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These 10 questions should take 12 minutes?

  1. Aug 8, 2014 #1
    Let me know how long it takes you to solve these 10 questions that are supposedly representative of what 10 Quantitative Reasoning questions on the GRE would be like: http://www.majortests.com/gre/problem_solving_test03. The first time I took the GRE (2 years ago), I remember most of the QR questions being no-brainers. Maybe I need to get back in the groove, or maybe I've become stupider.
     
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  3. Aug 8, 2014 #2

    verty

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    9m20 here, though for some reason I messed up on one question. I hate those "choose all the numbers that cannot be..." questions, one has to be so careful with them. The hardest question was the addition one, it had a little trick.
     
  4. Aug 8, 2014 #3
    That question didn't specify whether 0 can be one of the digits. You can't automatically assume D=1. That's what I assumed to get my answer.
     
  5. Aug 8, 2014 #4

    WWGD

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    But D has to be 1 , since B+C +k =4 , where k=0 (if you don't carry) or k=1 ( if you carry), so B must be 8 or 9, so you must carry a 1 , which is D; you cannot have 50-something + a 2-digit number be 40-something; if 5+B+k =4, then B is either 8, or 9, so you carry a 1.
     
  6. Aug 8, 2014 #5

    jbunniii

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    I got all ten within 12 minutes, but I had to spend several minutes trying to interpret what the 5A + BC = D43 question really meant. The way it is worded, one solution should be A=0, C=3, B=9, D=1. Then all the digits being summed (i.e., 5, A, B, and C) are distinct. But 13 wasn't one of the answers. Then I realized that by "all the digits in the sum," they must have meant that all of 5, A, B, C, D, 4, 3 must be distinct, and I got 22 as an answer under that interpretation. Also, they didn't specify base 10, so the question as written is ill posed.
     
  7. Aug 8, 2014 #6

    D H

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    Sure you can. The only way to get D=0 is if B is -1 (or -2 in the case A+C=13), and certainly neither -1 nor -2 counts as a digit.
     
  8. Aug 8, 2014 #7

    Matterwave

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    I got all 10 in 13 minutes and some seconds. I made a dumb error on the digits one and added the wrong digits up at the end after figuring out it could be 57+86=143 (I added 7,8,6 and 3...instead of 7,8,6 and 1) so I got none of the answers...took me 2 minutes to figure out my error! lol
     
  9. Aug 12, 2014 #8
    I need to come up with a more organized structure of how I approach those problems to solve them efficiently. Right now I'm using a "greedy" approach: try to search for the first digit whose value I know; narrow it down from there.
     
  10. Aug 12, 2014 #9

    D H

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    That problem isn't that hard. You know A+C is either 3 or 13. If A+C=3, that means B=9, D=1, and the sum is 13. If A+C=13, that means B=8, D=1, and the sum is 22. The first solution isn't on the list, but the second is.

    The problem would have been trickier if the problem was 5A+BC=D40 instead of 5A+BC=D43 and if both 10 and 19 were listed as choices. With this change, A+C is 0 or 10. A+C=0 means B=9, D=1, and the sum is 10. A+C=10 means B=8, D=1 and the sum is 19. Both solutions are listed! Which one is wrong? A+C=0 isn't valid because that means A=C=0, and this violates the clause that A,B,C, and D are different digits.
     
  11. Aug 15, 2014 #10

    cjl

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    But if A+C = 3, as you point out, D = 1, B = 9, but then you're stuck with A+C = 3 where A and C must be distinct digits not equal to 1 or 9 (or 5 or 4 or 3). This is not possible - the only A and C such that A+C = 3 are A = 1 and C = 2 (or A = 2 and C = 1). D is already equal to 1 though, so this fails the criterion that all the digits in the sum must be different. Thus, A+C = 13 is the only valid answer.
     
  12. Aug 15, 2014 #11

    D H

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    Who says A and C must be between 1 and 9? Zero is a digit. There's nothing wrong with A or C being zero. If the question had both 22 and 13 as answers and had I checked 13, I would most certainly protest that selection as being marked wrong.
     
  13. Aug 16, 2014 #12
    it took me 50 minutes for 9 questions (3 incorrect) and I still don't understand what the second question means yet
     
    Last edited: Aug 16, 2014
  14. Aug 16, 2014 #13
    A+C can not be 3 (3 is present in the sum, D is 1 already), so A or C can not be either 0,1,2,3,4,5
     
  15. Aug 16, 2014 #14
    ##\sqrt{5}## percent of ##5\sqrt{5}## is ##\frac{\sqrt{5}}{100}5\sqrt{5}##
     
  16. Aug 16, 2014 #15

    D H

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    The expression "a percent of b" means a·b/100. So in this case, with [itex]a=\sqrt 5[/itex] and [itex]b=5\sqrt 5[/itex], the answer is [itex]\frac {(\sqrt 5)\cdot(5\sqrt 5)}{100} = \frac{5\cdot 5}{100}=\frac {25}{100} = \frac 1 4[/itex].

    Sure they can. The question does not say that all of the digits in the problem are different from one another. It only says that "A,B,C and D represent different digits, and all the digits in the sum are different." The constraint against all of the digits in the sum being different just tells us that D is neither 3 nor 4, but we already know D has to be 1. The solution A=0, B=9, C=3, D=1 satisfies the sum problem and the constraints. A, B, C, and D are different digits. The only reason this is not the correct answer is because it is not one of the proffered choices.

    This is somewhat common on multiple choice questions such as the GRE. There might well be multiple solutions to a problem, but only one will be listed.
     
  17. Aug 16, 2014 #16
    Thank you a lot.
    Yes it is likely an issue with wording and comma; I think "all the digits in the sum are different" is also applied to 5A+BC=D43 not only to A+B+C+D
     
  18. Aug 17, 2014 #17

    OmCheeto

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    I only got 6 correct. Were we not supposed to do these in our head? I think I probably would have gotten most all correct if I'd just written them down.

    I missed:
    #2: Didn't read the question correctly. Picked the closest answer to "5 is what percent of 25" = 25. :rolleyes: Ha!
    #5: Can't divide in my head anymore. By this point I was wondering how anyone could do this quiz in 10 minutes. 668 / 4 = 100 with 268 left over. 268 / 4 = 60 with 28 left over. 28? 28 doesn't look divisible by 4. :tongue:
    #8: The carry threw me off by one. :mad:
    #10: Word problems.... Blech! I totally guessed on that one.
     
  19. Aug 17, 2014 #18
    Got 7 correct in 9 minutes 48 seconds.

    The 8th one was tricky.
     
  20. Aug 17, 2014 #19
    That's exactly where I went wrong.
     
  21. Aug 17, 2014 #20

    mfb

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    10 correct, 4:45.

    I missed the "A or C = 0" option, that saved some time.

    #5 can be solved as "6^5 - 6^4 = 6^4 (6-1) = 6^4 * 5" then you don't have to divide or multiply anything (because the right answer is given as 6^4 as well).
     
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