Arithmetic TRICKS, take two

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Hey Guys,

I was wondering what ARITHMETIC tricks you could implement int he MCAT...im just confused because there are a lot (ignoring the ones in EK), but I need to know if you have a list compiled after extensive practice that you would recommend knowing.

Rounding, no seriously rounding really helps.

I seriously went to the library and checked out a 5th grade text book. It has lots of stuff on unit conversions, rounding, fractions, and exponentials. I felt stupid doing problems out of it...but it helps you to do them fast and gives you practice. I hate how I've was so dependent on calculators! I couldn't even add fractions without the calculator until I looked at the text book. Just practice problems because you don't want to spend a lot of time on them on the test or to make stupid math errors. Good luck!

I so 100% agree with your comment. It's funny you mention fifth grade, because in class we discuss the highly advanced method of having no shame and doing it like a fifth grader.

Key things that I have found hurt people are:

Powers of ten: This is helped by decimal hopping and labeling numbers with "increased by factor of 10" or "decreased by factor of 100". It comes down to paying attention really.

Ratios: These are made easiest by making denominators easy to deal with. Also, if a ratio is hard to calculate as written, flip it and see if it is easier the other way. We somehow emotionally deal with bigger-over-smaller ratios better than smaller-over-bigger ratios.

Fractions and Equivalent Decimal-based Value: Learn the correlation between fractions and decimals 1/4 = .25, 1/5 = .2, etc... These can prove EXTREMELY useful on the exam.

There are many more strategies and techniques, but these are generally the big three for most people. A little practice goes a long, long way.

The funny thing I've found is that the supposedly more difficult math skills like logs, square roots of complex numbers, and exponential decay/growth are typically easier for people than basic division, fractions, and multiplication.

Learn how to estimate log calculations:

-log (1E-5) = pH 5
-log (1E-4) = pH 4

And -log(3E-5) =~ pH 4.5

If you remember that a '3' in front of the 10^-X, corresponds to the halfway point between two pH values then you can estimate whether all H+ (or OH-) concentration values will give a pH between X and X.5, X.5 or X+1, etc...

What textbook or SAT-like book could I use that would get these fundamentals into becoming a second language. I love ur advice Vihsdas, and its little bitty tricks like those I was wondering if you had a list of after taking the MCAT, I am pretty sure u may have had like 20-30 tricks in ur bag to tackle physics problems...other than that I really need book titles to do in a couple of days.

thanks

Wow, what a great little tidbit! Thanks much for this stuff!
"My log has a magic number. It's three."
A silly way to put it, but instantly memorable.

One of my favorite tricks is Left Add Right Subtract. This refers to exponents, for example, if you move the decimal point to the left one, you need to add one to the exponent. If you move the decimal point to the right one, you need to subtract one from the exponent.

1 x 10^5, move left one decimal point to the left and you got .1 x 10^6

I hope this helps.

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Long division was one thing I ended up using on the MCAT that I hadn't really expected. Since I rounded first ("Round, round, round" my pre med advisor used to say) the math wasn't too hard, but it was helpful to be familiar.
 
Here's an invaluable one that we use to help our students. Its a process more than a trick, but the end result is the trig function value of every important angle on the MCAT. Is seems long when written out, but its completely brainless and takes about 15 seconds to write down (tutorial anyone?)

1. create a table with 3 columns, one for angle, one for sin, one for cos.
2. write in the most important angles on the mcat in the left column: 0, 30, 45, 60, 90
3. in each cell in the sin and cos columns, put in "/2". Essentially every value has a 2 in the denominator.
4. now put a square root sign with nothing inside of it in the numerator of each cell fraction.
5. now start counting from the top. in the sin of 0, put a 0 inside the sqrt, in the sin of 30, put a 1 inside the sqrt, 45 gets 2, 60 gets 3, 90 gets 4. If you did it right, the sin of 0 should show up as sqrt 0 / 2, which is 0. sin 90 should show up as sqrt 4 / 2, which is 1.
6. do the same, but in reverse, for the cos column, cos 90 is (sqrt 0)/2, etc.

Like I said, it looks much worse written out than if you just create the table. Using this you'll have an elegant way to pull up the sin and cos of any important angle on the mcat.
 
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Here's an invaluable one that we use to help our students. Its a process more than a trick, but the end result is the trig function value of every important angle on the MCAT. Is seems long when written out, but its completely brainless and takes about 15 seconds to write down (tutorial anyone?)

1. create a table with 3 columns, one for angle, one for sin, one for cos.
2. write in the most important angles on the mcat in the left column: 0, 30, 45, 60, 90
3. in each cell in the sin and cos columns, put in "/2". Essentially every value has a 2 in the denominator.
4. now put a square root sign with nothing inside of it in the numerator of each cell fraction.
5. now start counting from the top. in the sin of 0, put a 0 inside the sqrt, in the sin of 30, put a 1 inside the sqrt, 45 gets 2, 60 gets 3, 90 gets 4. If you did it right, the sin of 0 should show up as sqrt 0 / 2, which is 0. sin 90 should show up as sqrt 4 / 2, which is 1.
6. do the same, but in reverse, for the cos column, cos 90 is (sqrt 0)/2, etc.

Like I said, it looks much worse written out than if you just create the table. Using this you'll have an elegant way to pull up the sin and cos of any important angle on the mcat.
I love that table! It works wonderfully...and you'll have it memorized if you use it enough.
Vihsadas log advice is great (like always). But always be careful. Make sure you're going in the right direction. ie log(3x10^-5)=4.5 NOT 5.5. Just practice a few logs.
The best way to deal with these little arithmetic problems is practice! Practice a lot of them...make some really stupid math mistakes, hit your head against the wall, and never make that mistake again. You'll never now what arithmetic rules are important until you practice!
 
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One important point that really should be raised is that often, you can get out of doing any calculations at all if you just stop for a few seconds and *think* about the kind of answer you need. Here's an example:

Q: What is the pH of an aqueous solution of 4 x 10^-9 HCl at 25 C?

A. 6.9
B. 7.4
C. 7.9
D. 8.4
E. 8.9

Note that the answer is *not* D!

People who plug and chug into formulas (even using V's excellent trick) get into trouble here, because they come up with a pH of eight point something. However, I told you in the question stem that this was a solution of *hydrochloric acid* at RT. So you know darn well that it can't have a pH of eight point anything! Even if you forgot all about water autoionization (which is the major source of protons here, not the HCl), you *still* should have ruled out any pH over 7 because this is an *acidic* solution. Thus, logically, using no calculation whatsoever, the only possible answer here is A.
 
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One important point that really should be raised is that often, you can get out of doing any calculations at all if you just stop for a few seconds and *think* about the kind of answer you need. Here's an example:

Q: What is the pH of an aqueous solution of 4 x 10^-9 HCl at 25 C?

A. 6.9
B. 7.4
C. 7.9
D. 8.4
E. 8.9

Note that the answer is *not* D!

People who plug and chug into formulas (even using V's excellent trick) get into trouble here, because they come up with a pH of eight point something. However, I told you in the question stem that this was a solution of *hydrochloric acid* at RT. So you know darn well that it can't have a pH of eight point anything! Even if you forgot all about water autoionization (which is the major source of protons here, not the HCl), you *still* should have ruled out any pH over 7 because this is an *acidic* solution. Thus, logically, using no calculation whatsoever, the only possible answer here is A.

VERY nice! Great point once again!

Make sure you're going in the right direction. ie log(3x10^-5)=4.5 NOT 5.5.

Just as a point of interest here, it's negative log you're considering here, not log. -log (3 x 10exp-5) = 4.5. As it were, log (3 x 10exp-5) = -4.5. You are correct in the magnitude, but preprogrammed to be thinking of pH.

QBy how much does the decibel level increase from when one engine is running after nineteen more engines are started, assuming a symmetric distribution?
A. 1.3 dB
B. 13 dB
C. 19 dB
D. 30 dB

The trick here is that the intensity is increased by a factor of 20 = 2 x 10exp1. If you know that log 2 = 0.3, then you can determine that the log of 20 is 1.3. If you don't know the log exactly, then using the Vishadas method of referencing log 3. We know that log 20 is less than 1.5 and we also know it's more than 1.0 (log 10 = 1.0). The decibel scale increases by 10 log (deltaI), so the decibel increase lies between 10 dB and 15 dB.

If you know log 2 and log 3, you can calculate nearly every log between 1 and 10
 
Another good one is just simply becoming comfortable with scientific notation.

This one is more just practice than anything. The basic rule is that you can add exponents of scientific notation and multiply the number in front, and, you can change the sign of exponents on the bottom and move them to the top.

Practice, practice, practice these tricks by turning every single number (even easy numbers) into scientific notation. You'll eventually be able to do most of these calculations in your head.
 
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VERY nice! Great point once again!



Just as a point of interest here, it's negative log you're considering here, not log. -log (3 x 10exp-5) = 4.5. As it were, log (3 x 10exp-5) = -4.5. You are correct in the magnitude, but preprogrammed to be thinking of pH.

QBy how much does the decibel level increase from when one engine is running after nineteen more engines are started, assuming a symmetric distribution?
A. 1.3 dB
B. 13 dB
C. 19 dB
D. 30 dB

The trick here is that the intensity is increased by a factor of 20 = 2 x 10exp1. If you know that log 2 = 0.3, then you can determine that the log of 20 is 1.3. If you don't know the log exactly, then using the Vishadas method of referencing log 3. We know that log 20 is less than 1.5 and we also know it's more than 1.0 (log 10 = 1.0). The decibel scale increases by 10 log (deltaI), so the decibel increase lies between 10 dB and 15 dB.

If you know log 2 and log 3, you can calculate nearly every log between 1 and 10

Great advice Berkrevteach, you could be my teacher any day :D
 
hey arithemetic pros, need your help... never been a fan of this but just wondering with these types of exponents how you come to this answer?

e^–(0.2)(1)/e ^ –(0.2)(6) = e ^ (0.2)(5)

thanks dudes!
 
Another good one is just simply becoming comfortable with scientific notation.

This one is more just practice than anything. The basic rule is that you can add exponents of scientific notation and multiply the number in front, and, you can change the sign of exponents on the bottom and move them to the top.

Practice, practice, practice these tricks by turning every single number (even easy numbers) into scientific notation. You'll eventually be able to do most of these calculations in your head.

Agreed. If the problem involves using scientific notation my rule of thumb was to first check the solution to see if the orders of magnitude are similar among the answer choices. So 10^-22, 10^-23, and 10^-24 are all similar. If they are similar this trick won't work. If they're more like 10^-12, 10^-15, 10^-19 then the trick will work. The trick is to deal only with the order of magnitude and not with the numbers in front. So if you have 4.8 x 10^-9 you just drop the 4.8 and use 1x10^-9. If you get the correct order of magnitude then you did the problem correctly. The problem with when the magnitudes get too close to each other in the answer choices is that the factors in front of the magnitude play an important role. :(
 
hey arithemetic pros, need your help... never been a fan of this but just wondering with these types of exponents how you come to this answer?

e^–(0.2)(1)/e ^ –(0.2)(6) = e ^ (0.2)(5)

thanks dudes!

When posting math on a forum you should be explicit about the order of operations.

In any case, e^–(0.2)(1) / e^–(0.2)(6) = e ^ (0.2)(5) can be rephrased as e^(-.2) / e^(-0.2*6). Subtract the value of the exponent in the denominator from the value of the exponent in the numerator and you get the exponent of the answer. This only works when you have the same base number, in this case 'e'. Exponential numbers are of the form base^exponent.

A simpler example would be: What is 10^3 / 10 ? Well, 10^3 / 10^1 = 10^2 because 3-1=2. You can check this easily as it's just asking what is 1000 divided by 10.

To rephrase it one more way: The rule is when you have an equation of the form: (base)^exponent1 / (base)^exponent2 or base^exponent3 * base^exponent4 where the base of both numbers are the same then you can respectively just subtract or add the exponents to get the answer. If you don't get it do a few examples to convince yourself of this.

2^2 * 2^3 = 2^5
2^5 / 2^3 = 2^2
 
after looking at it, I think I just made it seem more complicated because of the e function. honestly, sorta embarassed lol

thanks for doing it though
 
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hey arithemetic pros, need your help... never been a fan of this but just wondering with these types of exponents how you come to this answer?

e^–(0.2)(1)/e ^ –(0.2)(6) = e ^ (0.2)(5)

thanks dudes!

You could have also just thought of it as:

make e^-0.2 = x and therefore you now have x^1 / x^6 = x^-5 = e^(-0.2*-5) = e^(0.2*5)
 
Are we gonna have to manipulate natural logs? The prep books don't seem to suggest this, but they still have the formulas containing "ln". What do you guys think?
 
Are we gonna have to manipulate natural logs? The prep books don't seem to suggest this, but they still have the formulas containing "ln". What do you guys think?


Just use base e instead of base 10. Thats the only difference
 
Just use base e instead of base 10. Thats the only difference

I'm sorry, but perhaps someone can explain how to do that for me, because i am incapable of such things. Would someone please take pity on one so incompetent as myself to explain to me what the heck base e is or how that new information might help me solve natural logs. Do the same rules apply as in regular logs, because it doesn't seem like they should.

In advance, thanks to the taker
 
When you see log₁₀(x) = 3 you can isolate x by doing the following:

10^(log₁₀(x))=10^3 --->
x=10³=1,000

So, you can do the same thing with natural logs.

ln(x)=2
e^(ln(x)) = e^2 -->
x=e² = 2.7² (for MCAT purposes that's between between 4 and 9) = 7.38
 
Anyone remember that angle circle from pre-calc? Its where you draw a circle and write out the values for the various angles on the diameter. (for example, you would create lines going from the origin to the diameter and label them 30,45,60,etc. and where they met the diameter, you would put the x (cos) and y (sin) value of the things.)

Anyway, if you do this, you'll notice that going up from 30 to 45 to 60, the x and y values all have 2 in the denomenator, and a square root in the numerator. Furthermore, the numerators in the x values decrease from 3 (30) to 2( 45) to 1 (60). The y values aree reversed, so its 1 (30), 2(45), and 3(60).

Basically anytime I see a cos function or a sin function corresponding to one of these angles, I think of the wheel. I never really bothered to memorize the x, y values for the angle because it only takes 1/10 of a seecond to figuree it out through this method.

I hope that made sense to you guys. Good luck. PM me if you guys have any tips on shortening calculation time on PS section. I really need it!
 
Anyone remember that angle circle from pre-calc? Its where you draw a circle and write out the values for the various angles on the diameter. (for example, you would create lines going from the origin to the diameter and label them 30,45,60,etc. and where they met the diameter, you would put the x (cos) and y (sin) value of the things.)

Anyway, if you do this, you'll notice that going up from 30 to 45 to 60, the x and y values all have 2 in the denomenator, and a square root in the numerator. Furthermore, the numerators in the x values decrease from 3 (30) to 2( 45) to 1 (60). The y values aree reversed, so its 1 (30), 2(45), and 3(60).

Basically anytime I see a cos function or a sin function corresponding to one of these angles, I think of the wheel. I never really bothered to memorize the x, y values for the angle because it only takes 1/10 of a seecond to figuree it out through this method.

I hope that made sense to you guys. Good luck. PM me if you guys have any tips on shortening calculation time on PS section. I really need it!


Honestly it would take up less space in your brain to just plain memorize the sin and cos of 0, 30, 45, 60, 90, 180
 
Here is one. It might be a little much for the mcat, but it is cool nonetheless. It's the Babylonian Method (old school).

To calculate the square root of a number:

First guess roughly what you think it would be (number less than 1 guess bigger, for a number greater than 1 guess smaller)

Then divide your guess into the square root number.

Now take your answer and add it to your guess and divide by 2. Presto, you should be very close to the real number.

so for example:

sqrt of .78 = Guess .85

.78
-----= ~.9
.85

.85+.9
-------= ~.87 And this should be your answer (or close enough)
2

The real answer using a calculator is: 0.866 or rounded .87!!

Anyway, this method is quick and dirty. If you guess really wildly just use the answer from your first guess and run through the process again. You can do it in seconds once you get good at it.
 
Here is one. It might be a little much for the mcat, but it is cool nonetheless. It's the Babylonian Method (old school).

To calculate the square root of a number:

First guess roughly what you think it would be (number less than 1 guess bigger, for a number greater than 1 guess smaller)

Then divide your guess into the square root number.

Now take your answer and add it to your guess and divide by 2. Presto, you should be very close to the real number.

so for example:

sqrt of .78 = Guess .85

.78
-----= ~.9
.85

.85+.9
-------= ~.87 And this should be your answer (or close enough)
2

The real answer using a calculator is: 0.866 or rounded .87!!

Anyway, this method is quick and dirty. If you guess really wildly just use the answer from your first guess and run through the process again. You can do it in seconds once you get good at it.

Cool. I guess this is an iterative method then? I haven't learned this method, but from looking at it for a few seconds it looks as if you can guess any number and then if you do the process again and again you'll keep on getting closer to the real value.
So I guess a quick check to see if your guess was good is to do the process once, and then do it again with the answer from the first and see how different they are. If the difference is large, then your guess wasn't that good, if it's pretty small, then it was probably a good guess.
 
Cool. I guess this is an iterative method then? I haven't learned this method, but from looking at it for a few seconds it looks as if you can guess any number and then if you do the process again and again you'll keep on getting closer to the real value.
So I guess a quick check to see if your guess was good is to do the process once, and then do it again with the answer from the first and see how different they are. If the difference is large, then your guess wasn't that good, if it's pretty small, then it was probably a good guess.

Exactly. If you wanted to keep doing it you could always get to the precise answer, at least as precise as you want to make it. But typically for the basics, .01 to about 1000, 1 reasonable guess should be good enough.
 
One important point that really should be raised is that often, you can get out of doing any calculations at all if you just stop for a few seconds and *think* about the kind of answer you need. Here's an example:

Q: What is the pH of an aqueous solution of 4 x 10^-9 HCl at 25 C?

A. 6.9
B. 7.4
C. 7.9
D. 8.4
E. 8.9

Note that the answer is *not* D!

People who plug and chug into formulas (even using V's excellent trick) get into trouble here, because they come up with a pH of eight point something. However, I told you in the question stem that this was a solution of *hydrochloric acid* at RT. So you know darn well that it can't have a pH of eight point anything! Even if you forgot all about water autoionization (which is the major source of protons here, not the HCl), you *still* should have ruled out any pH over 7 because this is an *acidic* solution. Thus, logically, using no calculation whatsoever, the only possible answer here is A.

I didnt follow the reasoning there about autoionization. I get that it should be less than a pH of 7, but what is the twist in the question stem?

If I saw something like an "aqueous solution of 4 x 10^9 M HCl", why am I supposed to know that that number doesnt (shouldnt) "agree". Apart from the fact that strong acids have low pHs...

thx
 
I didnt follow the reasoning there about autoionization. I get that it should be less than a pH of 7, but what is the twist in the question stem?

If I saw something like an "aqueous solution of 4 x 10^9 M HCl", why am I supposed to know that that number doesnt (shouldnt) "agree". Apart from the fact that strong acids have low pHs...

thx

Well, how many H+ ions come as a result of auto-ionization? What is the order of magnitude of that number?
What is the order of magnitude of the number of HCl in QoQ's question?
So knowing those two things, which one (either the HCL or the autoioniz) is the greatest contributor to H+ ions?
 
Well, how many H+ ions come as a result of auto-ionization? What is the order of magnitude of that number?
What is the order of magnitude of the number of HCl in QoQ's question?
So knowing those two things, which one (either the HCL or the autoioniz) is the greatest contributor to H+ ions?

10^-7 (from auto-ionization) is bigger than 10^-9 (from HCl).

Okay if the H+ from auto-ionization is the bigger contributor, where does that leave HCl?
 
10^-7 (from auto-ionization) is bigger than 10^-9 (from HCl).

Okay if the H+ from auto-ionization is the bigger contributor, where does that leave HCl?

Remember that my pH trick only works with H+ concentration. So you have to take into account all of the contributors to H+. In this problem, you have the HCl and the Autoionization. Because the HCl contributes H+ at two orders of magnitude lower than the auto ionization, it would decrease the pH only very slightly below 7.
If you think about "Why is the pH of water 7??" You realize that the pH scale wasn't some arbitraty choice. In fact, the concentration of H+ ions from autoionization is 1x10-7 (as you knew). So -log[1x10-7] = ph 7.

So if you have -log[1x10-7 + 1x10-9] you have a slightly lower pH than 7.

My chem is really rusty, I may have left out some important detail somewhere? :O
 
this a biology textbook companion website site---registration is free. After you do that, scroll down below the contents, and there's a bunch of resources. The "math for life" pdf and the experiment links are helpful for the whole mcat, not just biology...there's helpful stuff on logs for chem. I used to be good at math, but got rusty after graduating a year ago, so I relate to the math stuff. This is helpful b/c it's written in a really down to earth way.
 
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i know i just posted but I want to share my arithmetic tricks lol...I've noticed my math improving (slowly, but it's improving nonetheless)! I was frustrated b/c I was missing such easy questions! So, if you're in that position, you can follow some of the things that worked for me:

I printed out this thread, and I also printed out this pre-med 411 Math tips file. The 411 thing is about 20 pages and it explains a lot of common mcat math concepts, like proportions, logs, sin/cos, vectors, geometry formulas (vol of sphere, for example), and so on. Anyways, I stapled them together and used them as my notes for all things MCAT math related. Anytime I encountered a little math trick, I'd take out my little packet and take note. I sat down and wrote out the rules for fractions, long division, square roots, sci notation, and other things I learned in junior high.



i drew a ton of graphs to help myself visualize the relationships between variables more clearly. Plus, the mcat is obssessed with graphs, so it's a good thing to get comfortable with. (As a side note, I found it super helpful to draw a few sin and cos graphs....)




Also, Berkeley Review has a great method for decimal and fraction conversions. This is straight from their stoichiometry chapter.

If you memorize common fractions and decimals, you can mutiply and divide fractions by quickly converting an easy denominator. I've memorized the decimal values of the all fractions from ½ to 1/12 and just wrote them down in a little table. I'd list them here, but it's a little too much time. lol it's the one time you can prob use a calculator for the next however many months!


Ex) 18/66=?
18/66=3/11----->>> 1/11 = 0.091; 3/11=
3 x (1/11)=
3 x (.091)=.273


Ex2) to estimate 11/12,
*1/12=.083


11/12--->(12-1)/12--->12/12 - 1/12----->1- (1/12)--->1-(0.083)=.917


It was painful to sit down and go back to the basics, b/c I'd been studying for a while and mastering "harder" concepts. I think I was protecting my ego a little too much. Honestly though, it felt good to have some humility about it and finally just get it over with. I would have felt a lot more dumb if I'd had the opportunity to fix this issue and hadn't b/c I didn't want to feel dumb...while I'm studying...all alone...not like I'm even in front of anyone.


anyways, that's my math trick. memorize decimals and be humble. :cool:
 
Although not really MCAT related, my favorite math trick is when you multiply two number, multi-digit in the from XA and X(10-A) (i.e 23x27, 35x35, 41x49), or in other words the ones digits add up to 10 and the rest of the digits are the same. So first multiply the ones digits normally and write that down. Then, add one to the rest of the digits and multiple those4 together and place them in front of the two digits you wrote down eariler.

and by the way if the ones digits are 1 and 9 you wriet down 09.

For exapmple 23x27

One digits - 3x7 = 21
add on to rest and multiply 3x2 = 6
place both halves together and 23x27= 621
and it works for every single case.

Oh and when multiplying two numbers (that are relativly reasonable) together without brute forcing it is to multplie two numbers taht are close to it and then add on the diffference. Say 52x18, you could do 50x18 (which can bee done in your head) then add 36, or 20x52 (which is also preety simple) minus 104. not too bad really.
 
Could you please explain the method again? like would I be able to solve 48*52 the same way? if yes then I didn't understand the method properly.
Thanks
 
You can use 0kazak1's method...but the fastest way to do 48*52 would be to to rewrite it as:

48*52 = (50-2)(50+2) = 50^2 - 2^2 = 2496.

If two numbers are close together, both even or both odd (so that their average is an integer), you can use this method if you know the square of the average.
 
Could you please explain the method again? like would I be able to solve 48*52 the same way? if yes then I didn't understand the method properly.
Thanks

This doesn't work with the first method I mentioned, the tens digits and beyond have to be exactly the same.

Bu t if you do the second way you could do 50x48=2400 + 96 (48x2)
or 50x52= 2600 - 104 which would give you the same thing. or up above works too, but if you get good at seeing things the way i sometime see them you can do in your head.
 
Here's an invaluable one that we use to help our students. Its a process more than a trick, but the end result is the trig function value of every important angle on the MCAT. Is seems long when written out, but its completely brainless and takes about 15 seconds to write down (tutorial anyone?)

1. create a table with 3 columns, one for angle, one for sin, one for cos.
2. write in the most important angles on the mcat in the left column: 0, 30, 45, 60, 90
3. in each cell in the sin and cos columns, put in "/2". Essentially every value has a 2 in the denominator.
4. now put a square root sign with nothing inside of it in the numerator of each cell fraction.
5. now start counting from the top. in the sin of 0, put a 0 inside the sqrt, in the sin of 30, put a 1 inside the sqrt, 45 gets 2, 60 gets 3, 90 gets 4. If you did it right, the sin of 0 should show up as sqrt 0 / 2, which is 0. sin 90 should show up as sqrt 4 / 2, which is 1.
6. do the same, but in reverse, for the cos column, cos 90 is (sqrt 0)/2, etc.

Like I said, it looks much worse written out than if you just create the table. Using this you'll have an elegant way to pull up the sin and cos of any important angle on the mcat.

Okay, I'm bumping this thread again because I've never seen this before and it is AMAZING. THANK YOU!
 
Okay, I'm bumping this thread again because I've never seen this before and it is AMAZING. THANK YOU!

This just goes to show you that some of the absolutely best tricks are the ones you get if you're lucky enough to have a really good high school teacher. I had a high school math teacher who taught that trick, as well as few other great ones. My favorite is the following:

3^2 - 2^2 = 3 + 2
4^2 - 3^2 = 4 + 3
5^2 - 4^2 = 5 + 4

This because x^2 - y^2 = (x + y)(x - y)

When x and y are consecutive integers, then (x - y) = 1.

It's useful for calculating large squares:

71^2
71^2 - 70^2 = 71 + 70
so 71^2 = 70^2 + 71 + 70
= 4900 + 141 = 5041

53^2
skipping a step here
53^2 = 50^2 + (51 + 50) + (52 + 51) + (53 + 52)
= 2500 + 309 = 2809

0.79^2
80^2 - 79^2 = 80 + 79
so 80^2 - (80 + 79) = 79^2
6400 - 159 = 6241 so 0.79^2 = 0.6241

I hope this thread keeps going. If we each toss in a few of the good math tricks we've picked up over the years, then it will be a great thread for those calculation-rich MCATs.
 
this isn't really a trick, but it's useful if you hate long division, and haven't thought of it before:

5500/62 = (5500/2) / (31) ---> [(5500/2)/2] / 15.5 etc.....

you can keep breaking it down in your head until you get something you're comfortable with.

also works with multiplication:

5500 * 62 = 5500 * 2 * 31-----> 5500 * 2 * 2 *15.5---->
5500 *2*2*2*7.75
 
Here's an invaluable one that we use to help our students. Its a process more than a trick, but the end result is the trig function value of every important angle on the MCAT. Is seems long when written out, but its completely brainless and takes about 15 seconds to write down (tutorial anyone?)

1. create a table with 3 columns, one for angle, one for sin, one for cos.
2. write in the most important angles on the mcat in the left column: 0, 30, 45, 60, 90
3. in each cell in the sin and cos columns, put in "/2". Essentially every value has a 2 in the denominator.
4. now put a square root sign with nothing inside of it in the numerator of each cell fraction.
5. now start counting from the top. in the sin of 0, put a 0 inside the sqrt, in the sin of 30, put a 1 inside the sqrt, 45 gets 2, 60 gets 3, 90 gets 4. If you did it right, the sin of 0 should show up as sqrt 0 / 2, which is 0. sin 90 should show up as sqrt 4 / 2, which is 1.
6. do the same, but in reverse, for the cos column, cos 90 is (sqrt 0)/2, etc.

Like I said, it looks much worse written out than if you just create the table. Using this you'll have an elegant way to pull up the sin and cos of any important angle on the mcat.

trig_30_60_90.gif


trig_45_45_90.gif
 
I've been able to calculate pretty well on some pretty wild rounding (takes about a second). For example the above,

48 x 52.

I say 50 x 50 = 2500

Calculated value = 2496

When numbers get past 1,000 I use scientific notation.

23215 x 3523 = 23x10^3 x 3.5x10^3 = 23 x 3.5 = ~ 80x10^6

Above, I made the assumption you can add the exponents of 3 + 3 = 6.

The calculated value us 81,786,445. Anyways my point is, this works for the MCAT cause they don't usually care for exact values. Many instances I've seen just knowing the correct power of ten gets you the right answer..

And btw, keeptouchingme, those triangles are MONEY!! I wish I would have remembered this when I first started studying, now all these values are just ingrained in my head from hundreds of practice problems. Great tips everyone.
 
Also, Berkeley Review has a great method for decimal and fraction conversions. This is straight from their stoichiometry chapter.

If you memorize common fractions and decimals, you can mutiply and divide fractions by quickly converting an easy denominator. I've memorized the decimal values of the all fractions from ½ to 1/12 and just wrote them down in a little table. I'd list them here, but it's a little too much time. lol it's the one time you can prob use a calculator for the next however many months!


Ex) 18/66=?
18/66=3/11----->>> 1/11 = 0.091; 3/11=
3 x (1/11)=
3 x (.091)=.273


Ex2) to estimate 11/12,
*1/12=.083


11/12--->(12-1)/12--->12/12 - 1/12----->1- (1/12)--->1-(0.083)=.917

I've never seen this before. What a great trick. That's not that many fractions fractions to know. Without a calculator, you can get really accurate answers. Are there others like this in the chapter?
 
a square root trick I learned recently that could be helpful for trig problems

sqrt 1 = 1 (as in 1/1 = New Year's Day)
sqrt 2 = 1.4 (as in 2/14 = Valentine's Day)
sqrt 3 = 1.7 (as in 3/17 = St. Patrick's Day)
sqrt 4 = 2(0) (need I explain...?)
 
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