Dyani and Tim are each walking at a constant speed along Front and Main Streets, respectively, which are perpendicular. When Dyani is at the intersection, Tim is still 500 yards away. In two minutes they are equidistant from the intersection. In eight more minutes they are again equidistant from the intersection. What’s the ratio of Dyani’s speed to Tim’s speed?
- 4:5
- 5:6
- 2:3
- 5:8
- 1:2
Let D’s speed = d and Tim’s speed = t
in yards per minute: yd/min.
Let T_{0} be Tim’s position at time = 0, etc.
T_{0} = 500 yd.
T_{0}T_{2} = 2t
So T_{2}O = 500 – 2t
T_{0}T_{10} = 10 t
So OT_{10} = 10t – 500
T_{0}T_{2} = 2t
So T_{2}O = 500 – 2t
T_{0}T_{10} = 10 t
So OT_{10} = 10t – 500
D_{0}O = 0
OD_{2} = 2d (= 500 – 2t)
OD_{10} = 10d (= 10t – 500)
2d = 500 – 2t → 2d + 2t = 500
10d = 10t – 500 → 10d – 10t = -500
2d + 2t = 500 | → x 5 → | 10d + 10t = 2500 | |
10d – 10t = -500 | |||
20d = 2000 | → d = 100 yd/min (≈ 3.4 mi/hr) |
2(100) + 2t = 500 → 2t = 300 → t = 150 yd/min (≈ 5.1 mi/hr)
So d:t = 100:150 = 2:3. The answer is (c).