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In each production lot for a certain toy, 25 percent of the [#permalink]
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24 Mar 2008, 13:36
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In each production lot for a certain toy, 25 percent of the toys are red and 75 percent of the toys are blue. Half the toys are size A and half are size B. If 10 out of a lot of 100 toys are red and size A, how many of the toys are blue and size B? (A) 15 (B) 25 (C) 30 (D) 35 (E) 40



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Re: PS:Toys % [#permalink]
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24 Mar 2008, 14:54
D35
25 are Red  10 are Size A  so 15 are Size B 75 are Blue  40 are Size A and 35 are Size B (because there 50 toys each with sizes A and B)



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Re: PS:Toys % [#permalink]
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24 Mar 2008, 14:58
100 toys total.
25 red, 75 blue. 50 A, 50 B.
10 red/A, therefore, 15 red/B.
We have 40 left in A and 35 left in B.
Therefore, 35 in blue/B.
I think.



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Re: PS:Toys % [#permalink]
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24 Mar 2008, 15:16
Given 100 Toys in total. So toal read toys = 25 Total blue toys = 75 Total size A = 50 Total size B = 50 Total Red and Size A = 10 So Remaining Red toys will be all size B = 25  10 = 15 Remaining Size B toys which must come from Blue toys = 50  15 = 35
Answer D.



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Re: PS:Toys % [#permalink]
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24 Mar 2008, 15:27
wow, i did it this way and got it wrong. Anyone know what i did wrong please explain to me.
1/4*1/2=1/8 sizeA and red.
3/4*1/2=3/8 sizeB and blue.
3/8*100=37.5
where did my logic go wrong here?



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Re: PS:Toys % [#permalink]
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24 Mar 2008, 15:32
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You cannot multiply as you did because It is not necessary that half of red toys are size A and remaining half are size B. Similarly it is also not necessary that half of blue toys are size A and remaining half are size B. But by your calculation you are assuming just that. Thereby the flaw.










