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amina.shaikh309
- Senior | Next Rank: 100 Posts
- Posts: 31
- Joined: Sat Apr 23, 2016 9:06 pm
We might be able to solve this one faster by first converting the fractions to decimals.
S1=0.25
S2=0.2
S3=0.17 (approx)
S4=0.15
S5=0.13 (approx)
S6=0.1
Statement 1: S2 and S4 were shipped on the first truck.
First truck has 0.2 + 0.15 = 0.35
Since the first truck holds more than 0.5, S3 may or may not be on that truck. For example, consider these two possible cases:
case a: first truck holds S2, S3 and S4, and second truck holds S1, S5 and S6,
case b: first truck holds S1, S2, and S4, and second truck holds S3, S5 and S6,
Statement 2: S1 and S6 were shipped on the second truck
Second truck has 0.25 + 0.1 = 0.35
Since the first truck holds more than 0.5, the second truck must have less than 0.5
Since S3 = 0.17, S3 cannot be on the second truck, otherwise the second truck would have more than 0.5
Since S3 cannot be on the second truck, we can be certain that it's on the first truck, in which case statement 2 is sufficient and the answer is B
Cheers,
Brent
