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A vending machine randomly dispenses four different types of [#permalink]
 21 Mar 2008, 23:18
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A vending machine randomly dispenses four different types of fruit candy. There are twice as many apple candies as orange candies, twice as many strawberry candies as grape candies, and twice as many apple candies as strawberry candies. If each candy cost $0.25, and there are exactly 90 candies, what is the minimum amount of money required to guarantee that you would buy at least three of each type of candy? A $3.00 B $20.75 C $22.50 D $42.75 E $45.00
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Re: PS: Vending Machine [#permalink]
22 Mar 2008, 00:04
Bfrom: "There are twice as many apple candies as orange candies, twice as many strawberry candies as grape candies, and twice as many apple candies as strawberry candies." we can conclude following ratio: 4:2:2:1 or 40:20:20:10 To satisfy the condition of buying at least three of each type of candy, we have to buy N=40+20+1=61 P=61*0.25$=15.25$ So, I picked B. BTW, 90*0.25$=22.5$, Therefore, C,D,E are out. A is too small. So B remains
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Re: PS: Vending Machine [#permalink]
22 Mar 2008, 02:30
prasannar wrote: A vending machine randomly dispenses four different types of fruit
candy. There are twice as many apple candies as orange candies,
twice as many strawberry candies as grape candies, and twice as
many apple candies as strawberry candies. If each candy cost $0.25,
and there are exactly 90 candies, what is the minimum amount of
money required to guarantee that you would buy at least three of
each type of candy?
A $3.00
B $20.75
C $22.50
D $42.75
E $45.00 I confused this: "three of each type of candy", if it means "each type three" I think A win. What do you think?
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Re: PS: Vending Machine [#permalink]
22 Mar 2008, 02:43
I've found my mistake at least three of each type of candy: 3 apple, 3 orange, 3 strawberry, and 3 grape candies. N=40+20+20+3=83 P=83*0.25=20.75$
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Re: PS: Vending Machine [#permalink]
22 Mar 2008, 19:41
walker wrote: I've found my mistake
at least three of each type of candy: 3 apple, 3 orange, 3 strawberry, and 3 grape candies.
N=40+20+20+3=83
P=83*0.25=20.75$ Walker, Why not N = 10 + 20 + 20 + 3 ? You are deserved to 51! 
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Re: PS: Vending Machine [#permalink]
22 Mar 2008, 20:46
sondenso wrote: walker wrote: I've found my mistake
at least three of each type of candy: 3 apple, 3 orange, 3 strawberry, and 3 grape candies.
N=40+20+20+3=83
P=83*0.25=20.75$ Walker, Why not N = 10 + 20 + 20 + 3 ? You are deserved to 51! Sondenso, What is the guarantee that the last 3 are Grape Candies, they could be very well Apple[since Orange and Strawberry are completely used as we picked 20,20 of them] thus to make sure there are 3 of each, we need to make sure, we pick up ALL of the rest of the candies but for the least number variant. Hope this helps
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Re: PS: Vending Machine [#permalink]
22 Mar 2008, 22:20
prasannar wrote: sondenso wrote: walker wrote: I've found my mistake
at least three of each type of candy: 3 apple, 3 orange, 3 strawberry, and 3 grape candies.
N=40+20+20+3=83
P=83*0.25=20.75$ Walker, Why not N = 10 + 20 + 20 + 3 ? You are deserved to 51! Sondenso, What is the guarantee that the last 3 are Grape Candies, they could be very well Apple[since Orange and Strawberry are completely used as we picked 20,20 of them] thus to make sure there are 3 of each, we need to make sure, we pick up ALL of the rest of the candies but for the least number variant. Hope this helps Hey, this concept have just come up to me. I tried to understand the logic. But Honestly, I did not understand. Do you guys mind writting it in more detail? Many thanks.
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Re: PS: Vending Machine [#permalink]
22 Mar 2008, 22:56
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we have: the number of apple candies = Na=40 the number of orange candies = No=20 the number of strawberry candies= Ns = 20 the number of grape candies = Ng = 10 our question: "what is the minimum amount of money required to guarantee that you would buy at least three of each type of candy?" If we get 40 candies, we can get only 40 apple candies and therefore N=40 does not guaranty that we will have always 3 of each type. If we get 60 candies, we may get 40 apple candies and 20 orange candies - no guarantees If we get 80 candies, we may get 40 apple candies, and 20 orange candies, and 20 strawberry - no guarantees If we get 82 candies, we may get 40 apple candies, and 20 orange candies, 20 strawberry, and 2 grape candies - no guarantees If we get 83 candies, we will always get at least three of each type of candy: 3 apple, 3 orange, 3 strawberry, and 3 grape candies, that is, we will get 3 of each type regardless chance.
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Re: PS: Vending Machine [#permalink]
23 Mar 2008, 00:11
walker wrote: we have:
the number of apple candies = Na=40
the number of orange candies = No=20
the number of strawberry candies= Ns = 20
the number of grape candies = Ng = 10
our question: "what is the minimum amount of money required to guarantee that you would buy at least three of each type of candy?"
If we get 40 candies, we can get only 40 apple candies and therefore N=40 does not guaranty that we will have always 3 of each type.
If we get 60 candies, we may get 40 apple candies and 20 orange candies - no guarantees
If we get 80 candies, we may get 40 apple candies, and 20 orange candies, and 20 strawberry - no guarantees
If we get 82 candies, we may get 40 apple candies, and 20 orange candies, 20 strawberry, and 2 grape candies - no guarantees
If we get 83 candies, we will always get at least three of each type of candy: 3 apple, 3 orange, 3 strawberry, and 3 grape candies, that is, we will get 3 of each type regardless chance. Walker, thanh you  
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Probabilty and Ratios [#permalink]
15 Sep 2011, 16:02
A vending machine randomly dispenses four different types of fruit candy. There are twice as many apple candies as orange candies, twice as many strawberry candies as grape candies, and twice as many apple candies as strawberry candies. If each candy cost $0.25, and there are exactly 90 candies in the machine, what is the minimum amount of money required to guarantee that you would buy at least three of each type of candy?
A $3.00
B $20.75
C $22.50
D $42.75
E $45.00
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Re: Probabilty and Ratios [#permalink]
15 Sep 2011, 16:19
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This question isn't that tough when you break it all down. We are given the following ratios: Code: A O S G
2 1
2 1
2 1
Consolidate these ratios: Code: A O S G
4 2 2 1
We have 90 items, so using the ratio above we have these pieces: Code: A O S G
40 20 20 10
Now we need at least three of each type. Think about the worst case, if we buy 40, we might get all of A. If we buy 60, we might get all of A and O, if we buy 80 we might get A, O and S, so we need to buy another three and now we have guaranteed we have A,O,S and G. 83 * $0.25 = $20.75 B.
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Re: Probabilty and Ratios [#permalink]
15 Sep 2011, 22:16
sameer1986 wrote: A vending machine randomly dispenses four different types of fruit candy. There are twice as many apple candies as orange candies, twice as many strawberry candies as grape candies, and twice as many apple candies as strawberry candies. If each candy cost $0.25, and there are exactly 90 candies in the machine, what is the minimum amount of money required to guarantee that you would buy at least three of each type of candy?
A $3.00
B $20.75
C $22.50
D $42.75
E $45.00 Given: Apple:Orange = 2:1 Strawberry:Grape = 2:1 Apple: Strawberry = 2:1 So if we have 1 Grape candy, we have 2 strawberry ones and 4 apple ones, which means we have 2 Orange candies. So in all we would have 1+2+4+2 = 9 candies Since we actually have 90 candies, we must have 10 Grape, 20 Strawberry, 40 Apple and 20 Orange candies. If we need at least three of each type and the machine dispenses them randomly, we have to take the worst case scenario (we will have to buy maximum number of candies). In the worst case, we will get the grape candies at the end. We will end up buying all other 80 candies and then get 3 grape candies because only grape candies will be left. So we will need to buy 83 ($20.75) candies.
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Re: PS: Vending Machine [#permalink]
18 Sep 2011, 12:57
B. (40+20+20+3)* .25 = 20.75
I also missed the wording 3 of each kind.
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Re: PS: Vending Machine
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18 Sep 2011, 12:57
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