A and B begin, at 6 a.m. on the same day, to walk along a road in the same direction, B having a start of 14 miles, and each walking from 6 a.m. to 6 p.m. daily. A walks 10 miles, at a uniform pace, the first day, 9 the second, 8 the third, and so on: B walks 2 miles, at a uniform pace, the first day, 4 the second, 6 the third, and so on. When and where are they together?
They meet at end of 2d. 6h., and at end of 4d.: and the distances are 23 miles, and 34 miles.
Let x = no. of days.
Then (2 × 10 − x − 1) ·
(x)/ |
(2) |
(x)/ |
(2) |
(21x)/ |
(2) |
(x²)/ |
(2) |
(19 ± 5)/ |
(6) |
(7)/ |
(3) |
Now the above solution has taken no account of the discontinuity of increase, or decrease of pace, and is the true solution only on the supposition that the increase or decrease is continuous, and such as to coincide with the above data at the end of each day. Hence ‘4’ is a correct answer; but ‘
(7)/ |
(3) |
Now in 2 days A has got to the end of 19 miles, B to the end of (14 + 6), i.e. 20.
∴ 19 + y ·
= 20 + y · (8)/ (12) (6)/ (12)
i.e. y ·
= 1 + y · ½; ∴ y = 6.
(2)/ (3)
Hence they meet at end of 2d. 6h., and at end of 4d.: and the distances are 23 miles, and 34 miles.
Q.E.F.