# MATH TIME......How do you solve this word problem?

Discussion in 'The Lounge' started by TX Mudder, Aug 8, 2004.

1. ### TX Mudder1/2 ton status

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A friend asked me for help with this math problem. I immediately got the answer by figuring it in my head, but that does not help my friend.
I need to be able to show the equation one would set up to solve this problem.
Again, I am not looking for the answer but HOW to get the answer.

The word problem:

How much of a 50% salt solution would you add to 40 liters of a 20% salt solution to get a 35% salt solution?

2. ### 84gmcjimmy1 ton status

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aah my brain!
someone should explain it so I can know too!

3. ### big83chevy4x43/4 ton status

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the answer is 40 liters right?
if so, this is what i did,
50%+20%=70%
70%/2=35%
it requires equal 50% and 20% to make 35%
/forums/images/graemlins/dunno.gif not sure if this is right but thats how i did it

4. ### 4X4HIGH1 ton statusPremium MemberGMOTM Winner

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I'm no expert but I don't think there is enough information to give an answer.

5. ### 84gmcjimmy1 ton status

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me too! I dont understand it!

6. ### Can CanPusher ManStaff MemberSuper Moderator

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Lucky for you my wife and I are smart. Hopefully this is right:

1. Take the original mixture of 20/80 and times it by 40(litres) which equals 800/3200.

2. Through process of elimination, we figured that 40(litres)of the 50/50 solution(= 2000/2000) added to our original 40 litres of the 20/80 solution(= 800/3200) divided by total amount of litres (80 litres) equals 35/65.

I have a headache now. Thanks very much.

7. ### mrk5The Sticker GuyModeratorVendorGMOTM WinnerAuthor

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[ QUOTE ]
How much of a 50% salt solution would you add to 40 liters of a 20% salt solution to get a 35% salt solution?

[/ QUOTE ]

x=amount of 50% solution
y=amout of 35% solution

x+40=y
(.5)x+(.2)40=(.35)y

In the second equation we know y=x+40 from the first equation. Substitute it into the second equation.

(.5)x+(.2)40=(.35)(x+40)

Now solve for x and you need to add 40L like already said you just don't have to use trial &amp; error.

8. ### clubba681/2 ton status

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[ QUOTE ]
[ QUOTE ]
How much of a 50% salt solution would you add to 40 liters of a 20% salt solution to get a 35% salt solution?

[/ QUOTE ]

x=amount of 50% solution
y=amout of 35% solution

x+40=y
(.5)x+(.2)40=(.35)y

In the second equation we know y=x+40 from the first equation. Substitute it into the second equation.

(.5)x+(.2)40=(.35)(x+40)

Now solve for x and you need to add 40L like already said you just don't have to use trial &amp; error.

[/ QUOTE ]

This is correct.

9. ### az-k51/2 ton status

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[ QUOTE ]
[ QUOTE ]
How much of a 50% salt solution would you add to 40 liters of a 20% salt solution to get a 35% salt solution?

[/ QUOTE ]

x=amount of 50% solution
y=amout of 35% solution

x+40=y
(.5)x+(.2)40=(.35)y

In the second equation we know y=x+40 from the first equation. Substitute it into the second equation.

(.5)x+(.2)40=(.35)(x+40)

Now solve for x and you need to add 40L like already said you just don't have to use trial &amp; error.

[/ QUOTE ]

Kinda BS that they start with 40 and you find 40 is the number to add. It rarely works like that. I got all pissed when I double checked the numbers. Then again I am a real dork for math (I can't resist a problem).

10. ### behemoth1/2 ton status

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I HATE word problems !!!!! /forums/images/graemlins/angryfire.gif /forums/images/graemlins/yikes.gif Give me an equation problem any day!!

11. ### spearchucker1/2 ton status

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Real life math problems are in the form of word problems. /forums/images/graemlins/shame.gif

12. ### behemoth1/2 ton status

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[ QUOTE ]
Real life math problems are in the form of word problems.

[/ QUOTE ]
I know - but I'm still allowed to HATE them !! /forums/images/graemlins/histerical.gif

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