Cube + Rate

Hi there,

This question requires a few steps. The best thing to do is to go step by step.

First.... find the dimensions of the smaller cubes.

The formula for volume of a cube is
(side)(side)(side) = volume

If the cubes have a volume of 125, they must have an edge length of 5, because (5)(5)(5) = 125.

Next... let's build the bigger cube.

If we stack four of the small cubes in a row....

|-----| |-----| |-----| |-----|
---5 --- 5 ---- 5 ---- 5

That makes a length of 20.

The problem says that this length becomes the length of the larger cube.

So we can find the volume of the large cube. The large cube will have a volume of 20 x 20 x 20.

= 8000

I hope that helps!

I'll take a look at your second question in a moment.

Again, this problem needs to be handled step by step.

First, let's forget about the issue that water is coming in and out. Let's make this question easier for a moment and pretend it just asked,

"At what rate would you have to fill the 128 gallon container so that it's full in 192 hours?"

That rate will be gallons/hours, so 128/192 = 2/3 = the rate at which you'd have to fill the container to make sure that it's full in 192 hours.

However, this easier version of the question isn't the whole issue. The water is flowing in and flowing out. We know it's flowing IN at the rate of 4 gallons per hour, so how many gallons would have to flow out again so that it fills at a rate of 2/3 gallons per hour.

Let's think about how flowing in, flowing out works. Imagine that it's flowing in at the rate of 4 gallons per hour and flowing out at the rate of 4 gallons per hour. Then nothing will stay in the container. Now imagine it flows in at 4 gallons per hour but flows out at 5 gallons per hour. Then definitely nothing will stay in the container!! Finally, imagine that it flows in at the rate of 4 gallons per hour and flows out at the rate of 3 gallons an hour. That's like a tub with a slow drain, that starts to fill up as you take a shower. Every hour 4 gallons go in and only 3 go out -- it fills up at the rate of 1 gallon per hour, in this scenario. So we see that the rate at which it fills up is "in rate" - "out rate" = "fill rate".

So basically, since it's flowing in at 4 gallons per hour, and we want it to fill up at the rate of 2/3 gallon per hour. If we call "x" the rate that it flows out, then
"in rate" - "out rate" = "fill rate"
4 - x = 2/3
x = [spoiler]10/3[/spoiler]

For the second question , I have followed these steps,

in 192 hours from x at 4 gallon per hour
It must given 192 * 4 = 768 gallons

Now capacity of tank is 128

so 768 - 128 = 640 { this is amount of solution that must flow out in 192 hours }

so 640/192 = 10/3 ( is the answer that we were looking for )

For the second question use the basic formula: D=R*T (D can be substituted with Capacity in this question... )

Capacity=128 gallons
Rate - in = 4 gallons per hr
Rate - out = x gallons per hr
Effective rate = (4-x) gallons per hr
Time = 192 hrs

subst. in formula:

128 = (4-x) * 192

x=10/3

hope it helps!!!

Hey laura, thanks for the crystal clear explanation! Appreciate it