Question
A cyclist travels from town P to town Q at 12 km/h and
returns from Q to P by the same route at a speed which is 25% higher. Due to the increase in speed on the return journey, he takes 1 hour less for the entire trip than he would have taken if he had travelled at 12 km/h both ways. What is the distance between P and Q?Solution
ATQ, Let distance = d km. Time if speed is 12 km/h both ways = 2d/12 = d/6 hours. Actual speeds: P→Q: 12 km/h ⇒ time = d/12 Q→P: 25% more ⇒ 12 × 1.25 = 15 km/h ⇒ time = d/15 Actual total time = d/12 + d/15 = (5d + 4d)/60 = 9d/60 = 3d/20 Given: 3d/20 = d/6 − 1 Multiply by 60: 9d = 10d − 60 ⇒ d = 60 km.
205     255     292     ?     335     345
...14Â Â Â Â Â 15Â Â Â Â Â Â Â 32Â Â Â Â Â Â 99Â Â Â Â Â Â 400Â Â Â Â Â Â ?
...32, ?, 41, 68, 132, 257
...If  2  4  6  x  14  28  30
Then, x³+ 2 x+1 = ?
5Â Â Â Â Â Â Â Â Â Â Â Â Â Â 9Â Â Â Â Â Â Â Â Â Â Â Â Â Â 30Â Â Â Â Â Â Â Â Â Â Â 115Â Â Â Â Â Â Â Â Â 582Â Â Â Â Â Â Â Â Â 3483
15Â Â Â Â ...
If   314     306      x      269       394    178    1480
Then, 1/3 x + 2x - x = ?
...29    51    95    183    ?     711
5     16     ?      66     119      200
...28         57           172           689          ?              20677
...21     84      168      172     ?     348
...
