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# KCC's Quizzes AQQ262 about conical bottle height

1. Quote of the week: "If each day is a "gift", I'd like to know where I can return Monday" - Unknown

2. New quiz AQQ262 about a conical bottle height

Good luck!

Kuo-Chang

P.S. Don't hesitate to share those weekly quizzes in EZ to colleagues or friends!

• Yes, you found it for both questions, congratulations  !

• I'm getting complex numbers after equaling two equations of the occupied volume. One is the left conical bottle, [pi/3 * R^2 * H]-[pi/3 * r1^2 * 8]. While the second one is, [pi/3 * r2^2 * (H-2)]. After using the radius height ratio to solve, the final equation i got is 512 = H^3 - H(H-2)^3. It's giving me complex number roots. May i Know what mistakes did i make?

• Time to publish the answers:

1. Let’s call H the full height of the cone, r the base radius of the first empty cone and r’ the base radius of the liquid cone:

The volume of the liquid must be equal for both bottle positions:

Liquid volume in first position is full cone volume (height H and radius r)  minus  empty zone cone volume (height 8cm and radius r)

Liquid volume on the second position is the cone of height (H-2) and base radius r’

πR²H/3 - πr²8/3 = πr’²(H-2)/3 or R²H – 8r² = (H-2)r’²

Then, using triangles similarity:

• On first position: 8/r = H/R giving r = 8R/H
• On second poistion: (H-2)/r’ = H/R

Conbinung the 3 relations, we can eliminate r and r’:

R²H – 8*64R²/H² = (H-2)3*R²/H²            or H – 8*64/H² = (H-2)3/H²

By simpligying: H3 – 512 = H3 – 3H²*2 + 3H*2² - 8 or 6H² - 12H – 504 = 0

And finally: H² - 2H – 84 = 0 that contains one positive solution:

H = 1+85 = 10.22 cm

1. Volume of the liquid is : πr’²(H-2)/3 with r’ = (H-2)/H * R

Volume is therefore πR²(H-2)²/H² * (H-2)/3 = [π(H-2)²/H²] * R² = 2.03 R²

With the data available, the volume depends on R

Congratulation to or 4 winners:  ,  ,  and Barry Kulp

Be prepared for the next challenge (will be a quite tough one...)

• Hello  , thnaks for your feedback.

I think you have probably made a typo somewhere. In effcet, you came with the equation 512 = H^3 - H(H-2)^3 which contains one abnormally : in the second term you have H^3 (thus a cube) subtracting H(H-2)^3 which has H^4 term in it. The 2 terms should be in the same exponent level, otherwise they cannot be added or substracted...

• Ok. Thank you for your response, will try to do it again.

{Found my mistake, Thanks}

• Great!