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Water fills a tank at a rate of $150$ litres during the $1$st hour, $350$ litres during the $2$nd hour, $550$ litres during the third hour and so on.

Find the number of hours necessary to fill a rectangular tank $16m \times 9m \times 9m$.

So far I have tried this... $16 \times 9\times 9= 1296 m^3$ .

Converted is $1296000$ litres.

The common difference is $200$ litres an hour as it show arithmetic progression $(a_1=150, a_2=350,a_3=550) a_3-a_2=200, a_2-a_1=200$.

Using the formula $S_n=\frac{n}2(2a+(n-1)d)$

I get $1296000=\frac{n}2(150 \times 150(n-1)200)$ which is then $2,592,000= n(22,500+200n-200)$


then $2,592,000=n(200n-22,500) = 2,592,000=200n^2-22,500n-2,592,000=0$.

I feel I am going wrong some where though. I'm unsure of the next step.

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$a=150, d=200, S_n=1,296,000, n=?$.

Now, $S_n=\frac{n}{2}[2a+(n-1)d] \iff 1,296,000=\frac{n}{2}[2(150)+(n-1)200].$

Expanding and simplifying, we get: $$100n^2+50n-1,296,000=0.$$ You now have a quadratic in $n$.

Use the quadratic formula (Try it as an exercise!) to give: $$\underbrace{n\approx-114.09 \ \text{hours}}_{\text{silly}}, \underbrace{\boxed{n\approx113.59 \ \ \text{hours}}}_{\text{sensible}}.$$

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  • $\begingroup$ Thanks for your help, but I’m confused as to how you got to that simplified equation and values. I have tried this below... 1296000=n/2(2(150)+(n-1)200 2,592,000=n(300+200-200) 2,592,000=n(200n-100) 2,592,000=200nSquared-100n 200nSqu-100n-2,592,000=0 Using Common 100 to simplify.... 2nsqu - n - 25920=0. $\endgroup$ – user154569 Jun 8 '14 at 10:00

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