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- October 13th 2009, 08:48 AM #1

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## GRE Question

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.

- October 13th 2009, 09:52 AM #2

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- October 13th 2009, 10:24 AM #3

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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

- October 13th 2009, 10:40 AM #4

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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

- October 13th 2009, 10:45 AM #5

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500 lbs ----- 20 pigs in 7 days

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

- October 13th 2009, 11:03 AM #6

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- October 13th 2009, 11:34 AM #7

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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.

- October 13th 2009, 11:59 AM #8

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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

We want:

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.