The outer circumference of a circular race-track is 528 m. The track is 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre (Use $\pi =\dfrac{22}{7}$)
Answer
623.4k+ views
Hint: First try to determine the outer radius of the track from the given outer circumference. Then the inner radius can be obtained as width of the track is given. Find the area of the track by subtracting the inner circular area from the outer circular area. Determine the cost of levelling the track by multiplying the total area and cost of levelling of one square metre.
Complete step by step answer:
Let the outer radius of the track is ‘R’ metre and inner radius is ‘r’ metre.
Outer circumference$=528m$
$\begin{align}
& \Rightarrow 2\pi R=528 \\
& \Rightarrow R=\dfrac{528}{2\pi }=\dfrac{528}{2\times \dfrac{22}{7}} \\
& \Rightarrow R=84m \\
\end{align}$
Given, width of the track$=14m$
Therefore inner radius, $r=R-14=84-14=70m$
Area of the track$=$ outer area $-$ inner area
As we know, the area of the inner track$=\pi {{R}^{2}}$ and the area of the outer track$=\pi {{r}^{2}}$
So, subtracting both we can get the required area as
$\begin{align}
& \Rightarrow \pi {{R}^{2}}-\pi {{r}^{2}} \\
& \Rightarrow \pi \left( {{84}^{2}}-{{70}^{2}} \right) \\
& \Rightarrow \dfrac{22}{7}\times 2156 \\
& \Rightarrow 6776{{m}^{2}} \\
\end{align}$
For levelling the track; one square metre costs 50 paise which $=\dfrac{50}{100}=0.5$ rupees.
6776 square meter will cost $=6776\times 0.5=3388$ rupees.
So, the total cost of levelling the track is 3388 rupees.
This is the required solution of the given question.
Note: The required area should be found by subtracting the area of the inner circle from the area of the outer circle. For levelling the track the cost of one square meter should be multiplied to the total area of the track.
Complete step by step answer:
Let the outer radius of the track is ‘R’ metre and inner radius is ‘r’ metre.
Outer circumference$=528m$
$\begin{align}
& \Rightarrow 2\pi R=528 \\
& \Rightarrow R=\dfrac{528}{2\pi }=\dfrac{528}{2\times \dfrac{22}{7}} \\
& \Rightarrow R=84m \\
\end{align}$
Given, width of the track$=14m$
Therefore inner radius, $r=R-14=84-14=70m$
Area of the track$=$ outer area $-$ inner area
As we know, the area of the inner track$=\pi {{R}^{2}}$ and the area of the outer track$=\pi {{r}^{2}}$
So, subtracting both we can get the required area as
$\begin{align}
& \Rightarrow \pi {{R}^{2}}-\pi {{r}^{2}} \\
& \Rightarrow \pi \left( {{84}^{2}}-{{70}^{2}} \right) \\
& \Rightarrow \dfrac{22}{7}\times 2156 \\
& \Rightarrow 6776{{m}^{2}} \\
\end{align}$
For levelling the track; one square metre costs 50 paise which $=\dfrac{50}{100}=0.5$ rupees.
6776 square meter will cost $=6776\times 0.5=3388$ rupees.
So, the total cost of levelling the track is 3388 rupees.
This is the required solution of the given question.
Note: The required area should be found by subtracting the area of the inner circle from the area of the outer circle. For levelling the track the cost of one square meter should be multiplied to the total area of the track.
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