Help an idiot solve simple math problems

[MrPlow]

Thx... but i found out something about myself. If i haven't solved a similar problem(and by similar i mean that it uses the same formula) i won't be able to solve any new problems. I found out that my ability to think and react accordingly by using logic is poor to say the least. I'm more like a bot that can only do something only when i know how. For example: now i know how to solve problems similar to the first and second question( and i have found a few in earlier tests and managed to solve them successfully) but if in lets say second question instead of thirds,halves and quarters they were replaced by percentages i wouldn't know how to solve it. I know the basics of percentages and i know how to solve the problem with halves,thirds,etc.. but when they combine i get lost.

Anyway here is another problem i got stuck at.

Question Three
For 120kn you could buy two chocolates more than after their prices got raised by 25%.

a)How many chocolates could you buy before their prices went up?

b)What is the price of one chocolate after the price went up?


Now in this one i'm kind of lost with the lack of information.
I know i need to find out the amount and price of chocolates. I tried setting it up like this
x-amount of chocolates
y-price

now in my head the basic formula looks like this:
x*y=120
but now i need to include the price raise and those 2 chocolates more. I thought that if i took 25% off of 120 and divided it by 2 that i'd get the price of one chocolate .. and well i think i got the answer to the b) but it bugs me because when i was solving it i had it in mind that i was finding out the price of one chocolate before the price went up... i confused myself with that one. Now for the a) one i'm a bit confused. I'm not sure if i solved it right. I tried by finding out that 25% of 15 is 3.75 and then got that the initial price was 11.25, then i divided 120:11.25 and i got 10.6666.... Now when i looked in the answer sheet it say that the answer to a) is 10 . I got 10.6666 i mean it's 10 and you can't buy .666666.... chocolates but still i'm kind of unsure if i did it right. help?

 

[Grega]

Actually all you need is X, which is the price of 1 chocolate. Everything else comes after.

So saying Y is the number of chocolates you can buy.

120/X = Y

120 = your money
X = price of one chocolate
Y = number of chocolates

Another thing you know is this.

120/(X*1.25) = Y-2 What this means is that after X goes up by 25% you will be able to buy Y - 2 chockolates.

Now switch it round a bit.
(120/(X*1.25) +2) = Y

Now you have 2 equations where you get the same amount which is Y.

So just erase Y from the equation and put both of those together.

120/X=(120/X*1.25)+2

All thats left is for is to get X out of that equation.so put everything that has an X in it on one side and everything else on the other.

120/X - 120/1.25X = 2

now to get rid of the 1.25X you have to multiply the entire thing with 1.25

(120*1.25)/X - 120/X = 2*1.25

Then just calculate it out.

150/x - 120/x = 2.5
30/x = 2.5

Now you have to get X out of the bottom part of the fraction so you multiply the entire equation with X

30 = 2.5x and finally to get X to stand alone you divide the equation with 2.5

30/2.5 = X
And the result is X = 12

So one chocolate costs 12kn BEFORE the price goes up. Now to check if this is correct.

Simply insert X into the first 2 equations.

120 / X = Y
120 / 12 = Y
Y = 10

120 / (X*1.25) = Y - 2
(120 / (12*1.25)) + 2 = Y
(120 / 15) + 2 = Y
8 + 2 = Y
Y = 10

In both cases Y = 10 after you finish the calculation so it means you have the correct answer.

So you have.
A.) 10 chocolates before the price went up and 8 after.
B.) 12kn before the price went up and 15kn after

But i said you dont need Y and thats because you can simply start with this

(120/x)-2 = 120/(x*1.25)

Its the same formula only that you skip a few steps, since they really arnt needed. Since Y doesnt matter. All you need is the fact that you can buy 2 more before the price went up. But i did the entire process as an explanation step.

And yes you can change the 1.25 into 5/4 and the 2.5 into 5/2, but i prefer decimals than fractions. The result is still the same though.

 

[harSens]

You were on the right track. As an alternative to Grega's explanation:
x: amount of chocolate
y: price of one chocolate (before prices were raised)

As you said before price raise: x * y = 120
A price raise of 25% means that the price of one chocolote is 1.25x the price it was, so the new price per chocolate is y*1.25 and you can buy 2 less chocolates for that:
So after price raise: (x-2)*(y*1.25) = 120

So you get
x * y = (x-2)*y*1.25 (since both are 120)
x = (x-2) * 1.25 = 1.25x + 2.5 (divide by y on both sides, allowed because y cannot be 0)
- 0.25x = 2.5 (substract 1.25 x on both sides)
x = 10 => you could buy 10 chocolates before prices went up
x * y = 120 => y = 12 => price per chocolate before prices went up is 12, after it is 1.25*12 = 15

 

[MrPlow]

You were on the right track. As an alternative to Grega's explanation:
x: amount of chocolate
y: price of one chocolate (before prices were raised)

As you said before price raise: x * y = 120
A price raise of 25% means that the price of one chocolote is 1.25x the price it was, so the new price per chocolate is y*1.25 and you can buy 2 less chocolates for that:
So after price raise: (x-2)*(y*1.25) = 120

So you get
x * y = (x-2)*y*1.25 (since both are 120)
x = (x-2) * 1.25 = 1.25x + 2.5 (divide by y on both sides, allowed because y cannot be 0)
- 0.25x = 2.5 (substract 1.25 x on both sides)
x = 10 => you could buy 10 chocolates before prices went up
x * y = 120 => y = 12 => price per chocolate before prices went up is 12, after it is 1.25*12 = 15

I've got one question. I'm feeling kind of tired now so i'm probably missing the obvious but

A price raise of 25% means that the price of one chocolote is 1.25x the price it was, so the new price per chocolate is y*1.25 and you can buy 2 less chocolates for that

how come it's 1.25x? Ain't it 0.25 since 25% of something is 25/100? I dunno

I've got less than 3 days before the test and it's supposed to be easy(IT IS EASY), i mean i almost managed to pass the advanced so this one shouldn't be as hard. But as each day passes i feel more and more tired and deconcentrated. I have a feeling that i'm going be average and i don't want that

 

[Dokutayuu]

It's 1.25 because it's 25% extra. So it's 1 with 0.25 added.

Don't worry, you did pretty well on the harder one. This shouldn't be too bad. Remember not to wear yourself out before exam though, take the day before off.

 

[Grega]

Simply said 100% = 1

a 25% increase means 1 + 25% which is 1.25

 

[MrPlow]

oh... i see... thanks
Well i still need to learn how to draw functions properly... I never was good at anything that involved drawing. When it comes to functions i could solve almost any type of problem i've encountered so far but when i need to draw those i get totally thrown off. At the end of this last year i had to solve a problem involving extremes and maximum and minimum of some random cubic function to pass math with grade 4(equivalent to B in american grade evaluation system or w/e it's called) and i actually managed to solve the whole function right... but my drawing was way off. Also need to work a bit on non-equations or however these are called(those equations with > or < instead of = for which i have to find their intervals or something(forgot how it's called) and present it on a numerical line). I've forgotten them since i haven't done them in a while(it's been general and quadratic equations mostly which are quite simple)

Also i would like to apologize if it was hard for you to understand my posts since english isn't my main language

Well here is another one but this one isn't a text question.
It's a non-equation or w/e and my result is a bit different so..

Problem one
Solve the non-equation(i wish someone would tell me how is this really called) ( (5x-3)/6 ) - ( 3x/2 ) > 1
*i've put fractions in brackets to avoid confusion
Now by the looks of it it's obvious that is not a quadratic non-equation i'm used to but a linear equation.
Now i know i need to find x but well i've messed up with the <,>

anyway first i reduced both fractions to their lowest common number which in this case is 6 so it ended up like this

( (5x-3-9x-6)/6 ) > 0 (i forgot to mention that i switched the 1 that was behind > over to the other side and included it into the fraction)

And after that i multiplied everything by 6 to get rid of the fraction so then i got

5x-3-9x-6 > 0

which when i switched it all around became like this:

-4x > 9

and then finally

x > -(9/4)

but according to the answer sheet the final answer is x < -(9/4)
So i'm 99% sure that i solved it right but i forgot the rule for those non-equations. When do those <,> characters switch?
Also since it's a linear non-equation i guess i don't have to draw it on the numerical line and look for inequalities and such

 

[Dokutayuu]

When multiplying by a negative number, the inequality sign switches.

e.g 3 > 2
but -3 < -2

You did well on this one, in a UK exam paper you'd have got most of the marks for that question.

 

[MrPlow]

When multiplying by a negative number, the inequality sign switches.

e.g 3 > 2
but -3 < -2

You did well on this one, in a UK exam paper you'd have got most of the marks for that question.

oh yea now i remember.... thx

EDIT: i'm stuck.... again -_-

Problem two

Function graph f(x)=2x-4 cuts(touches) the absciss(x-axis) axis in point A, and ordinate(y-axis) axis in point B. Which are the coordinates of points A and B?

I totally forgot how to do that. I'm thinking of making a small table and adding 2 random numbers instead of x(like 1 and 2 or something) and then drawing it but i don't believe that is the right way. Especially since it's a multiple choice question so it can't be just two random numbers. I suck at functions when i'm required to draw it in the cartesian coordinate system. Anyway i hope i translated the problem right so that you can understand me.

 

[Dokutayuu]

f(x) = 2x - 4

At A, it's at the x axis so y = 0. At B, x = 0.

a) So at A, 0 = 2x - 4
4 = 2x
2 = x

So A is at (2, 0)

b) At B, x = 0 so;
y = (2 x 0) - 4
y = -4

So B is at (0, -4)

 

[MrPlow]

f(x) = 2x - 4

At A, it's at the x axis so y = 0. At B, x = 0.

a) So at A, 0 = 2x - 4
4 = 2x
2 = x

So A is at (2, 0)

b) At B, x = 0 so;
y = (2 x 0) - 4
y = -4

So B is at (0, -4)

Yea that makes sense.

I have to remember that f(x) = y and the positions of the numbers in the answers (x,y)

 

[FalconFury]

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.

Man I am stupid, Stupid, stupid. I cannot believe that this guy is full smart and full educational. I have to do math better but first I have to do English. @_@ I am stupid and I am acting like some none sense freak. I made a terrible mistake in my whole life. But the worst part that was EVIL and very confusable is algebra, I hated it. So much and painful. If I don't finish math I will not be the smartest scientist in the whole world. *cry* why am I so stupid, I should do math better and don't get any help from someone. I let my self down.

 

[MrPlow]

Man I am stupid, Stupid, stupid. I cannot believe that this guy is full smart and full educational. I have to do math better but first I have to do English. @_@ I am stupid and I am acting like some none sense freak. I made a terrible mistake in my whole life. But the worst part that was EVIL and very confusable is algebra, I hated it. So much and painful. If I don't finish math I will not be the smartest scientist in the whole world. *cry* why am I so stupid, I should do math better and don't get any help from someone. I let my self down.

What?

 

[Rocky]

This thread is giving me a headache, I used to be so good at this **** but I'm failing at every single one.

 

[MrPlow]

Well tomorrow is the exam and i'm feeling way more pressure than i did when i had the advanced exam. Probably because this is my last chance, and i don't want to mess it up. Don't get me wrong i know i'll pass... just don't know how bad/good i'll do.
Anyway before i take a rest i want to recap and also make sure i know all i need to know. So answer me these questions.

i had a problem which i managed to solve on my own it went like this (2x/x^2-4)-(1/x-2)
well anyway i solved it by taking the x^2-4 and splitting it into two (x-2)(x+2) and then it was easy going on from there but i'm not quite sure about that rule of splitting so can someone explain me when can i use it. For example if it were x^2+4 would i still be able to use it(i believe that i could since + or - it doesn't matter because square is always = +) also if the second known(by that i mean the 4 in the previous example) number is 16,25 or any other number for which a square root is not a decimal can be split in two right? and what if there is a number like 4 in front? it can still split into 2 like (2x-4)(2x+4) or is it 4(x-4)(x+4) and what if any of the numbers is not a rootable number like let's say if it's x^2-13 or 3x^2-16 ?

I also found a problem which is something like this:
Draw a straight line using the given equation: 2x+3y=6 (on the cartesian coordiante system)
now i tried to solve it like this

2x+3y=6
2x=6+3y /:2
x=3+3y/2

2(3+3y/2)+3y=6
6+3y+3y=6
6y=0
y=0

2x+0=6
x=3

now there is something wrong with y...
i looked at the answer sheet and i saw the drawing of a straight line on the cartesian coordinate system passing through both +3 on y-axis and +3 on x-axis so i assume that the answer is supposed to be y=3 and x=3. As i already said i suck at anything that has to do with drawing. I can't find my way around drawing functions and equations on the cartesian coordinate system. So how do i draw according to coordinates here. I thought that you had to get both x and y and then mark them on the coordinate system and draw a straight line towards 0 from the point where these two touch but how do i draw these that don't go through 0 like this one. How do i discern which line should i draw according to x,y

 

[Dokutayuu]

About splitting a quadratic, you've been doing something called "the difference of 2 squares". This happens when you have an x^2 term taking a way a negative square number. So x^2 -9 can be (x+3)(x-3), for example. This doesn't have an x term, when expanded. It can't be done for positive terms. x^2 + 16 can't be split into (x+4)(anything), you can't use the quadratic formula to get a real answer.

If there's a coefficient (number) in front of the x^2 term, you can either take it out straight away, which makes it easier, or keep it.

For example, 4x^2 + 16x + 16 = 0
can go to 4(x^2 + 4x + 4) = 0
which you can then solve as
4(x+2)(x+2) = 0

However, going straight in to get
(2x+4)(2x+4) = 0 is acceptable but a little trickier to spot.

They shouldn't get you to deal with awkward numbers, don't worry. They want to see if you can spot things, not torture you.

With something like 2x + 3y = 6, you know it's a straight line as there are no ^2 or ^3 symbols. You can actually cheat on ones like these, all you need to find out is where it cuts each axis.

When it cuts the y axis, x = 0.
This means (2 x 0) + 3y = 6
y = 2

When it cuts the x axis, y = 0.
2x + (3 x 0) = 6
x = 3.

You can now mark both points on the graph and draw a straight line through them.

 

[MrPlow]

About splitting a quadratic, you've been doing something called "the difference of 2 squares". This happens when you have an x^2 term taking a way a negative square number. So x^2 -9 can be (x+3)(x-3), for example. This doesn't have an x term, when expanded. It can't be done for positive terms. x^2 + 16 can't be split into (x+4)(anything), you can't use the quadratic formula to get a real answer.

If there's a coefficient (number) in front of the x^2 term, you can either take it out straight away, which makes it easier, or keep it.

For example, 4x^2 + 16x + 16 = 0
can go to 4(x^2 + 4x + 4) = 0
which you can then solve as
4(x+2)(x+2) = 0

However, going straight in to get
(2x+4)(2x+4) = 0 is acceptable but a little trickier to spot.

They shouldn't get you to deal with awkward numbers, don't worry. They want to see if you can spot things, not torture you.

With something like 2x + 3y = 6, you know it's a straight line as there are no ^2 or ^3 symbols. You can actually cheat on ones like these, all you need to find out is where it cuts each axis.

When it cuts the y axis, x = 0.
This means (2 x 0) + 3y = 6
y = 2

When it cuts the x axis, y = 0.
2x + (3 x 0) = 6
x = 3.

You can now mark both points on the graph and draw a straight line through them.

oh i see.. i actually know about the difference of 2 squares but somehow i never connected it with this since i was taught for it to always be like (a^2 - b^2)=(a-b)(a+b) and always imagined it as two unknown terms with ^2. Pretty dumb of me actually, but that's me.. i always take everything literally

about the drawing thingy.. yea i wasn't talking about knowing if it is a straight line or not... i figured that quadratic equations end up as a parabola, and cubic as some sort reverse mirror like look and that the linear functions are .. linear. It's just that i always got confused with the drawing. Sometimes i had to draw the line that goes straight through the two points on the coordinate system and sometimes i had to find a spot in between the two points and draw a line through them all the way to 0(now that i think about it that's how parabola is drawn... i was probably thinkin about that /facepalm) as you can see i suck at those and i get all confused but i think i understand it now.. thanks.. i still need to get it in my head that it's always better to look for y(or f(x) in some cases) first and if it's a straight line then when y cuts x, x is naturally 0 and vice versa. It is really simple now that i think about it but i just don't get it why i get so confused when i look at it

EDIT: I need a bit of advice. I have a bit of trouble dealing with ratios. I mean i know how to solve them but sometimes i get a bit confused so i tend to waste time. Can anyone give me some pointers as to what to pay the most attention to and what never to do when dealing with ratios(w/ an example if possible since that's the best way i learn=

Anyway thanks a lot... taking your time to help me, i really appreciate it

 

[Dokutayuu]

If you find ratios tough, try to picture them as fractions.

For example, Doku brand pudding is made by mixing the pudding mix with dried milk and adding sugar in a 12 : 7 : 1 ratio. Water is added afterwards.

a) Grega's holding a pudding party, he needs to make 1kg of pudding. How much of each ingredient should he use?

As a ratio this can be hard to work out, but as a fraction it's easier. The total of the ratio is 12 + 7 + 1 = 20. So there are 20 "parts" in the mixture. 12 of these are pudding mix, 7 are milk and 1 is sugar. This means 12/20 (or 3/5) is mix, 7/20 is milk and 1/20 is sugar. Out of the kilogram we need to make, we know what it's made from.

3/5 x 1000g (1kg) = 600g of pudding mix
7/20 x 1000g = 350g of milk
1/20 x 1000g = 50g of sugar (the mix is really bitter)

There's another type of question they could ask.

b) Impressed with Grega's party, Sky decides he wants to make some pudding and buys 1500g of pudding mix. How much of each ingredient does he need to use all of it. He also wants to know how much pudding he can make.

We know that 12/20 is pudding mix which is equal to 1500g (1.5kg). If "x" is the total amount of pudding, we know;
3/5x = 1500g
x = 2500g
So we're making 2500g of pudding.

Then we can work out 7/20 and 1/20 of 2500g.
7/20 x 2500g = 875g
1/20 x 2500 = 125g

They can't really ask much else.

 

[harSens]

For the drawing the line task, you don't have to solve an equation.
Since the equation is 2x+3y=6, there are 2 variables and only one equation, so there is no way to solve this anyway. You made small mistake in trying your solution:
2x=6-3y rather than 2x=6+3y
so
x = 3 - 3/2y
2(3-3/2y)+3y=6
6-3y+3y=6
6=6
and you know absolutely nothing

The way to 'solve' these is by simply picking two different x positions and calculate y for them. Since it's a linear function, you can then simply draw a line. It is customary to find the points that cross the x axis (y=0) and the y axis (x=0), but really any 2 different points will do.

So
3y=6 => y = 2 for x-axis crossing
2x=6 => x = 3 for y-axis crossing

edit: nvm, missed Dokutayuu post.

 

[MrPlow]

Well less than an hour left until the exam. I just hope i won't at least mess up on the simple ones.
Well, wish me luck, and thank you, all of you who took a moment of their time to help me, i really appreciate it.

EDIT:Well i'm back from the exam. It wasn't hard but it wasn't too easy too. I didn't solve around 2.5 problems and for 2-3 of the multiple choice ones i had to take a guess. I know i'll pass... just don't know how good/bad. There were some problems similar to what you guys helped me and i knew how to solve some and i messed up and had to guess some.. Well now i have to wait a few weeks for the final verdict...i mean score..