| When a person cycled at 10 km/h, he arrived at his office 6 min late. He arrived 6 min early, when he increased his speed by 12 km/h. The distance of his office from the starting place is |
A) 6 km
B) 7 km
C) 12 km
D) 16 km
Correct Answer: C
Solution :
| Let the distance be D km, and actual time taken by man to reached the office at time = x |
| According to the question, |
| \[\frac{D}{10}=x+\frac{6}{60}\] (i) |
| and \[\frac{D}{12}=x-\frac{6}{60}\] (ii) |
| From Eqs (i) and (ii), we get |
| \[\frac{D}{10}-\frac{D}{12}=x+\frac{6}{60}-x+\frac{6}{60}\] |
| \[\frac{12D-10D}{120}=\frac{1}{10}+\frac{1}{10}\]\[\Rightarrow \]\[\frac{2D}{120}=\frac{2}{D}=12\,km\] |
| Alternate Method |
| Here, \[{{b}_{1}}=6,\]\[{{t}_{2}}=6,\] \[{{S}_{1}}=10\] and \[{{S}_{2}}=12\] |
| Distance \[=\,\,\frac{({{t}_{1}}+{{t}_{2}}){{S}_{1}}\,.\,{{S}_{2}}}{({{S}_{2}}-{{S}_{1}})\times 60}\,\,=\,\,\frac{(6+6)\,10\times 12}{(12-10)\times 60}\] |
| \[=\frac{12\times 120}{2\times 60}=\frac{24}{2}=12\,km\] |
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