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08 Apr 2019, 11:27
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
61% (02:35) correct
39%
(02:21)
wrong
based on 320
sessions
History
Date
Time
Result
Not Attempted Yet
Sam drove at a constant speed from City X along a highway to City Y. Did Sam reach City Y from City X in less than an hour?
(1) If Sam had driven at the same speed to City Z that was 15 kilometers further down the highway, he would have taken 50% more time
(2) If Sam's average speed for the drive had been 20 kilometers per hour lesser, he would have taken 50% more time
(1) If Sam had driven at the same speed to City Z that was 15 kilometers further down the highway, he would have taken 50% more time
(2) If Sam's average speed for the drive had been 20 kilometers per hour lesser, he would have taken 50% more time
18 Apr 2019, 07:57
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Sam drove at a constant speed from City X along a highway to City Y. Did Sam reach City Y from City X in less than an hour?
Question: is T<1 hours??
Solution: The question uses the same unit for distance, time and speed so no conversion is required.
1) If Sam had driven at the same speed to City Z that was 15 kilometers further down the highway, he would have taken 50% more time
X----T-----Y---(50%T)---Z
\(Speed1 = \frac{distance}{time} = \frac{d}{t}\)
\(Speed2 = \frac{d+15}{1.5t}\)
\(Speed1= speed 2 = \frac{d}{t} = \frac{d+15}{1.5t} => d+15 = 1.5d => 15=0.5d => d=30 Km\)
Speed and time are unknown, Not sufficient
(2) If Sam's average speed for the drive had been 20 kilometers per hour lesser, he would have taken 50% more time
\(Distance1 = Speed*Time = (S)T\)
\(Distance2 = Speed*Time = (S-20)(1.5)T\)
\(Distance1 = Distance2 = ST = (S-20)(1.5)T => S=(S-20)(1.5) => S=60 Km/hr\)
Only speed is available, distance and time are not... if the distance is more than 60 then time is greater than 1 hour, less than 60 time is less than 1 hour. Not sufficient
Using 1 and 2
From 1: distance=30 Km
From 2: Speed = 60 Km/Hour
\(Time = \frac{Distance}{Speed} = \frac{30}{60}=\frac{1}{2}hour\) = 30 mins < 1 hour sufficient. C
Question: is T<1 hours??
Solution: The question uses the same unit for distance, time and speed so no conversion is required.
1) If Sam had driven at the same speed to City Z that was 15 kilometers further down the highway, he would have taken 50% more time
X----T-----Y---(50%T)---Z
\(Speed1 = \frac{distance}{time} = \frac{d}{t}\)
\(Speed2 = \frac{d+15}{1.5t}\)
\(Speed1= speed 2 = \frac{d}{t} = \frac{d+15}{1.5t} => d+15 = 1.5d => 15=0.5d => d=30 Km\)
Speed and time are unknown, Not sufficient
(2) If Sam's average speed for the drive had been 20 kilometers per hour lesser, he would have taken 50% more time
\(Distance1 = Speed*Time = (S)T\)
\(Distance2 = Speed*Time = (S-20)(1.5)T\)
\(Distance1 = Distance2 = ST = (S-20)(1.5)T => S=(S-20)(1.5) => S=60 Km/hr\)
Only speed is available, distance and time are not... if the distance is more than 60 then time is greater than 1 hour, less than 60 time is less than 1 hour. Not sufficient
Using 1 and 2
From 1: distance=30 Km
From 2: Speed = 60 Km/Hour
\(Time = \frac{Distance}{Speed} = \frac{30}{60}=\frac{1}{2}hour\) = 30 mins < 1 hour sufficient. C
General Discussion
26 Jul 2023, 08:06
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rohan2345
Solution:
Pre Analysis:
- Let the distance between X and Y be D and speed of Sam be S
- Then time to reach Y will be \(\frac{D}{S}\) and we are asked if \(\frac{D}{S}<1\) or not
Statement 1: If Sam had driven at the same speed to City Z that was 15 kilometers further down the highway, he would have taken 50% more time
- According to this statement, to move extra 15 km, it would have taken 50% more time i.e., \(\frac{50}{100}\times \frac{D}{S}=\frac{D}{2S}\)
- Or, we can say \(\frac{15}{S}=\frac{D}{2S}\) or \(D=30\)
- But we do not get the value of S or \(\frac{D}{S}\)
- Thus, statement 1 alone is not sufficient and we can eliminate options A and D
Statement 2: If Sam's average speed for the drive had been 20 kilometers per hour lesser, he would have taken 50% more time
- According to this statement, time when speed \(= S-20\) would have been 50% more than \(\frac{D}{S}\)
- Or we can say \(\frac{D}{S-20}=\frac{3D}{2S}\) or \(S=60\)
- But we do not get the value of D or \(\frac{D}{S}\)
Combining:
- Upon combining we get the value of both D and S
Hence the right answer is Option C
