If a man and a half can eat a cake and a half in a minute

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In summary, the problem is asking how many men would it take to eat 60 cakes in 30 minutes if a man and a half can eat a cake and a half in a minute and a half. Using the formula R = 2/3 cakes/men minutes, we can solve for the number of men (Q) needed by setting up the equation Q * 2/3 * 30 = 60, which gives us Q = 3. However, there is a possibility of complex solutions if we allow for different arrangements of the cylinder and sphere in the second part of the conversation.
  • #1
ƒ(x)
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If a man and a half can eat a cake and a half in a minute and a half, allowing the same situation, how many men would it take to eat 60 cakes in 30 minutes?


Btw, I'm getting these off an app on my phone.
 
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  • #2


One cake per minute per man.

So, 2 men.

I want half the prize money.
 
  • #3


DaveC426913 said:
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


davee123 said:
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.
: faceplant :

I misread. I thought it was 1,1.5,1.5 not 1.5, 1.5, 1.5.
 
  • #5


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


ƒ(x) said:
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?

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


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


regor60 said:
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


davee123 said:
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


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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  • #11


are we allowed complex solutions?
 
  • #12


I'm not sure about complex solutions, but I just realized 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


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


regor60 said:
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


davee123 said:
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


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 :redface:).
 

1. What is the significance of "a man and a half" and "a cake and a half"?

The phrase "a man and a half" and "a cake and a half" is used to represent a ratio of 3:2. It implies that for every 3 units of one thing, there are 2 units of another thing. In this context, it means that for every 3 units of man, there are 2 units of cake.

2. How is the speed of eating a cake related to the ratio of "a man and a half" and "a cake and a half"?

The speed of eating a cake is directly related to the ratio of "a man and a half" and "a cake and a half" because it indicates the proportion of cake that can be consumed in a given amount of time. In this scenario, the ratio of 3:2 suggests that for every 3 units of man, 2 units of cake can be eaten in a minute.

3. Is this phrase meant to be taken literally or is it a hypothetical scenario?

This phrase is most likely meant to be taken as a hypothetical scenario rather than as a literal statement. It is used as a way to illustrate a proportion and is not based on any scientific evidence or research.

4. Is it possible for a man and a half to physically eat a cake and a half in a minute?

It is highly unlikely that a man and a half can physically eat a cake and a half in a minute. This phrase is used as a hypothetical scenario and is not supported by any scientific evidence. It is important to note that individual eating speeds and cake sizes may vary, making it difficult to accurately measure this scenario.

5. What is the purpose of using this phrase in a scientific context?

This phrase is commonly used as a way to explain and understand ratios and proportions in a more relatable manner. It can also be used as a thought experiment to challenge scientific reasoning and critical thinking skills. However, it should not be taken as a literal statement or used to support any scientific claims.

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