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You can directly select 80 in this particular Question, as the answer should be a multiple of 3 - 1 because we are selecting from 3 ranges i.e. 700-800, 800-900, 900-1000 and whatever applies to one range would apply to others as well. -1 because 700 is excluded.
@@skahmed4155 Since you are selecting from 3 ranges (700's 800's and 900's), you should have equal number of options in each of them. Say you have n options in each, the result should be equal to 3n and therefore be a multiplier of 3. However, you cannot select 700 therefore, the result becomes 3n-1 or a multiplier of 3 from which you deduct 1. 81 is a multiplier of 3 and therefore 81-1=80 works while 91 83 46 and 37 are not multipliers of 3.
The fastest way of solving this on the test would be using combinatorics: - our 3 digit number should be AAB - first digit I could choose in 3 ways(7,8,9) - second one in 1 way(must be the same one as first) - the third one in 9 ways(0 to 9 minus the one we’ve already used for the first two) - remember, we must permute them because order matter! In total: 3*1*9*3!/2!=81(now, we must subtract 1 because 700 should not be part of our range). We got 800
Once you learn the concept of looking for patterns, it's just a matter of practicing it until you get faster. I bet if you saw a question similar to this already, you'd be able to solve it more quickly simply because you know how to do it now.
+David Carvalho Hahahah, good catch! Looks like I forgot to change the slide template when I swapped out the question :) Don't worry, the real GMAT won't be so confusing....
If you would rate the difficulty of this question between 1-10 where would you consider it? Just getting started with GMAT and would like to have an overview
How did you find it personally? Easy or hard? It's a moderate-difficulty question, but the reasoning approach is something that makes it much easier and will help you with today's GMAT, which is increasingly reasoning-based.
Is there no mathematical way to solve this? I'm a quadriplegic and I would have to tell my assistant test note taker to write things down which in this case would be very redundant and would take a lot of time :-(
+Honza No, for this one you'd actually have to write out the possibilities. The good news is that you won't see too many questions like this, so hopefully your note taker won't have to work too hard :)
I don't think its correct because you are counting numbers thst have the first and last digits as same and the question didn't specify that so you may need to explain this better
I'm counting numbers that have the first two digits equal, the last two digits equal, and the first and last digits equal. So we are accounting for everything.
moossifer Not for this particular question, no. For some "how many ways..." questions, you can use straight permutations or combinations formulas. But this isn't one of those.
+Dominate the GMAT What if you just find an answer that can be divided by 3 if you add 1? Each 'Hundred' will have the same number of 'double digit' integers. There are 3 hundreds you are working with. And the 'add 1' is important because we do not include 700 as an included integer, yet we include 800 and 900. 91, no factor of 3. 83, no factor of 3. 81, yes factor of 3. 46, no factor of 3. 37, no factor of 3. Are there any downsides to this way, other than if all of the answers fit the rule?
+Dominate the GMAT So there are an equal amount of integers which follow the rule in each group of hundreds: 700-799, 800-899, and 900-999 each have the same amount of integers which follow the rule. I would call this number x. 3x would account for all integers which follow the rule from 700-999. Next I need to remove 700 because it is not included jn the range based on the question. So I have the answer 3x-1. What answer choice is divisable by 3 if I add 1?
@@tanushreeagarwal9401 Simple in the first case the first digit and second digit are the same. The number that can in the first 7,8,9 = 3. Then only 1 way for the second. And only 9 ways excluding the first digits. You can same argument first and last digit. Including th casee for tens and ones digits being the same. We have 27 * 3 = 81.
@@MrRobocop12 Great job explaining it. The math on the GMAT isn't supposed to go trig and up and this method looks like statistics? More power to you if you can solve these types of problems in a more efficient manner.
People ask how can i do this in exam... You cant You have to study each example and if you know the one in exam you solve it If you dont Leave it till the end
The GMAT is question-adaptive, meaning you can't get to the next question until you answer the one currently on your screen. So it would be nice if you could leave questions you're unsure about to the end, but unfortunately that's not an option on the current iteration of the GMAT.
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You can directly select 80 in this particular Question, as the answer should be a multiple of 3 - 1 because we are selecting from 3 ranges i.e. 700-800, 800-900, 900-1000 and whatever applies to one range would apply to others as well.
-1 because 700 is excluded.
Indian brain 👌👌
@@truthtalk3577 wish that applied for verbal too :(
Can you explain the answer please? @yash mehta
@@skahmed4155 Since you are selecting from 3 ranges (700's 800's and 900's), you should have equal number of options in each of them. Say you have n options in each, the result should be equal to 3n and therefore be a multiplier of 3. However, you cannot select 700 therefore, the result becomes 3n-1 or a multiplier of 3 from which you deduct 1.
81 is a multiplier of 3 and therefore 81-1=80 works while 91 83 46 and 37 are not multipliers of 3.
The fastest way of solving this on the test would be using combinatorics:
- our 3 digit number should be AAB
- first digit I could choose in 3 ways(7,8,9)
- second one in 1 way(must be the same one as first)
- the third one in 9 ways(0 to 9 minus the one we’ve already used for the first two)
- remember, we must permute them because order matter!
In total: 3*1*9*3!/2!=81(now, we must subtract 1 because 700 should not be part of our range). We got 800
*80
But why you divided it on 2!m
I understand the logic of how to do this if we have all the time in the world, but how do we figure this out under 2 minutes :(
Once you learn the concept of looking for patterns, it's just a matter of practicing it until you get faster. I bet if you saw a question similar to this already, you'd be able to solve it more quickly simply because you know how to do it now.
Exactly my question.
hmm yeah..
And what is the area of the shaded region after all the counting? ;)
+David Carvalho Hahahah, good catch! Looks like I forgot to change the slide template when I swapped out the question :) Don't worry, the real GMAT won't be so confusing....
If you would rate the difficulty of this question between 1-10 where would you consider it?
Just getting started with GMAT and would like to have an overview
How did you find it personally? Easy or hard? It's a moderate-difficulty question, but the reasoning approach is something that makes it much easier and will help you with today's GMAT, which is increasingly reasoning-based.
Super good stuff, always from you sir! Thanks man
You're welcome. Glad you found it helpful.
This helps more than school
Why did you multiply two 9s?...we can count manually from 701 to 799..then proceed same for 800 to 899 and 900 to 999
In less than two minutes?
thank you!
sometimes 'how many' questions can be solved easier by reverse plug in however reverse plug in wouldn't work in this case
Man you write like a gmat maker... frightening...
@@oxy204 😂😂😂😂😂😂
Thats a long way. Won't have so much time in the exam.
Is there no mathematical way to solve this? I'm a quadriplegic and I would have to tell my assistant test note taker to write things down which in this case would be very redundant and would take a lot of time :-(
+Honza No, for this one you'd actually have to write out the possibilities. The good news is that you won't see too many questions like this, so hopefully your note taker won't have to work too hard :)
See my answer..check if its right?
Great teacher ❤❤
I appreciate that!
I don't think its correct because you are counting numbers thst have the first and last digits as same and the question didn't specify that so you may need to explain this better
I'm counting numbers that have the first two digits equal, the last two digits equal, and the first and last digits equal. So we are accounting for everything.
This is really not a smart method of doing it
This takes too much time, is there a faster method?
moossifer Not for this particular question, no. For some "how many ways..." questions, you can use straight permutations or combinations formulas. But this isn't one of those.
+Dominate the GMAT What if you just find an answer that can be divided by 3 if you add 1? Each 'Hundred' will have the same number of 'double digit' integers. There are 3 hundreds you are working with. And the 'add 1' is important because we do not include 700 as an included integer, yet we include 800 and 900.
91, no factor of 3. 83, no factor of 3. 81, yes factor of 3. 46, no factor of 3. 37, no factor of 3.
Are there any downsides to this way, other than if all of the answers fit the rule?
+tylerbeardshow I'm not quite following how being a factor of 3 ensures that "2 of the digits will be equal to each other, but the 3rd will not."
+Dominate the GMAT So there are an equal amount of integers which follow the rule in each group of hundreds: 700-799, 800-899, and 900-999 each have the same amount of integers which follow the rule. I would call this number x. 3x would account for all integers which follow the rule from 700-999. Next I need to remove 700 because it is not included jn the range based on the question. So I have the answer 3x-1. What answer choice is divisable by 3 if I add 1?
There is a better method by just going with basics that takes less than 1 min
H-T-O
First two digits same:3*1*9=27
First and last same:3*9*1=27
Tens and ones same:3*9*1=27
Total=81
Answer=81-1(one digit that is 700)=80
@@adityaguptavishu hey I didnt understand this solution could u please explain it to me
@@tanushreeagarwal9401 Simple in the first case the first digit and second digit are the same. The number that can in the first 7,8,9 = 3. Then only 1 way for the second. And only 9 ways excluding the first digits. You can same argument first and last digit. Including th casee for tens and ones digits being the same. We have 27 * 3 = 81.
@@MrRobocop12 Great job explaining it. The math on the GMAT isn't supposed to go trig and up and this method looks like statistics? More power to you if you can solve these types of problems in a more efficient manner.
*Considering that we only have 2 min on avg per problem, any time saving way can go a very long way for this test.
People ask how can i do this in exam...
You cant
You have to study each example and if you know the one in exam you solve it
If you dont
Leave it till the end
The GMAT is question-adaptive, meaning you can't get to the next question until you answer the one currently on your screen. So it would be nice if you could leave questions you're unsure about to the end, but unfortunately that's not an option on the current iteration of the GMAT.
My answer is 82.
Because u didn't count 800 and 900
No, I think Brett is right because went about in a categorical manner and left almost no room for error.
No.. it s 80.. he count 800 and 900
.. but.you also count 888 and 999.. you should eliminate that.. 888 and 999 are not included
@@rezairdha2 Is seems he is not considering 800 nor 900 04:01 =/
It is 80,probably you included 777 ,because I just did..😂