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e.cartman

2nd Level
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Everything posted by e.cartman

  1. It's actually a simple problem if you are not distracted by the right-angle like I was. Just 1 step: Exterior angle = sum of interior opposite angles. PRS = PQR + QPR Stmt 1: PRS = PQR + 30 => PRS - PQR = 30. suff Stmt 2: We again get QPR=30 and same result. suff No need for PSR to be right angle. HTH
  2. 3/4 - 2/5 = ans. Smallbee, imo your final fraction should be 35/100
  3. Question translates to jfk = 1/2 miami; miami = 4 logan. It is not even ambiguous. I think OA is telling people what to think! :)
  4. Imo D. Either way it looks like median will be equidistant from smallest and biggest. OA please.
  5. Thanks for pointing that out silvanahv. I see it is essentially same as Bertrand's box; when the problem says drawn-out counter is white, it is analogous to eliminating the box with 2 silver coins, so it is really just reworded version of the same.
  6. Question: When positive integer n is divided by 3, the remainder is 2; and when positive integer t is divided by 5, remainder is 3. What is the remainder when product nt is divided by 15? 1) n-2 is divisible by 5 2) t is divisible by 3
  7. I took long time for this so I hope there is a short-cut. Together, Stmt 1: n=3i+2=5j+2 By trial and error, remainder of n by 15 would be 2. Donno about t, so insuff. Stmt 2: t=5p+3=3q by trying small values for p, remainder of t by 15 is 3. Donno about n so insuff. Together, n=15a+2 t=15b+3 So multiplying, remainder of nt by 15 has to be 6.
  8. That's why this is a good problem. Probability is not 1/2. You need to consider drawing w1 separate from w2 because they are equally likely different events. So probability is 2/3 as pointed out before.
  9. Multiply Nr and Dr by 2^4 2^4/10^something =16/10^something => 2 non-zero digits.
  10. good question, monty-hallish.
  11. Imo D. Let angle subtended at center be x. 4pi/3 = 2pi*4 * x/360 => x =60 If center is O, ORU is an equilateral triangle so RU=4.
  12. same method as thankont. This is weighted average where the weight on 20 is more than weight on 10 since y>x. So result will be greater than (10+20)/2 but less than 20.
  13. Question says factors, not prime factors.
  14. Aren't you listing all the ways in which couples can be selected? If so your list is correct. 24 ways in which a couple is included in committee. But question says there should be no couple. So just subtract that from all possible ways: 8C3 - 24 = 56 - 24 = 32.
  15. Imo Q1 is D. It says t+2 must be in set not t-2.
  16. no it's not said r is 3-digit number. ABC is only for convenience. tens digit of r/10 is always hundreds digit and hundreds digit of 10r is tens digit.
  17. question should say product of all the even integers...
  18. Imo C. Together probability is 1/6.
  19. Imo E. Together (p+1)(r+1) = 3(r+1) factors, but since r is not given, insuff.
  20. Let x be interest rate last year. x*1.1 = 10 => x=10/1.1 = 9.1%
  21. Answer is B. Stmt1 is insuff because x=3, y=-3 is possible.
  22. Imo B. Let r=ABC. We need to find digit B. 1) A=3. insuff 2) B=6. suff.
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