I think for the last question we can equate them first to test if we can find the root that can makes them equal. So in this case 1/(x+1) = x/2 has 2 roots -2 and 1. Because -2 is still
@@khalilshaik6161So you used only his videos to understand the concepts of something else as well? And which resources for practicing? Please mention man, I could really use some help
@16:26, you were give that a and b were two digit integers. And yes, C and D were contained in the interval of the max and min product. My question is: Why didn’t you check that the factors were between the constraints for a and b?
Thank you so much! We hired Adam Sandler's brother to do these videos. He's funny, but not nearly as funny as he thinks he is. ;) Have fun studying, Amanda!
Hi, GRE Ninja. Firstly, Big fan of your content. Keep up the honest work. But Correct me if I am wrong... Regarding Question #2: Although calculating range of values helps us eliminate 3 options, that itself isn't sufficient to ascertain that all the values within range are possible. This is because a & b are integers. We have to factorize the remaining two options to make sure they comply to the limitations of both a & b. For example, there are primes within the range which don't comply with the restrictions of a>60 & b
Thank you so much for the kind words, and apologies for my belated response! I think you're onto something here. We agree on the range as a starting point, I think: trueab must be less than 8000 (because b < 80, and a is a two-digit integer that must be less than 100), and ab must be greater than 600 (since a > 60 and b must be at least 10). That leaves us with just answers (C) and (D). In theory, it's not correct -- or at least not "safe" -- to stop there, for exactly the reasons you mentioned. A number could be in the correct range (between 600 and 8000), but still be impossible. Primes are one example, but there are plenty of others, since there's a limited number of possible integer values for a and b. So in theory, you'd want to verify that answers (C) and (D) are actually possible before moving on. It turns out that both ARE possible. So you're all good here, given the way this particular question is written. For simplicity, I didn't explicitly mention verifying those answers -- but you're correct that it would be wise to do so. :) Thank you for the thoughtful comment, and have fun studying!
Hi, quick follow up on the Arithmetic Mean problem. Could m or n be negative, that could be one of the edge cases right, they have not mentioned any constraints. In that case, answer would b D - Relationship cannot be determined from given information.
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!
For the second wording of q1, is there any time in a problem like that that you shouldn't just pick the two values that fall in the range? I see that you didn't do any double checking that 4800 and 5600 can be decomposed into products of two digit integers within the given ranges for a and b (though its obvious that they can), and that would be a helpful assumption to be able to make for that kind of question
For the average question you don't need to do all that algebra, right? The means are equal which means sum A/5 = sum B/4. Because sum B is only divided by 4 and 4A
We can say that because the sum of the items in list A divided by 5 is equal to the sum of the items in list B divided by 4, the sum of the items in list A is greater than the sum of the items in list B. The sum of the items in list A is 9 + 5m + n and the sum of the items in list B is 9 + 4n, so we can say 9 + 5m + n > 9 + 4n, which simplifies to m + n > 0. From this, we can't say whether n is greater than 4. It's possible that m is -5, so n needs to be greater than 4 to make m + n > 0 work. However, it's also possible that m could be 5, so n could be less than 4 and we still satisfy m + n > 0. All this shows that the algebra that Charles used in the video explanation is, sadly, necessary to reach the answer to this question. I hope that helps!
Thank you for asking! Most of us have taught SAT (and ACT) prep at some point or another, but at this stage, we don't have any plans to produce SAT videos. Maybe a in a few years, depending on how often we're asked for them. :) Thank you for watching, and have fun studying!
Charles didn't assume that every number from 610 to 7821 is the product of one integer greater than 60 and another less than 80. This question asked us which of the given answer choices could be the product of the two integers described in the question. We can rule out (A), (B), (E), and (F) by knowing that the product must lie between 610 and 7821. Now, we only need to worry about 4,800 and 5,600. We can check these numbers if we want so we don't have to assume they're a product of the integers described in the question. Since 4,800 = 80 * 60 and 5,600 = 80 * 70, these two can both be possible values of ab. I hope that helps!
Yes, definitely! The "new" GRE isn't all that different from the old one -- it's just shorter, but the quant content is completely identical. So all of our quant videos apply equally to the old and new versions of the exam. I hope that helps a bit, and have fun studying!
Oh wow, I see that you had the same experience with two of our videos. I'm answering on this particular video ("Episode 0") because it makes me wonder if you're struggling with carelessness more than anything else -- maybe you get complacent on the easier ones, but do a much better job of focusing when the questions start to feel difficult? If that's the case... well, the processes I babble about in this video might be more important for you than for many test-takers. Adaptive tests are annoying in that sense: if you slip up on questions that aren't actually hard for you, the test will punish you disproportionately. I might be way off here -- it's hard to diagnose test-takers using just a couple of brief comments. But if I'm barking up the right tree, hopefully this gives you some good food for thought. Have fun studying, and thank you for watching!
Haha! Nah, ETS isn't out to get you. The questions in these quant videos are our own, anyway -- they just test similar concepts in similar ways as the real thing. If you see identical questions on the actual exam, it's purely coincidence. :) Have fun studying!
you are my knight in shining headphones
Your videos are gem
Thank you so much, Garv! Have fun studying.
I think for the last question we can equate them first to test if we can find the root that can makes them equal.
So in this case 1/(x+1) = x/2 has 2 roots -2 and 1. Because -2 is still
Hi to everyone at GRE Ninja Tutoring.
I scored 170 on quant!
your videos helped me to build a routine around quant practice.
Thank you.
Congratulations!! Did you rely solely on these videos? I wanted to brush upon my basics as well as practice advanced questions
wow how much time you took to prepare? and what was the resources and books you used?
@@saiteja-bi9yq it was around 1.5 months .. but math is my fav subject ..
@@keke-u6i I used several resources to practice ...
@@khalilshaik6161So you used only his videos to understand the concepts of something else as well? And which resources for practicing? Please mention man, I could really use some help
i was really confused about the math section and oh my god i have Just watched this video and already a fan❤
Thank you so much! ❤️❤️❤️
Really Good Tutorial. I enjoyed it
Thank you, Joseph!
@@GRENinjaTutoring sir please tell the answer of 2th question its option C or D
@@LioPaul-qp7ks The question says Select "All" which is correct/applicable. And for this question, there are two answers.
@16:26, you were give that a and b were two digit integers. And yes, C and D were contained in the interval of the max and min product. My question is: Why didn’t you check that the factors were between the constraints for a and b?
I'm wondering this too. For example, if the numbers were prime then obviously they are not correct.
This guy is hilarious. Thanks for your videos.
Thank you so much! We hired Adam Sandler's brother to do these videos. He's funny, but not nearly as funny as he thinks he is. ;)
Have fun studying, Amanda!
Thank you for this!!
Thank you so much for watching!
so helpful thank you!
Hi, GRE Ninja. Firstly, Big fan of your content. Keep up the honest work.
But Correct me if I am wrong...
Regarding Question #2: Although calculating range of values helps us eliminate 3 options, that itself isn't sufficient to ascertain that all the values within range are possible. This is because a & b are integers. We have to factorize the remaining two options to make sure they comply to the limitations of both a & b. For example, there are primes within the range which don't comply with the restrictions of a>60 & b
Thank you so much for the kind words, and apologies for my belated response!
I think you're onto something here. We agree on the range as a starting point, I think: trueab must be less than 8000 (because b < 80, and a is a two-digit integer that must be less than 100), and ab must be greater than 600 (since a > 60 and b must be at least 10). That leaves us with just answers (C) and (D).
In theory, it's not correct -- or at least not "safe" -- to stop there, for exactly the reasons you mentioned. A number could be in the correct range (between 600 and 8000), but still be impossible. Primes are one example, but there are plenty of others, since there's a limited number of possible integer values for a and b. So in theory, you'd want to verify that answers (C) and (D) are actually possible before moving on.
It turns out that both ARE possible. So you're all good here, given the way this particular question is written. For simplicity, I didn't explicitly mention verifying those answers -- but you're correct that it would be wise to do so. :)
Thank you for the thoughtful comment, and have fun studying!
Hi, quick follow up on the Arithmetic Mean problem. Could m or n be negative, that could be one of the edge cases right, they have not mentioned any constraints. In that case, answer would b D - Relationship cannot be determined from given information.
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!
4 out 4 - But I guessed and moved on the one w/ the average. Lol.
Haha! Hey, guessing is a skill. I'll take it. ;)
Have fun studying, Joe! And please keep us posted on your progress.
- Charles
For the second wording of q1, is there any time in a problem like that that you shouldn't just pick the two values that fall in the range? I see that you didn't do any double checking that 4800 and 5600 can be decomposed into products of two digit integers within the given ranges for a and b (though its obvious that they can), and that would be a helpful assumption to be able to make for that kind of question
For the average question you don't need to do all that algebra, right? The means are equal which means sum A/5 = sum B/4. Because sum B is only divided by 4 and 4A
We can say that because the sum of the items in list A divided by 5 is equal to the sum of the items in list B divided by 4, the sum of the items in list A is greater than the sum of the items in list B.
The sum of the items in list A is 9 + 5m + n and the sum of the items in list B is 9 + 4n, so we can say 9 + 5m + n > 9 + 4n, which simplifies to m + n > 0.
From this, we can't say whether n is greater than 4. It's possible that m is -5, so n needs to be greater than 4 to make m + n > 0 work. However, it's also possible that m could be 5, so n could be less than 4 and we still satisfy m + n > 0.
All this shows that the algebra that Charles used in the video explanation is, sadly, necessary to reach the answer to this question.
I hope that helps!
Thanks ❤
Thank you so much for watching and for being so positive!
Hlo sir, I completed vocab words but I am finding hard in solving vocab questions. So pls do suggest a way to crack
Thank you... Please do you have videos for SAT prep too
Thank you for asking! Most of us have taught SAT (and ACT) prep at some point or another, but at this stage, we don't have any plans to produce SAT videos. Maybe a in a few years, depending on how often we're asked for them. :)
Thank you for watching, and have fun studying!
Now i can start my prep i guess.thank you
Re problem 2: Can we really assume, the way you have, that every number >= 610 (61*10) and
Charles didn't assume that every number from 610 to 7821 is the product of one integer greater than 60 and another less than 80. This question asked us which of the given answer choices could be the product of the two integers described in the question.
We can rule out (A), (B), (E), and (F) by knowing that the product must lie between 610 and 7821. Now, we only need to worry about 4,800 and 5,600. We can check these numbers if we want so we don't have to assume they're a product of the integers described in the question. Since 4,800 = 80 * 60 and 5,600 = 80 * 70, these two can both be possible values of ab.
I hope that helps!
Great stuff! Are the contents of these videos applicable to the new GRE equally?
Yes, definitely! The "new" GRE isn't all that different from the old one -- it's just shorter, but the quant content is completely identical. So all of our quant videos apply equally to the old and new versions of the exam.
I hope that helps a bit, and have fun studying!
Then there's me....
I got the last one correct and the easy ones incorrect....
Oh wow, I see that you had the same experience with two of our videos. I'm answering on this particular video ("Episode 0") because it makes me wonder if you're struggling with carelessness more than anything else -- maybe you get complacent on the easier ones, but do a much better job of focusing when the questions start to feel difficult?
If that's the case... well, the processes I babble about in this video might be more important for you than for many test-takers. Adaptive tests are annoying in that sense: if you slip up on questions that aren't actually hard for you, the test will punish you disproportionately.
I might be way off here -- it's hard to diagnose test-takers using just a couple of brief comments. But if I'm barking up the right tree, hopefully this gives you some good food for thought.
Have fun studying, and thank you for watching!
A COMPLETE COURSE IN QUANT , I WONDER IF ETS SAW THIS AND CHANGE THE QUESTION DISSCUSSED IN THIS SERIES COZ ETS IS MALIGN
Haha! Nah, ETS isn't out to get you. The questions in these quant videos are our own, anyway -- they just test similar concepts in similar ways as the real thing. If you see identical questions on the actual exam, it's purely coincidence. :)
Have fun studying!