| A man decides to travel 80 km in 8 h partly by foot and partly on a bicycle. If his speed on foot is 8 km/h and on bicycle 16 km/h, then what distance would be travel on foot? |
A) 20 km
B) 30 km
C) 48 km
D) 60 km
Correct Answer: C
Solution :
| Here, \[{{S}_{1}}=8\,\,km/h\]and \[{{S}_{2}}=16\,\,km/h\] |
| \[\therefore \] \[{{d}_{1}}={{S}_{1}}\times {{t}_{1}}=8\,\,{{t}_{1}}\] ... (i) |
| and \[{{d}_{2}}={{S}_{2}}\times {{t}_{2}}=16\,\,{{t}_{2}}\] ... (ii) |
| We know that, |
| \[{{t}_{1}}+{{t}_{2}}=8\] ... (iii) |
| and \[{{d}_{1}}+{{d}_{2}}=80\] (given) ... (iv) |
| From Eqs. (i) and (ii) put the value of di and da in Eq, (iv), we get |
| \[{{d}_{1}}+{{d}_{2}}=80\] |
| \[8{{t}_{1}}+16{{t}_{2}}=80\] |
| \[\Rightarrow \]\[8{{t}_{1}}+8{{t}_{2}}=8{{t}_{2}}=80\] |
| \[8\,\,({{t}_{1}}+{{t}_{2}})+8{{t}_{2}}=80\] |
| \[8\times 8+8{{t}_{2}}=80\] [from Eq. (iii)] |
| \[8{{t}_{2}}=80-64=16\] |
| \[\Rightarrow \] \[{{t}_{2}}=\frac{16}{8}=2\,\,h\] |
| \[\therefore \] \[{{t}_{1}}=8-2=6\,\,h\] |
| Distance travelled by foot \[={{d}_{1}}=8\times 6=48\,\,km\] |
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