Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question

  1. May 24, 2004 #1
    [​IMG]

    A = 13
    B = 24
    C = ?

    I'm curious how this question works. Does C equal 11 or...?
    This question has so many things to take into account I'm not sure what the answer is.
     
    Last edited: May 24, 2004
  2. jcsd
  3. May 24, 2004 #2
    Set these up as a system of equations:

    3S+1L=13
    5S+2L=24

    solve for S and L and then plug those values into 2S+L=?

    [edit]corrected 2S+L=?
     
    Last edited: May 24, 2004
  4. May 24, 2004 #3
    How do you factor in the unconnected small sphere? Are you sure it's a system of equations problem? It also has parallel lines and spheres that are connected. They don't have anything to do with the answer?
     
  5. May 24, 2004 #4
    I believe Faust9 is quite right. Setting up the equations

    3S+1L=13
    5S+2L=24

    then solving for S and L appears to be the right solution. The values for A and B fits, and the value for C is 11. I don't think whether the spheres are connected or not has anything to say.
     
  6. May 24, 2004 #5
    What parallel lines are you refering to? I chose the simple system because most of the other things you could throw into the problem to make it more difficult are subjective. For instance, color. How does that factor in? How do you see the colors (my laptop, and PC show A and B differently). Color is too subjective to act on the system. How does the combination of small spheres, and big shperes affect volume? Again, subjective because I may think the two big red spheres share 10% of their volumes you may say they share 12.5%. Surface area sufferes from the same subjectivity. Assigning the numbers 13,24,? as masses is the only logical thing (IMO) to do here. Now, we are shown to sphere sizes which differ drastically thus you can assume one doesn't represent a neutron and the other a proton. That means if we are looking at these as mass systems we can safely assume there is no change in mass whether there is a direct bond between the small and large spheres hence the system I presented.

    Also, I see this is for an IQ test. IQ test questions usually boil down to very easy solutions to seemingly complex problems.

    Give us the answer if you find it.

    Thank you and good luck.
     
  7. May 24, 2004 #6
    I originally determined the answer to be 11 without using Systems. Thanks for your help guys. I was referring to the non-existent but possibility of parallel lines when you look at the small circles. Amounts of diagonal lines as well made the problem more complex. I was looking at it in too much depth I guess. That seems to happen a lot with me. ;)

    Here's how I originally determined the answer. I wish I remembered Systems at the time. That's what were currently reviewing in class as well, oh the irony. lol.

    Diagram A: The amount of connections in the circle is 3. Add that to the big circle which equals 10.
    Diagram B: The left large circle plus it's connections = 14 plus the other big circle being connected equals 24. The left circle being the starting point actually makes the answer 23 if you go from there.
    Using that logic I got 13 for Diagram C, which happened to be correct. Perhaps another part of my mind was using Systems, lol.

    The answer was 11. Man, I hate when I miss the obvious solving method to a relatively easy question.
     
  8. May 24, 2004 #7
    So the answer is 11? when you subtract B = 24 to A = 13 you also get eleven
     
  9. May 24, 2004 #8
    Adding up all the spheres on A = 4 (including the largest one) + B = 7 equals 11
     
  10. May 25, 2004 #9
    That's funny. I also got eleven, but by the following logic:
    The first group, they tell you, is 13. That must be because a big sphere = 10 and the little ones =1 each. If the second group = 24 there is something about the fifth little sphere that makes it not count. I figured it wasn't in close enough proximity to the others. All the ones that seem to "count" in that one and the previous one, are within a certain proximity to each other. The third image shows one large, and two small, but one of the small ones isn't touching the large sphere and it doesn't fullfill the proximity requirement. Therefore the correct answer for that one had to be eleven.
     
  11. May 25, 2004 #10
    The way I saw this was similar to Zoobieshoe. Owing largely to group A it smelled like a base-10 numbering scheme. Glancing at each group from left to right my focus came to the detached ball, which I realized must be intended as a clue. Looking back from right to left caused me to appreciate the small red ball which seemed out of place, whereupon I recognized a pattern; tight, loose, separate. Immediately following I intuited that C = A-2 = 11, then looked to validate my hunch by studying the particulars of group B in greater detail.

    There is, incidentally, a plant which somewhat resembles group A (usually only having two smaller balls), I do not know the scientific name for it, but locally it is known as ‘young gal booby’.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Question
  1. Question (Replies: 2)

  2. Question (Replies: 9)

  3. A Question (Replies: 3)

  4. A Question (Replies: 20)

Loading...