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zuckerman

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Everything posted by zuckerman

  1. batty, Congratulations on a wonderful job! All the best with your applications. Regards. Zuck
  2. singhman, Congratulations! Splendid job in a very efficient timeline. All the best for your future plans! Zuck
  3. bmwhype, what's the OA, please? I am getting 31/45 as well, with similar logic as provided by others above. Thanks.
  4. krusta, thanks. Just to confirm... you meant 'unique combinations of two', am I right? Sorry, I am having a terrible week, and this has been quite common lately.
  5. I must be missing something since two people have already solved this one, but can someone throw some light on what the question is actually asking? "If we were to choose 1 member of each group, how many combinations are possible?" What type of combinations? Consisting of how many people? Thanks.
  6. Hi Bob, Can you please explain the underlined part of your comment? "It almost never happens that the answer to a yes-no question is no, and it absolutely NEVER happens that one statement gives you an answer of yes and the other statement gives you an answer of no." Thanks much! Regards. Z
  7. QUES 4. Ans C If k and n are integers, is n divisible by 7? 1) n-3 = 2k When n = 11, k = 4 11 - 3 = 2*4, n=11 is not divisible by 7 When n = 7, k = 2 7 - 3 = 2*2, n=7 is divisible by 7 Cannot say for sure. INSUFF 2) 2k-4 is divisble by 7. Does not say anything about n. INSUFF From 1) and 2) n-3 = 2k n-3-4 = 2k -4 n-7 = 2k-4 => n-7 is divisible by 7 => n is divisible by 7. SUFF, so C. QUES 7. Ans B If x+y+z > 0, is z > 1? 1) z > x+y+1 Try z = 0.75, x = y = - 0.25. Here z Try z = 2, x = y = - 0.25. Here z > 1 So INSUFF 2) x+y+1 0 > x+y+1 Also x+y+z > 0 x+y+z > x+y+1 => z > 1 SUFF QUES 9. Ans A ?? If x and y are integers between 10 and 99, inclusive, is (x-y)/9 is an integer? 1) x and y have the same two digits, but in reverse order. SUFF This is actually a well discussed condition for division by 9, as far as I know. 2) The tens' digit of x is 2 more than the units' digit, and the tens' digit of y is 2 less than the units' digit. I started to working on this one, and it seemed to lead nowhere. I am not 100% sure if this is INSUFF or not. Maybe someone else can chime in?
  8. QUES 8. Ans A If xy > 0, does (x-1)(y-1)=1? Rephrase the question as: Does xy -x -y + 1 = 1? Or, Does xy = (x+y) ? 1) x+y=xy SUFF. 2) x=y x*x = x + x x^2 = 2x This is true only for one value (since x, y are not equal to 0): x = y = 2 INSUFF
  9. QUES 5. Ans B 1) n > m+15 Say n = 20, m = 1 sqrt(n-m) = sqrt(19) is not an integer. INSUFF 2) n=m(m+1) n = m^2 + m n - m = m^2 => sqrt(n-m) = m, which is an integer SUFF QUES 6. Ans A If rs!=0, is (1/r)+(1/s)=4? 1) r+s = 4rs (1/r)+(1/s) = (r+s)/(r*s) = (4*r*s)/(r*s) = 4 SUFF 2) r=s (1/r)+(1/s) = (r+s)/(r*s) =(2*s)/(s*s) = 2/s INSUFF
  10. QUES 1. Ans D 1) x>r and y x - r > 0 and y - s x - r > y - s so x - y > r -s SUFF 2) y=2, s=3, r=5, x=6 x - y = 4 r - s = 2 4 > 2 SUFF QUES 2. Ans A 1) On the number line, z is closer to 10 than it is to x. SUFF We can draw a sketch to verify. 2) z=5x Test for x = 1 and x = 2, and we can is that this is INSUFF. QUES 3. Is 5^k less than 1,000? Note: k need not be an integer. 1) 5^(k+1) > 3,000 => k +1 > 4.975 => k > 3.975 But 5^k > 1000 for all k > 4.295 So INSUFF 2) 5^(k+1) = 5^(k) -500 Is this condition given correctly here? Can you please check? If it is correct the way you have it, I will got for E. We know 5^3 = 5^4 - 500, leading me to suspect the above condition. Or is it one of: 5^(k+1) = 5^(k) + 500 5^(k-1) = 5^(k) -500 5^(k) = 5^(k+1) -500 k = 4 or 3 depending on which condition we use. Here 5^K
  11. I can tell you how to go from step 2 to step 3, without considering the original question or the correctness of step 2. 5^21 * 2^22= 2 * 10^n (step 2) 5^21 * (2*2^21)= 2 * 10^n 2 * (5^21 * 2^21) = 2 * 10^n 2 * (5*2)^21 = 2 * 10^n 2 * (10)^21 = 2 * 10^n Is this what you were looking for?
  12. popatpandu, I don't know. I was not referring to any particular shortcut. I was wondering if you had used one.
  13. I got the same answers too. 1) 6 tables 2) 2.5 hrs popatpandu: How did you solve 2)? Did you solve for speed first? Or did you use a shortcut? Care to tell? Thanks.
  14. The n and n-4 are subscripts, denoting the nth term and the (n-4)th term of the series, respectively. The nth term equals the (n-4)th term for n> 4, meaning that the cycle repeats after every four terms.
  15. diya, what's the source of this question please, since you happen to have access to the OA? I would vote for A out of the choices given.
  16. 1) tells us that x and y share the same mutual relationship (sign wise) that a and b share. Thus implying that (-x, y) can lie either in quadrants II or IV. But there is nothing to tie a, b to x, y. Thus, we can have (-a, b) and (-b, a) in II, but (-x, y) can be in IV; or vice versa. 2) fills this gap by tying the signs of a and x. But 2) by itself is not sufficient, since now we lose the same-sign relationship of x and y that 1) offers. However, if we combine 1) and 2) then we have tied the points to the same quadrant. Therefore C.
  17. seriVolk, the point being inspected is (-x,y). That will affect the answers.
  18. 4 / (x-2) or, 2/(x-2) Then, I plugged in numbers. You can see that if x Also, if x > 4, the left hand side For x between 2 and 4 (inclusive) it fails. Is this the OA you have?
  19. seriVolk, thanks that explains. We don't get a unique number of factors by using 2) alone. Though I believe you meant to say either '3 or 4 factors' if we consider just 2); namely 1, 7 and 49 (if N=49) OR 1, 5, 7 and 35 (if N=35).
  20. I get x > 4 and x What's the OA, please?
  21. I have a major question here... Say N = 35 = 5*7 N is divisible by 7, and hence satisfies the question. So far so good. Let's look at condition 2) first. (Pretend the conditions were given to us in reverse order!) Condition 2) N/7 has only 2 different factors. N = 5*7 => N/7 = 5 = 1*5 N/7 has two different factors (1 and 5), and so we can say that N has four different factors, 1, 5, 7, 35 - SUFFICIENT What's wrong with the above reasoning? Condition 1) N is divisible by 49. Obviously 35 is NOT divisible by 49, so should we even pursue this condition any further? Since from 1) ALONE we got the total number of factors of N. My question is: In such questions, do we need to consider the validity of both statements even though just one ALONE apparently does the job?? What if the author of the question had the number N = 35 in mind? Is there any GMAT rule about some consistency/interdependence between conditions, such as in this case?
  22. Looking at the rule, we see that the series repeats itself after every four terms. Sum of first four terms = 2 - 3 + 5 - 1 = 3 Also, 97/4 = 24*4 [i.e. 24 complete cycles] + 1 [first term of 25th cycle] Thus the required sum = 3*24 + 2 = 74 So, IMO B
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