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 :)