Just some feedback and observations on this chapter which I wanted to throw out onto the forum to get your views.
1) Picking the best numbers was sometimes not as straightforward. Is there an easy way to quickly identify the smart numbers which should be picked.
2) Equation-based VICs: my concerns are around how long it will take to get to the solution based on the technique explained in the book -> pick values for one side of the equation, then solve the equation to figure out the remaining variables, then we find the target number, then we use the target number to test each answer choice. This will easily eat up 2 minutes per question ! Especially in the pressure environment of a test.
3) I need to figure out a way of quickly identifying some of the VICs type questions when in the midst of the GMAT exam ... from other types (basic equations, word translation type problems) ... by this I mean ... read a question, and be able to say ... right ... I can use the VICs technique to solve the question.
In other words, the VICs-type question looks very similar to a number of other types of problems (eg. Word translations etc).
Hope this makes sense. And would be grateful for any feedback or general comments on this.
Thanks.
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- IIL
- RonPurewal
- Students
- Posts: 19744
- Joined: Tue Aug 14, 2007 8:23 am
sure thing
(1)
you have to scan the problem for clues. here are a number of possible clues:
- if a problem involves percents of a number, then 100 is a great choice for that number.
- if a problem involves fractions of a number, then the least common denominator of those fractions is a great choice. (e.g., if the problem involves fifths and sevenths of a number, then 35 would be a good choice; this time, 100 would be a terrible choice)
- if a problem involves unit conversions, then pick a number that works well with such conversions. for instance, if the problem requires converting x minutes into hours, then let x = 60, 120, etc.
- DO NOT pick numbers that appear in excess elsewhere in the problem. (so don't pick 20 or 2000 in in-action problem #11, for example)
- DO NOT pick 0 or 1, unless you're verified ahead of time that all the answer choices are different with that choice of number.
if none of these hints is applicable, then just go ahead and pick numbers that are:
- 'ugly' enough to yield unique answers (for instance, shy away from 0, 1; also, if you're picking angles for a geometry problem, shy away from 30, 45, 60, 90, and the like)
BUT
- pretty enough to make computation easy
--
(2)
this is why the most important step of the VIC method is step zero: recognize that the problem is a vic problem in the first place.
as you have correctly observed, the vic method eats up A LOT of time. therefore, you need to recognize candidates for the vic method quickly, and commit to using the method at the very start of the problem.
you cannot afford to bang you head against the wall with a different method for 1-2 minutes and then begin using the vic method. instead, you need to start in on the vic method right away.
if you can recognize vic problems IMMEDIATELY, and get started on them within a reasonable amount of time, you should be able to arrive at a solution within the requisite two minutes. in fact, on harder problems, the vic approach can even save you time vis-a-vis the 'textbook' method.
--
(3)
you're right: vic problems look like problems from other categories.
and the reason is ... because vic problems really are problems from other categories!
it's not as though vic problems are their own special little class of problems. instead, a vic problem is ANY problem - whether a word translation, number properties, geometry, pure equation, etc. - that features undetermined variables that can be 'plugged into'. indeed, there are vic problems in each and every subject category.
the point is that the vic method is a backup, for use when you don't know how to do the primary method right away. (sometimes, for especially difficult problems, the vic method will actually be a primary solution method.) it is versatile enough to work on all kinds of different problems.
one point: vic problems are much more diverse than you might think by looking at the problems in the in-action section. those problems, to a certain extent, mostly look like each other, whereas the vic problems on the test feature greater variety. if you want a better sample of what vic problems look like, go to the vic problem set in the back (the og problems), look up those problems one at a time, and stare at them. look for common threads. you'll get the idea.
(1)
you have to scan the problem for clues. here are a number of possible clues:
- if a problem involves percents of a number, then 100 is a great choice for that number.
- if a problem involves fractions of a number, then the least common denominator of those fractions is a great choice. (e.g., if the problem involves fifths and sevenths of a number, then 35 would be a good choice; this time, 100 would be a terrible choice)
- if a problem involves unit conversions, then pick a number that works well with such conversions. for instance, if the problem requires converting x minutes into hours, then let x = 60, 120, etc.
- DO NOT pick numbers that appear in excess elsewhere in the problem. (so don't pick 20 or 2000 in in-action problem #11, for example)
- DO NOT pick 0 or 1, unless you're verified ahead of time that all the answer choices are different with that choice of number.
if none of these hints is applicable, then just go ahead and pick numbers that are:
- 'ugly' enough to yield unique answers (for instance, shy away from 0, 1; also, if you're picking angles for a geometry problem, shy away from 30, 45, 60, 90, and the like)
BUT
- pretty enough to make computation easy
--
(2)
this is why the most important step of the VIC method is step zero: recognize that the problem is a vic problem in the first place.
as you have correctly observed, the vic method eats up A LOT of time. therefore, you need to recognize candidates for the vic method quickly, and commit to using the method at the very start of the problem.
you cannot afford to bang you head against the wall with a different method for 1-2 minutes and then begin using the vic method. instead, you need to start in on the vic method right away.
if you can recognize vic problems IMMEDIATELY, and get started on them within a reasonable amount of time, you should be able to arrive at a solution within the requisite two minutes. in fact, on harder problems, the vic approach can even save you time vis-a-vis the 'textbook' method.
--
(3)
you're right: vic problems look like problems from other categories.
and the reason is ... because vic problems really are problems from other categories!
it's not as though vic problems are their own special little class of problems. instead, a vic problem is ANY problem - whether a word translation, number properties, geometry, pure equation, etc. - that features undetermined variables that can be 'plugged into'. indeed, there are vic problems in each and every subject category.
the point is that the vic method is a backup, for use when you don't know how to do the primary method right away. (sometimes, for especially difficult problems, the vic method will actually be a primary solution method.) it is versatile enough to work on all kinds of different problems.
one point: vic problems are much more diverse than you might think by looking at the problems in the in-action section. those problems, to a certain extent, mostly look like each other, whereas the vic problems on the test feature greater variety. if you want a better sample of what vic problems look like, go to the vic problem set in the back (the og problems), look up those problems one at a time, and stare at them. look for common threads. you'll get the idea.
- StaceyKoprince
- ManhattanGMAT Staff
- Posts: 9359
- Joined: Wed Oct 19, 2005 9:05 am
- Location: Montreal
By the way, if you get good at VIC, the technique is both fast and less prone to mistakes. I don't take any more time to do VIC than to do algebra, and often less.
Note certain things:
Get really good at recognizing, as Ron says (and this is easy! variable expressions in the answers!) AND at figuring out very quickly what numbers are likely to work well in the problem (this only comes from practice). As a general rule:
- don't pick zero or one
- don't pick a number that shows up in 4 or 5 of the anwer choices
- follow the constraints of the problem (eg, if it tells you the number is even, or positive, or whatever)
- pick different numbers if you have multiple variables
- pick something that makes your life easy! My default is 2, as long as it's still in the mix after following the above rules. If it's an age problem, I pick 10. Percents, 100. Etc.
Also, notice what doesn't work - eg, I wouldn't pick 50% on a percent problem. Let's say I were to pick 50% for the discount off the price of a TV set. Why wouldn't I want to do that? Well, now both the price I pay AND the discount I get are the same number! That's likely to mess things up somewhere.
Remember that you are picking a number for any variable that shows up in the answer choices. If a variable shows up in the problem but not the answers, DO NOT pick a number for it.
Now, plug those numbers into the problem and do the math - this should not take any more time than doing the algebra would have taken... and, really, it should take less, because we're almost always faster at arithmetic than we are at algebra. Additionally, you won't make as many mistakes with arithmetic - no matter how good you are at algebra, you're better at arithmetic.
Finally, when you test your answer choices - DO NOT actually calculate five numbers for the five answers. You have an answer that you're looking for - let's say it's 10. All you care about is finding a choice that equals 10; you don't care what an answer actually is unless it's close to or equal to 10. For at least two (sometimes three or four!) answers, you will be able to stop well before you finish the calculation because you can clearly tell the number won't be anywhere near 10.
So there's really not that much added work - IF you practice. Just remind yourself that all you're doing is turning algebra into arithmetic - and who wouldn't prefer arithmetic to algebra? :)
Note certain things:
Get really good at recognizing, as Ron says (and this is easy! variable expressions in the answers!) AND at figuring out very quickly what numbers are likely to work well in the problem (this only comes from practice). As a general rule:
- don't pick zero or one
- don't pick a number that shows up in 4 or 5 of the anwer choices
- follow the constraints of the problem (eg, if it tells you the number is even, or positive, or whatever)
- pick different numbers if you have multiple variables
- pick something that makes your life easy! My default is 2, as long as it's still in the mix after following the above rules. If it's an age problem, I pick 10. Percents, 100. Etc.
Also, notice what doesn't work - eg, I wouldn't pick 50% on a percent problem. Let's say I were to pick 50% for the discount off the price of a TV set. Why wouldn't I want to do that? Well, now both the price I pay AND the discount I get are the same number! That's likely to mess things up somewhere.
Remember that you are picking a number for any variable that shows up in the answer choices. If a variable shows up in the problem but not the answers, DO NOT pick a number for it.
Now, plug those numbers into the problem and do the math - this should not take any more time than doing the algebra would have taken... and, really, it should take less, because we're almost always faster at arithmetic than we are at algebra. Additionally, you won't make as many mistakes with arithmetic - no matter how good you are at algebra, you're better at arithmetic.
Finally, when you test your answer choices - DO NOT actually calculate five numbers for the five answers. You have an answer that you're looking for - let's say it's 10. All you care about is finding a choice that equals 10; you don't care what an answer actually is unless it's close to or equal to 10. For at least two (sometimes three or four!) answers, you will be able to stop well before you finish the calculation because you can clearly tell the number won't be anywhere near 10.
So there's really not that much added work - IF you practice. Just remind yourself that all you're doing is turning algebra into arithmetic - and who wouldn't prefer arithmetic to algebra? :)
Stacey Koprince
Instructor
Director, Content & Curriculum
ManhattanPrep
Instructor
Director, Content & Curriculum
ManhattanPrep
4 posts Page 1 of 1