2019 In-Training Examination Answers, Question-Breakdown, and Score calculator

[emt409]

Here is a link to the In-Training Answer key (as I and my co-residents have scored it). Yes, there are some terrible questions. Yes there are at a few with multiple correct answers, and possibly a few with no correct answer.

If you enter your answers in the User Answer Column, it will calculate your percentages for overall, and each of physics, radbio, biostats, and clinical categories. I suggest you copy the spreadsheet into another google doc, or excel sheet if you don't want people in real time seeing you enter your answers and your score.



As indicated in the pie chart, the breakdown of questions as I have categorized them was:
Radbio: 16.7%
Physics: 16.0%
Breast: 8.0%
CNS: 7.7%
H&N: 7.3%
GU: 7.0%
GI: 6.3%
GYN: 6.0%
PEDS: 5.7%
LYMPHOMA: 5.0%
LUNG: 4.7%
BIOSTATS: 4.7%
OTHER: 1.7%
SARCOMA: 1.3%

Enjoy! If you don't like the way I have categorized the question, or what I have keyed as an answer, then just change it yourself.


******edit*****

continuing to make corrections as they are pointed out to me, e.g.
-for a-bomb survivors, solid tumor risk is LNT for females and everyone, but LQNT for males
-pCR rates best for hormone negative, Her2+

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[emt409]

If someone will verify to me that everyone has taken the exam by now, I'll post scanned test booklet as well.

 

[noisyrope]

Thanks for your work.

I believe to officially take the exam you need to take it within 24 hours of March 1st. On that note. Does anyone have a blank copy they could scan?

 

[evilbooyaa]

I think OP has a scanned test booklet that was deleted last week as maybe some people hadn't taken it yet. EMT I'd probably just wait until this Friday afternoon (give people a week for any re-takes) although I'm not confident in the exact re-test timing.

 

[seper]

If you guys have a scanned 2019 booklet, I'd greatly appreciate posting.

 

[emt409]

I'll post it when I get back into my office tomorrow. I thought I had it saved in dropbox, but can't seem to find it.

 

[emt409]

I'll post it when I get back into my office tomorrow. I thought I had it saved in dropbox, but can't seem to find it.

Ok, ladies and gentlemen, long-awaited, and as promised -- here is a pdf of the 2019 In-Training Exam. If you want to venmo me a beer for the trouble, I won't say no (@emt409) -- and, thanks to one of my colleagues letting me scan his, as one of my favorite protagonists would say, the copy is:

 

[Chartreuse Wombat]

Ok, ladies and gentlemen, long-awaited, and as promised -- here is a pdf of the 2019 In-Training Exam. If you want to venmo me a beer for the trouble, I won't say no (@emt409) -- and, thanks to one of my colleagues letting me scan his, as one of my favorite protagonists would say, the copy is:

The first two questions break every rule about writing a good MCQ. Written by rank amateurs

 

[emt409]

Can you explain #107 and #133

I honestly don't know what they want from 107. For all else, the question is: Relative exposure rate of an x-ray generator is proprortional to:

A. kVp
B. kVp2
C. mAs2
D. mAs1/2

In reality, exposure is proportional to kVp (remember bremstrahhlung efficiency higher with higher energy), mA, and s. However, they ask for exposure rate. The simplified Larmor equation breaks down to P = 2/3(q^2*a^2)/c^3, so I see one could make a case for almost all of the answer choices. **** that question.

As for 133, anything near a line source will always get more exposure than being near a point source, since receiving radiation from linear array of points, so E(point)/E(line) < 1

 

[evilbooyaa]

I honestly don't know what they want from 107. For all else, the question is: Relative exposure rate of an x-ray generator is proprortional to:

A. kVp
B. kVp2
C. mAs2
D. mAs1/2

In reality, exposure is proportional to kVp (remember bremstrahhlung efficiency higher with higher energy), mA, and s. However, they ask for exposure rate. The simplified Larmor equation breaks down to P = 2/3(q^2*a^2)/c^3, so I see one could make a case for almost all of the answer choices. **** that question.

As for 133, anything near a line source will always get more exposure than being near a point source, since receiving radiation from linear array of points, so E(point)/E(line) < 1

I had no idea for 107. Maybe that's why all residents are such scrubs, for not knowing physics questions like that.

For 133 you're right for the question as written, with the caveat that at large distances for brachy (much higher than 1cm from source) a line source essentially acts like a point source. So if the question had said 100cm instead of 1cm, the ratio would be 1.

 

[emt409]

I had no idea for 107. Maybe that's why all residents are such scrubs, for not knowing physics questions like that.

For 133 you're right for the question as written, with the caveat that at large distances for brachy (much higher than 1cm from source) a line source essentially acts like a point source. So if the question had said 100cm instead of 1cm, the ratio would be 1.

Ratio asymptotically approaches one from above, so they'd have to use the physics buzzterm "at a very large distance", otherwise the d at which the ratio becomes essentially one depends on length of line source.

But yea, seriously people, you can't figure out 107??? You def to deserve to all fail your physics boards. Pretty much all dumber than DTJ.

 

[Ray D. Ayshun]

I honestly don't know what they want from 107. For all else, the question is: Relative exposure rate of an x-ray generator is proprortional to:

A. kVp
B. kVp2
C. mAs2
D. mAs1/2

In reality, exposure is proportional to kVp (remember bremstrahhlung efficiency higher with higher energy), mA, and s. However, they ask for exposure rate. The simplified Larmor equation breaks down to P = 2/3(q^2*a^2)/c^3, so I see one could make a case for almost all of the answer choices. **** that question.

As for 133, anything near a line source will always get more exposure than being near a point source, since receiving radiation from linear array of points, so E(point)/E(line) < 1

That said, wouldn't the answer be a? If we turn up the current, well get more exposure over a certain period of time, so directly proportional to mA, and if we turn up the voltage, more energetic photons, and slightly more photons, will create more charge in the detector, so proportional to kvp. mAs affects total exposure, but it would seem you'd divide out the s for exposure rate. Ultimately, wtf?

 

[emt409]

That said, wouldn't the answer be a? If we turn up the current, well get more exposure over a certain period of time, so directly proportional to mA, and if we turn up the voltage, more energetic photons, and slightly more photons, will create more charge in the detector, so proportional to kvp. mAs affects total exposure, but it would seem you'd divide out the s for exposure rate. Ultimately, wtf?

Perhaps it is, perhaps it isn't. That's the point, another epically bad question.

 

[Ray D. Ayshun]

Perhaps it is, perhaps it isn't. That's the point, another epically bad question.

The good news is that I was not able to Google relative exposure rate. Iow, it's not that I didn't know what they were talking about, it's that they didn't.

 

[seper]

Question: are all figures in the actual booklet black and white?

 

[emt409]

The good news is that I was not able to Google relative exposure rate. Iow, it's not that I didn't know what they were talking about, it's that they didn't.

After revised review and discussion with our head physicist, we think best answer is probably kvp^2.

Question: are all figures in the actual booklet black and white?

And yes, all the figures were black and white.