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Andrew [12]
4 years ago
6

The first barn contained 3 times more hay than the second one. After 20 tons of hay were removed from the first barn and 20 tons

were added to the second barn, the amount of hay in the second barn was 5/7 of the amount remaining in the first barn. How many tons of hay was there in each barn?
Mathematics
2 answers:
Vika [28.1K]4 years ago
8 0

The solution would be like this for this specific problem:

 

a + 20 = 5/7 (3a - 20) <span>
a + 20 = (15a - 100)/7 </span><span>
Multiply both side by 7, </span><span>
7a + 140 = 15a - 100 </span><span>
8a = 240<span> 
</span></span>a = 30

3a = 90

a + 20 = 50

3a - 20 = 70

<span>5/7 x 70 = 50</span>
OverLord2011 [107]4 years ago
7 0

Answer:

90 in the first and 30 in the second for before and for after there is

70 in the first and 50 for the second.

Step-by-step explanation:

Before solution:

a+20=5/7(3a-20)

a+20=(15a-100)7

multiply each side by 7

7a+140=15a-100

8a=240

a=30

3a=90  

After solution (with information from last equation):

a+20=5/7(3a-20)

30+20=5/7(90-20)

50=5/7(70)

50=50

a=50

3a times 5/7=70


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Grandpa and Grandma are treating their family to the movies. Matinee tickets cost $4 per child and $4 per adult. Evening tickets
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Complete Question

Grandpa and Grandma are treating their family to the movies. Matinee tickets cost $4 per child and $4 per adult. Evening tickets cost $6 per child and $8 per adult. They plan on spending no more than $80 on the matinee tickets and no more than $100 on the evening tickets. Could they take 9 children and 4 adults to both shows? Show your work. A yes or no answer is not sufficient for credit.

Answer:

Yes it is  possible to take the  9 children and 4 adults to both shows

Step-by-step explanation:

From the question we are told that

    The  cost of the Matinee tickets for a child is  z =  $4

    The  cost of the Matinee tickets for an adult is  a  = $ 4

     The cost of the Evening tickets for  a child is  k =  $6

      The cost of the Evening tickets for an adult is  b =  $8    

      The  maximum amount to be spent on Matinee tickets is  m = $80

       The maximum amount to be spent on Evening tickets is  e =  $100

       The  number of child to be taken to the movies is  n  = 9

        The number of adults to be taken to the movies is  j  =  4

Now the total amount of money that would be spent on Matinee tickets is  mathematically evaluated as      

           t = 4 n + 4 j

substituting values

           t = 4 *  9  + 4* 4

           t =  52

Now the total amount of money that would be spent on  Evening ticket is mathematically evaluated as      

      T =  6n + 8j

substituting values

     T =  6(9) + 8(4)

    T = 86

This implies that it is possible to take 9 children and 4 adults to both shows

given that

      t \le m

i.e  $56  \le$ 80

and  

      T \le e

i.e   $ 86 \le $ 100

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