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A vending machine randomly dispenses four different types of fruit can

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A vending machine randomly dispenses four different types of fruit can [#permalink]

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New post 18 Nov 2015, 01:44
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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: A vending machine randomly dispenses four different types of fruit can [#permalink]

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New post 18 Nov 2015, 02:35
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Let number of Apple , Orange , strawberry and grape candies be A, O ,S and G respectively .
A= 2O
S= 2G
A= 2S
A= 4G

A+O+S+G = 90
=>A + A/2 + A/2 +A/4 = 90
=> 9A/4 = 90
=> A = 40

O = 20
S = 20
G = 10

Cost of each candy = .25 $
Mininum amount of money required to guarantee that you would buy at least three of each type of candy
We can buy 40 Apple candies , 20 orange candies , 20 strawberry candies and 3 grape candies to ensure atleast 3 of each type of candies .
Total = 83 candies
Amount required = 83 * .25 = 20.75 $
Answer B
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Re: A vending machine randomly dispenses four different types of fruit can [#permalink]

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New post 18 Nov 2015, 02:58
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let a denote no of apple candies
o denote no of orange candies
s denote no of strawberry candies and
g denote no of grape candies

as per the question
a=2o
s=2g
a=2s

each candy cost $0.25
total candies=90
also total candies = a+o+s+g = a +a/2 +a/2 +a/4=90 > 4a+2a+2a+a=360 => a=360/9 => a=40
o=a/2=20
s=a/2=20
g=a/4=10

to gurantee atleast 3 of each type =40*(0.25) + 20*(0.25) + 20*(0.25) + 3*(0.25)=10 + 5 + 5 + 0.75 =20.75
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Re: A vending machine randomly dispenses four different types of fruit can [#permalink]

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New post 10 Apr 2017, 23:57
Skywalker18 wrote:
Let number of Apple , Orange , strawberry and grape candies be A, O ,S and G respectively .
A= 2O
S= 2G
A= 2S
A= 4G

A+O+S+G = 90
=>A + A/2 + A/2 +A/4 = 90
=> 9A/4 = 90
=> A = 40

O = 20
S = 20
G = 10

Cost of each candy = .25 $
Mininum amount of money required to guarantee that you would buy at least three of each type of candy
We can buy 40 Apple candies , 20 orange candies , 20 strawberry candies and 3 grape candies to ensure atleast 3 of each type of candies .
Total = 83 candies
Amount required = 83 * .25 = 20.75 $
Answer B


As stated in question that minimum 3 types of each candy and every thing i understand as per your solution but why only 3 grapes because you took 40 apple , 20 oranges , 20 strawberry and why 3 grapes only because its 10
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Re: A vending machine randomly dispenses four different types of fruit can [#permalink]

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New post 11 Apr 2017, 01:37
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Bunuel 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


Say number of grapes candies is G.
Number of strawberry candies is 2G.
Number of apple candies is 4G.
Number of orange candies is 2G.
G + 2G + 4G + 2G = 90
G = 10
Since it dispenses randomly, let's look at the worst case scenario. Say it dispenses only apple candies till all 40 are over. Then it dispenses only 20 strawberry candies and then 20 orange candies. At the end, we will be left with only grape candies and since we need at least 3 of them, we will take 3 so that we spend minimum money.

In this case, you would need to buy 40 + 20 + 20 + 3 = 83 candies which will cost 83*(1/4) = $20.75

Answer (B)
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Re: A vending machine randomly dispenses four different types of fruit can [#permalink]

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New post 18 Apr 2017, 16:29
1
Bunuel 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


We can let a = the number of apple candies, r = the number of orange candies, s = the number of strawberry candies, and g = the number of grape candies.

From the given information, we can create the following equations:

a = 2r

s = 2g

a = 2s

Since a = 2r and a = 2s, we see that 2r = 2s or r = s. Since s = 2g, r = 2g. Finally, since a = 2s and s = 2g, we see that a = 2(2g) = 4g.

Since there are 90 candies, we can create the following equation:

a + r + g + s = 90

Let’s get all of our variables in terms of g.

4g + 2g + g + 2g = 90

9g = 90

g = 10

Since g = 10, we have a = 40, r = 20, and s = 20.

We need to determine the minimum amount of money required to guarantee that we would buy at least three of each type of candy. To guarantee that, we would need to buy 83 pieces of candy. That is because if we only have 82 pieces of candy, it is possible that we have all 40 apple candies, all 20 orange candies, all 20 strawberry candies, and 2 grape candies. So, we don’t have at least three of each type of candy. Therefore, we need one extra piece of candy to make sure we have at least three of each type of candy.

Finally, since each candy costs $0.25, we need to spend 83 x $0.25 = $20.75 to guarantee that we would have at least three of each type of candy.

Answer: B
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Re: A vending machine randomly dispenses four different types of fruit can   [#permalink] 18 Apr 2017, 16:29
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