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nydimaria [60]
3 years ago
9

The Magic Dragon has 2013 heads. Leo the Brave is cutting off the Dragon's heads at a speed of 5 heads per minute. The Dragon's

heads are growing back at a speed of 2 heads per minute. How many heads will the Dragon have in an hour? How long will the "fight" last?
Mathematics
1 answer:
aivan3 [116]3 years ago
3 0

Answer:  The answer is 1833 heads  and 11 hours, 11 minutes.


Step-by-step explanation:  Given that the Magic Dragon has 2013 heads. The brave Leo is cutting off the Dragon's heads at a speed of 5 heads per minute. But, the Dragon's heads are growing back at a speed of 2 heads per minute.

So, the number of heads of Dragon that are decreasing per minute = 3.

Therefore, after 1 hour, i.e., 60 minutes, The number of heads of the Dragon

= 2013 - 60 × 3 = 2013 - 180 = 1833 heads.

Now, the fight will last until all the heads of the Dragon are cut. Let the fight lasts for 'x' minutes, then

2013-x\times 3=0\\\\\Rightarrow 3x=2013\\\\\Rightarrow x=\dfrac{2013}{3}=671.

Hence the fight lasts for 671 minutes, i.e., 11 hours and 11 minutes.

Thus, the Dragon will have 1833 heads after 1 hour and the fight will last for 11 hour and 11 minutes.


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<u>Answer: </u>

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

<u> Solution: </u>

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)  

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3)        ------(1)

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Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

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Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)    

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x = \frac{2}{3}

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Length = x+6 = (\frac{2}{3}) +6 = 6.667m

Width = x +3 = (\frac{2}{3}) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2(\frac{2}{3}) + 6 = \frac{22}{3} = 7.333m

Width = 2x + 2 = 2(\frac{2}{3}) + 2 = \frac{10}{3} = 3.333m

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Area of pool A = (\frac{20}{3}) x (\frac{11}{3}) = \frac{220}{9} = 24.44 meter

Area of pool B = (\frac{22}{3}) x (\frac{10}{3}) = \frac{220}{9} = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

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