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Norma-Jean [14]
3 years ago
11

A package of cookies, packed , ate the fourth part at breakfast , then 3/5 at tea time. there are still 18 cookies in the pakage

...
¿how many cookies were in the package ?
Mathematics
2 answers:
Kay [80]3 years ago
7 0
Break fast =  1/4
Tea time =  3/5
Still 18 left.

Fraction left =  1 - (1/4 + 3/5) = 

1/4 + 3/5 =  (1/4)*(5/5) + (3/5)*(4/4) = 5/20 + 12/20 = (5 + 12)/ 20

 = 17/20

Fraction left =  1 - (1/4 + 3/5) =  1 - 17/20 = (20 - 17)/20 = 3/20

Fraction left = 3/20

Assuming total number of cookies in the package is x.

(3/20)x = 18

3x/20 = 18

3x = 18*20

x = 18*20/3 = 120

There were 120 cookies in the package.  
Leno4ka [110]3 years ago
5 0
Ok so
1/4 of total eaten
3/4 left
3/5 of 3/4 eaten
9/20 left
18=9/20 of whole
multiply both sides by 20
360=9 of whole
divide both sides by 9
40=whole


40 total originlly

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Number of students that enjoy going to the movies, B = 12

Number of students that enjoy solving mathematical problems, C = 24

A∩B∩C = 8

Here we have;

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) -n(A∩C) + n(A∩B∩C)

= 38 + 12 + 24 - n(A∩B) - n(B∩C) -n(A∩C) + 8

Also the number of student that like only one activity is found from the following equation;

n(A) - n(A∩B) - n(A∩C) + n(A∩B∩C) + n(B) - n(A∩B) - n(B∩C) + n(A∩B∩C) + n(C) - n(C∩B) - n(A∩C) + n(A∩B∩C) = 30

n(A) + n(B) + n(C) - 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) + 3·n(A∩B∩C) = 30

38 + 12 + 24 - 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) + 24 = 30

- 2·n(A∩B) - 2·n(A∩C) - 2·n(B∩C) = -68

n(A∩B) + n(B∩C) + n(A∩C) = 34

Therefore, the number of students that like only two of the activities = 34.

8 0
3 years ago
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