Question

Kathleen had a bag of candy-coated chocolates. Of the 55 candies in the bag, 20% were red, 30% were orange, 30% were green, and 20% were blue. Of the blue ones, 20% had a damaged candy coating. Approximately how many of the blue ones had a damaged candy coating?
A.
2
B.
11
C.
22
D.
6

Answer 1

For this case, we find the amount of each colored candy in the bag, having a total of 55.

• Reds:

55 -------------> 100%

R ---------------> 20%

Where "R" represents the amount of red candies.

$R = \frac {20 * 55} {100}\\R = 11$

• Oranges:

$O = \frac {30 * 55} {100}$

O = 16.5, approximately 16 orange candies.

• Green:

$G = \frac {30 * 55} {100}$

G = 16.5, approximately 16 green candies.

• Blues:

$B = \frac {20 * 55} {100}\\B = 11$

Now, having a total of 11 blue candies, we must find 20% to know how many have the damaged candy coating.

11 ---------------> 100%

x -----------------> 20%

Where "x" represents the quantity of blue candies with the damaged candy coating

$x = \frac {20 * 11} {100}\\x = 2.2$

So, of 2 blue candies have the damaged candy coating

Answer:

two

Answer 2

Answer:

2

Step-by-step explanation:

55 x 0.2 = 11

11 x 0.2 = 2.2