MCAT Questions and Answers

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Hello,

I have signed up for the March 27th MCAT and I have been following SN2's plan and about two weeks into it, however I am noticing that when I get to doing the passages I am missing several more than I would like to be missing. I was wondering if this is typical and if you had any tips on how to work the passages? Do you spend under 45seconds reading the questions and then use P.O.E? Please let me know. Thank you.
 
why does it take so long to get your books after you mailed the money in?
 
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Hello,

I have signed up for the March 27th MCAT and I have been following SN2's plan and about two weeks into it, however I am noticing that when I get to doing the passages I am missing several more than I would like to be missing. I was wondering if this is typical and if you had any tips on how to work the passages? Do you spend under 45seconds reading the questions and then use P.O.E? Please let me know. Thank you.

Generally speaking, the typical student gets about 4.5 out of 7 on our passages. They also go on to get over 30 on the actual MCAT, so a score of 4.5/7 is actually a decent one. The downside to this is that it can be demoralizing when one is used to doing much better than that. The upside is that it forces you to review the answer explanation, which is where most learning occurs. Reading is a passive process, so while reading the section will expose you to the information, it's a different skill altogether to apply the information.

This is why we include multiple-choice questions as you read through the text portions of our books and why we have such lengthy answer explanations. What we also do is put in about ten to fifteen particularly challenging questions (polite phrasing for "trick" question) in each homework set with the idea people will miss those. When one misses a question, they read the answer explanation. Because this is the only place we know people will read for sure, we include lengthy explanations that summarize what was important in the text and on that particular question. These are learning tools. If we removed those, then the average score would probably be 5.5 to 6 out of 7, and while people would feel better about themselves, they wouldn't have certain key points hammered in.

The moral to the story is that you shouldn't take the score too seriously right now. Instead ask yourself, "would I get it right if I saw this question again (with altered wording) in under a miniute?"

As far as timing goes, we emphasize to our students in class to spend at most two minutes per science passage and one minute per question. We over-emphasize doing math quickly in your head and altering your equations into an easier to use form whenever possible. Calculation are the number one issue for people who run out of time.

And P.O.E. is HUGE with us, as you have likely seen in our answer explanations. It is helpful and instinctive to use it on all of the challenging questions.

Thanks for asking and more so thank you for using our materials. We take great pride in people's opinions of our materials and are honored that you chose us.
 
why does it take so long to get your books after you mailed the money in?

If we had an FAQ on our webpage, that would no doubt get the most frequent hits. The honest answer is that the man in charge of book distribution is also a teacher in Berkeley, a manager in Berkeley, and several other random jobs that present themselves. He dedicates a portion of each weekday and Saturday to processing the orders and sending out the books. On light weeks (with few orders and when he's not teaching) people get them in record time (two to three day turnaround). On busy weeks (tons of orders and he's teaching several classes) and around the holidays, things run much slower. Because he does all of the work, it's done by his schedule and preferences. This includes waiting a long time for checks to clear (learned lesson for him apparently) and using FedEx to ship. Hopefully with the credit card option available now and soon (defined by his terms) an on-line credit card click-to-order option, this will expedite things greatly. For now, think of it as bonding with the previous generation by using mail order methods so popular before this crazy internet thing.
 
Is there any discount for the actual class? special promotion or student discount?! This is a silly question since most applicants are also students, but it doesn't hurt to ask! :)
 
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Is there any discount for the actual class? special promotion or student discount?! This is a silly question since most applicants are also students, but it doesn't hurt to ask! :)

There are discounts for members of certain student organizations or volunteer programs. If your club gets a discount, then you probably got a global email to all members about it. Your best bet would be to contact the manager at the local center (either Todd or Dale) and see which clubs and/or oragnizations at your school get a discount. You could also ask the leadership people of your club or organization.
 
What is the average time one should spend reading a passage and doing the questions for the passages found in either the chem or physics book?
 
What is the average time one should spend reading a passage and doing the questions for the passages found in either the chem or physics book?

On the exam, given that there are typically seven passages to go with 52 questions, we recommend getting in the habit of spending 2 to 2 1/2 minutes per passage max and one minute per question. Our hope is that by learning tricks to do our questions quickly, you'll be even faster on the real MCAT and be able to invest any extra time (time leftover once you finished) finding and correcting careless errors. Our motto is "it's about speed, not precision." Practice finding a best answer quickly and not a perfect answer slowly.
 
How is Berkley Prep different from Kaplan? Does the program also come with computerized tests?
 
Also, can orders be faxed in instead of mail? I'm very happy SDN decided to have this forum. I've never heard of Berkley review and physics and gen chem are my problem areas on the MCAT. From the great reviews I know that when I work with these books and CBT's my score and understanding of the subjects will vastly improve.
 
How is Berkley Prep different from Kaplan? Does the program also come with computerized tests?

The program comes with 9 CBTs. There are both online explanations to each question as well as the option to discuss the questions live with the author or instructor.

As far as being different from Kaplan, you would need to sit in a lecture for each class and decide. The tangible differences are that we have more hours of live instruction and roughly half of our classes are taught by an instructor with more than ten years of experience teaching the MCAT. Expereince makes a huge difference in the insights they can share with class, because over time you learn what topics students have anxieties about, and you adapt and perfect your lectures to work through those anxieties. Perhaps the biggest difference is that we specialize in the MCAT and offer our class in only three cities, so we can cater the course to the location.
 
If someone signs up for your course do they also get the books or does one have to pay for the course and buy the books?
 
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I'm planning to take Berkeley Review's MCAT course this summer. Fortunately, I have the option to choose between Berkeley, Stanford, Irvine and LA locations. Which location would Todd and Dale be at the most? Which one would you recommend? Thanks!
 
I'm planning to take Berkeley Review's MCAT course this summer. Fortunately, I have the option to choose between Berkeley, Stanford, Irvine and LA locations. Which location would Todd and Dale be at the most? Which one would you recommend? Thanks!

I didn't even realize we had a Stanford class, so I can't comment on that site. Dale lives up north and handles the biology there. Todd comes up to teach some of the chemistry. In Westwood and Irvine it's the opposite, except Todd teaches physics and chemistry in Westwood and Irvine. You get more lectures total by them in southern California.
 
I'm doing Circuits now and had a question:

In a RC circuit, what is the relationship between the capacitor and the resistor?

And on page 180 in the Physics 2 book (same chapter), eq. 9-35:

I =(V/R) - (q/RC)

what is the q/RC term mean in relation to the example/figure 9-13?

And also, what do we have to understand about eq's 9-36 and 9-37?

Thanks! This part of physics confuses me!
 
I'm doing Circuits now and had a question:

In a RC circuit, what is the relationship between the capacitor and the resistor?

And on page 180 in the Physics 2 book (same chapter), eq. 9-35:

I =(V/R) - (q/RC)

what is the q/RC term mean in relation to the example/figure 9-13?

If you consider the circuit in the question, you'll note that it has a resistor and a capacitor in series. According to Kirckhoff's loop rule, the voltage is lost as it completes the circuit, so the resistor and the capacitor only get a fraction of the voltage (and hence Vtotal = Vdrop across the resistor + Vdrop across the capacitor). Across the resistor, the voltage drop is found using V = IR. Across the capacitor, because Q = CV, the voltage drop across the capacitor is Q/C. This means that Kirckhoff's loop rule cn be expressed as:

Vtotal = ItotalR + Q/C

They divided both sides by R and subtracted the Q/RC term from both sides to isolate the current.

It's important to look at this conceptually more than as an equation to use for a calculation.

Itotal = Vtotal/R - Q/RC

If the voltage of the battery is increased, then according to the equation, the current will increase. This is expected.

If the resistance is increased, then the current decreases, because the first term is larger than or equal to the second term (given that current must have some value or equal zero). Increasing resistance to lower the current seems feasible.

The important term to get straight is the charge on the capacitor, Q. Over time, the capacitor fills until it reaches it's maximum charge. Once it reaches its maximum charge, then no more current can flow through the capacitor. Because all of the circuit elements are in series, they share the same current, so when the capacitor is full, there is no current through the circuit. This is explained by the equation, because if you increase Q, you are subtracting a bigger value until at some point Q is large enough that I = 0.

Hopefully this explains the equation conceptually.


And also, what do we have to understand about eq's 9-36 and 9-37?

Thanks! This part of physics confuses me!

The equations tell us how fast a current can be released from a capacitor. For devices such as flash bulbs on disposible cameras or pacemakers, you want a rapid discharge of current so you design the wires and capacitor accordingly. In the case of a pacemaker, the wires connecting the device itself with the heart (SA node?) are generally thick, so that R is small and the discharge time is fast. You want a quick burst of current, not a slow burn of current.
 
Thanks for the quick reply and for clarifying things for me! TBR rocks!
 
I have been racking my brain over example 6.4a in Physics Part II: Sound, pg 18.

The solution states: ..."which means n and L are linearly related, and more importantly that a bigger sound speed v results in a smaller slope".

I don't understand the latter part of the explanation. I tried manipulating the formula to express it in terms of a straight line, y = mx + b, but I came up empty. So why exactly is the speed of sound lower for pipe A than for pipe B?

ps. I just ordered the TBR exams over USPS. I hope you have the time to help me out there as well! Thanks again!
 
I'm trying to find the General Chemistry/Physics TBR books, but I cannot get into contact with the Berkeley Staff for shipping charges (I'm from Canada). Has anyone ordered any of the BR books, had them shipped to Canada, and know the actual price of shipping? Or at least had any success in contact the company?
 
I have been racking my brain over example 6.4a in Physics Part II: Sound, pg 18.

The solution states: ..."which means n and L are linearly related, and more importantly that a bigger sound speed v results in a smaller slope".

I don't understand the latter part of the explanation. I tried manipulating the formula to express it in terms of a straight line, y = mx + b, but I came up empty. So why exactly is the speed of sound lower for pipe A than for pipe B?

ps. I just ordered the TBR exams over USPS. I hope you have the time to help me out there as well! Thanks again!

What they are saying is that it can be written as an equation of a line where y = mx + b becomes n = (2fn/v)L + 0, making the slope of the line equal to 2fn/v.

According to the graph, the line connecting the data points for Pipe A has a steeper slope than the line connecting the data points for Pipe B. This means that v (found in the denominator) must be less for Pipe A than Pipe B, so that (2fn/v) is greater for Pipe A than Pipe B.
 
I'm trying to find the General Chemistry/Physics TBR books, but I cannot get into contact with the Berkeley Staff for shipping charges (I'm from Canada). Has anyone ordered any of the BR books, had them shipped to Canada, and know the actual price of shipping? Or at least had any success in contact the company?

Someone earlier posted that the FedEx shipping charges to Canada are something like $157 US. You'd be better off either buying them used or having someone in the states buy them and ship them. Not sure why the shipping price is so different for a purchased item than a non-purchased item.
 
Hello. Hoping someone might be able to help...

I understand that launching a projectile at 45 degrees maximizes the range of the projectile IF the starting and ending heights are the same. I'm having a little trouble understanding how this changes, however, when the starting and ending heights are different like when you launch a projectile off a cliff.

Particularly, Physics Part I Chapter 1 (Translational Motion), Practice Passage IV, Question 27 (page 41). I read the explanation (page 55) which says "The best compromise of maximizing the angle to maximize the flight time, while simultaneously minimizing the angle to maximize the x-direction velocity, is achieved at a value of slightly less than 45 degrees." I get why a bigger angle increases flight time and a smaller angle increases x velocity (cos(theta)), but I don't get how we know that an angle less than 45 degrees is the right compromise between these competing factors.

Also, in the problem (page 41), each graph for answers A-D shows the range being the same for 0 degrees and 90 degrees. How is that possible? Isn't zero degrees straight ahead which would definitely have a greater range than 90 degrees which is shooting straight up in the air so the projectile would come right back down???


THANKS!!
 
Hello. Hoping someone might be able to help...

I understand that launching a projectile at 45 degrees maximizes the range of the projectile IF the starting and ending heights are the same. I'm having a little trouble understanding how this changes, however, when the starting and ending heights are different like when you launch a projectile off a cliff.

Particularly, The Berkeley Review (TBR) Physics Part I Chapter 1 (Translational Motion), Practice Passage IV, Question 27 (page 41). I read the explanation (page 55) which says "The best compromise of maximizing the angle to maximize the flight time, while simultaneously minimizing the angle to maximize the x-direction velocity, is achieved at a value of slightly less than 45 degrees." I get why a bigger angle increases flight time and a smaller angle increases x velocity (cos(theta)), but I don't get how we know that an angle less than 45 degrees is the right compromise between these competing factors.

It might be easier to think about what the situation allows different than when the launch point and landing point are at the same height. In this case, because you are landing at a lower elevation than you start, the object will be in the air longer than the standard scenario. This means that flight time is increased. If you lower the launch angle slightly, you can convert some of the extra flight time into extra speed, which leads to a balance between flight time and speed in the x-direction.

Another way to think about this involves your instincts when you throw an object onto a roof. You instinctively make the launch angle greater than 45 degrees, because with an elevated landing point, the flight time has been reduced. A greater launch angle will give back some of the lost flight time. Throwing off of a cliff is the opposite situation.

Also, in the problem (page 41), each graph for answers A-D shows the range being the same for 0 degrees and 90 degrees. How is that possible? Isn't zero degrees straight ahead which would definitely have a greater range than 90 degrees which is shooting straight up in the air so the projectile would come right back down???
THANKS!!

You are right. In the graphs in my picture it looks like the difference in range between 0 and 90 is minimal (I had to use a second piece of paper to see the height difference.
 
It might be easier to think about what the situation allows different than when the launch point and landing point are at the same height. In this case, because you are landing at a lower elevation than you start, the object will be in the air longer than the standard scenario. This means that flight time is increased. If you lower the launch angle slightly, you can convert some of the extra flight time into extra speed, which leads to a balance between flight time and speed in the x-direction.

Another way to think about this involves your instincts when you throw an object onto a roof. You instinctively make the launch angle greater than 45 degrees, because with an elevated landing point, the flight time has been reduced. A greater launch angle will give back some of the lost flight time. Throwing off of a cliff is the opposite situation.



You are right. In the graphs in my picture it looks like the difference in range between 0 and 90 is minimal (I had to use a second piece of paper to see the height difference.


Thanks for the answer! About the range at 90 degrees, it should be zero since the projectile should come straight down, right?
 
Hi, I'm having trouble finding equations at the end of Chapter 1, Physics Book 1. The body of the text refers to equations 1.28, 1.29, and 1.30 but I can't find them anywhere before or after that point. The last numbered equation I have is 1.27, h= ((v-nought-y)^2)/2g, and the last equation is the quadratic equation a couple pages after.

I bought my copy new ~2 months ago.

Thanks!
 
Hey Teach!

So, I totally got pawned by test 2 today. I have already completed the first 2/3 of all the passages in the books (just started the last 1/3s yesterday) and I have to say that those books are phenomenal. Great passages... and they really do kick the snot out of me sometimes, but I'm always learning.
Well, for the test today (which I will thoroughly review tomorrow) I only scored a 26 (9,8,9) after a 30 (10, 12, 8) on cbt 1. I definitely made a few easily corrected mistakes on test 1 that would have increased my score by about 3 points overall, and from what I can tell, the same happened on test 2.

My plan is to stick with my schedule of taking a cbt every few days (all 7 TBRs and the last 3 AAMCs bc I took the rest during my last go at the MCAT). On areas where I'm feeling week, I will reinforce with problems from other sources.

What do you think? Do you think it's for me to break that 30 mark and maybe even take it up a few points beyond? Less than 3 weeks to go. I feel I have a solid grasp of a majority of the content. But I've done some dumb $#!+ and missing easy points.

Any tips, pointers, input?

Much respect and thanks.
 
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