Math debate that you guys can help me with

[Tsaico]

Ok, the question is simple: A mechanic tests the current ratio of antifreeze to water and finds that it is 37.5%. How much does he need to drain and how much does he add of pure antifreeze to make the ratio 50% if the tank holds 20 litres total?

A lot of people are saying 2.5 litres, since a 37.5% will be 7.5 litres of antifreeze out of 20 total and to make 50%, you need 10 litres. 10-7.5=2.5 litres.

But I am thinking that is incorrect, since I am adding pure 100% to the 37.5% to try to get 50% is the real question. Or am I totally a wack job and it was good that I kept my mouth shut in study group? This is a basic algebra question, so I understand that 2.5 might be the "math book" right answer, but how would i set up the equation to solve this in real world?

 

[chuckywang]

Let x be the amount you have to drain of the mixture and then add of pure antifreeze.

Therefore, you need 10L of antifreeze after the drain/add process.

After draining, there will be .375*(20-x) L of antifreeze.

After adding, there will be .375*(20-x)+x L of antifreeze.

Therefore, you need .375*(20-x)+x = 10. Solve for x.

 

[dullard]

You can't just drain 2.5L of water from that mixture and add 2.5L of antifreeze. Why? Without a complex distillation process, you'd be withdrawing antifreeze as well when you drain it. Thus 2.5L is incorrect.

 

[QED]

This is a problem where there can be more than one answer based upon your assumptions.

If you drain FIRST, and then add the pure antifreeze, it's rather simple to calculate. Since you want a 50% mix in 20 litre tank, you need 10 litres of antifreeze total in the tank. Let x be the number of litres we need to first drain-- it will also be the number of litres of pure antifreeze we add later. Then we have 7.5 * ( 1 - x/20) litres of antifreeze in the tank after draining, and 7.5 * (1 -x/20) + x litres of antifreeze after adding the pure antifreeze. So just solve 7.5 * (1 - x/20) + x = 10 for x.

However, the problem gets a little more complicated if you drain and fill with pure antifreeze simultaneously-- in fact, finding the solution requires solving a differential equation. Nothing too difficult-- but a little beyond the scope of simple algebra.

 

[drinkmorejava]

EDIT: gahhh, I'm so confused, I give up

There appears to be some confusion with the ratio. A ratio of 50% would mean 1 anti:2 water a concentration of 50% would be 1 anti:1 water

same goes for the first part of the question, do you mean a concentration of 37.5% or a ratio of .375:1

If the concentration is 37.5% and you want to make it 50% the answer is 1.654 gallons.

It's 4 gallons if you mean the ratios of .375:1 and .5:1

 

[KK]

drain 4 gallons, add 4 gallons 100% antifreeze

37.5 antifreeze
62.5 water

16 * 62.5% = 10 gallons of water

 

[JujuFish]

1 and 1/3 liters to get from a ratio of 3:8 to 1:2.
4 liters to get from 37.5% anti-freeze to 50% anti-freeze.

 

[Tsaico]

Well, hmm.. we all are getting different answers here. And there are no gallons, so those are already wrong.

And while I understand there might be a difference with ratio and conecntration. I believe the question wants us to find out how much to drain and replace with 100% antifreeze to achieve a 50/50 of water and antifreeze mixture without having to dump the whole thing.

 

[chuckywang]

THE ANSWER IS 4!!!!!

Forget about gallons! EVERYBODY GOT 4!!!

 

[KK]

Originally posted by: Tsaico
Well, hmm.. we all are getting different answers here. And there are no gallons, so those are already wrong.

And while I understand there might be a difference with ratio and conecntration. I believe the question wants us to find out how much to drain and replace with 100% antifreeze to achieve a 50/50 of water and antifreeze mixture without having to dump the whole thing.

It's simple. gallon, litre, same difference.

you want 10 litres of water. if the ratio is 62.5% then times that by 16 litres and you get 10 litres of water. just fill it back up to the top with pure antifreeze. and that'd be 4 litres.

 

[6000SUX]

Originally posted by: Tsaico
Ok, the question is simple: A mechanic tests the current ratio of antifreeze to water and finds that it is 37.5%. How much does he need to drain and how much does he add of pure antifreeze to make the ratio 50% if the tank holds 20 litres total?

A lot of people are saying 2.5 litres, since a 37.5% will be 7.5 litres of antifreeze out of 20 total and to make 50%, you need 10 litres. 10-7.5=2.5 litres.

But I am thinking that is incorrect, since I am adding pure 100% to the 37.5% to try to get 50% is the real question. Or am I totally a wack job and it was good that I kept my mouth shut in study group? This is a basic algebra question, so I understand that 2.5 might be the "math book" right answer, but how would i set up the equation to solve this in real world?

You didn't give enough information to work the problem, sorry. All the other answers are wrong. The fact that a tank has a capacity of 20 litres doesn't mean it is full.

 

[chuckywang]

Originally posted by: 6000SUX

Originally posted by: Tsaico
Ok, the question is simple: A mechanic tests the current ratio of antifreeze to water and finds that it is 37.5%. How much does he need to drain and how much does he add of pure antifreeze to make the ratio 50% if the tank holds 20 litres total?

A lot of people are saying 2.5 litres, since a 37.5% will be 7.5 litres of antifreeze out of 20 total and to make 50%, you need 10 litres. 10-7.5=2.5 litres.

But I am thinking that is incorrect, since I am adding pure 100% to the 37.5% to try to get 50% is the real question. Or am I totally a wack job and it was good that I kept my mouth shut in study group? This is a basic algebra question, so I understand that 2.5 might be the "math book" right answer, but how would i set up the equation to solve this in real world?

You didn't give enough information to work the problem, sorry. All the other answers are wrong. The fact that a tank has a capacity of 20 litres doesn't mean it is full.

LOL at your post.

 

[Tsaico]

Ok ok!! You guys crack me up... Thanks...

 

[6000SUX]

Originally posted by: chuckywang

Originally posted by: 6000SUX

Originally posted by: Tsaico
Ok, the question is simple: A mechanic tests the current ratio of antifreeze to water and finds that it is 37.5%. How much does he need to drain and how much does he add of pure antifreeze to make the ratio 50% if the tank holds 20 litres total?

A lot of people are saying 2.5 litres, since a 37.5% will be 7.5 litres of antifreeze out of 20 total and to make 50%, you need 10 litres. 10-7.5=2.5 litres.

But I am thinking that is incorrect, since I am adding pure 100% to the 37.5% to try to get 50% is the real question. Or am I totally a wack job and it was good that I kept my mouth shut in study group? This is a basic algebra question, so I understand that 2.5 might be the "math book" right answer, but how would i set up the equation to solve this in real world?

You didn't give enough information to work the problem, sorry. All the other answers are wrong. The fact that a tank has a capacity of 20 litres doesn't mean it is full.

LOL at your post.

No, d!ckweed. You're just embarrassed because you're too stupid to realize what was staring you in the face.

 

[thesurge]

The answer is 4.

 

[chuckywang]

Originally posted by: 6000SUX

Originally posted by: chuckywang

Originally posted by: 6000SUX

Originally posted by: Tsaico
Ok, the question is simple: A mechanic tests the current ratio of antifreeze to water and finds that it is 37.5%. How much does he need to drain and how much does he add of pure antifreeze to make the ratio 50% if the tank holds 20 litres total?

A lot of people are saying 2.5 litres, since a 37.5% will be 7.5 litres of antifreeze out of 20 total and to make 50%, you need 10 litres. 10-7.5=2.5 litres.

But I am thinking that is incorrect, since I am adding pure 100% to the 37.5% to try to get 50% is the real question. Or am I totally a wack job and it was good that I kept my mouth shut in study group? This is a basic algebra question, so I understand that 2.5 might be the "math book" right answer, but how would i set up the equation to solve this in real world?

You didn't give enough information to work the problem, sorry. All the other answers are wrong. The fact that a tank has a capacity of 20 litres doesn't mean it is full.

LOL at your post.

No, d!ckweed. You're just embarrassed because you're too stupid to realize what was staring you in the face.

What was "staring" me in the face was a well-formed, unambiguous algebra problem. Everybody else understood it, and everybody else solved it correctly. You, however, seem to not know how to use common sense when reading.

 

[6000SUX]

Originally posted by: chuckywang

Originally posted by: 6000SUX

Originally posted by: chuckywang

Originally posted by: 6000SUX

Originally posted by: Tsaico
Ok, the question is simple: A mechanic tests the current ratio of antifreeze to water and finds that it is 37.5%. How much does he need to drain and how much does he add of pure antifreeze to make the ratio 50% if the tank holds 20 litres total?

A lot of people are saying 2.5 litres, since a 37.5% will be 7.5 litres of antifreeze out of 20 total and to make 50%, you need 10 litres. 10-7.5=2.5 litres.

But I am thinking that is incorrect, since I am adding pure 100% to the 37.5% to try to get 50% is the real question. Or am I totally a wack job and it was good that I kept my mouth shut in study group? This is a basic algebra question, so I understand that 2.5 might be the "math book" right answer, but how would i set up the equation to solve this in real world?

You didn't give enough information to work the problem, sorry. All the other answers are wrong. The fact that a tank has a capacity of 20 litres doesn't mean it is full.

LOL at your post.

No, d!ckweed. You're just embarrassed because you're too stupid to realize what was staring you in the face.

What was "staring" me in the face was a well-formed, unambiguous algebra problem. Everybody else understood it, and everybody else solved it correctly. You, however, seem to not know how to use common sense when reading.

You won't get far in math, or any other logical pursuit, assuming things that are not stated. Have a nice life, simpleton.

 

[chuckywang]

Originally posted by: 6000SUX

You won't get far in math, or any other logical pursuit, assuming things that are not stated. Have a nice life, simpleton.

1) I have gotten very far in math.

2) Problems are written by PEOPLE. And sometimes those people make little ambiguities (to say nothing about the OP who transcribed the problem to an online forum). It doesn't take a rocket scientist to know what someone meant given the context if there is just some tiny wording that seems out of place or missing. The only way this problem makes sense given what he wrote is if the tank is full. The context of this problem is a simple algebra class. This is not a freaking math journal paper.

 

[everman]

.99999 != 1 :Q

 

[TuxDave]

Originally posted by: everman
.99999 != 1 :Q

But 0.9999.... = 1!!!

 

[xcript]

No, I will not help you math debate.

 

[silverpig]

4

 

[HomeBrewerDude]

Originally posted by: everman
.99999 != 1 :Q

i clicked in here just to see how many posts before this reference was made

 

[BrownTown]

lol, maybe since we are computer geeks we should make a reference to the Pentiums FDIV bug, according to Intel .9999 is close enough to 1 and we are stupid for thinking otherwise .

 

[6000SUX]

Originally posted by: chuckywang

Originally posted by: 6000SUX

You won't get far in math, or any other logical pursuit, assuming things that are not stated. Have a nice life, simpleton.

1) I have gotten very far in math.

2) Problems are written by PEOPLE. And sometimes those people make little ambiguities (to say nothing about the OP who transcribed the problem to an online forum). It doesn't take a rocket scientist to know what someone meant given the context if there is just some tiny wording that seems out of place or missing. The only way this problem makes sense given what he wrote is if the tank is full. The context of this problem is a simple algebra class. This is not a freaking math journal paper.

No, you haven't gotten far in math-- or you were passed incorrectly. It takes someone smarter than you to read a problem correctly, or at least better-trained. This proves your error very clearly. In the context of a simple algebra class, the correct answer would be "Not enough information given." Have a great life with your, um, advanced math skills. (Or should I say "skilz"?)