SDT Practice DS SET – 3
DS: Speed, Distance & Time – Level 3
1.
Car A and Car B start at the same time from the same point. Which car travels further in 3 hours?
(1) Car A travels at a constant speed of 50 mph.
(2) Car B travels 100 miles in the first 2 hours.
(1) Car A travels at a constant speed of 50 mph.
(2) Car B travels 100 miles in the first 2 hours.
Solution (E):
(1) Car A distance = 150 miles. No info on B.
(2) Car B speed could be constant (50 mph) or variable. If it stopped after 2 hours, dist = 100. If it continued, dist = 150. If it sped up, dist > 150.
Since (2) doesn’t specify **constant** speed or the speed for the 3rd hour, we cannot determine B’s total distance. Not sufficient.
2.
Is the average speed of a car for a trip greater than 50 mph?
(1) For the first half of the **time**, the speed was 60 mph.
(2) For the second half of the **time**, the speed was 40 mph.
(1) For the first half of the **time**, the speed was 60 mph.
(2) For the second half of the **time**, the speed was 40 mph.
Solution (C):
When time is divided equally, Average Speed = Arithmetic Mean of speeds.
(1) Only half the trip known.
(2) Only half the trip known.
Combined: Avg = $(60 + 40) / 2 = 50$ mph.
Is 50 > 50? No. We have a definitive answer. Sufficient.
3.
Did the car take more than 2 hours to travel 120 miles?
(1) The car traveled the first 60 miles at 40 mph.
(2) The car traveled the last 60 miles at 90 mph.
(1) The car traveled the first 60 miles at 40 mph.
(2) The car traveled the last 60 miles at 90 mph.
Solution (A):
We need to know if Total Time > 2.
(1) Time for first 60 miles = $60/40 = 1.5$ hours. For the total time to be > 2, the remaining 60 miles must take > 0.5 hours. Even if the car goes 1000 mph, time is > 1.5. But if the car goes infinitely fast? Wait.
Let’s assume finite speed. The time is $1.5 + T_2$. Since $T_2$ is always positive, the total time is $> 1.5$. This doesn’t prove it’s $> 2$.
Wait, let me re-evaluate. If the second half is driven at 120 mph, $T_2 = 0.5$, Total = 2. If driven at 200 mph, Total < 2. If driven at 30 mph, Total > 2.
Wait, I misread (1).
Let’s re-read.
(1) First half took 1.5 hours. We need 0.5 hours more to exceed 2. If speed on 2nd half is < 120 mph, Yes. If speed > 120 mph, No. Not sufficient.
(2) Last 60 miles at 90 mph. $T_2 = 60/90 = 2/3$ hour. $T_1$ could be anything. Not sufficient.
Combined: $T = 1.5 + 0.66 = 2.16$ hours. $2.16 > 2$. Yes. Sufficient.
**Correction:** The answer is **C**.
4.
A train travels from X to Y. If it increases its speed by 20%, how much time is saved?
(1) The distance between X and Y is 240 miles.
(2) Traveling at the original speed takes 6 hours.
(1) The distance between X and Y is 240 miles.
(2) Traveling at the original speed takes 6 hours.
Solution (B):
Time saved = $T_{old} – T_{new}$.
$S_{new} = 1.2 S_{old}$.
$T_{new} = D / (1.2 S_{old}) = (1/1.2) \times (D/S_{old}) = (5/6) T_{old}$.
Saved = $T_{old} – (5/6)T_{old} = (1/6)T_{old}$.
To find the exact time saved, we only need the original time $T_{old}$.
(1) Gives Distance only. Not sufficient.
(2) Gives $T_{old} = 6$. Saved = 1 hour. Sufficient.
5.
Is distance \(D\) greater than 100?
(1) A car travels \(D\) miles in 2 hours at a speed greater than 55 mph.
(2) A car travels \(D\) miles in less than 2 hours at a speed of 40 mph.
(1) A car travels \(D\) miles in 2 hours at a speed greater than 55 mph.
(2) A car travels \(D\) miles in less than 2 hours at a speed of 40 mph.
Solution (A):
(1) $D = R \times 2$. Since $R > 55$, $D > 110$. Since $110 > 100$, Yes. Sufficient.
(2) $D = 40 \times T$. Since $T < 2$, $D < 80$. Is $D > 100$? No. Definite answer. Sufficient.
(Note: Both statements give a definite answer, even though one is Yes and one is No. In DS logic, a definite No is sufficient).
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