mikemagoosh

pemdas: We have to be very careful here. The probability asked in the question is not "the probability of being a woman and bringing dessert". The probability asked in the question is not "the probability of being a woman and not bringing dessert". The probability asked in the question is not "the probability of being someone who does not bring dessert". The probability asked in the question is "the probability of being a woman or being a man who doesn't bring dessert." It's an idiosyncratic group -- "success" for this probability would involve picking any of the six women (regardless of whether they bring dessert) or any of the five men who don't bring dessert. Of the 12 people present, 11 of them would constitute a "yes" to this question. P = 11/12 What I suggested in the second half of my post was --- one alternative way to think about the problem is to calculate the probability of the complement. Let A = person picked is a woman, and B = man not bringing dessert P(A or B) = 1 - P[not (A or B)] = 1 - P[(not A) and (not B)] = 1 - P(person picked is not a woman and is a man bringing dessert) We can't multiply probabilities because they are not independent ---- if you are a man bringing dessert, you can't be a woman!! We have to consider, in this group of 12 --- how many individuals fit this description? Only one does, the one dude bringing dessert --- that's 1/12, so the probability we want is 1 - 1/12 = 11/12 Again, you cannot just multiply probabilities whenever you see the word "and". You can only multiply probabilities when the two events are independent. A and B are independent if and only if whether A happens has absolutely no bearing whether B happens. Again, this A (or not A) and B (or not B) are not independent, because if I know the person I pick is a woman, that automatically means it can't be a man bringing dessert; if the person I pick is a man bringing dessert, it automatically means it can't be a woman. The fact that the outcome of one has direct implications for the possible outcomes of the other means the two events are not independent, which in turn means it's mathematically incorrect to multiply the probabilities under any circumstances. Does that make sense? Let me know if you have any further questions. Mike :)

I'm happy to help with this. :) So, first of all, there are 6 women. There are also 6 men, and of those, one brings a dessert and five don't. The set of (women) + (men who don't bring dessert) is 6 + 5 = 11. Remember that, as a rough & ready rule, the word "or" means add in probability. Thus, probability is P = 11/12 Another slightly more sophisticated way to look at this is the following. The negative of (A or B) is given by: not (A or B) = (not A) and (not B). The "or" changes to an "and" when negated. The negation of (woman) or (man without dessert) is someone who is a man and who is bringing dessert --- well, that's just one person. the one man bringing a dessert. If the negation of the set is one person, the set must be the other 11 --- again, probability is 11/12. Does all this make sense? Let me know if you have any further questions. Mike :)

With all due respect, no professionally produced book will have p. 1 on the left. All professionally printed books have odd pages on the right (the "recto") and all even numbers on the left (the "verso"). If the GRE gives a problem about pages in a random unspecified book, you can definitely assume it follows the odd-recto/even-verso convention. Also, if you go back to look at the post, it was not about 189 sheets of paper making the book, but rather 189 digits printed for the page numbers, and the question was about sorting out how many one-digit vs. two-digit page numbers. Does that make sense? Mike :)

So for that first question, see this diagram: [ATTACH=CONFIG]6457[/ATTACH] If we construct right triangle ADO, notice that AD bisects the 60 degree angle at A, so triangle ADO is a 30-60-90 triangle. The ratio of AD:DO is sqrt(3):1. Call OD = r, because it's the radius of the circle. Then, AD = sqrt(3)*r. D is the midpoint of AC, so AC = 2sqrt(3)*r, and therefore the perimeter of the equilateral is p = 6sqrt(3)*r. That's always true of a circle inscribed in an equilateral triangle. What's messy about this problem is that the area of the circle not expressed in terms of pi --- that makes this a much messier problem than it would be otherwise. A = (pi)r^2 = 154 r^2 = 154/(pi) r = sqrt(154/(pi)) So, perimeter of equilateral triangle is p = 6sqrt(3)*r = 6sqrt(3)*[sqrt(154/(pi))] Yuck. Wherever you got that, it's a butt-ugly problem because of the numbers chosen. 2. Five bells ring together and thereafter, ring at the intervals of 5, 6, 15, 18 and 24 seconds respectively. How many times will they ring together? This is a poorly worded question. How long will all the bells go on ringing? Indefinitely? In perpetuity? Then, there will be an infinite number of occasions on which they will ring together. I think the question is really asking -- when is the next time all five will ring together? Or, given that number, how many time will they ring together in some specified interval of time. In that guise, the problem is really asking for the LCM of 5, 6, 15, 18, and 24. Well, 5 is a factor of 15, so it's automatically included in 15; likewise, 6 in 18 or 24. To find the LCM of 18 and 24, use the fact that their GCF = 6, and use the formula 18*24 = GCF * LCM 18*24 = 6 * LCM 18*24/6 = LCM 18*4 = LCM 72 = LCM of 18 and 24. We know 15 = 3*5. We know 72 already has a 3 in it, so it's simply missing a factor of 5. Therefore 72*5 = 360 is the LCM of all five numbers. Every 360 second (= six minutes), all five bells will ring once again in synch. After the first time, they will ring together 10 times in an hour. 3. Find the least number which when divided by 2,3,4,5,6 leaves in each case a remainder of 1, but when divided by 7 leaves no remainder. For this one, first we need to find the LCM of 2, 3, 4, 5, and 6. Since 2 and 3 are already factors of 6, we can simply find the LCM of 4, 5, 6. The LCM of 4 and 6 is 12, and times 5 is 60. 60 is the LCM of 2, 3, 4, 5, and 6, but 60 + 1 = 61 is not divisible by 7 --- in fact, it's prime. We need a multiple of 60, plus 1, that is divisible by 7. 120 + 1 = 121 = 11^2, not divisible by 7 180 + 1 = 181, prime 240 + 1 = 241, prime 300 + 1 = 301 = 7*43 So, 301 is the lowest number which, when divided by 2, 3, 4, 5, or 6, gives a remainder of 1, but which is divisible by 7. The next numbers which satisfy that property are 721 and 1141. Does all this make sense? Please let me know if you have any other questions. Mike :-)

I'm not sure I understand your approach. There are nine one digit numbers. Two digit numbers start at 10 and end at 99 --- inclusive counting gives us 90 two-digit numbers. There are not 91 two-digit numbers in the base ten system. (Of course, if we went to base 12, there would be 132 two-digit numbers --- but that brings us well outside the pale of GRE math.) Mike :)

This is a fun question. :) In printing pages 1-9, nine digits were printed. Then, we get to the two digit numbers. The 180 digits are 180/2 = 90 two-digit numbers. Nine one-digit numbers + 90 two digit numbers = 99 pages, which is less than 100. Let me know if you have any questions about that. Mike :)

Hi, there. I'm happy to help with this. :) This is a fine example of a rate problem. If you are a Magoosh member, I highly recommend watching the "Work Problems" lesson in the "Word Problem" section of the Math lessons. The basic idea is: we have to convert everything to information in terms of rate of work. First of all, when people or machines work together, we add rates. Let's say that one of these machines has a rate of R. Then 18 of them have a combined rate of 18R. Work rate is given in the form of (how much work)/(how much time). The rate at which these 18 machines work is (1 job)/(40 days), or (1/40) job/day. Thus, 18R = (1/40) job/day. We want to know what would happen if 6 more machines had been working from the beginning --- in other words, we want to know 24R. Notice that we can multiple 18 by 4/3 to get 24, so we will multiple both sides of the equation above by 4/3 (4/3)*18R = (4/3)*(1/40) job/day 24R = (1/30) job/day = (1 job)/(30 days) If 24 machines had been working from the beginning, their combined rate is 1/30, which means in a single day they would complete 1/30 of a job, and in 30 days, that would complete a full job. Does that make sense? Let me know if you have any further questions on this. Mike :)

I'm happy to help with this one as well. I love geometry! :) These diagrams may well make things clear: [ATTACH=CONFIG]6447[/ATTACH] [ATTACH=CONFIG]6448[/ATTACH] [ATTACH=CONFIG]6449[/ATTACH] So, as you see, four distinct lines can intersect at six different points. Actually, this is identical to the question: in a group of four people, everyone shakes hands with everyone else. How many handshakes are there? Mathematically, that's the number of two-person combinations taken from a pool of 4. That equals 4C2 = (4!)/[(2!)(2!) = 6 So, those are two ways to go about getting the answer. Does all that make sense? Please let me know if you have any further questions. Mike :)

I'm happy to help with this. This is a tricky one, because it involves something called inclusive counting. Suppose the first day of Harry's vacation is the 8th and the last day is the 26rd. How many days long is Harry's vacation? The tempting mistake is to say 26 - 8 = 18, but that's not right. That would be right if we were counting the 26rd but not the 8th as part of the vacation. If they are both part the vacation, we have to subtract the numbers and then add one. His vacation is nineteen days long. Inclusive counting from a to b = b - a + 1 What does this have to do with this problem? Well, consider the simpler scenario --- suppose there's just a 90 ft line, and we are spacing the trees 30 feet apart. 90/30 = 3, so you might naively thing there would be three saplings planted, but we have to consider the first planted at "zero' before we started spacing. As the diagram shows, we can plant four saplings along the 90 ft line. [ATTACH=CONFIG]6446[/ATTACH] This is another example of inclusive counting --- here, we divide to find the number of spaces, and then add one to find the number of trees. Now, to this problem. 455/30 = 15 and 1/6 Obviously, if we have some fraction of 30 feet left, we are not going to plan a fraction of a sapling! The answer 15 1/6 is there for people doing the problem on snooze-control automatic pilot. Obviously, we round down --- we have 15 full intervals of 30 ft. It is tempting here to say that we have 15 intervals so we can plant 15 trees, but that's not correct. We need to use inclusive counting. We add one, and find that we can plant sixteen trees. Does all this make sense? If you are a Magoosh customer, I recommend watching the "Sum of Sequences" videos for more on inclusive counting. Please let me know if you have any more questions. Mike :)

I'm happy to help here. With all due respect, you stated the first problem incompletely. You said "a record of 6 inches" --- on a websearch, I found it was suppose to be "a record with a radius of 6 inches." The fact that you missed that crucial detail when you typed the problem in actually gives me some insight into why math is a challenge for you. In math, every little detail matters. It's never enough to have the "gist" of a problem --- you have to scour the problem for fine details that will reveal further clues. You can never be too careful with math. The details are everything. [ATTACH=CONFIG]6445[/ATTACH] I've attached a diagram of this problem. The point on the circumference, 6 in from the center, is on a larger circle than the point 5 inches away from the center. Each time they make a complete revolution, the outer point moves further than the inner point. It doesn't matter how many times the circle turns. The outer point covers more distance. Cost of any quantity of anything = (number of units)*(price per unit) The cost of x pounds of meat at y $ per pound = (x lb)*($y/lb) = $ x*y The cost of y yards of material at x $ per yard = (y lb)*($x/lb) = $ x*y Does all that make sense? Please let me know if you have any further questions. Mike :)

HI, there. I'm happy to help with these 1) This one shows a line with negative slope, going through the origin and the point (8, -8). The shaded region is above that line and to the left of the y-axis. The question asks: Of the following pair of coordinates, which represents a point in the shaded region? (A) (3, -5) (B) (-3, -5) © (-3, 5) (D) (-5, 3) (E) (-5, -3) Well, first of all, the shaded region is in Quadrant II, which means that x-coordinates are negative and the y-coordinates are positive. Right there, that eliminates (A), (B), and (E). Because you're a Magoosh customer, you can watch Magoosh GRE - The Coordinate Plane to get clear on that. This line happens to be the line y = -x. This line makes a 45 degree angles with both the x- and y-axes. The points on this line in Quadrant II have x- and y-coordinates of equal absolutely value, but the x is negative and the y is positive --- for example, (-1, 1), (-2, 2), (-3, 3), (-4, 4), etc. If we start from one of those points and go down (decrease y), then we will not be in the shaded region. If we start at one of those points and go up (increase y), then we will be in the shaded region. So, examples of points in the shaded region are . . . (-1, 2), (-1, 3), (-1, 4), etc., (-2, 3), (-2, 4), (-2, 5), Those are all points in the shaded region. Notice the pattern is: the x-coordinate is a negative number with a smaller absolute value, and the y-coordinate is a positive number with a larger absolute value. The point that fits this pattern is (-3, 5). Answer = 2) (For this one, I made a scaled diagram, in the jpg attached.) ACEF is a square region, and B, D, and G are the midpoints of AC, CE, and BD respectively. The fraction of ACEF that is shaded | 7/16 In the second diagram of the jpg, the upper half of the square is subdivided into little triangles. There are 8 triangles in the upper half, so there would be 16 triangles in the square overall. The shaded region occupies 7 of those triangles, so it is 7/16 of the square. Answer = 3) This one shows two overlapping triangles, which may or may not be equilateral, and may or may not be congruent. The altitude of ABC from B to AC | The altitude of DEF from E to DF. In the GRE, we know that figures are not necessarily drawn to scale. We are given zero information about the size/shape of the triangles other than the diagrams. Since all we have to go on is the diagram, which we know is not necessarily drawn to scale, we can't conclude anything. Answer = Does all this make sense? Please let me know if you have any further questions. Mike :)

Dear smuf, As Chris explained, 20 is a possible area. In fact, 20 is the maximum area -- that's when we have the side of 5 and the side of 8 at right angles (which is always when you will get maximum triangle area). If you make the angle less than 90 degrees, you will get areas of less than 20. In particular, suppose in slow motion you decrease the angle between the side of 5 and the side of 8 from 90 degrees to zero --- in that process, the area of the triangle will decrease continuously from 20 to zero, and in that process, it will pass through every positive integer less than 20. Once you know the maximum area, you know it's possible for the triangle to have an area equal to any positive integer less than the maximum area. Therefore, of course 5 is a possible area. No additional calculation is needed. Does that make sense? Please let us know if you have any further questions about this. Mike :)

Hi, I'm happy to help with these, :) Question #1: This shows a point P with coordinates (sqrt(3), 1), and asks: "In the rectangular coordinate system, segment OP is rotated counterclockwise through an angle of 90 degrees to position OQ (not shown). the x-coordinate of Q | -1 Here, we need to remember our 30-60-90 triangles. Call K the point directly below P on the x-axis. We know that OK = sqrt(3) and that PK = 1. This means that angle POK = 30 degrees. Now, rotate that 90 degrees ccw. That will result in a segment OQ that is 30+90=120 degrees ccw from the positive axis. That means the angle between OQ and the y-axis is 30 degrees. If we call W the point on the y-axis at the same height as Q, we notice that triangle QOW is a congruent rotation* of triangle POK. That means QW = PK = a length of 1, and if Q is a length of 1 to the left of the y-axis, it has a y-coordinate of -1. *yes, yes, for the all the geometry nerds out there, I realize that "congruent rotation" is redundant --- a rotation is an isometry, so by definition, its image is always congruent to its preimage. I was being redundant for pedagogical clarity. Question #2 This shows what I presume must be a circle. RS is the diameter of the circle. Point T is on the circle, so angle RTS is an inscribed angle. Point X is outside of the circle, so angle RXS is the angle between two tangent lines. There are two ways to do this (a) the rough-and-ready way, and (b) the proper geometric way. (a) the rough-and-ready way X is clearly further from segment RS than T is. Suppose you started with the third vertex at a triangle at T, and dragged it back to X. As you drag the third vertex from a triangle further away from the opposite side, the angle has to get smaller. This is precisely the reason why objects "look smaller" (i.e. occupy a smaller visual angle) when you are further from them. Hence, RXS must be smaller. (b) the proper geometric way Of course, in the original diagram, we didn't know the measure of arc BC. But we do know So, Does all this make sense? Please let me know if have any questions on this. Mike :)

Hi, there. I'm happy to help with this. :) Given 2a + 2b = 6c, 2a -2b = 2c and b – c = 0. How many unique solutions are there for b? (1) 2a + 2b = 6c --> a + b = 3c (2) 2a -2b = 2c ---> a - b = c (3) b – c = 0 ---> b = c Plug the b=c equation from (3) back into (2) (4) a - b = c ----> a - b = b -----> a = 2b At this point, we can express every variable in terms of b, so plug these expression into (1) to get a single equation we can solve for b: (5) a + b = 3c ----> 2b + b = 3b ---> 3b = 3b ---> b = b Unfortunately, we don't get an equation we can solve. We get a degenerate equation. What b's satisfy b = b? All real numbers. The variable b can take on a continuous infinity of real number solutions. So, that's how many solutions b has --- infinity. Does that make sense? Please let me know if you have any questions on this. Mike :)

Hi, there. I'm happy to help. :) The question: Charlotte Salomon's biography is a reminder that the currents of private life, however diverted, dislodged, or twisted by____________ public events, retain their hold on the _________ recording them. a)transitory..culture b)dramatic..majority c)overpowering..individual d)conventional..audience e)relentless..institution It's easy to imagine several possibilities for the first blank. This is an interesting Sentence Completion question, insofar as I believe the second blank really is quite decisive. Consider this shortened version . . . . The currents of private life retain their hold on the _________ recording them. Well, the currents of private life --- the emotionally impactful things that happen to you or me in our most private and intimate spaces in life --- say, a troubling dream about a lover or coworker, something like that ----- If I have that troubling dream, that will have a hold on me, and maybe even on the one or two people with whom I share it, but it's not going to have a hold on a (A) culture, a (B) majority, a (D) audience, or an (E) institution. It's just the nature of the emotionally powerful things that happen in our private lives --- we don't go running to the newspapers to broadcast that stuff. It simply doesn't make sense for the second blank in the question to be anything involving a large number of people --- people don't share their private lives with large numbers of people. Therefore, the only word that works in the second blank is © "individual". Then, the completed sentence is: Charlotte Salomon's biography is a reminder that the currents of private life, however diverted, dislodged, or twisted by overpowering public events, retain their hold on the individual recording them. The word "overpowering" to describe the "public events" is a contrast to the powerful currents of private life. This sentence works well. The answer is ©. Does that make sense? Please let me know if you have any questions. Mike :)

With all due respect to smuf, I would say that this answer is not correct. It's a tricky question. If A and B both attempt a solution, then if either one of them comes up with the solution, the problem will be solved. In order for the problem to be solved, we don't need both of them to come up with solutions, only one of them. So this is fundamentally an "OR" question, not an "AND" question. So, P(A or B) = P(A) + P(A) - P(A and B) The first two terms, in green, are given in the problem. They're fine. The third term is tricky. If we know that A solving the problem and B solving the problem are independent, then we can say that P(A and B) = P(A).P(B). In the actual real world, with real world problems, that's a highly dubious assumption, because if the problem is easier, it's likely both A and B will solve it, and if the problem is more intractable, it's likely that both A and B won't solve it Any factor that makes A & B likely to succeed or fail together necessarily implies that they are not truly independent. It would be a very strange scenario of problems such that whether A could solve it and whether B could solve it would be truly independent. Nevertheless, we have absolutely no way to get a numerical answer to the question without making the assumption of independence. If this were a QC, the argument could be made that we simply do not have enough information, because nothing about independence is specified, and the answer would be D. If this were a PS question, then the real GRE would be very good about specifying, quite explicitly, that we could assume independence. So, for the sake of argument, let's assume independence. Then: P(A or B) = P(A) + P(A) - P(A and B) = P(A) + P(A) - P(A)*P(B) = (1/4) + (1/3) - (1/4)*(1/3) = 3/12 + 4/12 - 1/12 = 6/12 = 1/2 Does this make sense? Please let me know if anyone has any questions on what I've said. Mike :)

So, if I understand the nature of the question, it is: Given square PQRS, can you have a rectangle completely enclosed in the square, that has a greater perimeter than the square? To cut to the chase, the answer is: no, the square always has a larger perimeter. To see the argument why, see the attached pdf. Does this make sense? Please let me know if you have any more questions. Mike :) [ATTACH]6379[/ATTACH]

OK, some cool questions here. :) 1) On a number line , the distance b/w the 2 points with co-ordinates -5 and 1 is how much less than the distance b/w the 2 points with coordinates 2 and 14 ? (A) 6 (B) 8 © 10 (D) 12 (E) 16 Distance between x = -5 and x = 1 is d = 6. Distance between x = 2 and x = 14 is d = 12 !2 is 6 more than 6. Answer = A. 2) [diagram of a right-ish triangle, with legs of 5 & 11, and hypotenuse of 13 ] x | 90 No x appears in the diagram, but I assume that the right-ish angle is labeled x. Think about it this way. If this were a right triangle, the Pythagorean Theorem would apply. Let's say we have a right triangle with legs 11 and 5. Then the hypotenuse h is given by: h2 = 112 = 52 = 121 + 25 = 146 So, h is an irrational number, the square root of 146. Now, instead of that h, we remove that hypotenuse, and stick in a side of length 13, as we have in the diagram. The side of 13 (which equals the square root of 169) is bigger than the square root of 146, so we have to take our erstwhile right triangle and crank the angle wider, separate the legs more, to be able to fit a side of 13 where the hypotenuse was. That means, we had to make x is larger than 90 to fit in the side of 13, which is larger than the hypotenuse would be if x were equal to 90. Thus, x is larger than 90. Answer = A. 3) [unclear diagram] Perimeter of square pqrs | perimeter of the shaded rectangular region I'm a bit perplexed by this. In the diagram I see square pqrs, but I don't see a shaded region. I see what appears to be a smaller square/rectangle inside the square pqrs, and to all appearances, it has a smaller perimeter than square pqrs. What's not at all clear to me are the geometric constraints of the problem. Do we know that the inner square/rectangle is at a 45 degree angle? Do we know that exactly two vertices of the square/rectangle lie on sides of square pqrs? I suspect there is more information given in the original question that is missing here. Let me know what the original question is. I'm curious about this, and I would like to opportunity to help you. 4) 3x3xn = 2x2xp np isn't equal to 0 n/p | 2/3 3x3xn = 9n and 2x2xp = 4p. 9n = 4p Divide by p ----> 9n/p = 4 Divide by 9 ---->n/p = 4/9 Now, the question: can we reduce 4/9 to a simpler form? NO! This is a HUGE and very common fraction mistake. If both the numerator and the denominator are squared, we can't "cancel" the squares. If both the numerator and the denominator are raised to any power, we can't "cancel" the powers. We have to leave 4/9 as 4/9 --- it's already in lowest terms. Now, the question, how does 4/9 compare in size to 2/3? Well, 2/3 = 6/9, and 6/9 is clearly bigger than 4/9. Answer = B. Does all this make sense? Mike :)

Hi, there. I'm happy to help with this. :) When it was $9/share, we bought $4500, or 4500/9 = 500 shares When it was $10/share, we bought $3000, or 3000/10 = 300 shares So, we now own 500 + 300 = 800 shares, and we have spent a total of $7500 on them. So, our average is 7500/800 = $9.375/share, which is answer D. Does that make sense? Please let me know if you have any questions. Mike :)

Hi, there. I'm happy to help with this. :) What is the source of this problem? It seems to me a tad more fiendish :evil: than what one would actually see on the GRE. So, first of all, five digits, none repeated, so there are 5! = 120 possible numbers. (Are you cool with permutations and factorials?) In each place (the ones place, the tens place, etc.), each one of digits will appear the same number of times --- 24 times each in the 120 numbers on the list. Thus, if we scan down the ones place on the list, we will have twenty-four 1's, twenty-four 3's, twenty-four 5's, twenty-four 7's, and twenty-four 9's. That's 24*(1 + 3 + 5 + 7 + 9) = 24*25 = 600 The sum of digits in the ones place is 600 The sum of digits in the tens place is 6000 The sum of digits in the hundreds place is 60,000 The sum of digits in the thousands place is 600,000 The sum of digits in the ten-thousands place is 6,000,000 Add those, for a sum of 6,666,600. Answer = Does that make sense? Let me know if you have any questions at all about what I've said here. Mike :)

Hi, there. I'm happy to help with these. :) The toll for a certain bridge is $0.15 or 1 token. Tokens are sold in pack of 40 for $4.00. % saved on 40 trips across the bridge if a token , rather than $0.15, is used to pay each toll | 66 2/3 % Cost of 40 trips with tokens = $4. Cost of 40 trips with coinage = $6 $4 is 33 1/3 % less than $6. (Incidentally, $6 is 50% more than $4, but that's not relevant to the problem). Therefore, % saved is less than 66 2/3 %. Answer = 2) (a+b)^2 = 49 ab = 12 a + b | 7 This one is tricky tricky tricky. (a + b)2 = 49, so a + b = 7, right? WRONG! a + b = +7 or -7. Two values are possible. (3 + 4)2 = (-3 + -4)2 =7 Now, we look at the second one. The product of a and b is positive, which could mean (i) both positive, or (ii) both negative. (3)(4) = (-3)(-4) = 12 Neither statement allows us to distinguish between positive and negative. Therefore, we do not have sufficient information to determine whether a + b is equal to 7, or equal to -7 and thus less than 7. Answer = Does all that make sense? Please let me know if you have any questions. Mike :)

Dear Smuf You asked: "In the second question , 4x5 parallel intersecting lines result in 12 parallelograms. But these small parallelograms combine to form larger ones, How could we determine the total # of these? Explanation by dash method would help." For the sake of argument, let's say we have 4 vertical lines and 5 horizontal lines. _ _ _ |_|_|_| |_|_|_| |_|_|_| |_|_|_| Now, pick any parallelogram, big or small. Say, we pick . . . _ _ _ |_|_|_| |_|_|_| |_|_|_| |_|_|_| That parallelogram is formed by one pair of vertical lines and one pair of horizontal lines. In fact, any parallelogram we could pick would be formed uniquely by one pair of vertical lines and one pair of horizontal lines. Therefore, if we just counted how many pairs of vertical lines we could have, and how many pairs of horizontal lines we could have, that would be in every way equivalent to counting the number of parallelograms. # of possible pairs of vertical lines = combination of 2 from 4 = 4C2 = 6 # of possible pairs of horizontal lines = combination of 2 from 5 = 5C2 = 10 Let (VL1) equal the first vertical line, (VL2) equal the second vertical line, etc. The six possible pairs of vertical lines are (VL1)(VL2) (VL1)(VL3) (VL1)(VL4) (VL2)(VL3) (VL2)(VL4) (VL3)(VL4) Let (HL1) equal the top horizontal line, (HL2) equal the next horizontal line down, etc. The ten possible pairs of horizontal lines are (HL1)(HL2) (HL1)(HL3) (HL1)(HL4) (HL1)(HL5) (HL2)(HL3) (HL2)(HL4) (HL2)(HL5) (HL3)(HL4) (HL3)(HL5) (HL4)(HL5) Every possible parallelogram is a combination of one pair of vertical lines and one pair of horizontal lines. For example, the red parallelogram "selected" in the diagram above could be expressed as (VL2)(VL4)(HL2)(HL5). Since 6 vertical pairs each can be matched with each of 10 horizontal pairs, the total number of combinations (by the FCP) is 6*10 = 60. Does all that make sense? Mike :)

Dear smuf: My explanation in the immediate preceding post in this thread should make clear where the 15 came from. As for the 6C4, that's pemdas' contribution, and to be honest, I am not completely clear on how 6C4 applies to the problem. It's true that 6C4 = 15, but beyond that, I don't see the connection. Mike :)

Dear Praveen This is a response to your question: "In the first question, I still don't get it how you arrive at the 15 arrangements without listing! listing seems to be quite a difficult task on this question. I would be happy if you could elaborate." Think about it this way. You have four women, who need to be separated. _____F_______F_______F_______F_______ Of the five guys, you need to put three of them as female-buffers, in the three places between the females . . . . _____FM_______FM_______FM_______F_______ Now, no matter where the other two men sit, we are guaranteed that the women will be separated, which is what we need. As for where the other two men sit, we have options. __1__FM___2___FM___3___FM___4___F___5___ That's five slots where we could put the two men. We have two options: a) We could stick both extra men in the same slot --- there are five slots where we can put them. ==> 5 possibilities. b) We could put the two extra men in two different slots ---- so, we would choose a combination of two slots from the five --- 5C2 = 10 possibilities. Add those --- 5 + 10 = 15 possible arrangements of the M's & F's. Does that make sense? Mike :)

Hi, there. I'm happy to help with these. :) 1) The sociologist responded to the charge that her new theory was __________ by pointing out that it did not in fact contradict accepted sociological principles. a) banal b) heretical c) unproven d) complex e) superficial The sociologist needed to defend herself by saying her theory didn't contradict accepted principles, so the charge must have been that her theory did contradict accepted principles. The only word that that means anything like "contradicting accepted principles" is (b) heretical. That's a very strong work which connotes attacking & contradicting what was most precious in the previous accepted order. 2) People should not be praised for their virtue if they lack the energy to be_________; in such cases, goodness is merely the effect of __________. a) depraved..hesitation b) cruel..effortlessness c) wicked..indolence d) unjust..boredom e) iniquitous..impiety The first blank could be any word that means "bad" --- really, any of the five answer choices would work for the first word. The key to this is the second blank. These folks are virtuous because they lack the energy to be bad people. What seems like virtue is just a by-product of lack of energy, of inertia, of not wanting to get off their lazy butts to do anything at all. Lack of energy, inertia, laziness -- the word with all those connotations is © indolence. Choice (a) "hesitation" implies the lack of action is due to an intellectual quandary rather than a lack of energy/effort. Choice (b) "effortlessness" is actually a word we use for folks who are so skilled at their action that it looks like the results occur as if by magic; for example, Yo-Yo Ma plays the cello with effortlessness. It is definitely not a word that means anything like laziness. Choice (d) "boredom" often goes along with lack of energy & laziness, but technically is not the same as laziness; one can easily be bored and restless -- that is, have a lot of energy and nowhere to direct it. That's not the case with the folks here. Choice (e) is right out --- it's almost impossible to imagine how goodness could result from "impiety." Of these answers, © wicked . . . indolence is by far the best. 3) No longer________ by the belief that the world around us was expressly designed for humanity, many people try to find intellectual_______ for that lost certainty in astrology and in mysticism. a) satisfied..reasons b) sustained..substitutes c) reassured..justifications d) hampered..equivalents e) restricted..parallels The first blank has to be a happy word --- folks had the happiness of a happy belief, and now that we've lost that happiness, we need the comfort of astrology and mysticism. Choice (a) & (b) & © all work for the first blank --- satisfied & sustained & reassured -- are all happy things a belief can do to one. The last two, "hampered" and "restricted", are negative words that don't fit the context. The belief was a happy thing that ended. The second blank is tricky --- "intellectual reasons", "intellectual substitutes", and "intellectual justifications" are all phrase that ring familiar --- none of those are wrong idiomatically. What's crucial is what follows the blank --- "for that lost certainty". If the belief was a happy thing, then losing the certainty was a big yucky unhappiness. We don't want reasons for the lost certainty, nor do we want justifications for the lost certainty --- providing an intellectual framework for it is in no way going to take away the fundamental unhappiness here, and providing this framework is not what "mysticism and astrology" do anyway. The only thing that will take away the unhappiness would be to plug something else into that existential slot --- we want "intellectual substitutes" for that lost certainty, and the passage is suggesting that "mysticism and astrology" are what humans use to plug that hole. We lost the certainty, and where the uncertainty was, we plug in astrology and mysticism -- that is to say, we use them as substitutes. That's why (b) sustained . . . substitutes, is the best answer,. Does all that make sense? Please let me know if you have any questions on this. Mike :)