Question

Hey, guys, this is not a question, but a quick tutorial on word problems. I noticed that people have trouble with word problems, and they're actually really simple. :)

Enjoy!

Answer 1

HERE ARE SOME EXAMPLES:
Of problems that lead to simultaneous equations.

Example 1.Andre has more money than Bob. If Andre gave Bob $20, they would have the same amount. While if Bob gave Andre $22, Andre would then have twice as much as Bob. How much does each one actually have?
Solution. Let x be the amount of money that Andre has. Let y be the amount that Bob has.
Always let x and y answer the question -- and be perfectly clear about what they represent!
Now there are two unknowns. Therefore there must be two equations. (In general, the number of equations must equal the number of unknowns.) How can we get two equations out of the given information? We must translate each verbal sentence into the language of algebra.

Here is the first sentence:
"If Andre gave Bob $20, they would have the same amount."
Algebraically:
1) x − 20 = y + 20.

(Andre -- x -- has the same amount as Bob, after he gives him $20.)
Here is the second sentence:
"While if Bob gave Andre $22, Andre would then have twice as much
as Bob."

Algebraically:
2) x + 22 = 2(y − 22).
(Andre has twice as much as Bob -- after Bob gives him $22.)
To solve any system of two equations, we must reduce it to one equation in one of the unknowns. In this example, we can solve equation 1) for x --
x − 20 = y + 20


implies: x = y + 40
-- and substitute it into equation 2):
y + 40 + 22 = 2(y − 22).
That is,
y + 62 = 2y − 44,

y − 2y = − 44 − 62,


−y = −106

y = 106.
Bob has $106. Therefore, according to the exression for x, Andre has
106 + 40 = $146.

Answer 2

The last one failed :(

Answer 3

The reason people have problem is because they don't know how to solve the problem after it's been set up. The most difficult part about a world problem should be setting it up but even then after it's been set up people have trouble doing the problem.

Answer 4

It's actually really simple once you have it setup.

Answer 5

hah nice. i've always wanted to make a tutorial like this but i lacked the spirit and the idea. so congrats for doing it :D

Answer 6

I love these problem. There are a lot of fun. Good job

Answer 7

good tutorial .. @Compassionate
as romero said : they mainly have problem in converting the word problem to expressions

Answer 8

example :
X is older than Y by 2 years and their sum of ages is 9 ..Find the ages

Answer 9

:D Good job!

Answer 10

4 medals are enough

Answer 11

Bumpidty bump bump. Lol.

Answer 12

where are my medals

Answer 13

^ no medals for you. x)

Answer 14

plx comp it was my idea :D

Answer 15

yeah i got one :)

Answer 16

Thanks for your sharing your insight, i really like the way you took it through.

Answer 17

How would you set a problem like this up: A cone with a height of 6m and base radius of 2m has watered poured into the open spout and the water in the cone rises at 0.5m/s. How fast is the water being poured in when the cone has been filled up to 2m?

Answer 18

I'll make a tutorial on it later. Talking about it on this thread is against the rules and considered spam.

Answer 19

I thought the purpose of the thread was to discuss on how to set up world problems?

Answer 20

The purpose was how to set up word problems involving $$$. Not geometric word problems. They're different.

Answer 21

It wouldn't make much sense to ask a word problem concerning an integral on here. Only a certain type of word problems are accepted.

Answer 22

i think every word problem is a problem. and every question have and answer. Making tutorials on it later can solve these problem. thank u.

Answer 23

It wasn't explicitly stated at the beginning; it just states "word problems" in general, so how is one supposed to know the discussion was solely on algebraic problems?

Answer 24

I made a typo, sorry. This one is concerning cost.

Answer 25

main problem with word problems is that teachers act as if there is one way to do it
here is quick example

Answer 26

@Compassionate you can edit the question. Near the top inside the box your question is, you can click the edit question button to fix the typo. Good tutorial though. It's funny how people have trouble with these when they can be a lot harder.

Answer 27

@Calcmathlete They can be very hard. I've gone as far as integrals in word problems.

Answer 28

Yeah. I mean as in word problems of this level. THey can get a lot harder later on, but these are pretty simple.

Answer 29

I enjoy keeping it simple because most people on here are students.

Answer 30

I wasn't suggesting that you make it harder? I was just mentioning how questions at this level can seem like a piece of cake later on.

Answer 31

adam and eve have a total of 50 apples
adam has 4 more apples than eve

method one
solve \(A+E=50\) \(A-E=4\)

method two
solve \(A+(50-A)=4\)

method 3
solve \(A+E=50\) \(E=A-4\)

method 4
solve \(A+(A-4)=50\)

method 5
guess (easiest method)

Answer 32

ok method two was wrong, should be \(A-(50-A)=4\) but you get the idea

Answer 33

A + E = 50
A = 4 + E

(4+ E) E = 50
4 + 2E = 50
2E = 56
E = 28

E = 28 and A = 22.

Easiest method is substitution.

Answer 34

i can give the reason for each of these in simple english, but the point is that there is no one way to solve a "word problem"
you pick one variable to solve for (say the length) i pick the other (say the width) and we should all come up with the same answer
all too often it is assumed that "i have to come up with the one equation to solve" but that is nonsense
sometimes you need no equation at all

Answer 35

Sometimes it's really simplicity.

Answer 36

@Compassionate not to be argumentative (it is a very nice tutorial) but i completely disagree with your assertion that substitution is in any way "simpler' than any other method

Answer 37

Please explain.

Answer 38

they are all equally "simple" in that they all take exactly the same number of steps in the long run

Answer 39

if it was up to me i would guess
that would be simplest for this problem
evidently adam has more than eve, and 26 and 24 is wrong, try 27 and 23 (which is the right answer)

Answer 40

but that aside, if i was going to do this problem i would not use two variables
i would say adam has \(A\) and the total if 50 so eve has \(50-A\) but i am not saying that is easier, i am saying it is a matter of taste

Answer 41

Ah, my mistake on solving that question. You'll have to pardon me, I've been up for 16 hours and running low on energy.

All though your methods work well for you, they might not work well for others. It seems like getting a good knowledge of the elimination and substitution method first helps, and then learn some alternative ways that make it easier. True, it is not the easiest way to do these problems, but I'd say that statement is more subjective than anything.