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iogann1982 [59]
3 years ago
8

The farm keeps 20 gallons of the milk produced each day to use on the farm and sells the rest of the milk to a distributor to be

sold in stores. The farm packages 10 gallons of the milk it keeps into gallon containers. The remaining 10 gallons are divided so that 5 gallons will be divided among one-quart containers, and 5 gallons will be divided among one-pint containers. A. Find the number of one-quart containers needed each day. B. Find the number of one-pint containers needed each day. C. Explain how you would find the number of one-quart and one-pint containers needed each week.
Mathematics
1 answer:
Elena-2011 [213]3 years ago
5 0

Answer:

A. 20 containers

B. 40 containers

C. 140, 280 containers

Step-by-step explanation:

What we should do is find the equivalences of a quarter and a pint, in gallons.

We have to:

1 gallon = 4 quarts.

1 gallon = 8 pints.

If we have 5 gallons of each.

A. 5 gallon * 4 quarts / 1 gallon = 20 one-quart containers per day are required.

B. 5 gallon * 8 pints / 1 gallon = 40 one-pint containers per day are required.

C. To find out what is needed each week, multiply by 7, since one week is 7 days.

So:

For one-quarter containers:

20 * 7 = 140 containers per week.

For one-pint containers:

40 * 7 = 280 containers per week.

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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>

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So, For \triangle ABD,

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• b = 100 [AD]

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\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) }

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\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

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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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