My Math Forum GED assistance #2

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 June 14th, 2014, 04:10 PM #1 Member   Joined: Jun 2014 From: pennsylvania Posts: 45 Thanks: 0 GED assistance #2 Hello, So there were a few questions I had trouble on, and with your help I hope to master all the math I can handle and to become the smartest I can be. How do I solve this? I want to make sure I got this one right based on something I learned from the first assistance post I made. Is the answer $17 ? I did 400+15(200) divide by 200 = 3400/200=17 Ok so for this one I just redid on paper and wanted to make sure I am doing it right. When I did it last time, I multiplied by the full height of the flask. 0.33333 x 3.14 x 34 x 34 x 80 = 96795..... rounded to the nearest hundredths is answer c)96,800 Thanks Everyone. Last edited by skipjack; June 17th, 2014 at 01:16 AM.  June 14th, 2014, 06:14 PM #2 Member Joined: Jun 2014 From: pennsylvania Posts: 45 Thanks: 0 Anyone have any assistance to offer? My last post was answered pretty fast, I hope I did not ask too hard of questions? Last edited by skipjack; June 17th, 2014 at 01:14 AM.  June 14th, 2014, 07:45 PM #3 Math Team Joined: Dec 2013 From: Colombia Posts: 7,116 Thanks: 2369 Math Focus: Mainly analysis and algebra 1) Treat it as two separate inequalities $$90 \le \frac{84 + 94 + 91 + h}{4} \qquad \text{and} \qquad \frac{84 + 94 + 91 + h}{4} \le 95$$ Solve each inequality for$h$. 2) Looks OK to me. 3) The volume of a cone is $$V = \frac{\pi r^2 h}{3}$$ We want to know the volume of liquid in the flask when the height of the liquid is 80mm. We are told to approximate the volume in a full flask by using a cone of the same base and height. So our cone holds$$V = \frac{\pi {\frac{68}{2}}^2 \cdot 117}{3}$$ But in this case, only the bottom 80mm is filled. That is the same as saying that the flask is full except for the top$117-80=37mm$. So if we can find the volume of the top 37mm of the flask, we can subtract that from the capacity of the flask to get our answer. In order to do this, we need to work out the diameter (or radius) of the flask 37mm from the top. Can you do that? June 14th, 2014, 09:10 PM #4 Member Joined: Jun 2014 From: pennsylvania Posts: 45 Thanks: 0 Quote:  Originally Posted by v8archie 1) Treat it as two separate inequalities $$90 \le \frac{84 + 94 + 91 + h}{4} \qquad \text{and} \qquad \frac{84 + 94 + 91 + h}{4} \le 95$$ Solve each inequality for$h$. 2) Looks OK to me. 3) The volume of a cone is $$V = \frac{\pi r^2 h}{3}$$ We want to know the volume of liquid in the flask when the height of the liquid is 80mm. We are told to approximate the volume in a full flask by using a cone of the same base and height. So our cone holds$$V = \frac{\pi {\frac{68}{2}}^2 \cdot 117}{3}$$ But in this case, only the bottom 80mm is filled. That is the same as saying that the flask is full except for the top$117-80=37mm$. So if we can find the volume of the top 37mm of the flask, we can subtract that from the capacity of the flask to get our answer. In order to do this, we need to work out the diameter (or radius) of the flask 37mm from the top. Can you do that? Thanks v8archie, I appreciate you stepping up the the challenge and helping me out this bunch. I still don't understand how I come up with (h) in the first problem. I am probably doing it wrong, but I was adding the 3 values in the numerator and then dividing by the denominator for the two separate inequalities and then add the two values. Am I supposed to turn the inequality into a fraction over 1 and then make both denominators the same? Please help me understand this a little better. As for the 3 question, can you help me understand better? What am I dividing it all by 3 for? Maybe I'm just used to doing the formula and am not used to reading it anymore. Was the way I was doing it correct? Up to what point was I messing up? Thank you. Last edited by skipjack; June 17th, 2014 at 01:21 AM.  June 14th, 2014, 09:36 PM #5 Math Team Joined: Dec 2013 From: Colombia Posts: 7,116 Thanks: 2369 Math Focus: Mainly analysis and algebra 1) Manipulate the inequalities algebraically to make$h$the subject. That is, make them read$h \ge A$and$h \le B$where$A$and$B$are numbers. Your first move for each of them should be to multiply both sides of the inequality by 4. 2) You divide by 3 because that is the volume of a cone. Without dividing by three, it's the volume of a cylinder. The 80 in your attempt is wrong, because the flask is not 80mm tall. Look at this diagram: The flask is 117mm tall, but the blue part is only 80mm tall. Clearly, the blue part is not a cone. But the whole flask is, and the white part is too. If we take the white part away from the whole flask, we'll be left with the volume of the blue part, which is what we want. But to do that, we need to know the width of the bottom of the white part. Can you work that out? June 14th, 2014, 11:30 PM #6 Member Joined: Jun 2014 From: pennsylvania Posts: 45 Thanks: 0 Quote:  Originally Posted by v8archie 1) Manipulate the inequalities algebraically to make$h$the subject. That is, make them read$h \ge A$and$h \le B$where$A$and$B$are numbers. Your first move for each of them should be to multiply both sides of the inequality by 4. 2) You divide by 3 because that is the volume of a cone. Without dividing by three, it's the volume of a cylinder. The 80 in your attempt is wrong, because the flask is not 80mm tall. Look at this diagram: The flask is 117mm tall, but the blue part is only 80mm tall. Clearly, the blue part is not a cone. But the whole flask is, and the white part is too. If we take the white part away from the whole flask, we'll be left with the volume of the blue part, which is what we want. But to do that, we need to know the width of the bottom of the white part. Can you work that out? I appreciate you help. I am having trouble the way you are explaining this one. Could you please explain to me step by step how to solve this one. A bit more simpler. I'm having trouble understanding how much I actually got right, are you saying what I wrote was correct except for the 117mm? Like I said, I'm used to solving the volume for a cone and pyramid but not reading the formulas. So basically what I wrote was correct except for the switching the 80 and the 117. How do I find the difference that I need to remove? DO I do the same problem 2 times once with 80 and once with 117 and then subtract the difference? Thanks again for taking the time and patience with me. Last edited by skipjack; June 17th, 2014 at 01:24 AM.  June 15th, 2014, 02:38 PM #7 Member Joined: Jun 2014 From: pennsylvania Posts: 45 Thanks: 0 Anyone?  June 15th, 2014, 10:26 PM #8 Math Team Joined: Dec 2013 From: Colombia Posts: 7,116 Thanks: 2369 Math Focus: Mainly analysis and algebra You've used the formula in a way that makes some sense, but you need to step back and think about what you are calculating. We do not want the volume of a cone 80mm high. We are told that our flask/cone is 117mm high, but we also know that it isn't full. That means that the volume we need isn't a cone. It's a cone with the top cut off. But the only formula we have is that for a complete cone. So the question we need to answer first, is "how do I find the volume of a cone that has had the top cut off using only the formula for the volume of a cone?" Now go back to the diagram. We know the whole cone is a cone (pretty obviously), so we can work out it's volume, and we know that we can't directly calculate the bottom part (which is what we need). So what about the other part that is left? The top part of the cone, the part that is cut off? What shape is that? Hopefully you can see that it's another cone, obviously a smaller one that the whole cone. Now think about how the two cones and the piece we want are related. We could take the whole flask and cut the top off to leave the volume we require. Now, we know the volume of the whole cone. What we need is the volume of the piece we are going to cut off. To do this we need its height and the radius. The height is easy enough. The whole cone is 117mm high and the piece that we are going to leave behind is 80mm high. So the height of the piece must be the difference: 37mm. What about the diameter? Well, you can see that the diameter of the cone decreases as you move from the bottom to the top. It decreases at a constant rate from 68mm at the bottom, to 0mm at the top. Now comes the clever bit. The height of our piece is 37mm which is$\frac{37}{117}$of the 117mm of the whole cone. So the diameter of our piece is$\frac{37}{117} \cdot 68 = 21.504mm\$. Now, can you take the volume of our top piece away from the volume of the whole cone?
 June 15th, 2014, 11:16 PM #9 Member   Joined: Jun 2014 From: pennsylvania Posts: 45 Thanks: 0 Ahhh, Yes, Thanks for explaining that for me v8, But don't you mean the top part is a cylinder not a cone? I just want to make sure. Does the dotted line represent the where the 80mm mark is from the top 37mm? Either way, I can work from what you wrote. At the end tho the formula to solve the volume of the 37mm cylinder would be. 37(difference in height of what needs to be removed from the whole flask) / 117(the total height of the flask) x 68(original diameter of the flask)? So now we know how much volume is in the part of the flask we do not need? And we can do what now, subtract from what volume? Now that I look at the flask, it's not a full cone is it? A cone normally has a pointed top, so I'm not exactly sure? Last edited by skipjack; June 17th, 2014 at 01:26 AM.
 July 13th, 2014, 03:01 PM #10 Member   Joined: Jun 2014 From: pennsylvania Posts: 45 Thanks: 0 I'm not sure whether this question was very full answered, but if it was I'm having trouble reviewing it. Ok, so we got the volume of the top portion of the flask that is not filled, which is 21.504mm. I actually got 215.02mm when I did the calculation? So say one of those is the volume of the non-filled part. When I subtract this from the full volume of the cone, I still don't get the right number. The number seems to be pointing in the last answer (D), but I need to know how to do it, not just the answer. Last edited by skipjack; July 15th, 2014 at 07:12 PM.

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