I felt bad for him when he didn't see in question 9 the constraint that p must be odd in evaluating option B. I felt bad cause I know that it happens to everyone :(
Amazing video Arnold Reed!! Thank you so much, it helped me to see how it will be the GMAT. I havent done any practice test yet but now I'm confident with the type o questions I need to study. It was incredible to see this video for the last couple of days. Thank you!
I love this video! This is more like vlogging! this prepares potential test takers to be prepared mentally to face questions and gather the courage to pick a random answer and move on (esp the hard ones or the ones that you dont know) Need more tet taking videos like this one!
Unfortunatelly at 37:36 you missed a short cut. If y+p = 150, the angle QPR must be 30. This is equal to statement 1, which you proved sufficient already, thus answer D would be correct.
Yep. The stress of the test sometimes keeps you from seeing things right in front of your face! Oh well--the nice thing is that you don't have to be perfect to get a great score.
I don't get question 7. If C was correct (as he picked), then the question would need to give the constraint that all numbers in that sequence are > 0. Otherwise, if negative numbers are allowed, the sum and the average don't determine the number of terms in the sequence. That's why I would have chosen E. Can someone please help me out? What am I missing? Thanks in advance guys!
Actually, having negative numbers in a set doesn't affect the average calculation. Take the numbers 2,3,4,5,6. The average is 4, right? The sum is 20, there are five numbers. Okay, so how about the set -4, 0, 4, 8, 12. The average is still 4. The sum is 20, there are five numbers. -25, -10, -5, -1, 61... the sum is 20, there are five numbers, the average is 4. If you have an average, and you have the sum, you know how many numbers are in the set. Some of them might be negative, but that's not relevant to this question, as you know how many there.
Because one car sold for a 4000 dollar gain (bought for 16K, sold for 20K) and one car sold for a 5000 dollar loss (bought for 25K, sold for 20K). A 4000 dollar gain and a 5000 dollar loss amounts to a 1000 dollar loss.
Because that is mixing up concrete vs. relative numbers. Saying the tree was 1/5 taller in year 6 than in year 5 is a ratio. The tree grew by 20% in those two years. But I can't say that means 'the tree grew by 10% each year,' because each percent growth is based on the *new height.* If it said the tree was 1/5 *an inch* taller year 6 compared to year 4, in *that* case we could say, "Oh, the tree is growing 1/10 an inch every year."
Anything above 2 wrong would give you a 50. To get a 51, you need a near flawless paper. I got 1 incorrect and got a 51 on the actual thing. But getting a 51 is definitely getting harder imo
@@reedarnold5468 Not possible on the actual test. In one of the official mock I got 2 wrong but still got 50. 51 in my opinion is very difficult nowadays. Also, it depends which questions you got wrong. If you got the experimental questions wrong then I think its still possible to score a 51 with 3-4 wrong but out of the marked questions if you get more than two wrong then you would not get 51 come what may.
Don't lose hope! Don't forget I *teach* this stuff. This is what an expert looks like taking the test. When I first took the GMAT, I didn't move nearly as fast as I do here (and I still got a 760.) What do you mean by 'mental math?' Do you mean just, like, arithmetic? You can practice arithmetic, and find new ways to do arithmetic. There are loads of tricks out there (come to one of our free "Foundations of GMAT Math" workshop. The first thing we do is talk about some fast mental tricks for arithmetic). If you mean quantitative reasoning--again, it can be practiced. It's often a skill of recognition--I see certain things in a question that immediately 'trigger' a certain line of reasoning or a certain set up (like the overlapping set tables). You can study and collect such 'triggers' as well, so you know more quickly what strategy will get you through a question.
@@reedarnold5468 thanks so much for the response - you are certainly an expert at this! I’ll check out the Manhattan workshop you mentioned. And yes I meant arithmetic. Like that car dealership question - you solved for the 125% of x = 20k before I even set it up. Funnily, I’m pretty confident about my quantitive reasoning and solved some of the data sufficiency questions or that ugly chemical reaction proportions question within the time you took to solve them. It’s the arithmetic that stumps me. Hopefully speed will come with practice! Thanks again
@@CanonNikonMan241 So stuff like that comes from lots of repetition and a change in how I think about percents. If you check out the "All about percents part 1" free prep hour, you'll get some insight into how I 'see' percents. I know that '125%' of some number is a 4:5 ratio. (since 5 is 125% of 4). In that problem, the '5' part of the ratio corresponded to the 20k. So I know I need to multiply the ratio by 4k to get the actual values. Basically, 4:5 = 16k : 20k. This is, again, not something I could do when I first took the test, but I do find it very convenient. The GMAT tends to use percentages that correspond to 'nice' ratios and ratios that correspond with 'nice' percents. E.g. 1:1, 1:2, 1:3, 2:3, 1:4, 3:4, 1:5, 2:5, 3:5, 4:5, 5:6, 9:10...
@@reedarnold5468 thats helpful, I’ll check the free prep hour out. Yes, already I’m getting more used to the GMAT percentages and it’s encouraging to hear that repetition will help
Yep! Made a silly mistake "1000 - 20 = 880." However, this is a good example of one of the advantages of Data Sufficiency... You don't actually have to *do the math right*. You just need to have the right set up and logical process, such that if you *did* do the math right, you'd get one single answer.
Yeah, it's definitely a very advanced divisibility question. I also made things unclear with the letters I chose. I should have written 3^n * 7 ^m, or something, since by re-writing 'k,' it looks like I'm referring to the 'k' in the question. During the test, I was aware enough to treat the 'k' exponent as a totally different value. In fact, I stopped thinking about the 'k' in the question itself, because when I was told it was a number with only prime factors 3 and 7, I just only thought about that number as a multiplication of 'some power of 3' and 'some power of 7.' So k = 3^n * 7^m For some integers 'n' and 'm.' Then the problem tells me that that number has 6 positive prime factors, including 1 and itself. There's a rule out there to determine how many factors a number has based on prime factors. It's a little complicated to write out, but basically consider that you can 'build' the factors of a number by choosing which 'prime factors' to include. For instance, 60 = 2^2 * 3 * 5 I can build a factor of 60 by taking some of these numbers. So I can build 15 out of 3*5 I can build 12 out of 2^2 * 3 I can build 20 out of 2^2 * 5 I can build 6 out of 2*3... etc. Notice, for each of these factors, it's like I'm 'choosing' a number of each prime factor to include. When I build '15,' I choose zero '2s,' the one 3, and the one 5. When I build '12' I choose both '2s,' the one 3, but not the 5. When I build '6,' I choose one of the '2's,' the '3', but not the 5. If I build '10' I choose one of the 2s, not the three, and the 5... Okay, so basically I have a 'choice' for every prime factor when building a factor. I can include zero, one, or two of the 2s, zero or one 3, and zero or one 5. So that's THREE choices for 2, TWO choices for 3, and TWO choices for 5, or THREE*TWO*TWO total choices. That means I have 12 overall options to 'build,' which represent all 12 factors fo 60 (count'em! 1,2,3,4,5,6,10,12,15,20,30,60) 1 is choosing 'ZERO' for each prime: 2^0 * 3^0 * 5^0 2 is choosing ONE 2 and ZERO 3 and 5: 2^1 * 3^0 * 5^0... 20 is choosing TWO 2s, ZERO 3s, and ONE 5: 2^2 * 3^0 * 5^1 Okay, so that can be generalized. When we break a number down to its prime factorization, if we ADD 1 to each exponent and MULTIPLY that, we figure out HOW MANY FACTORS THAT NUMBER HAS. So back to the number 3^n * 7^m... If that number has SIX factors, then I know that (n+1)(m+1) = 6. Now there aren' that many numbers that multiply to 6! (n+1)(m+1) 1 6 2 3 3 2 6 1 This would mean the possibilities for n and m are: n m 0 5 1 2 2 1 5 0 So notice I can't actually have the 0,5 or the 5,0 options, because that would actually would mean the number (3^n * 7^m) does NOT have a prime factor of either 3 or 7, respectively. So I actually know up front the number is either 3^1 * 7^2 (=98) or 3^2 * 7^1 (=63). Each statement tells me it must be 3^2 * 7^1. As you can tell... This is a very hard question.
Q24 is 1 ...at least 1 house sold for above 165k it can't be 3 alone, because suppose one house was sold for 129,999. Then to make up for this drop, another house would have to be sold for at least 170,001 to get an average of 150k for that pair. Similarly, if the median is 130k, there needs to be at least one house which is at 170k to balance this ((170k + 130k)/2 = 150k, the mean). 2 is also insufficient, because it still doesn't address the 130k median or any house potentially below that, which would have to be made up by a house that is greater than or equal to 170k. In all scenarios, you need one house that is at least 170k, and if there are houses below 130k, you need even more expensive houses. suppose we had 130k, 130k ,130 k, 130k ,130k, 130k, 130k, (130k), ----, -----, -----, -----, ------, ------, ------ (all those blanks need to be at a minimum of 170k to balance the 7 lots of 130k's on the left of the median (the median is in the 8th position) for example. Even then, one house would have to be even more than 170k to account for the median... so 6 houses of 170k on the right, and 1 which is 190k. This is just one possible scenario. If any of the houses on the left are lower than 130k, then even more expensive houses will be required...
Great video! For those following along... questions 9, 13, and 24 are the ones he got wrong.
I felt bad for him when he didn't see in question 9 the constraint that p must be odd in evaluating option B. I felt bad cause I know that it happens to everyone :(
Amazing video Arnold Reed!! Thank you so much, it helped me to see how it will be the GMAT. I havent done any practice test yet but now I'm confident with the type o questions I need to study. It was incredible to see this video for the last couple of days. Thank you!
I love this video! This is more like vlogging! this prepares potential test takers to be prepared mentally to face questions and gather the courage to pick a random answer and move on (esp the hard ones or the ones that you dont know) Need more tet taking videos like this one!
It's actually encouraging to me that someone so experienced in this exam still has the "ugh" response to some questions. hahahah.
Absolutely!!!
This is so great! Really helps show the reality of the GMAT test and why you should always draw out your thought process for each question.
Yay! Glad you enjoyed it!
it's actually refreshing to see that he got stuff wrong too!
Took class with him..real cool guy
Unfortunatelly at 37:36 you missed a short cut. If y+p = 150, the angle QPR must be 30. This is equal to statement 1, which you proved sufficient already, thus answer D would be correct.
This is correct. Was falling a little behind, and it's just impossible to catch everything when you're in the thick of it and the pressure is on!
Thank you sir.Loved it!!!
Thanks for the great video.... This increased my confidence by a huge margin! Would love to see more such videos. Thanks again.
Wow! Amazing video. Thank you!
Thank you! Very helpful. Keep doing more of these.
This was super helpful. Thank you!!!
This is amazing! Thank you!!
You're so welcome!
Great video i learnt a lot from it
This is like the Eye of the Tiger of GMAT prep videos
Haha, love that.
in question 13, you have x=30+y written on the screen. You didn't need to go further than that since it means x-y=30.
Yep. The stress of the test sometimes keeps you from seeing things right in front of your face! Oh well--the nice thing is that you don't have to be perfect to get a great score.
Thank you. It was really helpful.
Glad you enjoyed it!
Great video. Loved the takeaways.
Glad you enjoyed it!
I just got half of them right, should I give up the GMAT ? 🤔
17:48 and 28:12 me in front of every single Word Problem
Where are you taking these practice
tests ?
These are the official practice tests from MBA.com
I don't get question 7. If C was correct (as he picked), then the question would need to give the constraint that all numbers in that sequence are > 0. Otherwise, if negative numbers are allowed, the sum and the average don't determine the number of terms in the sequence. That's why I would have chosen E. Can someone please help me out? What am I missing? Thanks in advance guys!
Actually, having negative numbers in a set doesn't affect the average calculation. Take the numbers 2,3,4,5,6. The average is 4, right? The sum is 20, there are five numbers.
Okay, so how about the set -4, 0, 4, 8, 12. The average is still 4. The sum is 20, there are five numbers.
-25, -10, -5, -1, 61... the sum is 20, there are five numbers, the average is 4.
If you have an average, and you have the sum, you know how many numbers are in the set. Some of them might be negative, but that's not relevant to this question, as you know how many there.
why isn't the answer to number three a 1k gain?
Because one car sold for a 4000 dollar gain (bought for 16K, sold for 20K) and one car sold for a 5000 dollar loss (bought for 25K, sold for 20K). A 4000 dollar gain and a 5000 dollar loss amounts to a 1000 dollar loss.
45m Q 7) Why can't we say 1/5 is the rate in two years so 1/10 is the change in one yr and so the height increased by (4/10) 2/5 per yr?
Because that is mixing up concrete vs. relative numbers. Saying the tree was 1/5 taller in year 6 than in year 5 is a ratio. The tree grew by 20% in those two years. But I can't say that means 'the tree grew by 10% each year,' because each percent growth is based on the *new height.*
If it said the tree was 1/5 *an inch* taller year 6 compared to year 4, in *that* case we could say, "Oh, the tree is growing 1/10 an inch every year."
Anything above 2 wrong would give you a 50. To get a 51, you need a near flawless paper. I got 1 incorrect and got a 51 on the actual thing. But getting a 51 is definitely getting harder imo
On a practice test recently, I got a 51 and missed four total. But never on an official test.
@@reedarnold5468 Not possible on the actual test. In one of the official mock I got 2 wrong but still got 50. 51 in my opinion is very difficult nowadays. Also, it depends which questions you got wrong. If you got the experimental questions wrong then I think its still possible to score a 51 with 3-4 wrong but out of the marked questions if you get more than two wrong then you would not get 51 come what may.
How do you solve so fast? Like your mental math is way too fast compared to mine. Makes me feel I have no hope!
Don't lose hope! Don't forget I *teach* this stuff. This is what an expert looks like taking the test. When I first took the GMAT, I didn't move nearly as fast as I do here (and I still got a 760.)
What do you mean by 'mental math?' Do you mean just, like, arithmetic? You can practice arithmetic, and find new ways to do arithmetic. There are loads of tricks out there (come to one of our free "Foundations of GMAT Math" workshop. The first thing we do is talk about some fast mental tricks for arithmetic).
If you mean quantitative reasoning--again, it can be practiced. It's often a skill of recognition--I see certain things in a question that immediately 'trigger' a certain line of reasoning or a certain set up (like the overlapping set tables). You can study and collect such 'triggers' as well, so you know more quickly what strategy will get you through a question.
@@reedarnold5468 thanks so much for the response - you are certainly an expert at this! I’ll check out the Manhattan workshop you mentioned. And yes I meant arithmetic. Like that car dealership question - you solved for the 125% of x = 20k before I even set it up. Funnily, I’m pretty confident about my quantitive reasoning and solved some of the data sufficiency questions or that ugly chemical reaction proportions question within the time you took to solve them. It’s the arithmetic that stumps me. Hopefully speed will come with practice! Thanks again
@@reedarnold5468 and just to add - it’s very encouraging to hear you were slower initially and still pulled a 760. Thanks!
@@CanonNikonMan241 So stuff like that comes from lots of repetition and a change in how I think about percents. If you check out the "All about percents part 1" free prep hour, you'll get some insight into how I 'see' percents.
I know that '125%' of some number is a 4:5 ratio. (since 5 is 125% of 4). In that problem, the '5' part of the ratio corresponded to the 20k. So I know I need to multiply the ratio by 4k to get the actual values. Basically, 4:5 = 16k : 20k.
This is, again, not something I could do when I first took the test, but I do find it very convenient. The GMAT tends to use percentages that correspond to 'nice' ratios and ratios that correspond with 'nice' percents. E.g. 1:1, 1:2, 1:3, 2:3, 1:4, 3:4, 1:5, 2:5, 3:5, 4:5, 5:6, 9:10...
@@reedarnold5468 thats helpful, I’ll check the free prep hour out. Yes, already I’m getting more used to the GMAT percentages and it’s encouraging to hear that repetition will help
39:09 u meant 980
Yep! Made a silly mistake "1000 - 20 = 880." However, this is a good example of one of the advantages of Data Sufficiency... You don't actually have to *do the math right*. You just need to have the right set up and logical process, such that if you *did* do the math right, you'd get one single answer.
I did not understand Question 15 at all!
Yeah, it's definitely a very advanced divisibility question. I also made things unclear with the letters I chose. I should have written 3^n * 7 ^m, or something, since by re-writing 'k,' it looks like I'm referring to the 'k' in the question. During the test, I was aware enough to treat the 'k' exponent as a totally different value. In fact, I stopped thinking about the 'k' in the question itself, because when I was told it was a number with only prime factors 3 and 7, I just only thought about that number as a multiplication of 'some power of 3' and 'some power of 7.'
So k = 3^n * 7^m
For some integers 'n' and 'm.'
Then the problem tells me that that number has 6 positive prime factors, including 1 and itself. There's a rule out there to determine how many factors a number has based on prime factors. It's a little complicated to write out, but basically consider that you can 'build' the factors of a number by choosing which 'prime factors' to include.
For instance, 60 = 2^2 * 3 * 5
I can build a factor of 60 by taking some of these numbers.
So I can build 15 out of 3*5
I can build 12 out of 2^2 * 3
I can build 20 out of 2^2 * 5
I can build 6 out of 2*3...
etc.
Notice, for each of these factors, it's like I'm 'choosing' a number of each prime factor to include.
When I build '15,' I choose zero '2s,' the one 3, and the one 5.
When I build '12' I choose both '2s,' the one 3, but not the 5.
When I build '6,' I choose one of the '2's,' the '3', but not the 5.
If I build '10' I choose one of the 2s, not the three, and the 5...
Okay, so basically I have a 'choice' for every prime factor when building a factor. I can include zero, one, or two of the 2s, zero or one 3, and zero or one 5.
So that's THREE choices for 2, TWO choices for 3, and TWO choices for 5, or THREE*TWO*TWO total choices. That means I have 12 overall options to 'build,' which represent all 12 factors fo 60 (count'em! 1,2,3,4,5,6,10,12,15,20,30,60)
1 is choosing 'ZERO' for each prime: 2^0 * 3^0 * 5^0
2 is choosing ONE 2 and ZERO 3 and 5: 2^1 * 3^0 * 5^0...
20 is choosing TWO 2s, ZERO 3s, and ONE 5: 2^2 * 3^0 * 5^1
Okay, so that can be generalized. When we break a number down to its prime factorization, if we ADD 1 to each exponent and MULTIPLY that, we figure out HOW MANY FACTORS THAT NUMBER HAS.
So back to the number 3^n * 7^m... If that number has SIX factors, then I know that (n+1)(m+1) = 6.
Now there aren' that many numbers that multiply to 6!
(n+1)(m+1)
1 6
2 3
3 2
6 1
This would mean the possibilities for n and m are:
n m
0 5
1 2
2 1
5 0
So notice I can't actually have the 0,5 or the 5,0 options, because that would actually would mean the number (3^n * 7^m) does NOT have a prime factor of either 3 or 7, respectively. So I actually know up front the number is either 3^1 * 7^2 (=98) or 3^2 * 7^1 (=63).
Each statement tells me it must be 3^2 * 7^1.
As you can tell... This is a very hard question.
Q24 is 1 ...at least 1 house sold for above 165k
it can't be 3 alone, because suppose one house was sold for 129,999. Then to make up for this drop, another house would have to be sold for at least 170,001 to get an average of 150k for that pair. Similarly, if the median is 130k, there needs to be at least one house which is at 170k to balance this ((170k + 130k)/2 = 150k, the mean). 2 is also insufficient, because it still doesn't address the 130k median or any house potentially below that, which would have to be made up by a house that is greater than or equal to 170k. In all scenarios, you need one house that is at least 170k, and if there are houses below 130k, you need even more expensive houses.
suppose we had 130k, 130k ,130 k, 130k ,130k, 130k, 130k, (130k), ----, -----, -----, -----, ------, ------, ------ (all those blanks need to be at a minimum of 170k to balance the 7 lots of 130k's on the left of the median (the median is in the 8th position) for example. Even then, one house would have to be even more than 170k to account for the median... so 6 houses of 170k on the right, and 1 which is 190k. This is just one possible scenario. If any of the houses on the left are lower than 130k, then even more expensive houses will be required...
Yep yep. I mixed up the 'median' and 'average' on that question. Led to a miss!
epic true
I am sorry but is he getting sums wrong? It is just been 3 questions till now, I think 2 were already wrong
explanation is not good
that ain't the point of the video, genius!