Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Approximately how many seconds per year does the average 18 – 21 year women factory worker work?

Choices are...

A] 2,526
B] 80208
C] 5700
D] 3,443
E] 3958

assuming Accident per person per year is 0.057

assuming Accident per million seconds is 14.4

I did it 0.057 * 14. = 0.8208. so would the answer be 80208. I am not sure exactly how to calculate it Thanks in advance for your help?

share|improve this question
1  
This question doesn't make sense as written. It can't be asking for the number of seconds worked per year, as the choices (A)-(E) are all far too small. You need to give some clarification. –  Chris Taylor Nov 22 '11 at 15:57
    
Why did this get upvoted? o_O? –  Phonon Nov 22 '11 at 17:58
    
@Phonon - I am sorry that I could not make to clearer. –  DiscoDude Nov 22 '11 at 22:58
add comment

2 Answers

up vote 3 down vote accepted

I'll try using the data you give:

assuming Accident per person per year is 0.057

assuming Accident per million seconds is 14.4

Then this would mean that $$\frac{\text{number accidents}}{\text{number people}}=0.057$$ and $$\frac{\text{number accidents}}{\text{million total seconds worked}}=14.4$$ We want $\frac{\text{total seconds work}}{\text{number people}}$ which is the amount of seconds per person, so we divide $0.057$ by $14.4$, and then multiply by $1000 000$. This gives $3958$ seconds total, so the answer is $E$.

Keep in mind however, that none of the numbers in the question make any sense. As outlined by Phonon's answer, we expect the number of seconds for the average person to work in the year to be around $7200000$, which is almost $2000$ times larger then the answer to this question.

Edit: How did I know to divide? Remember, we are looking at the quantities $$A=\frac{\text{number accidents}}{\text{number people}}$$ and $$B=\frac{\text{number accidents}}{\text{million total seconds worked}}$$ And we want $$C=\frac{\text{total seconds work}}{\text{number people}}.$$ If I look at $A\times B$ I get

$$A=\frac{(\text{number accidents})^2}{(\text{number people})(\text{million total seconds worked})}$$ which is no good. If I look at $\frac{A}{B}$ I get

$$\frac{\frac{\text{number accidents}}{\text{number people}}}{\frac{\text{number accidents}}{\text{million total seconds worked}}}=\frac{\text{number accidents}}{\text{number people}}\times \frac{\text{million total seconds worked}}{\text{number accidents}}$$ $$=\frac{\text{million total seconds worked}}{\text{number people}}$$ which is what I want, but upside down. So then we look at $\frac{B}{A}$ and we get the answer.

share|improve this answer
    
thank you for a swift response. I was multiplying when divison should have used, as you clearly stated. Very difficult for me to know which one to use. –  DiscoDude Nov 22 '11 at 22:56
    
@DiscoDude: I added in an edit explaining how I knew to divide rather then multiply, and how I new which ones to divide. It is the reason I wrote the fractions in such a strange way. –  Eric Naslund Nov 22 '11 at 23:14
    
thank you for your time and help –  DiscoDude Nov 23 '11 at 18:00
add comment

Assuming factory workers usually belong to unions, we should assume that they work exactly 40 hours per week. Assuming the current state of the economy, the answer in fact approaches 0, but that's for a different SE web site.

If we subtract ~10 days off everyone gets and weekends we get around 250 working days, which multiplied to 8 hours/day gives us 2000 hours a ear. We multiply that by 3600 seconds/hour to get 7,200,000.

All of your other data makes no sense in context of this question, and obviously none of your choices look like my answer, so something wasn't communicated correctly.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.