A vending machine randomly dispenses four different types of fruit can

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

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

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

A:O=2:1
S:G=2:1
A:S=2:1

Hence, A:S:G:O= 4:2:2:1

Total number of candies=A+S+G+O=90

4O+2O+2O+O=90
O=10

A=40; S=20; G=20; and O=10

minimum number of candies required to guarantee that you would buy at least three of each type of candy= 40+20+20+3=83
Amount you gonna spend= 83*0.25=20.75$

B

OR

Total amount of money required to guarantee that you would buy at least three of each type of candy=X

[{(3/4)*90}+3]*0.25 < X < 90*0.25

Only B lies in our range

Hi karishma Bunuel

Can you please explain why are we assuming that it could dispense till all 40 apple candies are over, followed by strawberry and orange ones. When it is given that 4 different candies are dispensed in question. I am also confused that why can't we choose 3 (as least is covered) of each as cost of all 4 flavour candies are same. So that would make total cost to be 0.25*3*4 = 3 in that case then.

Your help would be appreciated.

We need the minimum number of candies to guarantee that we have at least three of each type of candy. For any number less than 83, there is no guarantee that we have at least three of each type of candy. For example, for 82 candies, it's possible to pick 40 apple candies, 20 orange candies, 20 strawberry candies and ONLY 2 grape candies, so we would not have at least three of each kind.


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