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Math Help - Crazy percent problem

  1. #1
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    Crazy percent problem

    It's almost midnight, I am about to finish my studies for the day (yup, from morning till now) and this is the problem I get:

    If 125% of j is equal to 25% of k, 150% of k is equal to 50% of l, and 175% of l is equal to 75% of m, then 20% of m is equal to what percent of 200% of j ?

    It's got to be a cruel joke, I have no idea how to tackle this, unless I substitute some numbers here and do the arduous work of calculating everything. But there must be a faster way. Will you please help? Last one for today, I promise.
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  2. #2
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    Quote Originally Posted by crazirani View Post
    It's almost midnight, I am about to finish my studies for the day (yup, from morning till now) and this is the problem I get:

    If 125% of j is equal to 25% of k, 150% of k is equal to 50% of l, and 175% of l is equal to 75% of m, then 20% of m is equal to what percent of 200% of j ?

    It's got to be a cruel joke, I have no idea how to tackle this, unless I substitute some numbers here and do the arduous work of calculating everything. But there must be a faster way. Will you please help? Last one for today, I promise.
    Hello,

    if you use fractions instead of percents the problem shrinks to normal size

    You may have notice that the second value is allways greater than the first one:
    k=5 \cdot j
    l=3 \cdot k= 3\cdot 5 \cdot j
    m=2 \cdot l=2 \cdot 3 \cdot k= 2 \cdot 3\cdot 5 \cdot j

    Therefore j=\frac{1}{30}m \Longleftrightarrow m = 30 \cdot j.

    20 \% m = \frac{1}{5} m = 6 \cdot j and 200% j = 2 j. Now calculate the percentage:

    p=\frac{6 \cdot j}{2j} \cdot 100= 300\%


    EB
    Last edited by earboth; January 19th 2007 at 01:27 AM.
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  3. #3
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    Hi earboth,

    I don't understand how you get those numbers. For example:

    k = 5 . j

    How did you get 5?

    Thanks.
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  4. #4
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    Quote Originally Posted by crazirani View Post
    Hi earboth,

    I don't understand how you get those numbers. For example:

    k = 5 . j

    How did you get 5?

    Thanks.
    Hi,

    according to your problem

    125\%j=25\%k divide the equation by 25% and you'll get: 5j=k

    But it is easier to use fractions instead of percents:
    25\%=\frac{1}{4}
    125\%=\frac{5}{4}
    150\%=\frac{6}{4}
    175\%=\frac{7}{4}

    Now you have some simple equations and by substitution you'll get those results I have posted in my first reply.


    EB
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  5. #5
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    Quote Originally Posted by crazirani View Post
    It's almost midnight, I am about to finish my studies for the day (yup, from morning till now) and this is the problem I get:

    If 125% of j is equal to 25% of k, 150% of k is equal to 50% of l, and 175% of l is equal to 75% of m, then 20% of m is equal to what percent of 200% of j ?

    It's got to be a cruel joke, I have no idea how to tackle this, unless I substitute some numbers here and do the arduous work of calculating everything. But there must be a faster way. Will you please help? Last one for today, I promise.
    Here is one way.

    Since in the end the j is there, or a percent of j is what is asked for, then carry over the j from beginning to end.

    "125% of j is equal to 25% of k,..."
    1.25j = 0.25k
    k = 1.25j /0.25 = 5j

    "...150% of k is equal to 50% of l,..."
    1.50k = 0.50L
    1.5(5j) = 0.5L
    L = 1.5(5j) /0.5 = 15j

    "...and 175% of l is equal to 75% of m,..."
    1.75L = 0.75m
    1.75(15j) = 0.75m
    m = 1.75(15j) /0.75 = 35j

    "...then 20% of m is equal to what percent of 200% of j ?"
    0.20m = x*2.00j
    0.2(35j) = x(2j)
    x = 0.2(35j) /2j = 3.5 = 350% ------------answer.
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  6. #6
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    Quote Originally Posted by ticbol View Post
    Here is one way.

    ..
    Hello ticbol,

    thanks for repairing my calculations: Of course you got the right result. I misread the problem.

    So sorry crazirani.

    EB
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  7. #7
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    Quote Originally Posted by earboth View Post
    Hello ticbol,

    thanks for repairing my calculations: Of course you got the right result. I misread the problem.

    So sorry crazirani.

    EB
    Good morning, earboth.
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  8. #8
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    Thanks guys.

    When I was in high school and college, I used to be able to "figure out" such problems. Right now I just don't know where to start, I guess I panic or something. With the test being tomorrow, I must take it up a notch, but how?
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  9. #9
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    Quote Originally Posted by crazirani View Post
    Thanks guys.

    When I was in high school and college, I used to be able to "figure out" such problems. Right now I just don't know where to start, I guess I panic or something. With the test being tomorrow, I must take it up a notch, but how?
    I see.

    Math needs practice. That's why I'm here in this forum.
    While in school, we may be good at the Math topics we are studying because they are "fresh" in our minds. Then out of the studying ib school, we tend to forget those topics. [It takes only about a month before I forget how to do most Math topics, by the way.]

    So, you need to take it up a notch?
    Practice, practice, and practice.
    See and try to solve more of similar problems in the Exercises portion of your textbook.

    I wonder why they call those "Exercises"?
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  10. #10
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    You are absolutely right. And thanks for your input.

    I am facing another problem though. I bought 4 GMAT books 3-4 months ago and I studied every single one of them every day. I had planned to study until 3-4 days before the exam, and the last 3-4 days, I had planned to take practice tests. So I studied these 4 books religiously and learned them inside and out, but now that I take the practice tests, I find the math much harder than the ones in the book. I am practicing as much as I can with the math that is in the practice tests, but in some ways, it's a bit too late.

    Am not asking for sympathy of course, just wanted to put that in there just in case someone else might benefit from it.
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