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Everything posted by coyote
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Statement 1 : X > Y^2 Since y^2 > 0 for all Y, we can conclude that X > 0 We do not know whether X and Y are integers or fractions. Hence we cannot compare X and Y. Statement 2 : X and Y are positive integers. Obviously insufficient. Combining both, X > |Y| and X>0 Thus sufficient to see that X>Y IMHO it is C.
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Merci, Bucky! I'll make it point to study the Princeton strategies well. :)
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That's a fantastic score, buckykatt! Congratulations. Could you please tell us a little more about how you prepared for the verbal section? Did you take any special efforts for verbal before the 2nd test besides the practice tests?
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25 and 26 do not have any common factors. So their LCM is 25*26 Hence the LCM of 2*25 and 2*26 is 25*26*2
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:2cents: If M is a positive integer, for sqrt(M) > 25 we need M>25*25 Statement 1 : M is divisible by 50 M >= 50 Insufficient. Statement 2 : M is divisible by 52 M >= 52 Insufficient. Statement 1 + Statement 2 : M is divisible by 51 and 52. M >= LCM (50,52) M >= (2*25*26) i.e. M >= (25*52) Hence we can conclude that M > 25*25 IMHO it's C.
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:2cents:IMHO it's B. 25^sqrt(2) = square of 5^sqrt(2) = [5^sqrt(2)]^2 = [5^(2*sqrt2)] = [5^2] ^ sqrt(2) Discussing other options, 5^2 = 25 [A = C] D. 25^2sqrt(2) = {[5^sqrt(2)]^2}^2 E. 5^sqrt(2)^2 = 5^[sqrt(2)^2] = 5 raised to square of sqrt(2)
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The population increase is a GP. a,ar,ar^2.... where r=2 Hence on 6th day, population = 3*2^5 Since no.of days IMHO the answer is 96.
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Of course we should. as Olm rightly said, 0 is divisible by any integer. Let's see it the other way, 1/2 = 0.5 = 0 + 5/10 = 0+(1/2) = (0+1) / 2 i.e remainder 1.
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Welcome. Glad I could be of some help. :)
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12^10 = (3*4)^10 Thus 12^10 = 3^10 * (2^2)^10 = 3^10 * 2^(10*2) = 3^10 * 2^20. [ (a^b)^c = a^(b*c) ] Since 3^10 will not contain any even factor , the maximum possible power of 2 in 12^10 will be 2^20.
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Just a glimpse will tell you that 7/22 = 0.3xxx As we know 22/7 = pi = 3.142 Thus 22/7 - 7/22 = 3.142 - 0.3xxx The nearest integer is 3.
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Choose 1 even integer from a set of 5 even integers --> 5C1 Choose 1 odd integer from a set of 5 odd integers --> 5C1 Total selections = 5C1 * 5C1 = 25
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710 GMAT - (Q49 (89%),V38 (84%) :: Overall 93%)
coyote replied to vbhup2's topic in Just Finished My GMAT
A detailed analysis in the debrief! [clap] Applause for the same. Your score of course, was not unexpected. Great show mate! We'll miss you in TM! Best wishes for your applications! -
:2cents:IMHO it's B Since the strip fits around the rug, length of strip = perimeter of rug. Say length and breadth of rug are L and B. 2(L+B) = 30 LB = 54 Solve to get L=9 and B=6 Thus col A = 9 Col B = 10
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yup. an error.
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Whenever you want to factorize such expressions, always try to express m^2-n^2 as (a-b)(a+b) where a and b are some convenient numbers. Preeet has already explained the solution. :)
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Confused about the first one. :hmm: This is my understanding about the second statement. (x+1)C4 = 2* (xC4) Hence this statement is individually sufficient.
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Jot down the sequence : 2,3,5,10,20 ....... From the third term every term is twice the earlier term. Hence the ratio of any two terms will be 2^d where d is the difference between the number of terms . So if we want the ratio between the 30th and 25th terms, we can first note that the difference is 5. Hence the ratio will be 2^5 = 32. IMHO the answer will be 32.
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I guess I've read this question elsewhere on the forum :hmm: IMO the answer is B (63/20) sqrt (x^2 + 6x + 9) - sqrt (y^2 - 2y +1) = |x+3| - |y-1| =|3/4 + 3| - |2/5 - 1| = |15/4| - |-3/5| = 63/20
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Terrific morpheus! An applause for this! [clap]
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Actually we do not have to find N. We only need to know its prime divisors. What confuses me is that if two statements individually yield answer to a question but those answers are contradictory then do we consider the answer as D or E? :hmm:
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Substitute values of co-ordinates of any one point.
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:hmm:confusing! Statement 1 says 2N has one prime divisor. IMO this means that 2 is the only prime divisor of N. Statement 2 says 3N has one prime divisor. IMO this means that 3 is the only prime divisor of N. But the two conditions cannot co-exist! Is it E? :confused: