Question

A company makes similar cylindrical containers in two sizes, as shown in the table below.

Containers
Radius (in centimeters)
Volume (in cubic centimeters)
Type 1
4
?
Type 2
7
1231.66

What is the volume, in cubic centimeters, of Type 1 containers?

2111.42

2155.41

402.17

229.81

Answer 1

Please help! Pretty please?

Answer 2

The answer I get is different than the answers provided, and I don't know what I am doing wrong.

Answer 3

Cylinder container is like a circle with a skirt. To compute volume you would find the area of the circular base and multiply by the height. So lets look at type I fpr which I see a problem, I don't have the height. How tall are these cylinders?

Answer 4

They don't provide the height.

Answer 5

I think we are to use the relationship between the two types of cylinder assuming they are the same height, but different radius.

Answer 6

Oh.... ok.

Answer 7

What I tried to do was (pi x 4^2 / pi x 7^2) = (x/1231.66)
But it didn't exactly work out, for the answer I got was 1226.29 = x. And that isn't labeled as one of the options.

Answer 8

So..... It is quite a conundrum.

Answer 9

So, may anybody please give me a clue, as what to do?

Answer 10

So let us calculate the height of the Type 2.
Volume=2pi r^2 h
1231.66 =2 pi 7 squared h\[1231.66 = 2\pi \times49\times h\]\[1231.66 \over 98 \pi = 4cm\] we can assume a height of 4 cm. Now for the volume of the type 1

Answer 11

V(Type 1)=2 pi r^2 * 4
Volume would then equal:\[2 \pi \times 16 \times 4 =402.1234 cu. cm. \]

Answer 12

OH... I never thought about it in that manner... Thank you... I, by myself would have never thought of finding the height through a reverse process. So then, by finding that out, you can plug it in to the formula to find out the answer we are seeking. It is more wise to view a problem through different views, so you could find all the different possibilities the problem in question may behold.

Answer 13

My satellite link failed because of a thunderstorm that we are having. Sorry for delay.

Answer 14

Did that value come close to one of your options?

Answer 15

Hold on, we may be getting some new input from Callisto.

Answer 16

Well, yes, it did come close to one of the answers.very close, in fact.

Answer 17

I have a Pi key on calculator which makes it very accurate.

Answer 18

Yes. The answer I have as one of the options provided, is rounded, so it came quite close. Only about a few hundredths off.

Answer 19

@iluvmangos , remember this was based on an assumption that the cylinder containers were equal height. Hopefully the assumption was a valid one.

Answer 20

Thank you. I am sure that it is. I will come on again in a few moments, and tell you if it is right or not.

Answer 21

And unfortunately it is marked wrong... but at least we tried. That's what counts.

Answer 22

@Callisto , did you have a comment?

Answer 23

I was hoping Callisto could straighten me out, obviously I have gone astray here.

Answer 24

I got some different actually..

Answer 25

*something

Answer 26

Well, this is certainly something I will ponder over..... and @Callisto , may you present your approach to the situation?

Answer 27

Did I err on the calculations or on the assumption?

Answer 28

I use property of similar figures to do it..

(ratio of sides)^3 = ratio of volume
\[(\frac{4}{7})^3 = \frac{V}{1231.66}\]
Solve V.

Answer 29

@radar I saw the problem.... ''Volume=2pi r^2 h'' It should be volume= pi r^2 h

Answer 30

Oh yes, I really boo booed! A=pi*r^2

Answer 31

I think I combined the formula for circumfurence and area lol

Answer 32

Oh goodness.... I guess that is where we went awry. Oops! ;)

Answer 33

And as usual there was a distractor that was close to my incorrect answer. Sorry about that iluvmangos.

Answer 34

Thanks Callisto for clarifying that for me. I will try not to do that any more.!!

Answer 35

I like your method Callisto as it eliminates having to use Pi.

Answer 36

Thanks for all the help you two have given me!

Answer 37

Did you get around 230 cu. cm for volume and was that an option.

Answer 38

You are welcome and sorry about the miscue.

Answer 39

The followings are quite useful:

For similar figures,
(ratio of sides)^2 = ratio of area
(ratio of sides)^3 = ratio of volume

Welcome and welcome and welcome :)