| An aeroplane first flew with a speed of 440 km/h and covered a certain distance. It still had to cover 770 km less than what it had already covered, but it flew with a speed of 660 km/h. The average speed for the entire flight was 500 km/h. Find the total distance covered. |
A) 3250 km
B) 2750 km
C) 4400 km
D) 1375 km
Correct Answer: B
Solution :
| Let x be the first lap of the flight, |
| Then, the second lap will be \[(x-770)\,\,km.\] |
| Total distance \[=\,\,(2x-770)\,\,km\] |
| Let be the total time of journey |
| \[\Rightarrow \]\[\frac{x}{440}+(x-770)/660=t\] (i) |
| \[500t=2x-770\] (ii) |
| From Eq. (i), |
| \[660x+440x-770\times 440=660\times 440t\] |
| \[\Rightarrow \]\[6x+4x-770\times 4=6\times 440t\] |
| \[\Rightarrow \]\[6x+4x-3080=2640t\] |
| \[\Rightarrow \]\[3x+2x-1540=1320t\] (iii) |
| From Eq. (ii), |
| \[2x-770=500t\] (iv) |
| On subtracting Eqs. (iii) and (vi), we get |
| \[3x-770=820t\] (v) |
| From Eqs. (iv) and (v), (v) and (iv), then \[x=320t\] |
| Putting the value of x in Eq. (iii), we have |
| \[5\times 320t-1540=1320t\] |
| \[\Rightarrow \] \[280t=1540\]\[\Rightarrow \]\[t=5.5\] |
| \[\Rightarrow \] \[x=320\times 5.5=1760\] |
| Total distance |
| \[=2x-770=2\times 1760-770=2750\,\,km\] |
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