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valentina_108 [34]
3 years ago
15

A candle shop sells scented candles for $16 each and unscented candles for $10 each. The shop sells 28 candles today and makes $

400.
Write a system of linear equations that represents the situation.
Please also include how many scented candles were sold and how many unscented were sold.
Thanks!!!
Mathematics
1 answer:
Rudiy273 years ago
7 0
<h3><u>The system of linear equations that represents the situation is:</u></h3>

a + b = 28

16a + 10b = 400

<h3><u>20 scented candles and 8 unscented candles were sold</u></h3>

<em><u>Solution:</u></em>

Let "a" be the number of scented candles sold

Let "b" be the number of unscented candles sold

From given,

Cost of 1 scented candles = $ 16

Cost of 1 unscented candles = $ 10

<em><u>The shop sells 28 candles today</u></em>

Therefore,

a + b = 28

b = 28 - a ------- eqn 1

<em><u>The shop sells 28 candles today and makes $400</u></em>

Therefore,

number of scented candles sold x Cost of 1 scented candle + number of unscented candles sold x Cost of 1 unscented candles = 400

a \times 16 + b \times 10 = 400

16a + 10b = 400 ------ eqn 2

<em><u>Substitute eqn 1 in eqn 2</u></em>

16a + 10(28 - a) = 400

16a + 280 - 10a = 400

6a = 400 - 280

6a = 120

a = 20

<em><u>Substitute a = 20 in eqn 1</u></em>

b = 28 - 20

b = 8

Thus 20 scented candles and 8 unscented candles were sold

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  • Question 1a. i) x=1.8m

  • Question 1a. ii)   Volume=27.6m^3

  • Question 1b) Volume=65.9m^3

Explanation:

<u><em>Question 1 a. i) Find the value of x.</em></u>

         tan(\theta )=\dfrac{opposite\text{ }leg}{adjacent\text{ }leg}

For the smalll triangle you can write:

        tan(\theta )=\dfrac{x}{1m}

For tthe big triangle:

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        \dfrac{x}{1m}=\dfrac{x+2.7m}{2.5m}

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        2.5x=x+2.7m\\\\2.5x-x=2.7m\\\\1.5x=2.7m\\\\x=2.7m/1.5\\\\x=1.8m

<u><em>Question 1a ii) Find the volume of the frustrum</em></u>

  • Find the volume of a cone with height = 2.7m + 1.8m = 4.5m, and radius = 2.5m

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         Volume=(1/3)\pi \times radius^2\times height

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  • Subtract the volume of the small cone from the volume of the big cone

        Volume\text{ }of\text{ }frustrum=9.375\pi m^3-0.6\pi m^3=8.775\pi m^3\approx 27.6m^3

<u><em>Question 1b. Calculate the volume of the bin</em></u>

<u>i) Upper frustrum</u>

This is the same frustrum from the equation of above, thus ist volume is 27.6m³.

<u>ii) Lower frustrum</u>

            \dfrac{x}{2.0m}=\dfrac{x+2.4m}{2.5m}

           2.5x=2(x+2.4m)\\\\2.5x=2x+4.8m\\\\0.5x=4.8m\\\\x=9.6m

        Volume=(1/3)\pi \times (2.5m)^2\times (9.6m+2.4m)-(2.0m)^2\times (9.6m)

       Volume=38.3m^3

<u>iii) Add the volume of the two frustrums</u>

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