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All of 115 different pills contain at least one of the vitamins A, B and C. Eighteen have

vitamin A only, fifteen have vitamin B only, and twenty have vitamin C only. If ten have

all three vitamins and there are 8x having vitamins A and B only, 2x-8 having vitamins

B and C only and 3x-5 having vitamins A and C only.

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Answer:

115 - all pills

A = 18

B = 15

C = 20

A, B, C = 10

A, B = 8x

B, C = 2x-8

A, C = 3x-5

18 15 20 10 8x (2x-8) (3x-5) = 115

8x 2x 3x = 115 - 18 - 15 - 20 - 10 8 5

13x = 65

x = 56/13

x = 5

this implies:

A, B = 8x = 8 * 5

A, B = 40

B, C = 2x-8 = 2*5-8 = 2

B, C = 2

A, C = 3x-5 = 3*5-5 = 10

A, C = 10

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