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aniked [119]
3 years ago
11

Water fills a tank at a rate of 150 litres during the first hour, 350 litres during the second, 550 litres during the third and

so on. Find the number of hours necessary to fill a rectangular tank 16m x 7m x 7m.

Mathematics
2 answers:
Lubov Fominskaja [6]3 years ago
5 0
Volume of tank = (16m)(7m)(7m) = 784 m³

Conversion of m³ to L:
(784 m³) × (1000L / 1m³) = 784,000 L

Rate in the 1st hour:
150 liters/hr

Rate in the 2nd hour:
350 liters/hr

Rate in the 3rd hour:
550 liters/hr

It is apparent that the Fill Rate is increasing by 200 liters/hr every subsequent hour . . . so that can be represented by the following equation

where:
t = number of hours

Fill rate (i.e. volume of water filled into tank within the specified hour) = 150 + 200(t - 1)

For t = 1 . . . Fill rate = 150 L/hr
For t = 2 . . . Fill rate = 350 L/hr
For t = 3 . . . Fill rate = 550 L/hr

Because after every hour there has been more water added to the tank, this problem can be represented as a geometric sequence in order to account for the compounding of the volume after each time step, but it can also be tabulated (which seems to me to be the more direct/simple approach), so I will build a table that accounts for the increasing Fill Rate and the compounding of water volume after each time step . . .

(see attached)

The answer (after all of this) is . . .  t = 88 hrs 17 1/2 mins (approx)



Alborosie3 years ago
3 0
Putting this as an arithmetic sequence gives:

u_n = 150+200(n-1)

The sum of the series = 16 x 7 x 7 = 784 m^3 = 784 000 L

The sum of an arithmetic series can be written as:

S_n=n/2 [2a+(n-1)d] = 784 000
\\n/2[2(150)+(n-1)200] = 784 000
\\n[300+200(n-1)=1 568 000
\\300n+200n^2-200n = 1 568 000
\\200n^2+100n- 1 568 000 = 0
\\2n^2 +n- 15680 = 0

\\n= 88.2...,-88.7

n has to be positive, so we get

n = <u>88.2 hours (3 s.f.)</u>
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Answer:

\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}

Step-by-step explanation:

The <u>width</u> of a square is its <u>side length</u>.

The <u>width</u> of a circle is its <u>diameter</u>.

Therefore, the largest possible circle that can be cut out from a square is a circle whose <u>diameter</u> is <u>equal in length</u> to the <u>side length</u> of the square.

<u>Formulas</u>

\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}

\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}

\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}

If the diameter is equal to the side length of the square, then:
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Therefore:

\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}

So the ratio of the area of the circle to the original square is:

\begin{aligned}\textsf{area of circle} & :\textsf{area of square}\\\sf \dfrac{1}{4}\pi s^2 & : \sf s^2\\\sf \dfrac{1}{4}\pi & : 1\end{aligned}

Given:

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  • radius (r) = 6 ÷ 2 = 3 in

\implies \sf \textsf{Area of square}=6^2=36\:in^2

\implies \sf \textsf{Area of circle}=\pi \cdot 3^2=28\:in^2\:\:(nearest\:whole\:number)

Ratio of circle to square:

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