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29 Jan 2021, 07:45
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C
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A cyclist is travelling on a square path having curved corners as shown in the figure. Each curved corner is a quarter circle of radius 42 m. If he travels on the curved path at a speed that is 2/3 rd of his speed on the straight path and traces the complete path once in 15 minutes, what is his speed on the straight path?
(A) 2.20 km/hr
(B) 6.65 km/hr
(C) 8.24 km/h
(D) 10 km/h
(E) 12.48 km/hr
Attachment:
2021-01-28_22-01-22.png [ 22.15 KiB | Viewed 3510 times ]
Let the speed be \(x\) \(\frac{m}{s}\).
Distance covered on the straight path \(= 4 \times (500-84)=(4 \times 416)\) m
Distance covered in curved path \(=2 \pi \mathcal{r}= 2 \times \frac{22}{7} \times 42 = 12 \times 22\) m
Therefore,
\(\frac{4 \times 416}{x} + 12 \times 22 \times \frac{3}{2x}=15\times 60\)
\(\frac{1}{x} \times (1664+396)=15 \times 60\)
\(x=\frac{2060}{15 \times 60}=\frac{206}{9} \times \frac{18}{5}=8.24 \frac{km}{hr} \)
Ans C
Distance covered on the straight path \(= 4 \times (500-84)=(4 \times 416)\) m
Distance covered in curved path \(=2 \pi \mathcal{r}= 2 \times \frac{22}{7} \times 42 = 12 \times 22\) m
Therefore,
\(\frac{4 \times 416}{x} + 12 \times 22 \times \frac{3}{2x}=15\times 60\)
\(\frac{1}{x} \times (1664+396)=15 \times 60\)
\(x=\frac{2060}{15 \times 60}=\frac{206}{9} \times \frac{18}{5}=8.24 \frac{km}{hr} \)
Ans C
08 May 2024, 18:22
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meanup
How did you get subtract 84 from 500 in order to get the straight path's distance?
IMO, since each curved path is a quater of a circle and since each curved path is shared by 2 of the sides of the straight paths, thus we should subtract (1/8*2)*circumference of the cicle which is equal to 66.
Thus the straight path's distance should be 4*(500-66) = 1736.
Please let me know where am I getting this wrong !!
Bunuel meanup
