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

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

  1. TX Mudder

    TX Mudder 1/2 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. 84gmcjimmy

    84gmcjimmy 1 ton status

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    aah my brain!
    someone should explain it so I can know too!
     
  3. big83chevy4x4

    big83chevy4x4 3/4 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. 4X4HIGH

    4X4HIGH 1 ton status Premium Member GMOTM Winner

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    I'm no expert but I don't think there is enough information to give an answer.
     
  5. 84gmcjimmy

    84gmcjimmy 1 ton status

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    me too! I dont understand it!
     
  6. Can Can

    Can Can Pusher Man Staff Member Super 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. mrk5

    mrk5 The Sticker Guy Moderator Vendor GMOTM Winner Author

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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 & error.
     
  8. clubba68

    clubba68 1/2 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 & error.

    [/ QUOTE ]

    This is correct.
     
  9. az-k5

    az-k5 1/2 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 & error.

    [/ QUOTE ]

    I agree as to this being the correct way to go about this.

    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. behemoth

    behemoth 1/2 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. spearchucker

    spearchucker 1/2 ton status

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

    behemoth 1/2 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
     
  13. TX Mudder

    TX Mudder 1/2 ton status

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    [ QUOTE ]
    (.5)x+(.2)40=(.35)y

    [/ QUOTE ]

    Thanks, that's exactly what my brain was not getting.
     

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