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Novay_Z [31]
2 years ago
10

Tickets for the baseball games were $2.50 for general admission and 50 cents for kids. If there were six times as many general a

dmissions sold as there were kids tickets, and the total receipts were $7750, how many of each type of ticket were sold?
Mathematics
1 answer:
Makovka662 [10]2 years ago
4 0
There was 3000 general admission tickets sold and 500 kid ticket sold.

How did I get this?

First, we need to see what information we have.
$2.50 = General admission tickets = (G)
 $0.50 = kids tickets =  (K)
There were 6x as many general admission tickets sold as kids. G = 6K

We need two equations:
G = 6K  
$2.50G + $.50K = $7750
Since, G = 6K we can substitute that into the 2nd equation.

2.50(6K) + .50K = 7750
Distribute 2.50 into the parenthesis

15K + .50K = 7750
combine like terms

15.50K = 7750
Divide both sides by 15.50, the left side will cancel out.

K = 7750/15.50
K = 500 tickets
So, 500 kid tickets were sold.

Plug K into our first equation (G = 6k)

G = 6*500
G = 3000 tickets

So, 3000 general admission tickets were sold,

Let's check this:

$2.50(3000 tickets) = $7500 (cost of general admission tickets)
$.50(500 tickets) = $250 (cost of general admission tickets)
$7500 + $250 = $7750 (total cost of tickets)




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Hello from MrBillDoesMath!

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The pattern  1,8,27, 64... is immediately recognizable as the the cube of the positive integers. But this question has a minus sign appearing before  each entry, suggesting we try this:


 - 1  =  (-1)^3

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Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and pro
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{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ Sanya has a piece of land which is in the shape of a rhombus.

★ She wants her one daughter and one son to work on the land and produce different crops, for which she divides the land in two equal parts.

★ Perimeter of land = 400 m.

★ One of the diagonal = 160 m.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Area each of them [son and daughter] will get.

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Let, ABCD be the rhombus shaped field and each side of the field be x

[ All sides of the rhombus are equal, therefore we will let the each side of the field be x ]

Now,

• Perimeter = 400m

\longrightarrow  \tt AB+BC+CD+AD=400m

\longrightarrow  \tt x + x + x + x=400

\longrightarrow  \tt 4x=400

\longrightarrow  \tt  \: x =  \dfrac{400}{4}

\longrightarrow  \tt x= \red{100m}

\therefore Each side of the field = <u>100m</u><u>.</u>

Now, we have to find the area each [son and daughter] will get.

So, For \triangle ABD,

Here,

• a = 100 [AB]

• b = 100 [AD]

• c = 160 [BD]

\therefore \tt Simi \:  perimeter \:  [S] =  \boxed{ \sf \dfrac{a + b + c}{2} }

\longrightarrow \tt S = \dfrac{100 + 100 + 160}{2}

\longrightarrow \tt S =  \cancel{ \dfrac{360}{2}}

\longrightarrow \tt S = 180m

Using <u>herons formula</u><u>,</u>

\star \tt Area  \: of  \: \triangle = \boxed{\bf{{ \sqrt{s(s - a)(s - b)(s - c) } }}} \star

where

• s is the simi perimeter = 180m

• a, b and c are sides of the triangle which are 100m, 100m and 160m respectively.

<u>Putt</u><u>ing</u><u> the</u><u> values</u><u>,</u>

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(180 - 100)(180 - 100)(180 - 160) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(80)(80)(20) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180 \times 80 \times 80 \times 20 }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{9 \times 20 \times 20 \times 80 \times 80}

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{ {3}^{2} \times  {20}^{2}  \times  {80}^{2}  }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  3 \times 20 \times 80

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} = \red{   4800  \: {m}^{2} }

Thus, area of \triangle ABD = <u>4800 m²</u>

As both the triangles have same sides

So,

Area of \triangle BCD = 4800 m²

<u>Therefore, area each of them [son and daughter] will get = 4800 m²</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

{\underline{\rule{290pt}{2pt}}}

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