/sci/ - Science & Math

[Anonymous No.15312824]

Am I just low IQ or is this test unreasonably hard for a job interview? It says you get 35 minutes and if you don't get more than 6/9 questions correct you won't get accepted.
(1/2)

[Anonymous No.15312826]

[Anonymous No.15312829]

>>15312824
Anon, I... wish you all the best.

[Anonymous No.15312837]

>>15312826
>>15312824
kek, they administer IQ tests for jobs where you live?

[Anonymous No.15312847]

>>15312824
Not really an iq test as much as a Math test.

[Anonymous No.15312848]

>>15312824
> is this test unreasonably hard for a job interview
For shelf-stacking, yes.

[Anonymous No.15312850]

>>15312837
Every internship I've applied to required me to complete a sort of IQ test (diagrammatic/spatial reasoning) as well as a coding test. I generally score decently on them but I've never come across one as hard as this. Maybe it wouldn't be that bad except you only get 35 minutes... I've got a test booked with them on the 11th but I don't think I'm smart enough.

[Anonymous No.15312856]

>>15312824
This test is racist

[Anonymous No.15312879]

>>15312824
https://youtu.be/1sFyrfqTdcg

[Anonymous No.15312904]

>>15312824
I got #2!!!
its 416723

[Anonymous No.15312926]

>>15312824
Hint for #1: p = 3 and q = 5

[Anonymous No.15312948]

>>15312824
#5 is asking you to do modulo 12 arithmetic

[Anonymous No.15312949]

>>15312824
>>15312826
and for 7. its 617
to lazy for the rest

[Anonymous No.15312951]

>>15312824
So here's the deal. 2-7 are all quite solvable. 1 and 9 I think are legit faulty/impossible to solve and 8 requires Math Olympiad type knowledge.

I think this is a test where you're supposed to realize something along those lines without getting discouraged or wasting too much time on the 3 remaining problems.

Go to the test with that in mind and you should be fine

[Anonymous No.15312985]

>>15312949
There's a fast way to get this one:
>bottom left corner square must contain a 4 (unique row/column)
>upper left corner square must contain a 3 (unique row/column)
>upper left corner letter must be D (A eliminated by row, B eliminated by column, C eliminated because C3 is already in the grid)

>>15312951
9 is a well-known puzzle: number the stacks N=1,2,...,11 and put N weights from stack N simultaneously onto the scale. Then the excess weight (from what you expect them to be) is exactly N kg (+1kg from each of the N weights), which tells you the stack number.

8 is just adding the areas of the 4 semicircles and then subtracting the area of the square.

1 is easy once you know the trick >>15312926: write n1 + n2 + n3 = (n2 - 2) + n2 + (n2 + 2) = 3 * n2 and you'll see that p^3 is a multiple of 3, so p has to be 3. Likewise for q = 5.

[bodhi No.15313034]

>>15312824
1. k=62, p=3, q=5, n1=7, nk=7+2(k-1), k=62, nk=129 (simple arithmetic progression)
2. 416723 (logic, write down the prime numbers from 13 to 97, determine the minimum and maximum sum of a sequence of 6 distinct digits, find EF, find AB with the sum, complete with CD with the total sum)
3. D (4th letter of alphabet, replace letters with entries in the English alphabet)
4. D3 (just a simpler variation of sudoku, you can reduce everything to C4, D3, D4, complete the other squares)
5. 6 PM (47 999 996 is divisible by 24 because it's divisible by 2, 3 and 4, it denotes the number of days, subtract 1 hour)
6. 40%, 20% (simple probability, they can either be all in the same class, all in different, or 3 other variations in which 2 of them are in one class and the other one in one of the other 2 classes)
7. 617 (just logic)
8. 2x^2(4 - pi) (hint: find the area of the full square, than the one of a semicircle, than the one of a circular pointy triangle, then the final surface)
9. divide them into groups of 5. if the balance is equal: the leftover stack is faulty. if it is uneven: heavier stack is faulty.

time taken: 27 minutes.
skeptical about: 9

[bodhi No.15313041]

>>15313034
fuck i fake posted as bodhi again JFL

[bodhi No.15313048]

>>15313034
ERRATA FOR 5:
48 000 000 is divisible by 24
the corect tine is 2 PM
sorry for stupidity

[Anonymous No.15313060]

>>15313034
Those are the answers.
You got 5/9.
Still better than me. Guess I'm just not smart enough for this job lel.

[Anonymous No.15313079]

>>15313034
>7. 617
this violates condition 3.

[Anonymous No.15313083]

>>15313079
wait i cannot read nvm

[Anonymous No.15313097]

>>15313060
i don't think those probabilities are correct

[Anonymous No.15313122]

>>15313097
The probabilities are fine.
Imagine that we assign the classes by drawing up a 20*3 grid of squares, then going through a list of student names and writing them down, one by one, into a randomly chosen empty square. (Then we take the three columns to be the assigned classes.)
We also assume that the list of names is sorted alphabetically, so Alice gets assigned first, then Bob, then Caroline.

In the beginning, Alice is assigned to any one of the 60 empty squares. Then, when Bob is assigned, there are 59 squares remaining, of which exactly 19 of them are in the same column as Alice's square. This yields the part 1 probability of 19/59.

Now suppose that Bob is assigned a square in a different column from Alice (this happens with probability 40/59). Then when Caroline is assigned, there are 58 squares left to choose from, of which 19 are in Alice's column, 19 are in Bob's, and the remaining 20 are "fresh". So the probability that Caroline is assigned a square in the "fresh" column is 20/58.
Multiplying together, we get the part 2 probability of 40/59 * 20/58 = 800/3422.

[Anonymous No.15313128]

>>15313060
Probability is wrong. It should be 19/59 * 18/58

[Anonymous No.15313133]

>>15313128
Fuck misread the question, thought it said in the same class

[Anonymous No.15313172]

>>15313122
aaaa kek
thanks probabilityanon
i think i am somewhat mentally retarded because I've always struggled with probabilities while other branches of math came naturally to me

[Anonymous No.15313186]

Can someone explain #3 to me?

[Anonymous No.15313241]

>>15313186
In each example pattern:
>the two letters that make up the top edge are exactly 1 letter apart
>the two letters on the left edge are exactly 8 letters apart (wrapping around from Z to A, e.g. V WXYZABCD E)
>the right-edge letters are 11 letters apart
>the bottom-edge letters are 4 letters apart
Either of the last two facts will let you deduce the missing letter.

[Anonymous No.15313251]

>>15313186
...and right after posting >>15313241 I just noticed the faster and cleaner solution (which would explain what the middle square is doing there)

[Anonymous No.15313260]

>>15313251
Ah! Thanks. That's the pattern. I just didn't see it.

[Anonymous No.15313278]

For number 1 you use a disgustingly clever technique by expressing the terms in each sum by the middle value so you can eliminate them.

Look:
[math]n_1 + n_2 + n_3 = (n_2 - 2) + n_2 + (n_2 + 2) = 3n_2 [/math]

You can do this because we are given that they are consecutive odd integers, and we know that consecutive odd integers differ by 2. You have to express it by the middle value so you can eliminate the +2 and -2.

So now [math]3n_2 = p^3[/math] implies that [math]p^3[/math] is divisible by 3.

However, one definition of primality is if you have a prime number (call it q) and q divides any product, then it must divide one or both of the constituent elements of that product. Ie: [math]q|ab \Rightarrow q|a[/math] or [math]q|b [/math] or both.

So applying this principle to what we have. We know 3 is a prime number, so if 3 divides [math]p^3[/math] it follows that 3 divides one of the products. But this means that 3 divides p.

But if 3 divides p, p is a multiple of 3, which doesn't make sense given our assumption that p is prime. The only option then is p=3.

Now with the other sum, our middle element this time is [math]n_{k-2}[/math]:
[math](n_{k-2} - 4) + (n_{k-2} - 2) + n_{k-2} + (n_{k-2} + 2) + (n_{k-2}+ 4)=q^4[/math]

So, by the same argument, q=5.

We know now that q=5, p=3. How do we find k?

Well, since it's a sequence of consecutive odd integers, we can express [math]n_k = 2k - 1[/math]. This takes some thinking to understand.

Now, since we have a formula, let's find [math]n_k[/math].
[math]n_k + (n_k - 2) + (n_k - 4) + (n_k - 6) + (n_k - 8) = 5n_k - 20 = q^4 = 5^4 = 625[/math]

So we know [math]5n_k = 645 \Rightarrow n_k = 129[/math]

Lastly, [math]129 = 2k -1 \Rightarrow 2k = 130 \Rightarrow k = 65[/math].

So in fact finding the value of p wasn't even necessary. I guess they just put that in to throw you off?

Either way, a very tricky problem which requires some knowledge of number theory and a lot of thought. Too complex for a coding job.

[Anonymous No.15313285]

>>15313278
Ah, looking at the answers it seems I got it wrong nevermind xD

[Anonymous No.15313288]

Fast(-ish) solution for #2:
>From [1] and [3], the last two digits EF are at most 9+8+7+6+5+4 = 39.
>From [4] and [5], EF is one of these 2-digit primes: {31, 23, 17, 13}. (It cannot be a 1-digit prime, because [3] would leave no space for the other digits.)
>From [2], AB is a 2-digit prime with the same digit sum as EF (but different digits, from 1). This rules out EF=31 and EF=13, leaving 41xx23 and 53xx17 as the only remaining possibilities.
>From [3], rule out 53xx17 leaving 416723 as the only correct answer.

[Anonymous No.15313301]

>>15313278
Ah ok I see where I went wrong. Instead of [math]n_k = 2k-1[/math] it should be [math]n_k = 2k + 5[/math] since we don't start at n_1 = 1 but at n_1 = 7. You can confirm this because 7 + 9 + 11 = 27 which is exactly 3^3. So in fact finding p was necessary. Sorry.

[Anonymous No.15313566]

>>15312824
Depends on the job. This is definitely too hard for most jobs, but might be appropriate for a quant or data scientist role.

1. you can rewrite them as 3*n1+6 = p^3 and 4*nk-20 = q^4. For the first one, note that 6 more than a multiple of 3 must be a multiple of 3, and for the second, note that a multiple of 4 minus 20 must be a multiple of 4. You should be able to go from there.

[Anonymous No.15313579]

>>15312824
>6/9 questions
Nice.

[Anonymous No.15313581]

>>15313034
Are you sure about 6? Remember that if Alice is in a class, that leaves only 19 spots for Bob or Caroline (vs 20 for the other two classes).

[Anonymous No.15313646]

>OP can literally trick /sci/ into doing his elementary school homework by calling it an "IQ test"
The absolute state of this board ...

[Anonymous No.15313727]

>>15313034
The area of the clover problem is not right. The area of the white space is 2 x the area of the square minus the inscribed circle.

So the area of the black space is the area of the area of the square less that.

[Anonymous No.15313755]

>>15312824
>It says you get 35 minutes
calculator though? For example the final problem is just a big divide-and-remainder, its not testing your intelligence any further than the first or second qs which require the same operations on smaller integers.

[Anonymous No.15313758]

>>15313251
These puzzles are such bullshit imo because if they were just using numbers it would be immediately obvious. The next obvious approach is indeed to just index the letters and make them into numbers, but there is always a chance that this is a red herring and you’re wasting critical time doing it. People think about the alphabet like a “linked list” so it always takes a stupid amount of time to try and figure out how far apart letters are from each other. So the reasonable approach is to try and gauge the intended difficulty of the problem and guess if the designer is simply wanting you to index the letters or is baiting you into wasting your time.

[Anonymous No.15313760]

>>15313755
ehh I take that back, I guess it's asking you to make the difference up to 48M hours and then divide it trivially, take the 5 back out. I guess the rest are appropriate difficulty for pen and paper too.

[Anonymous No.15313767]

>>15313755
IYou’re supposed to notice that it’s a few less than 48,000,000 which is 2,000,000 days. And it’s cyclic or whatever so you just wind the click back however many hours less than 48,000,000 it is.

You’re also supposed to recognize that this is the Trick and not think too hard about how the calendar is actually adjusted over time with respect to sidereal time which makes this problem almost impossible to actually solve.

[Anonymous No.15313768]

>>15312826
>#11

You really cannot tell simply by hand if a stack is 11kg overweight?

[Anonymous No.15313795]

>>15312985
Thanks for the help with the ones I had the least clue on.

for 8 I thought to just construct an integral to find the area of one, and multiply by 4.

[Anonymous No.15313804]

>>15312848
>>15313646
I honestly don't understand what people like you get out of making comments like this. Like do you think this "shows them" or something? Like that you come off looking so smart and you put them down because they belong down there.

Do you really think that's how an intelligent person acts? That they feel a need to put down others at every opportunity? You suck.

[Anonymous No.15313808]

>>15313767
>recognize that this is the Trick and not think too hard about how the calendar is actually adjusted over time
yeah I got that as soon as I actually looked at the digits instead of the lengths of the numbers. It also obviously doesn't take into account leap seconds or anything, I guess it's just about seeing the ratio.

> It says you get 35 minutes
I still think this is far more than enough time to figure them all out, even if you weren't familiar with these styles or structures of problems already, with an AP/honors highschool math education.

[Anonymous No.15313819]

>>15312824
1. hard (for me)
2. easy
3. easy
4. literally sudoku
5. easy
6. easy
7. easy
8. easy
9. no fucking idea

[Anonymous No.15313891]

>>15313804
I am intelligent and I hate unintelligent plebs. It is my right and my moral obligation to assert my intellectual superiority.

[Anonymous No.15313899]

>>15313891
No intelligent or successful person talks or thinks like this. You're a midwit at best.

[Anonymous No.15313919]

>>15313899
Intelligence comes with no obligation to be emotionally stable, mature, benevolent or modest. I can perfectly use my high IQ to belittle others in order to compensate for my social disabilities, my traumatic childhood and the bitterness caused by a chronic disease. What are you gonna do about it, midwit? Post more irrelevant platitudes about how you erroneously assume smart people ought to behave?

[Anonymous No.15313934]

>>15313919
if you're so smart then why does your life suck?

[Anonymous No.15313939]

>>15313934
Because we live in a soiciety that hates and discriminates against smart people.

[Anonymous No.15314141]

>>15313939
why do you care about the opinions of stupid people? That seems pretty dumb.

[Anonymous No.15314188]

>>15312850
Wait what? I've been asked once to take a test for a job and I turned them down. I always turn them down along with those retarded video record interviews.

It's like ultimate wagie bending. You're a JOB not an education, I'm not here to be certified by you.

[Anonymous No.15316120]

>>15312824
3=D
4=D3
>>15312826
7=697
9=faulty stack will weigh 11kg more

[Anonymous No.15316142]

>>15316120
shit, meant 617

[Anonymous No.15316153]

>>15313581
It would be 1/9th and 1/3rd anyway, even if you didn't bother taking that into account.

[Anonymous No.15316286]

>>15312826
For 9 the best solution I could come up with was taking 1 block from each stack and weighing them together, then removing them 1 by one to determine which one was from the faulty stack. Would this count as using the scale more than once?

[Anonymous No.15316319]

>>15313566
> Note obvious shit that doesn't help and go from there

Thanks

[Anonymous No.15316444]

>>15312824
Some of them require some rational calm thoughts, like q5. mod 24. you're 5 short. so 7pm - 5 hours = 2pm.
That's done in under a minute.
The rest aren't too hard either I don't think. Not sure if I'd have done it in 35 mins now, as I'm getting very rusty with these problems.
I think a few took me longer than 5 mins to work out. Add interview environment to it, and maybe I'd have failed too.

Not gonna lie though, this looks like a shitty interview question set. If you've seen these problems before, you can do them in substantially less time. Good example is q9, where you either write down answer, or spend time working it out. Having a time constraint on it just biases your candidates to the ones that have seen the problems before.
Best example of this bias might be q8 though
https://www.teachoo.com/1885/549/Example-6---Find-area-of-shaded-design--ABCD-is-a-square-10-cm/category/Examples/
Video explanation is 3 mins long. Seeing the problem afresh, 5 mins doesn't give you a lot of time to derive the idea from scratch, unless you've done a lot of these problems before.

But still, if it's a technical job, solving 6 out of 9 in 35 mins sounds ok if you want good candidates. All of these problems would have been covered in any king of mathematics competitions in primary/secondary schools.

[Anonymous No.15316466]

>>15316286
yes.
solution here
>>15313060
also, your method relies on remembering which coin is which.
now that you know the answer, you won't easily forget it.

[Anonymous No.15316493]

>>15312826
For (9), Take 1 weight from stack 1, 2 from stack 2, 3 from stack 3, etc. Then the weight should be 11x+y where x is the weight of each block and y is the stack with faulty blocks.

[Anonymous No.15316525]

>>15312824
I would give up after 35 seconds, hand in the paper, and go back to the lousy job I have now. I'm too old for this.

[Anonymous No.15316745]

>>15312948
Literally just subtracting 5.

[Anonymous No.15317141]

I got 7.5/9, but took closer to 1.5 hours rather than 35 mins.
Here is python for 4

def four():
arr = [
['', '', 'a1', ''],
['', '', '', ''],
['b2', '', '', ''],
['', 'c3', '', 'b1']
]
possible = [x + y for x in ['a', 'b', 'c', 'd'] for y in ['1', '2', '3', '4']]

def validate(mat):
# validate rows
for row in mat:
all_letters = [x[0] for x in row if len(x) == 2]
if len(all_letters) > len(set(all_letters)):
return False
all_numbers = [x[1] for x in row if len(x) == 2]
if len(all_numbers) > len(set(all_numbers)):
return False

# validate columns
for i in range(4):
col = [row[i] for row in mat]
all_letters = [x[0] for x in col if len(x) == 2]
if len(all_letters) > len(set(all_letters)):
return False
all_numbers = [x[1] for x in col if len(x) == 2]
if len(all_numbers) > len(set(all_numbers)):
return False

all_placed = [x for row in arr for x in row if len(x) == 2]
return len(all_placed) == len(set(all_placed))


def f(i, j):
if i == 4:
return True

if j == 3:
next_i = i + 1
next_j = 0
else:
next_i = i
next_j = j + 1

if len(arr[i][j]) == 2:
ans = f(next_i, next_j)
return ans

for val in possible:
arr[i][j] = val
if validate(arr):
ans = f(next_i, next_j)
if ans:
return True
arr[i][j] = ''

return False

f(0, 0)
for row in arr:
print(row)
return arr[0][0]

[Anonymous No.15317925]

>>15312824

[Anonymous No.15318029]

>>15312824
The answer to the first question is 61.
Second is 471,329.
I haven't attempted the others.

[Anonymous No.15318337]

>>15314188
so what jobs did you get that dont have this type of interviews and tests?

[Anonymous No.15318462]

sol to q2

[math]0+1+2+3+4+5 \leq \overline{EF} \leq 4+5+6+7+8+9 [/math]
[math]\Rightarrow 15 \leq \overline{EF} \leq 39 [/math]
[math]\Rightarrow \overline{EF} \in \{ 19, 23, 29, 31, 39 \} [/math]

Cross out [math]19, 29, 39[/math] since they are all [math]\geq 10[/math]. Also cross out [math] 31 [/math], since this implies [math] \overline{AB} \in \{ 04, 22, 40\} [/math] none of which are valid. So, [math] \overline{EF} = 23 [/math]

Hence [math]\overline{AB} \in \{ 05, 41 \}[/math] But since [math] A \neq 0 [/math], [math] \overline{AB} = 41 [/math]

Finally, [math] C+D = \overline{EF} - A - B - E - F[/math]
[math]\Rightarrow C+D = 13[/math]
[math]\Rightarrow \overline{CD} \in \{ 49, 58, 67, 76, 85, 94 \} [/math]

Then cross out invalid ones and get [math] \overline{CD} = 67[/math]

[math]\overline{ABCDEF} = 416723[/math]

[Anonymous No.15318466]

>>15318462
Where did you get the idea for the first inequality?

[Anonymous No.15318477]

>>15318466
>>15318462
Ignore it, I get it now.
It's between the lowest possible sum of 6 different digits and the highest possible sum of 6 different digits.

[Anonymous No.15318479]

ok but anyone know the sauce to OP's pic??

[Anonymous No.15318493]

>>15318462
**[math] \require{color} \overline{EF} = \{ 19,23,29,31, \textbf{37} \} [/math], not 39

[Anonymous No.15319710]

>>15318493
I don't think you can rule out [math]\overline{EF}=17[/math] at this stage either.

[Anonymous No.15319715]

>>15312824
Seems totally unreasonable to me.

[Anonymous No.15319941]

Very cool stuff! Took me about 40 minutes. Not so sure about #6, though.