} \frac{job}{day} \; = \; \frac{1}{4} \frac{job}{day}
\)

or 1 job per 4 days.

That's my fault, @WBahn -- I've been too lenient as regardes skipped steps, etc... (Case in point: reciprocation)

@RRITESH KAKKAR

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

 

That's my fault, @WBahn -- I've been too lenient as regardes skipped steps, etc... (Case in point: reciprocation)

@RRITESH KAKKAR

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

Oh, not to worry.

I consider three different categories of errors at play -- each deserving a different kind of response.

Units -- For ME this is a cardinal sin, but I realize that most people don't agree. So while I will harp about it as much as I can -- particularly when not tracking them results in mistakes not getting caught -- I generally don't jump and shout when others ignore them.

Missing steps -- This is definitely in the eye of the beholder and is also very context sensitive. If someone gives the general impression that they are mathematically competent and if their work is otherwise correct, I seldom say anything. But if the result isn't correct and there's not enough work shown to begin to spot where the error is, then it quickly becomes an issue, particularly if the person continues to not show their work despite continuing to make mistakes.

Sloppy math -- This I harp on whenever I see it (and, a few times, I've harped on it about a post that I made where I got sloppy -- I generally leave the error in place but add an edit note indicating that is it wrong and which post to look at for the correction). There are lots of common places where people get sloppy and reciprocation is certainly one of them. Another is writing expressions that violate order of operations, such as saying 15/3+2 = 3, and expecting everyone to just "know what I mean". This is the kind of sloppiness that really gets people into trouble -- and it's a kind of sloppiness that tracking units will almost always catch.

 

@WBahn Keep harping!

I used to harp at my kids the same way when helping with their math homework. Units ("Do them!"), Missing Steps ("Paper is cheap, show your work!"), Sloppy Math ("Recheck your answers - its easy if you pay attention to the first two rules").
It worked. My daughter's math-fu got her several research fellowships on her way to an MD. My son graduated USNA in economics of all things and wound up in Nuclear Power School for subs. He made it through when about 20 classmates washed out -many of those graduated in engineering. When I asked him how he managed to do it he said "Math, dad. I'm better at math then they were. You taught me how." I didn't teach him any of that but he did remember my harping on how to approach math and the insistence on actually getting to the correct answer, no partial credit for DadMath. It turns out there is no partial credit for nuclear reactor math either.

Harp on, friend! Someone will get it.