SDT Practice DS SET – 2
DS: Speed, Distance & Time – Level 2
1.
Two cars, A and B, are traveling towards each other from 300 miles apart. How long until they meet?
(1) Car A travels at 60 mph.
(2) Car B travels at 40 mph.
(1) Car A travels at 60 mph.
(2) Car B travels at 40 mph.
Solution (C):
To find the meeting time, we need the relative speed ($A + B$).
(1) Gives A only.
(2) Gives B only.
Combined: $60 + 40 = 100$ mph. Time = $300/100 = 3$ hours. Sufficient.
2.
Is the average speed of a round trip greater than 45 mph?
(1) The speed going was 60 mph and the speed returning was 40 mph.
(2) The distance of the one-way trip was 120 miles.
(1) The speed going was 60 mph and the speed returning was 40 mph.
(2) The distance of the one-way trip was 120 miles.
Solution (A):
(1) Avg Speed = $\frac{2xy}{x+y} = \frac{2(60)(40)}{100} = 48$ mph. $48 > 45$. Sufficient.
(2) Distance alone doesn’t give speed. Not sufficient.
3.
Car X leaves City A for City B at 1 PM. Car Y leaves City A for City B at 2 PM. At what time will Car Y catch Car X?
(1) Car X travels at 50 mph.
(2) City B is 400 miles away from City A.
(1) Car X travels at 50 mph.
(2) City B is 400 miles away from City A.
Solution (E):
To find the catch-up time, we need the speeds of BOTH cars.
(1) Gives Speed X. No info on Speed Y.
(2) Gives total distance. No speeds.
Combined: We know X is 50 mph. We still don’t know Y’s speed. Y might never catch X if Y is slower. Not sufficient.
4.
If Bob drove from home to work, how many miles is the trip?
(1) If he drove at 40 mph, he would arrive 15 minutes late.
(2) If he drove at 60 mph, he would arrive in 1 hour.
(1) If he drove at 40 mph, he would arrive 15 minutes late.
(2) If he drove at 60 mph, he would arrive in 1 hour.
Solution (B):
(1) We don’t know the scheduled arrival time or usual time. “15 minutes late” is relative to an unknown variable. Not sufficient.
(2) $D = R \times T = 60 \times 1 = 60$ miles. Sufficient.
5.
A boat travels upstream. What is the speed of the current?
(1) The boat’s speed in still water is 20 mph.
(2) The boat takes 5 hours to travel 80 miles upstream.
(1) The boat’s speed in still water is 20 mph.
(2) The boat takes 5 hours to travel 80 miles upstream.
Solution (C):
Let $B$ = boat speed, $C$ = current speed. Upstream rate = $B – C$.
(1) $B = 20$. No info on $C$.
(2) Upstream rate = $80/5 = 16$ mph. So $B – C = 16$. Can’t find $C$ without $B$.
Combined: $20 – C = 16 \implies C = 4$. Sufficient.
Score: 0 / 0