/sci/ - Science & Math » Thread #13038330

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Can you solve this math problem?

Anonymous Tue 27 Apr 06:42:50 2021 No.13038330 View Reply Original Report

Quoted By: >>13038380 >>13038464 >>13038528 >>13038550 >>13038606

I just started IB (International Baccalaureate) mathematics AA HL today, and I'm stuck. My peers said it was an easy problem, but I'm still having trouble solving it. I got the solution 9 cows, 3 pigs, 2 sheep, but my teacher said it's not correct. Please help me out here, thanks.

Anonymous

Anonymous Tue 27 Apr 2021 06:44:29 No.13038339 Report

Quoted By:

Funny

IB student 14

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IB student 14 Tue 27 Apr 2021 06:45:56 No.13038346 Report

Quoted By:

I think a viable solution is 9 cows 3 pigs 2 sheep.

Anonymous

Anonymous Tue 27 Apr 2021 06:46:03 No.13038347 Report

Quoted By:

10i+20-38i+80+28i = 100
10i*10+(20-38i)*3+(80+28i)*0.5 = 100

Anonymous

Anonymous Tue 27 Apr 2021 06:46:45 No.13038352 Report

Quoted By:

hello, i think 10 cows is $100. please tell me if that is not true. these cows may be undervalued. check with your teacher.

IB student 15

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IB student 15 Tue 27 Apr 2021 06:47:21 No.13038357 Report

Quoted By:

10 cow

impostor from anognus

impostor from anognus Tue 27 Apr 2021 06:48:13 No.13038361 Report

Quoted By:

94 chicken one cow and 4 sheep

monster.triffle113

monster.triffle113 Tue 27 Apr 2021 06:49:43 No.13038371 Report

Quoted By:

9 cows 3 pigs and 2 sheep don’t add up to 100 animals faggot.

imposter from among us

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imposter from among us Tue 27 Apr 2021 06:49:47 No.13038372 Report

Quoted By:

After careful consideration I have reached the answer 1 cow 1 pig and 174 sheep. I believe this is the optimal answer at the vertex of the parabola

IB San Francisco Mathematics department Mr. Shefield

IB San Francisco Mathematics department Mr. Shefield Tue 27 Apr 2021 06:50:35 No.13038377 Report

Quoted By:

I can confirm that its 9 cows, 3 pigs, 2 sheep as the total adds up to $100. This is an introductory question into the start of the IB 11 course. Your teacher should have said that you are correct. Please confirm back once you get into contact with your teacher.

Anonymous

Anonymous Tue 27 Apr 2021 06:52:00 No.13038380 Report

Quoted By: >>13038398 >>13038462

>>13038330
Your solution doesnt add to 100 animals. You need to set up a system of equations. C= number of cow, p=number of pig, s=number of sheep
C+p+s=100 (100 total animals)
10c+3p+.5s=100 (100$ spent)
Solve for a natural number solution for each variable

monster.triffle113

monster.triffle113 Tue 27 Apr 2021 06:52:51 No.13038383 Report

Quoted By:

Answer is 94 sheep 1 pig 4 sheep. Google search it faggot why are you wasting time on this shithole site retard

IB San Francisco Mathematics department Mr. Shefield

IB San Francisco Mathematics department Mr. Shefield Tue 27 Apr 2021 06:55:50 No.13038394 Report

Quoted By:

Monster.triffle113 please refrain from using such violent and offensive language. I will have to report your account as anyone should be allowed to ask a viable math questions on this thread. Also, you can't trust all google sites and as the head of the math department I can truly confirm it is indeed 10 cows 3 pigs 1 sheep

Anonymous

Anonymous Tue 27 Apr 2021 06:56:40 No.13038398 Report

Quoted By: >>13038416 >>13038499

>>13038380
OP here, how am I supposed to solve a system with 3 variables given 2 equations? There's an infinite number of solutions right?

cumsex

cumsex Tue 27 Apr 2021 06:57:43 No.13038406 Report

Quoted By: >>13038420

I don't think planet earth has infinite animals

Anonymous

Anonymous Tue 27 Apr 2021 06:58:14 No.13038409 Report

Quoted By:

kys fag no one gives a shit

Anonymous

Anonymous Tue 27 Apr 2021 06:59:48 No.13038416 Report

Quoted By:

>>13038398
Yes but you can maybe find a solution in the infinitely many solutions where all variable are integers between 1 and 100

monster.triffle113

monster.triffle113 Tue 27 Apr 2021 07:00:16 No.13038420 Report

Quoted By:

>>13038406

I think you are mentally retarded. the answers are obviously any multiple of 94 sheep 1 pig 4 sheep up to infinity. go to elementary school kid

IB San Francisco Mathematics department Mr. Shefield

IB San Francisco Mathematics department Mr. Shefield Tue 27 Apr 2021 07:03:49 No.13038433 Report

Quoted By:

monster.triffle11 I have been monitoring this thread to watch out for inappropriate behaviour and I must say as a math teacher with a bachelors degree in mathematics, you are incorrect. I also do not appreciate you slandering the mentally ill and using vulgar language once again. I will give you one final chance but be aware I am monitoring this thread and watching what you are posting.

Anonymous

Anonymous Tue 27 Apr 2021 07:07:29 No.13038444 Report

Quoted By:

Why even ask this question on here u retarded subhuman piece of shit

Anonymous

Anonymous Tue 27 Apr 2021 07:12:00 No.13038459 Report

Quoted By:

solve the question then u bigot. Lets see who the subhuman piece of shit is after you get the answer

Anonymous

Anonymous Tue 27 Apr 2021 07:14:10 No.13038462 Report

Quoted By: >>13038474 >>13038693 >>13038792

94 sheep
1 pig
5 cow
Solved with >>13038380 Method

Anonymous

Anonymous Tue 27 Apr 2021 07:14:21 No.13038464 Report

Quoted By:

>>13038330
He buys 100 animals dimwit. Does 9 cows, 3 pigs and 2 sheep sound like 100 animals to you?

This is the problem:
10x + 2y + 0.5z = 100 (cost)
x + y + z = 100 (number of animals)

There are an infinite amount of solutions to this however we have a further constraint that
x, y, z are natural numbers (i.e. positive integers > 0) such that the solution set should be sparse

Anonymous

Anonymous Tue 27 Apr 2021 07:15:48 No.13038474 Report

Quoted By: >>13038480 >>13038486

>>13038462
OP here, how?
I get the equation 19c+5p=100 but what do I do with it?

Anonymous

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Anonymous Tue 27 Apr 2021 07:17:16 No.13038480 Report

Quoted By:

>>13038474
ngmi

Anonymous

Anonymous Tue 27 Apr 2021 07:18:42 No.13038486 Report

Quoted By:

>>13038474
I was lazy and used a matrix calculator, put the solution for the cow variable to be 0<c<10, since its obvious we need more than 1 cow and less than 10 cows from the initial requirements, then it was easy to figure out what numbers would yield natural number solutions. If you want less hostility, go to /sqg/. I use it a lot because I'm a brainlet

Anonymous

Anonymous Tue 27 Apr 2021 07:23:32 No.13038499 Report

Quoted By: >>13038512

>>13038398
Using the equations we're given and letting the number of sheep be the free variable x.
The number of cows is $-\frac{200}{7} + \frac{5}{14}x$ .
The number of pigs is $\frac{900}{7} - \frac{19}{14}x$ .
These values must be at least 1 and at most 98. From that, we have that $83 \leq x \leq 94$ . From here you could brute force it.

Anonymous

Anonymous Tue 27 Apr 2021 07:27:10 No.13038512 Report

Quoted By: >>13038559

>>13038499
OP here,
Thanks man!

?????<3

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?????<3 Tue 27 Apr 2021 07:27:49 No.13038514 Report

Quoted By: >>13038555

????????????,??????????????????????????,???????????,??????,?????????????????????????????!???????,???????????,???????

????????????100??,?????????????????,?????????????,???????????!????!

Anonymous

Anonymous Tue 27 Apr 2021 07:30:07 No.13038528 Report

Quoted By: >>13038536 >>13038551

>>13038330
$ c + p + s=100 (1) 10c + 3p + 1/2s = 100 -> 20c+6p+s=200 (2) 2000-14p-19s=200 14p+19s=1800 $

dis be a diofantine equation so I use bezout monika,

$ gcd(14,19)=1 14y+19z=1 y=-4, z=3 p=-4*1800 + \lambda * 19 -> \lambda=ceil(4*1800/19)=379 s=3*1800 - \lambda * 14 p=1, s=94 (1) c = 100-94-1 = 5 $

so the solution be 5 cows, 1 white pig and 94 sheeeeeeeeeeeiiiiiiiippp

Anonymous

Anonymous Tue 27 Apr 2021 07:31:22 No.13038536 Report

Quoted By:

>>13038528
oh nono no newlines o shit nigga

Anonymous

Anonymous Tue 27 Apr 2021 07:35:36 No.13038550 Report

Quoted By:

>>13038330
>a cow costs $10, a pig $3, and a sheep costs 504
what the fuck?

Anonymous

Anonymous Tue 27 Apr 2021 07:35:42 No.13038551 Report

Quoted By: >>13038927

>>13038528
$ c + p + s=100 (1 )\\ 10c + 3p + 1/2s = 100 -> 20c+6p+s=200 (2) \\ 2000-14p-19s=200 14p+19s=1800 \\ $

$ gcd(14,19)=1 \\ 14y+19z=1 \\ y=-4, z=3 \\ p=-4*1800 + \lambda * 19 -> \lambda=ceil(4*1800/19)=379 \\ s=3*1800 - \lambda * 14 \\ p=1, s=94 \\ (1) \\ c = 100-94-1 = 5 \\ $

Anonymous

Anonymous Tue 27 Apr 2021 07:36:29 No.13038555 Report

Quoted By:

>>13038514
wu mao

Anonymous

Anonymous Tue 27 Apr 2021 07:37:19 No.13038559 Report

Quoted By: >>13038579

>>13038512
You can refine it even more, for $-\frac{200}{7} + \frac{5}{14}x$ to be an integer, $5x-400$ must be divisible by $14$ .
Then, $5x \equiv 8 \pmod{14}$ or $x \equiv 10 \pmod{14}$ . Only $94$ works given this constraint.

?????<3

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?????<3 Tue 27 Apr 2021 07:41:27 No.13038578 Report

Quoted By:

??!??!??!???!!!

Anonymous

Anonymous Tue 27 Apr 2021 07:42:07 No.13038579 Report

Quoted By:

>>13038559
no need for mods and such, substitute s and you get 19c + 5p = 100, so c must be a multiple of 5. The only multiple of 19 and 5 under 100 is 95, so c is just 5.

Anonymous

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Anonymous Tue 27 Apr 2021 07:50:00 No.13038606 Report

Quoted By: >>13038627 >>13038680

>>13038330
#include<iostream>

int main(){
for(int c=1; c<=100; c++){
for(int p=1; p<=100; p++){
for(int s=1; s<=100; s++){
if(10*c + 2*p + 0.5*s == 100 && c + p + s == 100) break;
}
}
}
std::cout << "c: " << c << std::endl;
std::cout << "p: " << p << std::endl;
std::cout << "s: " << s << std::endl;
}

Anonymous

Anonymous Tue 27 Apr 2021 07:58:00 No.13038627 Report

Quoted By: >>13038693

>>13038606
Doesn't compile and even if it did wrong solution the break just breaks the last loop. Write some unit tests next time

Anonymous

Anonymous Tue 27 Apr 2021 08:15:27 No.13038680 Report

Quoted By:

>>13038606
filter(x -> sum(x) == ([10 3 .5]*[x...])[1] == 100,collect(product(1:100,1:100,1:100)))
1-element Array{Tuple{Int64,Int64,Int64},1}:
(5, 1, 94)

Anonymous

Anonymous Tue 27 Apr 2021 08:17:35 No.13038693 Report

Quoted By: >>13038695

>>13038627
You're right, was to lazy to test it and wrote it up on the fly. This compiles though, there is more than one solution:

#include<iostream>

int main(){
for(int c=1; c<=100; c++){
for(int p=1; p<=100; p++){
for(int s=1; s<=100; s++){
if(10*c + 2*p + 0.5*s == 100 && c + p + s == 100){
std::cout << "c: " << c << std::endl;
std::cout << "p: " <>13038462
This answer is wrong, check it for yourself:
94*0.5$ = 47$
47$ + 2$ + 50$ = 99$

The correct answers are
>1 27 72 (10$ + 54$ + 36$)
and
>4 8 88 (40$ + 16$ + 44$)

Anonymous

Anonymous Tue 27 Apr 2021 08:18:30 No.13038695 Report

Quoted By: >>13038698

>>13038693
you are so fucking retarded.

Anonymous

Anonymous Tue 27 Apr 2021 08:20:25 No.13038698 Report

Quoted By:

>>13038695
>3$ instead of 2$
I am indeed blind

Anonymous

Anonymous Tue 27 Apr 2021 08:43:38 No.13038792 Report

Quoted By: >>13038927

>>13038462
Pls explain process sensai

Anonymous

Anonymous Tue 27 Apr 2021 09:11:56 No.13038927 Report

Quoted By:

>>13038792
this
>>13038551

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