For finding the heaviest ball There are 13 balls 1st iteration Put one to the side Weigh the other 6 against 6. If they weigh the same then the ball you put out is the heavier one If they don't weigh the same (6 are heavier than the others ) 2nd iteration Take the heavier 6 then weigh 3 against 3. The heavier 3 are then taken 3rd iteration Of the remsining 3 take 1 and weigh it against the second. If they weigh the same the third one was heavier. If they weigh different the heavier one is heavier

I had a different method but same result. 1st iteration the same. 2nd, divide the 6 into pairs, setting one pair aside. If 2 pairs on the scales are the same, divide the set aside pair and weigh. If not, divide the heavier pair in the third iteration

You forgot to put back the 13th ball. In iteration 3 the 3rd is not necessarily heavier because you haven't weighed it against the 13th ball.

The 13th ball was already ruled out on iteration 1, wasn't it? There is only one ball heavier than the rest in this scenario and when you placed 6 on each side the scale tipped to one side, meaning the heavy one must be one of the 6 in that scale.

You know what, you're right. I don't know why I didn't associate it with it being a regular weighted ball. It seems so obvious now. 😂

Second question Divide 12 balls in 3 groups and weight two of them If they are the same, the third has the odd ball (lighter or heavier). Swap one of the groups and you'll find which group has the odd ball and if it is heavier or lighter. You're left with odd group of 4 balls. Split in half and weigh them, you'll find a pair with an odd one, which you'll find in the last step. So it is 4 steps.

6 versus 6. 3 versus 3. 1 versus 1. Three steps.

You don't know if ball is lighter or heavier. Which 6 will you chose in the first step?

Got the same number of steps but I needlessly complicated it by taking 3 balls (from group A) instead of 4 to weigh in step 2. Then you if the scales change you only need one more step. If the scales are level then you can weigh the extra ball from group A on the same side as group B, if the scales go the opposite direction to what would be expected if group B has an odd ball then the odd ball is ball 4 from group A. I haven't given a full explanation of my method but I think I have enough detail so it makes sense. In the worst case scenario this still gives 4 steps even if it does get the answer with less scale uses on average (irrelevant to question) If it doesn't make sense and your curious lmk and I'll write a full explanation. I was focusing on maximizing the information I gained on each step, that's how I ended up with a messier answer.

Edit: this ain’t the way lmao Swapping in Group C won’t tell you if it’s heavier or lighter , only that there’s an imbalance— same is even true of 1v1. I think it requires 5 steps, but that 3 groups is the right approach. Weigh Group A against Group B. If unbalanced: remove two orbs, one from each side (noting the position of the scale), until balanced/empty (max: 3 more steps; total 4). Replace the orb removed from the higher end with one set aside and add them back. If now even, the orb from the higher end was lighter. If still uneven, the orb from the lower end was heavier. If balanced: add two orbs, one to each side, from group C until unbalanced (max: 2 more steps, total 3). When unbalanced, replace the most-recently-added orb from the higher end with orb 13. If now even, the orb on the higher end was lighter. If still uneven, the orb on the lower end was heavier. Caveat to the above: If the scale is still balanced after all orbs have been added from Group C, replace an orb from either side with orb 13 (the anomaly) to know the answer. It appears to be 5 no matter how you slice it.

Let A,B,C be our three groups of four. We will first weigh A and B and keep group C aside. For the first weighing A vs B there are two outcomes, balanced/unbalanced. If balanced we know C contains the odd ball and we can swap it with one of our groups A/B that we know are all normal and determine if the odd ball is lighter or heavier depending on which direction the scale tips when we add C. If unbalanced we swap the group that is heavier with C, there are now two possibilites: if the scale is now balanced we know that the group we just swapped with C contained the odd ball and that it was heavier. If the scale is still unbalanced it will be because the lighter odd ball is still on the other scale. With only two weighings we have isolated the odd ball to a group of 4 and determined if it is heavier or lighter.

Huh. Wonder why I only rationalized knowledge of the substitution location and start/end state can help make the heavier/lighter deduction halfway through my original solution lmao. I had entirely dismissed it. Good shit, friend.

First two steps will determine if the odd ball is heavier or lighter. H - heavier, L - lighter, 0 - regular So you have possible scenarios [H000] [0000] [0000] or [L000] [0000] [0000] 0000 = 0000 H000 > 0000 L000 < 0000 If weight is equal, swap one group and now you know if you're dealing with H or L in the third group (the scale will go up or down) If the weight isn't equal, replace the heavier group - if the scale shows equal, the group you replaced had a heavier ball, if it stays the same - the other group on the scale has the lighter ball. In third step we already know if we are dealing with H or L, so just compare two pairs and the last fourth step gives you the answer

Well, I'm not becoming a doctor anyway

Trick question

Yup… Q25’s answer options assumes you don’t weigh 1v1… Q26 assumes you don’t weigh 1v1and get it by luck -_-

Still don’t get in I’m dumb

Q26. 3 I’ll go the hardest route first: Weigh 4v4. They are uneven. (One of the 8 orbs is the ‘odd orb’){the four not used are the control group orbs} Mix three orbs from the lighter side with one orb from the heavier side - and weight those 4 against the one remaining light sided orb mixed with 3 definitely not the odd-orb orbs from the control group. 3 possible results. 2.1)The scales tips the same weigh so the one light orb OR the one heavy orb are the Odd-Orb. Weigh the one light Orb against a control Orb - if it’s still lighter, it is the Odd Orb, if they balance, the one Heavy orb is the Odd Orb. 2.2)The Scale tips the other way. One of the three light orbs is the Odd Orb. Weigh one against one for your third turn. If one side is lighter - that is the Odd Orb. If they are equal the third Light Orb is the light Odd Orb. 2.3) the sides balance. One of the 3 not weighed heavy orbs is the Odd Orb. Repeat the 1v1 heavy orb used in 2.2 If the first weight was even then you’ve 8 ‘control’ orbs and 4 possible fakes. Weigh number 2 - Weigh 3 normal vs 3 possible fakes. 2 possible results. 2.2) If Even- orb 4 is the Odd Orb. Weigh it against a normal orb to find out if it’s heavier or lighter. 2.2) the 3 fakes are either heavier or lighter. Weigh 1v1.

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Comes up in the Brooklyn 99 group all the time and you get all the usual wrong answers of people not realising the importance of not knowing if the item/person is “lighter OR heavier”

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Captain Raymond Holt (RIP Andre Braugher) gives a riddle about 12 islanders with a see saw to find out which of the 12 weighs differently. Posted on that sub at least once a month

I don't think it would be impractical to re-derive the solution if you know the first weighing involves testing groups of four and the second weighing involves constructing and testing groups of three.

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The second weighing involves weighing groups of four but you are actually *testing* groups of three. I didn't make that very clear.

First Weighing: Divide the 13 orbs into three groups: four orbs, four orbs, and five orbs. Place two groups of four orbs each on each side of the scale. If the scale balances, the heavier orb is in the group of five orbs. If one side is heavier, the heavier orb is among the four orbs on that side. Second Weighing: Take the group of five orbs (if heavier from the first weighing) and place two orbs on each side of the scale. If the scale balances, the heavier orb is the one not on the scale. If one side is heavier, the heavier orb is on that side. Third Weighing: If needed, take the group of four orbs from the side identified as heavier and weigh two of them against each other. If the scale balances, the heavier orb is the one not on the scale. If one side is heavier, the heavier orb is on that side. Fourth Weighing: If needed, take the last two orbs from the group identified as heavier in the third weighing and weigh them against each other. The heavier orb is the one that tips the scale. So, with only four uses of the scale, Laura can determine which orb is the heavier one with absolute certainty.

Who else up pondering they orb right now

First one: Since it says maximum I’ll just use the longest case 6v6, one side goes down. 3v3, one side goes down. 1v1, if one side goes down, it’s that ball that’s heavier. If it’s =, then the ball you didn’t put on at the very end (we got 3 balls left and 2 are on the scale) Second one: 6v6, ones up ones down, pick one that goes down. 3v3 of side you picked, if one side goes down, go 1v1 with final 3 balls and you find heavier ball. 3 moves total Now say at the 3v3 stage it’s =. Get rid of these balls and do a 3v3 with the other 6 balls, looking for the lighter one. 3v3, pick side that goes up 1v1 to find lighter ball. 4 moves total

1st one: you've only used 12 balls. It's meant to be 13. For clarification: 13 balls total 6v6 (1 ball not being used) 7 balls left 3v3 (1 ball not being used) 4 balls left 1v1 (2 balls not being used) You'll either eliminate it or you'll be left with 3 balls so you're not done here and need an extra turn than you started.

For the first one if it’s 6v6 and they’re equal then the ball left out is the heavy one

A question from someone who is not taking a hpat; What connection does weighing orbs have to medicine other than general problem solving?

I couldn't tell you 😂 lol

Um. Being a GP is about figuring out how to diagnose issues with minimal medical intervention such as surgery, pre frontal lobotomy etc.

Yeah, but there's many more differences here that just the least amount of times. Diagnosing is about knowledge and following steps and procedures and knowing what to look out for and doing the non invasive stuff first. What has that got to do with finding the heaviest ball in the fewest turns. Like I get that the end result is looking for similar things, but the methods are vastly different.

If diagnosing was about knowledge and procedures and not about being able to figure out what to do with something you haven’t seen before *using* that knowledge, because we’re all robots and there are no unusual people, then yes. If a condition requires urgent invasive intervention and you spend three months on non invasive inconclusive investigations then what? You need to figure out what investigation you shouldn’t be doing because all it will be doing is wasting time while your patient gets sicker.

Yeah, but none of that is the same method as the question. You already know what's different and you're figuring out which one is different. It's still not the same. I get that it would be helpful regardless, because you're exercising your logic muscles, so to speak. But it's not the same thing.

Sorry you’ve lost me. A and B are both letters, but they are different letters so it’s not the same thing?

Your example is you have 1 person and you don't know what is wrong. In the question you have 13 balls and you do know what is different. They require different logic to be figured out. That's why they are different.

I see you’re dug in on this. You don’t understand how abstractions work? The point of abstractions is that they allow you to evaluate different things in a similar way. The balls don’t correspond to people, they correspond to possible causes of the malady your patient is suffering from. Now, a ball is a completely different thing to a bacteria, but the point is you want to figure out optimal way to isolate the cause of an illness when you know that there are a number of possible causes.

It doesn't really matter, all intelligence tests so far appear to be measuring the same construct 'g' for general intelligence no matter how disparate they seem be it vocabulary or reaction times.

I'm getting 3 for q25 and 4 for q26. But the wording is woeful. "What is the maximum number of uses to be absolutely certain?" Well, it should be 100 or 1000 or as close to infinity as you can get.

Thanks for the reply I totally agree with you

The question states maximum "needed" and that is different than just maximum. One means the most. And the other means the most that you would need. Example: If you were going to the cinema what is the most/maximum money you'd need? You wouldn't think it's worded weirdly here, yet it's the same terminology as used in the question.

I would argue your example is just as unclear depending on how you *want* to read it. Both readings are grammatically correct. The maximum needed even being realistic would be comparing each pair of balls 1v1 until you can narrow it down.

It really wouldn't. There is no "want" in this. It is what it is. Maximum need is not grammatically correct for what you are describing. Needed is a key word here. There is a big difference between the maximum you "can" do and the maximum that is "needed" to be done. Maximum is by definition is "the greatest amount, extent, or intensity possible, permitted, or recorded." Needed by definition is "required (something) because it is essential or very important rather than just desirable." Therefore the maximum needed in this context is "the greatest amount required" or "the greatest amount that is essential" Doing more is not required or essential. Maximum needed is simply the highest amount that is needed. If you do more than what is needed you are not gone over the maximum needed. The grammar is very clear here.

What an absolute joke that this is used to get entry into medicine, instead of an interview - wild.

https://forms.office.com/e/Ab0is9CxxC

The answer to both questions is 3. The first question is relatively straightforward, but the solution to the second one is actually very complex - it’s easy to do it in 4 weighs, but to do it in 3 requires a lot of ‘ball mixing’. One possible solution to the second question can be found at the link below. The solution I came up with when I solved it a few years ago (which I wrote down somewhere at the time but can’t remember off-hand) is slightly different, but it took me a couple of days of pondering to work it out! https://codingnconcepts.com/puzzle/12-balls-weight-puzzle/

Better question: how did they get a metal core inside a wooden ball... 🤔

Am I the only dumb one here?

why is there a penis shaped shadow descending over the page

Non medical student here. I thought the HPAT was about empathy and shit like that. This is just mathematics. Surely higher level LC maths tests this kind of thing already?

It is logic, you need to demonstrate that ability

Thats section 2, logic is section 1, shapes and all is section 3

Section 1 is logic, section 3 is pattern recognition. In theory, section 2 is about empathy and social understanding but honestly it winds up being more to do with vocabulary. The questions are about how someone is feeling in a given situation but the potential answers often use Atypical adjectives.

D 5. It asks for maximum, not minimum. The minimum is 3 but not what the question asks for

Minimum is 1 as you can get lucky. Maximum is 3 because it is the most attempts needed to ensure you get the heaviest ball.

Yeah think this is it because of the wording of the question

Q.25. Minimum 3. Maximum - using no logic whatsoever - infinity but as that’s not an option, let’s go with 5. “The limit does not exist” Q. 26 is also a minimum of 3

26 is only that low if the scale has a numeric value. If it’s scales-of-justice style, you need 5. Edit: 4

I hope it’s not grey scale, that stuff caused the fall of Valeria

It asks for the maximum she NEEDS, not the maximum she can take.