If we multiply c/2 by 3/3, we'll get (c*3)/(2*3) = 3c/6. I hope that helps!

@@GRENinjaTutoring yes, it does. thank you!

That's another great way to solve this question, thank you for sharing! If you're comfortable with when to substitute 3p for X, then you're absolutely right: factoring X² - X - 42 is much nicer than factoring 3p² - p - 14. Thank you again for commenting!

Question 6 in this video is a Quantitative Comparison question. In one of these questions, answer choice (D) means we cannot tell which of the two quantities is greater. In this case, Alex showed that Quantity A is greater than Quantity B for some values of x and Quantity B is greater than Quantity A for other values of x. This means we cannot say for certain which quantity is greater for all values of x, so the answer to this question is (D). For more on QC questions, check out the first video in this series: kzbin.info/www/bejne/gpDZeYKgedObbKssi=JzDZ3cRvJGLQfd2B I hope that helps!

It might be easier to think about this question in terms of the units of each quantity. At the start, the company has enough fuel to run 20 trucks for 36 days, so instead of dividing 36 by 20, we can multiply these numbers to get 720 truck-days of fuel. Here, we're saying that the total amount of fuel the company has in its stock is equal to the number of trucks multiplied by the number of days the company can keep that truck running. To represent this figure, we can use the made-up unit of a truck-day, where one truck-day represents the amount of fuel one truck uses per day. This means we can divide the number of truck-days of fuel we have available by the number of trucks we want to run. If we think of the units as variables, we'd be doing truck-days / trucks and we could cancel the trucks on the top and bottom of this fraction to leave us with the number of days. Doing this with the numbers involved in the question means we'd do 720 truck-days / 16 trucks to get 45 days of fuel available. I hope this helps!

That's not wrong at all. At about 37:05, Dana showed that 3x/5 = y (using your variables of x for the price in Jan and y for the price in Feb.) If we make x the subject of this equation, we get x = 5y/3. If we convert the 5/3 to a decimal, it should read x = 1.66666....y, so x = 1.6y is a very close approximation of the actual value. I hope that helps!

The question stem tells us that x is the value we get when we add one randomly chosen element from set S with one randomly chosen element of set T. For example, if we chose 2 from set S and 3 from set T, then x would take the value of 5. Quantity A is "the probability that x is greater than 9." From the work Dana did in the video, we can show that this probability equals 9/20. I hope that helps!

Thank you! I'm glad that you're not completely tired of watching us yet. :) Is there anything in particular that you'd like to see in our verbal videos? We'll likely film a video on harder and stranger vocab-based questions at some point. Is there anything else that jumps to mind? Thank you for watching, and have fun studying!

harder and stranger vocab-based questions Would be great ..lot of thanks

Thank you so much!

Thank you so much for the kind words, and have fun studying!

If p takes the value of -1, then Quantity B is -1. We also know that x > 14p = -14 and y < 9p = -9, so let's assume that x = -10 and y = -10. In that case, Quantity A is (x - y) / 5 = (-10 - -10) / 5 = 0/5 = 0 which is greater than Quantity B. We can repeat this for any combination of x and y values we like, and we'll still find that Quantity A is greater than Quantity B. The first explanation provided in the video, the one for which Harry uses the blue pen, doesn't assume that p is a positive number. This way of looking at the question works whether p is positive or negative, so Quantity A will always be greater than Quantity B, no matter which value of p we choose. This means the answer to this question is (A). I hope that helps!

Yes, that's exactly what it means. There were F friends who were all going on this vacation together, but G of them dropped out of the trip. This means that (F - G) people are left to go on the trip and have to pay the full amount of the rental. I hope that helps!

Thank you so much for watching, and thank you for the kind words!

Thank you! I'm glad that you enjoyed it. Have fun studying!

Thank you so much for watching!

I'm sorry if I'm missing something really obvious, but I don't see a 45-45-90 triangle in the last question that would help us reach the answer. Which diagonal that has length 12 are you referring to? If you let me know a bit more about the path you're using to find the solution, I might be able to answer your question. Thank you!

​@@GRENinjaTutoring Actually the side length of the big square is 12 (2+10) when we cut the small square from the diagonal.It has the same length that has a big Square side so the diagonal length is 12. In short (side length of the big square =diagonal length of short square)

@@kashafbajwa9286 In the video, we found the side length of the small square was sqrt(104). We can use a 45-45-90 triangle to find the length of the diagonal of the short square as sqrt(2)*sqrt(104) or sqrt(208). This is not equal to 12, so we have shown that the side length of the big square is not equal to the diagonal of the short square. I'm sorry that doesn't work out, but thank you for posting!

As Charles showed in the video, we can simplify Quantity A to x/3 + 2/3 and Quantity B to x/3 + 4/9. This means that we can take whatever x value we like, positive or negative, and put it into these expressions. The x/3 is in both Quantity A and Quantity B, so that will be the same across both quantities no matter which value of x we choose. We then add 2/3 in Quantity A and 4/9 in Quantity B. Since we're adding a greater number in Quantity A, Quantity A will be greater than Quantity B. This means it doesn't matter what value x takes, Quantity A will always be greater than Quantity B. I hope that helps!

This is the sort of algebra that's awkward to explain in a KZbin comment, but I'll try! If I give you something like this, you'd be able to "expand" it, hopefully: (x + 2)(x + 3) = x^2 + 5x + 6. Different people might approach this slightly differently, but I'm basically multiplying the first term in each parentheses (x * x = x^2), then multiplying the outer terms (x * 3 = 3x), then the inner terms (x * 2 = 2x), then the last term (2 * 3 = 6). Add those up, and we get x^2 + 5x + 6. Same deal when we expand (x + y)^2. That's the same as (x + y)(x + y), and the process is the same as for the previous example: we get x^2 + xy + xy + y^2 = x^2+2xy+y^2. Apologies if that's still unclear -- algebra doesn't render beautifully in KZbin comments. Similar expressions appear in our algebra GRE videos, so maybe check this one out, in case it helps reinforce the concepts: kzbin.info/www/bejne/hInbq2Wrds50ec0. I hope that helps a bit, and have fun studying!

Thank you so much for the kind words, and have fun studying!

This is going to be very hard to discuss using text only, but I'll do what I can. The formula you're describing creates triangles by drawing diagonals across a hexagon. All these diagonals start from the same vertex. You can see examples of that through the link at the bottom of this message. If we draw as many diagonals as we can from one vertex of a hexagon, we split that hexagon into four triangles. In this example, Alex is not creating triangles by drawing diagonals from one vertex of the hexagon. He joins each vertex of the hexagon to its centre, creating six equilateral triangles. Each of these two processes is helpful, depending on what the question asks. In this question, splitting the hexagon up this way is the best way to find an answer. I hope that helps! Link: mathbitsnotebook.com/JuniorMath/Polygons/PYpolygonangles.html#:~:text=In%20a%20regular%20hexagon%2C%20four%20triangles%20can%20be%20created%20using%20diagonals%20of%20the%20hexagon%20from%20a%20common%20vertex.%20Since%20the%20interior%20angles%20of%20each%20triangle%20totals%20180%C2%BA%2C%20the%20hexagon%27s%20interior%20angles%20will%20total%204%E2%80%A2(180%C2%BA)%2C%20or%20720%C2%BA

I'm really sorry if I'm missing something obvious, but I can't see where Harry tries to do -16-(-36) in Q7. There is a point where he does -96-(-36) = -60, is that the part of the question you're asking about? Please let me know and I'll do what I can to help. Thanks!

​@@GRENinjaTutoringoh my bad ! I'm really sorry. You're right. Thank you for the response.

No worries at all. Thank you for commenting, I hope you're enjoying the videos, and good luck with your studies!

You get to use an on-screen calculator on the GRE quant section. Check out the link below for more details on what the calculator can do. I hope that helps! www.ets.org/pdfs/gre/on-screen-calculator-guidelines.pdf

Thank you for watching! ❤️

Yes! You've followed the method Harry described in blue at the start of the explanation. It's a perfectly valid way of doing this question and if you're comfortable with finding the values of the choose functions, it won't take too long. It's just not quite as efficient as the method shown in the explanation. If your method was the first one that came to mind and you're comfortable with solving it that way, you'll do just fine in this question. Thank you for adding an alternative solution path! I hope that helps!

@ alright thx!

In this case, you're totally right that it would have been easier to just do 75*4 to get to 300. However, it won't be the case in every scenario. For example, if we were looking at 6 and 8 (instead of 75 and 4) then multiplying them together would give 48 but the LCM of 6 and 8 is 24. This means that we might think we're looking for a multiple of 48 when we're really looking for a multiple of 24. We'd think the list of numbers we're looking for was 48, 96, 144, 192... when the actual list we'd be looking for was 24, 48, 72, 96, 120, 144, 168, 192... and we'd miss a few potential correct answers. We can't assume that the LCM of any two numbers is given by multiplying them together. If we default to finding the LCM in the way shown in the video, it might take us a bit longer but it will probably mean we make fewer errors. I hope that helps!

To use 6C3 as the method to find the number of teams, we've looked at the 3 engineers, put one of them in the team, and put the other two in a group with the 4 sales personnel so we can do the 6C3 operation. By doing things this way, we've effectively said that the first engineer we chose and put in the team before doing 6C3 with the others MUST be in the team. We build the rest of the team around them and they MUST be in the team. However, there can be teams that do not involve this first engineer that we'd miss if we only did 6C3 to find the solution. To get to the right answer, we also need to find the number of teams that include at least one engineer but do not involve this first engineer. We could repeat the process by excluding the first engineer, choosing a second engineer that MUST be in the team, putting the final engineer with the sales personnel, and doing 5C3 to find the other members of the team. But there's still more to do: we need to repeat this process one more time to find the number of teams that include at least one engineer but do not include the first two. To do this, we'd say the final engineer MUST be in the team, exclude the first two, and choose 3 sales personnel from the available 4 to make up the rest of the team. There are 4C3 ways of doing this. This means we'd have a total of 6C3 + 5C3 + 4C3 = 20 + 10 + 4 = 34 possible teams, giving the same answer that Harry found in the video. I hope that helps!

The equation Charles set up in the first line of his solution is the same whether m and n are positive or negative, so we could still use the same method. You're absolutely right that there aren't any constraints on m and n being positive or negative, so it's absolutely possible that m could be a negative number. Thankfully, the terms with m cancel each other out and the value of m doesn't have any effect on the final answer. On the other hand, Charles found the value of n at the end of the solution. Again, the method wouldn't change if m or n were positive or negative, so since Charles found that n = 9/4 at the end of the solution, there's no way n could be negative. I hope that helps!

Between the solid black lines representing 10% and 20%, there are four lighter lines. This means each line represents 2%, so we can split the 10-20% interval into 10, 12, 14, 16, 18, and 20. For the second six month period, City A sees an increase in subway passengers that is halfway between the first and second light lines above the 10% mark. The first light line represents a 12% increase and the second light line represents a 14% increase. Since we're looking halfway between 12 and 14%, we can say the number of subway riders increased by 13%. I hope that helps!

I like this way of looking at things and thank you for commenting! The only question I'd ask is whether this works all the time. We know that quantity B is greater than quantity A when x is at its maximum, but how about when x is far from y? My hesitation in doing this question the way you suggest is I'm not sure I could say for certain what would happen when x is not at its maximum. Can we be sure that the answer is (B), or might it be (D) because we're not sure whether quantity B is always greater?

Our GRE video series is pretty darned close to comprehensive, even if you're shooting for a 170 on the quant section. In all honesty, when we created these videos, we worried that we might have included too much detail for most test-takers -- very few people have a compelling admissions-related reason to score above a 160 on the GRE quant, and in most of these videos, some of the questions are beyond what you'd need for that score. That's a long-winded way of saying that these videos cover nearly every significant quant-related principle that you'll see on the GRE. If you deeply understand everything in the series, you'll be in good shape for a super-elite score, as long as your execution (good time management, avoiding careless errors, etc.) is great. The full series is available here: kzbin.info/aero/PL0rijwfA1veSQZE5bbAZ3lqajipQcy_g1. I hope that helps a bit, and have fun studying!

Thank you for watching! ❤️❤️❤️

There's nothing in Q7 that says p, x, or y must be integers, but I'm not 100% sure what you mean when you say we cannot pick anything. The explanation works whether or not these numbers have to be integers. I hope that helps, but please let me know if you have any other questions!