Math, Help!!!!

[CannonD]

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My Dad has a miniature Pyramid of [COLOR=blue ! important][COLOR=blue ! important]Egypt[/COLOR][/COLOR]. It is 3 inches in height. Dad was invited to display it at an exhibition. Dad felt it was too small and decided to build a scaled-up model of the Pyramid out of material whose density is (1 / 10) times the density of the material used for the miniature. He did a "back-of-the-envelope" calculation to check whether the model would be big enough.

If the mass (or weight) of the miniature and the scaled-up model are to be the same, how many inches in height will be the scaled-up Pyramid? Give your answer to two places of decimal.

 

[smile101]

My Dad has a miniature Pyramid of [COLOR=blue ! important][COLOR=blue ! important]Egypt[/color][/color]. It is 3 inches in height. Dad was invited to display it at an exhibition. Dad felt it was too small and decided to build a scaled-up model of the Pyramid out of material whose density is (1 / 10) times the density of the material used for the miniature. He did a "back-of-the-envelope" calculation to check whether the model would be big enough.

If the mass (or weight) of the miniature and the scaled-up model are to be the same, how many inches in height will be the scaled-up Pyramid? Give your answer to two places of decimal.

So...I am so proud of myself that I solved it right...but it took some time...say 3ish min
Answer is 4.33
This is really hard to explain for me, but I'll try.
Okay, so lets assume that the small pyramid has a density of d, so the big pyramid has a density of d/3. They have the same mass, so considering the equation D=m/V, we will set the equation D1V1=D2V2 (where D1 is the density d of small pyramid and D2 is the density d/3 of the big pyramid; V1 is the volume of small pyramid which would be 1/3 area of base * height. Area is (9*square root of 3)/2, since pyramid is equilateral triangle. The V2 would be H squared * squareroot of 3/2).

If you work out the algebra, the squareroot of 3/2 terms cancel out on both sides, which leaves you with 27=H cube/3, again doing some math, you should end up with cuberoot of 81=H, which should be a number between 4 and 5. I actually googled it here for exact number, but it should help in choosing from multiple choice in exam.

They have an explanation on website to help you understand if this is too much.

 

[CannonD]

i don't understand how did you get the base area, because the base it's a square right? so each side should be the same. i don't have the answer? anyway how do you find the Area of the base?

 

[raiden117]

I hope there aren't any questions this complex on the DAT...

 

[Streetwolf]

This is how I interpreted it.

D = M/V so M = D*V. You keep M constant and the new D is 1/10 of the old D. So if you divide D by 10, you need to multiply V by 10 to keep M the same.

The volume of a pyramid is (1/3)BH (B = base, H = height). If the Dad wants a scaled up model then every dimension needs to be proportional to the old model. In other words you need to multiply the length and width of the base, as well as the height of the new pyramid, by the same number (just a note here that this works for any shape base since area consists of some form of length * width). So if he wanted to double the length of each dimension he'd end up with a pyramid of 2^3 = 8 times the original volume.

Since we're working with 3 dimensions here and the Dad wants to increase the volume of the model 10-fold, each dimension needs to be increased by the cube root of 10 = approximately 2.1544.

Thus the height will be 2.1544 * 3 inches = approximately 6.46 inches. That's my answer.

 

[Ranelar]

This is how I interpreted it.

D = M/V so M = D*V. You keep M constant and the new D is 1/10 of the old D. So if you divide D by 10, you need to multiply V by 10 to keep M the same.

The volume of a pyramid is (1/3)BH (B = base, H = height). If the Dad wants a scaled up model then every dimension needs to be proportional to the old model. In other words you need to multiply the length and width of the base, as well as the height of the new pyramid, by the same number (just a note here that this works for any shape base since area consists of some form of length * width). So if he wanted to double the length of each dimension he'd end up with a pyramid of 2^3 = 8 times the original volume.

Since we're working with 3 dimensions here and the Dad wants to increase the volume of the model 10-fold, each dimension needs to be increased by the cube root of 10 = approximately 2.1544.

Thus the height will be 2.1544 * 3 inches = approximately 6.46 inches. That's my answer.

This answer looks right, the other answer using d/3 does not 😛

 

[CannonD]

Streetwolf, Can't the W and L of the new pyrimad varies? so are you telling that the height, width,and length have the same rate of increasing?

 

[Streetwolf]

If you make a scale model of something, it MUST be proportional in all dimensions.

You've heard of scale models where 1 inch represents 1 mile I'm sure. That means every dimension has been scaled down by the same value - whatever value takes 1 mile and turns it into 1 inch.

So if the volume needs to be increased 10-fold, and volume is dependent upon length, width, and height, then each of those 3 need to be increased by the same number - the cube root of 10. When you take the cube root of 10 and raise it to the third power, you get 10.

 

[CannonD]

that makes sense thank you a lot

 

[smile101]

So...I am so proud of myself that I solved it right...but it took some time...say 3ish min
Answer is 4.33
This is really hard to explain for me, but I'll try.
Okay, so lets assume that the small pyramid has a density of d, so the big pyramid has a density of d/3. They have the same mass, so considering the equation D=m/V, we will set the equation D1V1=D2V2 (where D1 is the density d of small pyramid and D2 is the density d/3 of the big pyramid; V1 is the volume of small pyramid which would be 1/3 area of base * height. Area is (9*square root of 3)/2, since pyramid is equilateral triangle. The V2 would be H squared * squareroot of 3/2).

If you work out the algebra, the squareroot of 3/2 terms cancel out on both sides, which leaves you with 27=H cube/3, again doing some math, you should end up with cuberoot of 81=H, which should be a number between 4 and 5. I actually googled it here for exact number, but it should help in choosing from multiple choice in exam.

They have an explanation on website to help you understand if this is too much.

Sorry for the confusion, but it turns out that my answer was apparentely right. Everytime I click on the link, it gives me different numbers. When I was solving ur problem, the website gave me that the height was 3 in and the density of new pyramid was 1/3 the density of miniature. Using these numbers, the answer is indeed 4.33.

 

[Streetwolf]

Hmm I didn't realize you could click a link there.

If the density was 1/3 then 4.33 is correct.