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Need help with 10th grade math problem.


IndieMate

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There are 6 barrels of root beer in the store.

In the picture below it is shown how much liters of root beer is in each barrel.

2 customers come to the store and one of them bought 2 full barrels, while the other one bought 3.

It is also known that the second customer bought twice as much root beer than the first one.

There was only 1 barrel left, which one?

 

Need an explanation. :(

post-14031-0-68314100-1489078272_thumb.png

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There are a few restrictions here.

- Since customer 1 bought twice as much as customer 2, and all barrels contain whole numbers of liters, the amount customer 1 buys must be an even number, limiting your options.

- Since customer 1 bought twice as much as customer 2 with 3 barrels versus 2 of customer 2, it means that customer 1 must have bought the 31 barrel, as with the other remaining barrels, it is impossible to pick 3 that get an even number you can evenly spread over the remaining 2 barrels (excluding the 31 as that alone will be more than half of the maximum aggregate total you can make with 3 of the remaining barrels).

- Since customer 1 bought the 31 barrel, it means he needs to buy one odd-numbered and one even-numbered aside from that, as per the first statement, he needs an even aggregate volume.

I am not sure if there are more ways to rule out options than the above, as after that, the first combination I tried worked. I can reveal the answer, but I think with the above, your options will have been limited so that it should be easier to get to the proper solution

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The 20 liter barrel is left. 

 

The first customer bought 2 barrels, the 15 and 18

 

That totals to 33

 

The second customer bought the 16, 31, and 19 barrels, which total to 66.

 

The only barrel left is the 20 barrel

 

 

I just guessed and checked til I got it. That's my usual approach to problems like this. It also helps to look for patterns that make sense in a problem like this: The second customer had to buy an even number of liters so it could be split evenly in half, there were no decimal barrels there.

 

31+19 make an even 50, and a round number like that usually means that's part of the way to the answer, so I played around with the 50 until I found the answer. I'm not sure what mathematical rules would be used to approach this problem from an algebraic standpoint though lol

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There are a few restrictions here.

 

- Since customer 1 bought twice as much as customer 2, and all barrels contain whole numbers of liters, the amount customer 1 buys must be an even number, limiting your options.

 

- Since customer 1 bought twice as much as customer 2 with 3 barrels versus 2 of customer 2, it means that customer 1 must have bought the 31 barrel, as with the other remaining barrels, it is impossible to pick 3 that get an even number you can evenly spread over the remaining 2 barrels (excluding the 31 as that alone will be more than half of the maximum aggregate total you can make with 3 of the remaining barrels).

 

- Since customer 1 bought the 31 barrel, it means he needs to buy one odd-numbered and one even-numbered aside from that, as per the first statement, he needs an even aggregate volume.

 

I am not sure if there are more ways to rule out options than the above, as after that, the first combination I tried worked. I can reveal the answer, but I think with the above, your options will have been limited so that it should be easier to get to the proper solution

 

 

The 20 liter barrel is left. 

 

The first customer bought 2 barrels, the 15 and 18

 

That totals to 33

 

The second customer bought the 16, 31, and 19 barrels, which total to 66.

 

The only barrel left is the 20 barrel

 

 

I just guessed and checked til I got it. That's my usual approach to problems like this. It also helps to look for patterns that make sense in a problem like this: The second customer had to buy an even number of liters so it could be split evenly in half, there were no decimal barrels there.

 

31+19 make an even 50, and a round number like that usually means that's part of the way to the answer, so I played around with the 50 until I found the answer. I'm not sure what mathematical rules would be used to approach this problem from an algebraic standpoint though lol

 

I need to write the solution and I can't write that I'm guessing all of this. I get what you mean, but can't there be an equation system or something like that done?

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I need to write the solution and I can't write that I'm guessing all of this. I get what you mean, but can't there be an equation system or something like that done?

Guessing and trying works best with these things

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Guessing and trying works best with these things

 

I know this, but writing your guessing process on a piece of paper doesn't do it for the teacher.

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I assumed you couldnt just guess, as trial and error would be an easy way out here. Are there any indications as to what you need to do in the assignment? and what math subject is this a part of? I have never been required to solve a similar problem, so if there is any form of generic equation that would work, then I do not know of it.

None of the things I mentioned are guesses, they're all definitions or derived from definitions of known generic sums (or information you directly get from the given problem). I tried to stick to the basics as much as possible as I do not know the assignment level, or the kind of rationale they require in this situation.

- You cannot divide an odd number by two in order to get an integer, hence the first statement is a requirement
- Removing the 31 barrel from the problem impossible, as the largest remaining barrel is 20 and the lowest barrel is 15. All other numbers are between that, and since 3x20 = 60, you know the highest aggregate volume you can get is by definition less than 60. 60/2 = 30 and 15x2 is also 30, so the lowest aggregate volume you can get by adding 2 of the barrels is by definition larger than 30, and thus larger than the maximum possible half-value you are required to work with. Therefore, you know that the 31 barrel needs to be involved in order to get a total volume high enough to be equal to a possible combination of the two remainders. You also know it has to be customer 1 that buys it because otherwise, you still have the same problem as indicated above.
- Again, statement 3 is solely based on definitions after deducting the above, as the sum of two odds is an even number, as is the sum of two even numbers. Since you know the first number to be an odd number, you therefore know that the sum of the two next barrels also has to be an odd number in order to satisfy the requirement of an even number total.

This is as systematic and elemental as I can project the abovementioned to my capability. I hope it helped, and if not, I wish you good luck finding the solution :P

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Simple system of equations.

 

x - liters first customer purchased

y - liters second customer purchased

z - liters in barrel left at the store

a - liters in barrel 1 (customer 1)

b - liters in barrel 2 (customer 1)

c - liters in barrel 3 (customer 2)

d - liters in barrel 4 (customer 2)

e - liters in barrel 5 (customer 2)

 

What is given:

 

x + y + z = 119

y = 2x

z belongs to { 31, 20, 19, 18, 16, 15 }

 

 

Solution:

 

x = y/2

z = 119 - x - y

 

=>

 

x + 2x + z = 119

3x + z = 119

 

Let's assume z = 31

 

3x = 119 - 31

3x = 88

x = 29.333 

 

x must be an integer, so x â‰  29.33

 

=>

 

Let's assume z = 20

 

3x = 119 - 20

3x = 99

x = 33

 

The first customer purchased barrels 1 and 2, or

 

a + b = x

 

a + b = 33

 

31 + 15 > 33

19 + 15  > 33

18 + 15 = 33

 

=>

 

y = 2x = c + d + e = 66

 

66 = 2 * 33 = 16 + 19 + 31 = 66

 

z = 20 

 

Answer:

 

The remaining barrel z had 20 liters in it.

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create a system of equations and reduce it or use substitution. Either that or create a system of equations and plug in values which make your matrix consistent with one of the 6 possible values.

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If you still are having trouble, try slipping a $20 note in with the homework. It may work :P

Haha, i would if i weren't broke af. Anyway, just explained the solution using some equations but mostly by words. Asked my math teacher for advice and she said that there no way of solving this by not guessing

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