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A vending machine randomly dispenses four different types

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by BTGmoderatorLU » Wed Aug 15, 2018 1:59 pm

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Source: Princeton Review

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

The OA is B.
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by GMATGuruNY » Wed Aug 15, 2018 4:24 pm
BTGmoderatorLU wrote:Source: Princeton Review

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

The OA is B.
There are twice as many strawberry candies as grape candies.
If G=1, then S=2.
There are twice as many apple candies as strawberry candies.
If S=2, then A=4.
There are twice as many apple candies as orange candies.
If A=4, then O=2.
Resulting ratio:
G:S:A:O = 1:2:4:2.

Since the sum of the parts of the ratio = 1+2+4+1 = 9, and the actual number of candies -- 90 -- is 10 times as great, all of the values in the ratio must be multiplied by 10:
G=10, S=20, A=40, and O=20, for a total of 90 candies.

To guarantee that we purchase 3 of each flavor, we must consider the WORST-CASE SCENARIO.
Here, the worst-case scenario occurs when we purchase all of the candies in red -- the three most numerous candies -- with the result that none of the grape candies are purchased.
If we purchase the 80 candies in red, we must then purchase 3 grape candies -- so that we purchase at least 3 of each flavor -- yielding a total purchase of 83 candies.
Since each candy costs $0.25, the cost of these 83 candies = (83)(0.25) = $20.75.

The correct answer is B.
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by Scott@TargetTestPrep » Sun Aug 26, 2018 5:54 pm
BTGmoderatorLU wrote:Source: Princeton Review

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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