| A is twice as fast as B and B is thrice as fast as C. The journey covered by C in \[1\frac{1}{2}h\] will be covered by A in |
| [SSC (10+2) 2014] |
A) 15 min
B) 20 min
C) 30 min
D) 1 h
Correct Answer: A
Solution :
| Let distance covered by A, B and C be d. |
| Now, let speed of A = 6x and speed of C = x |
| \[\therefore \] speed of B = 3x |
| \[\therefore \] This distance is covered by C in |
| \[1\frac{1}{2}h=\frac{3}{2}h=9\min \] |
| \[\therefore \] Time taken to covered this distance by |
| \[\text{C}\,\,\text{=}\,\,\frac{\text{Distance}}{\text{Speed}}\] |
| \[\Rightarrow \] \[\text{Time}=\frac{d}{x}\]\[\Rightarrow \]\[90=\frac{d}{x}\] |
| \[\Rightarrow \] \[x=\frac{d}{90}\] (i) |
| Now, time taken to covered this distance by A. |
| \[\Rightarrow \] \[\text{Time}\,\,=\,\,\frac{\text{Distance}}{\text{Speed}}\] |
| \[\Rightarrow \] \[\text{Time}\,\,=\,\,\frac{d}{6x}=\frac{d}{6\times \frac{d}{90}}\] [from Eq. (i)] |
| \[\Rightarrow \] \[\text{Time}=\frac{d\times 90}{6\times d}=\frac{90}{6}=15\min \] |
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