Well I got some homework and I don't really know how to figure it out. If anyone could lend a hand, it'd be greeatt. Question: In a factory, skilled workers are paid $300 a day and unskilled workers are paid $150. If there are 30 workers and their wages amount to $6000 a day, how many skilled workers are there? Q2) A student cycles home for lunch at 12km/ph, takes half an hour for his lunch and cycles back to school at 10km/ph. He is absent from school for 57.5 minutes altogether. How far from his home is school. All help will be appreciated EDIT: I need help ASAP, this is due tomorrow

I'll give you the system of equations for the first question. Let x be skilled workers Let y be unskilled workers x + y = 30 300x + 150y = 6000 Solve from there.

Thanks for your help, but I haven't learnt equations with 2 pronumerals, time to surf the net to find out

Just use the substitution method. x + y = 30 can be changed to y = -x + 30 So now that you know what y equals, substitute that into the other equation. 300x + 150(-x + 30) = 6000 Once you find what x is, substitute that value for x and then you can find y also.

Somewhat like the first one. He eats lunch in half an hour. So his cycling took him 57.5 minutes - 30 minutes = 27.5 minutes, or 0.4583 hours. We can assume he took the same route from and to school (aka, same distance). You also know he took more time going back to school, as he was cycling slower. So, if x and y are hours he spent cycling 12km/h and 10km/h repsectively, x*12 will equal the distance to his house, and y*10 will equal the distance to school. These distances are the same, so x*12km/hr = y*10km/hr. We also know that he spent 0.4583hours cycling (remember, his speed is in kilometers per HOUR, so do not use 27.5 MINUTES). Thus, x + y = 0.4583hours. You should be able to get it from there.

iffy525 took the first one the way I would do it. Q2) A student cycles home for lunch at 12km/ph, takes half an hour for his lunch and cycles back to school at 10km/ph. He is absent from school for 57.5 minutes altogether. How far from his home is school. First step is get everything in workable units i.e. all the same. 57.5 minutes will be a pain to work with in hours (57.5/60= 0.95833333........ and accounting for lunch makes no difference) so change to km/minute). 12/60= number of kilometres per minute, 10/60 = kilometres per minute for the return journey. A couple of ways to go here. You say 2 variable equations had not been done which probably discounts most linear acceleration equations/equations of motion. http://www.antonine-education.co.uk/Physic...s_of_motion.htm None the less they are based on Distance = average speed × time which in this case is s (distance) = ((v+u)*t)/2 Of course this is both legs so assuming the same route and it is as the crow flies or it does not matter then you have to divide by 2 to get the individual distance. Assuming I did not mess up somewhere along the way it is just over 2.5km edit: must remember to hit reply quicker.

Yep. And it's from private tutor. In normal school, I'm doing advanced placement maths (Year 9 Book) and in normal school, it's easy. So far, we've done in order: Number Skills (-2 + 6), Introductory Algebra, Expanding, Factorizing, Pythagoras and now Geometry. Seems to me Australia is behind in education, but it's okay, I'm only struggling with tutor