GRE Question

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October 13th 2009, 07:48 AM
fifthrapiers

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, 08:52 AM
aman_cc

Quote:

Originally Posted by fifthrapiers

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.

1 worker, 1 day does ------ 18/(15*24) driways

So, let x days be required.

40 workers in x days will do --------- 40x18/(15*24) = 22
solve for x , we get 11

This is typically how you can do such proportion problems
October 13th 2009, 09:24 AM
fifthrapiers

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, 09:40 AM
Defunkt

In 7 days, the "work" 20 pigs will do is 500. So in 7 days, the "work" 1 pig will do is $\frac{500}{20}=25$ . So the "work" one pig will do in one day is $\frac{25}{7}$ . So now you have 14 pigs, then their work in 1 day is $\frac{25}{7}\cdot 14 = 50$ . So we want to know how many days they have to eat 50 lbs of mush until they eat 200 lbs. So $50x = 200 \Rightarrow x = 4$
October 13th 2009, 09:45 AM
aman_cc

Quote:

Originally Posted by fifthrapiers

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

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, 10:03 AM
Soroban

Hello, fifthrapiers!

Quote:

If 15 workers can pave 18 driveways in 24 days,
how many days would it take 40 workers to pave 22 driveways?

It helps if you understand the "algebra" of these work problems.

$\begin{array}{cccc}\text{We have:} & \text{15 workers} & \text{18 driveways} & \text{24 days} \\ \\ \text{Divide by 15:} & \text{1 worker} & \frac{6}{5}\text{ driveways} & \text{24 days} \\ \\ \text{Multiply by 40:} & \text{40 workers} & \text{48 driveways} & \text{24 days} \\ \\ \text{Multiply by }\frac{11}{24}\!: & \text{40 workers} & \text{22 driveways} & \text{11 days}\end{array}$
October 13th 2009, 10:34 AM
fifthrapiers

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, 10:59 AM
fifthrapiers

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.