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Btw, I'm getting these off an app on my phone.

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- #1

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Btw, I'm getting these off an app on my phone.

- #2

DaveC426913

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One cake per minute per man.

So, 2 men.

I want half the prize money.

- #3

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One cake per minute per man.

Not quite-- If 1 man could eat 1 cake in 1 minute, then the one man in the example would've eaten 1 cake in 1 minute, while the other half-a-man would've eaten his half of a cake in 1 minute. So if it were 1 cake per minute per man, it'd be gone in 1 minute, rather than a minute and a half.

If R is the number of cakes per minute that 1 man can eat, then:

1.5 men * R * 1.5 minutes = 1.5 cakes

1.5 men * R minutes = 1 cakes

R = 1 cakes / (1.5 men minutes) = 2/3 cakes/men minutes

Hence, we now want:

Q men * 2/3 cakes/men minutes * 30 minutes = 60 cakes

Q men * 2/3 = 2 men

Q men = 3 men

Q = 3

DaveE

- #4

DaveC426913

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: faceplant :Not quite-- If 1 man could eat 1 cake in 1 minute, then the one man in the example would've eaten 1 cake in 1 minute, while the other half-a-man would've eaten his half of a cake in 1 minute. So if it were 1 cake per minute per man, it'd be gone in 1 minute, rather than a minute and a half.

I misread. I thought it was 1,1.5,1.5 not 1.5, 1.5, 1.5.

- #5

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I have another one:

If you have a sphere and you remove a cylinder with a length of six inches from it (assume that the cylinder is arranged in such a way so as to go through the maximum amount of the sphere), what is the new volume?

- #6

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If you have a sphere and you remove a cylinder with a length of six inches from it (assume that the cylinder is arranged in such a way so as to go through the maximum amount of the sphere), what is the new volume?

Maybe I'm missing something-- I get:

2*pi*((2/3)*R^3-3*R^2-27)

(where R is the radius of the sphere.) I'm guessing that some factors are supposed to cancel out nicely?

DaveE

- #7

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The answer is independent of radius because as the radius of the sphere increases the radius of the cylinder also increases, so the volume percentage remains the same. Kind of like why rain water depth is independent of the the volume used to measure it

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- #8

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The answer is independent of radius because as the radius of the sphere increases the radius of the cylinder also increases, so the volume percentage remains the same. Kind of like why rain water depth is independent of the the volume used to measure it

I think there's some sort of stipulation that's being implied. Here's what I get using two extremely different values for the radius of the sphere (see attachment). Am I missing some way of arranging the cylinder such that it takes up more volume?

DaveE

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- #9

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I think there's some sort of stipulation that's being implied. Here's what I get using two extremely different values for the radius of the sphere (see attachment). Am I missing some way of arranging the cylinder such that it takes up more volume?

DaveE

It looks like you've set the length to be >6 in the larger sphere, close to the diameter at 18 (?). Length is specified at 6, radius can vary

- #10

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What about differentiating the expression for the volume difference to find the value of radius\diameter that minimises it?

So I get for the volume difference V = 4/3*pi*(d/2)^3 - 6*(d^2-36)*pi/4 where d is the diameter of the sphere, which upon differentiating and setting to zero gives either a diameter of 9 or -15. Using 9 gives the difference as 54*pi.

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are we allowed complex solutions?

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I'm not sure about complex solutions, but I just realised I factorised the quadratic wrong. I should have got a diameter of ~7.82 and a volume difference of ~43*pi.

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- #13

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Okay I've completely messed up that differentiation, I should've got d = 6 or 0 and that would maximise the volume difference so no good.

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- #14

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It looks like you've set the length to be >6 in the larger sphere, close to the diameter at 18 (?). Length is specified at 6, radius can vary

I've done nothing of the sort. The cylinder in the larger sphere resembles a disc: it's 6" tall, and extremely wide (big radius). The cylinder in the smaller sphere resembles a rod: it's also 6" tall, but is extremely narrow, in order to fit within the sphere. In both images, the circle is the sphere, and the rectangle is a sideways view of the cylinder, with the left and right sides being the straight "length" of the cylinder, and the top and bottom sides representing the diameter of the cylinder.

DaveE

- #15

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I've done nothing of the sort. The cylinder in the larger sphere resembles a disc: it's 6" tall, and extremely wide (big radius). The cylinder in the smaller sphere resembles a rod: it's also 6" tall, but is extremely narrow, in order to fit within the sphere. In both images, the circle is the sphere, and the rectangle is a sideways view of the cylinder, with the left and right sides being the straight "length" of the cylinder, and the top and bottom sides representing the diameter of the cylinder.

DaveE

yes, I see, I didn't look carefully enough. I agree with your earlier analysis...I was remembering a strikingly similar problem in which the answer was a constant, not related to either radius

- #16

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So to repeat arithmetix' question, are we allowed complex solutions? I get two possible complex ones: (45+i*9*(27)^1/2)*pi or (45-i*9*(27)^1/2)*pi (assuming I haven't made anymore silly mistakes, which is a big assumption ).

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