Here are two different problems that I need help understanding.
If 15 workers can pave 18 driveways in 24 days, how many days would it take 40 workers to pave 22 driveways?
And a similar question:
If 500 pounds of mush will feed 20 pigs for a week, how many days will 200 pounds of mush feed 14 pigs?
They give the solutions but I don't understand them. Perhaps someone can give me THEIR way of solving it... by the way, the answers are 11 and 4.
Grr I don't understand why this is so hard to grasp.
For the next one..
500 lbs mush ..... 20 pigs ..... 7 days
200 lbs mush ..... 14 pigs ..... x days
Isn't there a way I can just put this into an equation
I tried 200*x*(20/(500*7)) = 14
But that didn't work out
In 7 days, the "work" 20 pigs will do is 500. So in 7 days, the "work" 1 pig will do is . So the "work" one pig will do in one day is . So now you have 14 pigs, then their work in 1 day is . So we want to know how many days they have to eat 50 lbs of mush until they eat 200 lbs. So
1 lb in 1 day will feed ------------ 20*7/500 pigs
Let num of days be x
200 lbs in x days will feed ...........200/x * 20*7/500 = 14
which gives x = 4
you need to ensure whether the variables are directly or inversely proportional.
Let num of days pigs can be fed=d
num of pigs = p
qty of mush = q
so d is directly proportional to q
d is inversely proprtinal to p
so d*p/q is a constant
Apply this under both cases (500lbs and 200lbs) you should get your ans
THANK YOU ALL!!
Soroban, I especially like your method. I tried it for the second one...
20 pigs ... 500 lbs mush ... 7 days
We want to get to:
14 pigs ... 200 lbs mush ... x days
So, we alter the first set to try get the second set (my question is can I alter any 2 numbers? In this case it worked):
First we will get down to 14 pigs (multiply by 14/20):
14 pigs ... 350 lbs mush ... 7 days
Then, we get 350 lbs mush down to 200 lbs mush (multiply by 200/350):
14 pigs ... 200 lbs mush ... 4 days
So I can always just alter 2 numbers? Like I could multiply the above and alter days/pigs? That's awesome.
I think I realized something. Another problem was: If 4 boys can shovel a driveway in 2 hours, how many minutes will it take 5 boys to do the same job?
So we have:
4 boys ... 1 driveway ... 120 minutes
5 boys ... 1 driveway ... x minutes
So I multiplied the first by
5/4 to get:
5 boys 1 driveway 120 minutes. Then I multiplied this initially by (5/4) and got the wrong answer. I guess you have to see whether it proportional or inversely proportional. So in this case I would have to multiply by (4/5).
= 96 days.
Ok I think I am getting it.