Help an idiot solve simple math problems

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A small brick of text. If you don't wanna read just look at the bottom of the post there is the question.

Well there is this "new" thing my beloved country of Croatia came up with for students who finished high school and want to go to college called "State matura" which is basically a series of tests on various subjects such as math,english,physics,etc... and the three main subjects involve 1.Croatian(our language),2.Mathematics,3.Foreign language for which the students can choose either the standard or advanced test depending on what the college requires. Anyway i took all the tests and among them advanced math believing that i would pass... but well... i didn't .... i failed by 1% under the required score to pass. So i had to wait 2 months to register for the 2nd try and now i took the standard level, but in the meantime instead of studying i've been screwing around and now i'm back to being a dummy. So instead of going to a math dedicated forum where i'd get laughed at i've decided to get laughed at here. I need some help with .... well standard math... funny ain't it?
Well i found the earlier copy of the 2010 standard level test and i only managed to solve around 70% ... so i'm askin you for help.

First this one.

The first question
A milk product comes in packages of 330g or 500g(g being grams). The merchant received the amount of 55 550g of that milk product in total of 140 packages. How many smaller packages did he get?

Now this seems really easy to me but i've never been good with these types where you have to figure the formula yourself(ya i know i'm THAT dumb) but i just can't seem to solve it.
 
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60.5 500g packages
91.7 330g packages
 
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according to the answers sheet that is not the correct answer. It's supposed to be 85 330g packages. Also i'm not looking for an answer i'm looking for a "how-to" . Thx anyway
 
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Nevermind, hold on, I misread the problem. Ok, the way you've written the problem is kinda confusing. Did he receive 55,550g in 140 packages? Because that wouldn't make sense.
 
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Nevermind, hold on, I misread the problem. Ok, the way you've written the problem is kinda confusing. Did he receive 55,550g in 140 packages? Because that wouldn't make sense.
well that's the way the question is written... it's 55550g total
 
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Basically the question is as follows spunky.

The guy received 55550g in 140 packages. The packages are either 330g or 500g. The question is how many packages of 330g did he receive.

Basically 330x + 500y = 55550 where x+y=140 The question is: How much is X?

---------- Double Post below was added at 04:39 PM has been merged with this post created 04:28 PM at ----------

I just did a quick calculation though i did cheat a bit.

The answer is 85 packages of 330g.

the real way to solve this would be to put the 2 equations next to each other and then eliminate Y. After that you are left with an equation where X is the only variable.

How exactly to do that. Dont ask me. Been years since i did that sort of math ^^;
 
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The total is 55,550g. With packages coming in 500g and 330g, and there being 140 of them all together.

Let's call the number of 330g packages "x" and the number of 500g packages "y". As Grega's said;

1)x + y = 140 (as the total number of packages is the number of 330g and 500g packs added together)

We also know that the total mass is;
330x + 500y = 555500 (330g times how many there are of that one and likewise for 500g)

This cancels to;
2)33x+50y=5555 (divide through by 10)

If we rearrange the one at the top;
y= 140-x

Replace "y" in 2) with this new term;

33x + 50(140-x)= 5555
33x + 7000 - 50x = 5555
7000- 17x = 5555
7000- 17x - 5555 = 0
1445 - 17x = 0
1445 = 17x
85 = x

Grega's spot on. 85 of 330g and 55 of 550g.
 
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The total is 55,550g. With packages coming in 500g and 330g, and there being 140 of them all together.

Let's call the number of 330g packages "x" and the number of 500g packages "y". As Grega's said;

1)x + y = 140 (as the total number of packages is the number of 330g and 500g packs added together)

We also know that the total mass is;
330x + 500y = 555500 (330g times how many there are of that one and likewise for 500g)

This cancels to;
2)33x+50y=5555 (divide through by 10)

If we rearrange the one at the top;
y= 140-x

Replace "y" in 2) with this new term;

33x + 50(140-x)= 5555
33x + 7000 - 50x = 5555
7000- 17x = 5555
7000- 17x - 5555 = 0
1445 - 17x = 0
1445 = 17x
85 = x

Grega's spot on. 85 of 330g and 55 of 550g.
wow thanks. It really is easy , it's just that i'm an idiot so i can't figure these things right away
 
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Don't worry, it's one of those questions where you need to apply principles, the content isn't too bad once you see it ^^.
 
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Omg it's been so long since I did algebra, that was fun and helpful hahaha
 
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Ah systems of equations. It gets fun when they assign those massive 10 variable problems :p And by fun I mean not particularly difficult, but time consuming :p
 
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Here's another one i can't seem to figure out right now(might be because i'm sleepy or dunno)
Second question:
Martin went to buy school accessories. He spent third of his money on notebooks, then, he spent a quarter(one fourth) of what was left on pens and in the end he spent half of what was left on a pencil case. He was left with 18kn(croatian currency). How much money did he have at the beginning?

now this, like the last problem this is a multiple choice question so i managed to solve it by using the answers one by one and then finally getting the right one but that is not how i want to solve this. The way i did it takes more time (unless the first choice you take is the correct one) and i also want to know which type of a arithmetical problem this is. I tried to go backwards... multiplying 18x2 then 36x4 and that's when i figured that what i'm doing doesn't really make sense. And i just can't picture the problem in my head.
I tried imagining it like this:
1/3x=y
1/4y=z
1/2z=18
now i tried replacing those terms with equals similar to what Dokutayuu showed in the previous question and well... it's the same as multiplying it in reverse just a different approach

I wish i had my books but unfortunately i had to sell them and the internet is useless(most of the tutorials i found that are in croatian either take the weirdest and unnatural way to explain things or actually leave me with more questions... but that's probably just me).Anyway i suck the most with these type of problems where i can't figure what i'm supposed to do. Well i guess that just shows how limited and deconcentrated i am.
 
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Let me check

Your main error is equating them like that.

1/2z=18 thats true

But 1/4y = z is false

If he spent 1/4 of his cash, then z=3/4 of what he had left And y = 2/3 of what he started with.

so

Y = 2/3X
Z = 3/4Y

And you know that before he bought the last item ha had 36kn

So 36 = 3/4Y

36/3=1/4Y
12=1/4Y
12*4= Y
Y=48

Y=2/3X
48=2/3X
48/2=1/3X
24=1/3X
24*3=X
X=72

that should be the answer.
 
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Okay, this seems easy. You have x, which is the original amount. Then you have amount a (which is after buying notebooks) b (after pens) and c (after pencil case).

a=2/3 x
b=3/4 a
c=1/2 b = 18kn

->> B= c*2, a=b*4/3, x=a*3/2

->>> b= 36kn a=48kn

->>>>>x=72kn



As to how to solve it when you have a brain fart, simply translate the text into a math formula, this is pretty much what I did.
 
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Or without introducing all them dummy variables:
after the first purchase he has 2/3 of his money left.
after the 2nd 3/4th of 2/3 = (2/3)*(3/4) = 6/12 = 1/2
after the 3rd half of that: (1/2)*(1/2) = 1/4

So 1/4 of his money is 18kn, 4 * 18kn = 72kn
 
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the coder comes in and owns the thread!
 
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Hes working in a university that doesnt count >.>
 
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Or without introducing all them dummy variables:
after the first purchase he has 2/3 of his money left.
after the 2nd 3/4th of 2/3 = (2/3)*(3/4) = 6/12 = 1/2
after the 3rd half of that: (1/2)*(1/2) = 1/4

So 1/4 of his money is 18kn, 4 * 18kn = 72kn
WOW!! Never would have thought of that....

Anyway do you guys have any advice for me as for how to recognize what the problem is and how to solve it. I mean i practice a lot but whenever i do i always have a clear problem which can be solved using a number of formulas but when i get text problems like these most of the time i lose myself in it. I'm not looking for a magic spell which will make me aware of every single thing but just some basic tips on what to pay special attention to. You see during all 4 years of high school i never had to solve problems like these... we always got straight up numbers and unknown terms and i didn't have that much trouble. Hell even in the Advanced math test the only text based problem was the last question... yet i still failed it... by 1% ....
 
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Finding information in context can be a little tricky. Remember, most of Maths can be expressed as Algebra. Dummy variables can be helpful but they can often confuse you further. You need to identify all relationships between variables and you can work from there.

You got a lot further in the second question though, with a little more practice, I think you'll get it soon.
 
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If you assign variables, start out with assigning variables to the thing you want to know (number of 330g packages, number of 500g packages in the first question). Sometimes you can then just go through the text and write all information on those variables in formula form (e.g. x+y=140, x *330+ y*550=55 550). In the case of question one this leads to a set of equations that you can then simply solve.

Key to the second question is that the information given is in different 'types' the 1/3, 1/4, 1/2 are on proportions spend, while the 18kn is on money left. In this kind of questions, it's often beneficial to see if you can convert one into the other. In this case you can convert the proportions spend into proportions left.
 

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