Let the distance travelled by slower train be ‘x’ km So, distance travelled by faster train = ‘x + 100’ km ATQ, (x/30) = {(x + 100)/40} Or, 40x = 30x + 3000 Or, 10x = 3000 So, x = 300 Total distance between station ‘A’ and station ‘B’ = (300 + 300 + 100) = 700 km
23 48 98 198 ? 798
...The question below is based on the given series I. The series I satisfy a certain pattern, follow the same pattern in series II and answer the question...
49 121 169 289 361 ?
...130 114 106 102 100 ?
3 15 35 63 99 ?
9 10 6 15 -1 ?
...35 70 105 140 175 ?
299, 302, 308, 317, ?, 344
14 15 32 99 400 ?
...56. 23 45 89 177 363 705
...
