Question

Word Problem Practice... @Hero

Answer 1

I am stealing this exact problem... from another member... I want to see if I can do this one correctly without screwing up.

Answer 2

Jimmy has 30 chocolates and wants to give them to Harry and Eric. If Jimmy gives Harry 2 times the number of chocolates that he is gives to Eric, how many chocolates does Eric get?

Answer 3

I would set this equal to 30... since that is how much Jimmy has in total. Plus, he has to give out of that 30

Answer 4

H + E = 30

Answer 5

Since it says Harry gets twice as many as Eric.. I would change that to...

2H + E = 30

Answer 6

Then... I was just going to simply change the H to an E...

2E + E = 30

Answer 7

3E = 30
E = 10

Making H = 20

Answer 8

Please, tell me I made no mistake...

Answer 9

Well, you got the final answer correct so that's the most important part, but basically, when you replaced H with E, that's essentially saying that H = E

Answer 10

But we know that is not true

Answer 11

So you are telling me that is not a legal way to do this. lol

Answer 12

How would you do it?

Answer 13

Well, like you, I would have started with

H + E = 30

Jimmy gives 2 times as much chocolate to Harry as he does to Eric. In other words

H = 2E

I know it seems weird to write that, but it is true. Whatever Eric gets, Harry will have twice as many. So in order to get what Harry has, we have to take whatever Eric has and multiply it by two.

So now you can replace properly:

2E + E = 30
3E = 30
E = 30/3
E = 10

Answer 14

Oh, I want to be like you. :D

Answer 15

Try to be like @satellite, @amistre or @hartnn...they are all very good.

Answer 16

Oh, I love @hartnn and @amistre64 :D they are awesome like you :)

Answer 17

Thanks again

Answer 18

Jimmy has 30 chocolates and wants to give them to Harry and Eric. If Jimmy gives Harry 2 times the number of chocolates that he is gives to Eric, how many chocolates does Eric get?

the chocolate is split in a ratio of 1:2

scaling this by some integer k we have 1k+2k = 30

3k = 30, k = 10

the chocolates are split up: 10:20

Answer 19

just another way to approach it is all ...

Answer 20

@amistre64,

The system I came up with begins with ratios:
H = amount of chocolate Harry gets
E = amount of chocolate Eric gets
C = total amount of chocolate

The ratio of Harry's chocolate to Eric's chocolate:
\[\frac{H}{E} = \frac{2}{1}\]

Cross multiply to get:
\[H = 2E\]

The ratio of Harrys chocolate to the total chocolate + the ratio of Eric's chocolate to the total chocolate = 1

\[\frac{H}{C} + \frac{E}{C} = 1\]
\[\frac{H + E}{C} = 1\]

C = 30 so

\[\frac{H + E}{30} = 1\]
\[H + E = 30\]

So even using ratios we still end up with the system:

H = 2E
H + E = 30

Answer 21

yes, they are intertwined. the system is a fine way to approach it. FI i had done rations I would have shortened the work load as:
\[\frac HE=\frac {2k}{1k}\]
\[H+E = 30 = 2k+k\]
\[30 = 3k\]
\[10 = k\]
\[\frac HE=\frac {20}{10}\]

Answer 22

if .. not "FI"

Answer 23

a ratio does not have to be in "simplest" form to be a ratio is all

Answer 24

I called myself shortening the workload by avoiding the ratios altogether when I began with

H + E = 30
H = 2E

The ratios are preliminary to the system, but it can be setup directly. But the ratios are definitely a good way to start off as well.

Answer 25

satellite needs to address this with a heaviside function, then ill be content lol

Answer 26

I don't think it is necessary to go that far.