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sets

by Lattefah84 » Fri Dec 25, 2009 10:47 am
Example from OG:

In a certain production lot, 40 percent of the toys are red and the remaining toys are green. Half of toys are small and half are large. If 10 percent of the toys are red and small, and 40 toys are green and large, how many of the toys are red and large?

On the given information it is easy to compute that there is 30% toys that are large and red. But what I don't understand is how can I know n - number of the toys n=200 (if the only number in task is 40 toys)?
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by truplayer256 » Fri Dec 25, 2009 12:13 pm
Let x represent the total number of toys in the production lot.
.60(x)=Green
.40(x)=Red

x/2= Small
x/2= Large

.10(x)= Small and Red
.30(x)= Large and Red

40 toys are green and small while .60(x)-40 toys are green and large. So:

.60(x)-40+.10(x)=x/2

x=200

Total number of toys that are red and large= 3/10*200=60 toys.

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by Lattefah84 » Fri Dec 25, 2009 1:15 pm
truplayer256 wrote:Let x represent the total number of toys in the production lot.
.60(x)=Green
.40(x)=Red

x/2= Small
x/2= Large

.10(x)= Small and Red
.30(x)= Large and Red

40 toys are green and small while .60(x)-40 toys are green and large. So:

.60(x)-40+.10(x)=x/2

x=200

Total number of toys that are red and large= 3/10*200=60 toys.
how can you compute 60% green toys and 40 green toys ...?

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by truplayer256 » Fri Dec 25, 2009 1:37 pm
how can you compute 60% green toys and 40 green toys ...?
I don't really know what you mean by this but .60(x) represents the total number of green toys that are both green and large and green and small. Since the problem tells us that there are a total of 40 green and small toys, subtracting these from .60(x) will give us the total number of green and large toys.

In case you were wondering how I computed .60(x), since the total number of toys is x and 40% of the toys are red, x-.40(x) represents the total number of green toys.

40 green and large toys are given to you in the problem.
I hope this helps.

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by Lattefah84 » Sat Dec 26, 2009 3:46 am
truplayer256 wrote:
how can you compute 60% green toys and 40 green toys ...?
I don't really know what you mean by this but .60(x) represents the total number of green toys that are both green and large and green and small. Since the problem tells us that there are a total of 40 green and small toys, subtracting these from .60(x) will give us the total number of green and large toys.

In case you were wondering how I computed .60(x), since the total number of toys is x and 40% of the toys are red, x-.40(x) represents the total number of green toys.

40 green and large toys are given to you in the problem.
I hope this helps.
oh, yes, yes

i got it now, thanks very much!

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by valleeny » Sun Dec 27, 2009 12:18 am
I guess the easiest to illustrate this is to make a table like below. Let total toys in production be x.

small large
red 0.1x 0.3x
green y 40

A few equations can be formulated.

1) y + 40 = 0.6x
2) 0.1x + y = 0.5x
3) 0.3x + 40 = 0.5x

Solving equation 3, 0.3x = 60