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Anika [276]
3 years ago
10

Tracy had a bag of candies, and none of the candies could be broken into pieces. She ate $\frac{1}{3}$ of them and then gave $\f

rac{1}{4}$ of what remained to her friend Rachel. Tracy and her mom then each ate 15 candies from what Tracy had left. Finally, Tracy's brother took somewhere from one to five candies, leaving Tracy with three candies. How many candies did Tracy have at the start?
Mathematics
1 answer:
Ymorist [56]3 years ago
7 0

Answer:

72

Step-by-step explanation:

Let the number of candies = n

Tracy ate = 1/3 of n = n/3 remaining 2n / 3

she gave 1/4 of the remainder to her friend = 1/4 of ( n - n/3) = 2n / 12

the new remainder = (2n / 3) - (2n / 12) = n /2

she and her mom then ate altogether = 30

and the brother took from 1 to five, the number he took = x

and she has 3 left

( n/2) - 30 - x = 3

n = 2 ( 33 + x)

now we know that the candies could not be broken  and x is between 1 to 5

n = 2 (33 + 1), 2 (33+2), 2(33 +3), 2 (33+ 4) and 2 (33 + 5) this are the possible values of n, ( 68, 70, 72, 74, 76) and the multiple of 3 is 72

n therefore = 72

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