The Mammoth Corporation has just completed hiring nine new workers: B, C, D, E, F, G, H, I, and J.
In no time at all you should see that this set asks you to arrange the entities in order of hiring—a sequence set. Moreover, it’s a special type of sequencing set, one we call a “free-floating” sequencing set because all of the entities are described relative to each other, as opposed to in a standard sequencing set in which the entities are slated for definite slots (Monday to Friday, 1 to 7 in a line, etc.). In contrast, in “free-floating” sets, the important aspect is where each entity is relative to the others—H is before D, D is after I, etc. On the basis of these types of rules, there are no specific slots we’re asked to fill in, although sometimes, as in this case, we’re able to deduce much of the relative ordering. That is, in easy sets of this type (and this is one of the easier ones), we get so much relative information that we’re able to fit entities into specific time slots, but that’s not always the case.
Each of the nine new workers—B, C, D, E, F, G, H, I, and J—was recently hired. The Key Issues will be:
1) What worker can, must, or cannot have been hired before or after what other worker?
2) What worker can, must, or cannot be hired on the same day as what other worker?
The Initial Setup:
Many people find it helpful to visualize a “free-floating” sequencing set in a vertical fashion. Let’s put the workers who are hired earlier above the workers who were hired later. Remember to list the workers’ letters off to the side, and we’re ready to hit the rules: B C D E F G H I J The Rules: A good way to start is to find a worker or workers who might come at the top of the sketch; that is, anyone who doesn’t seem to be hired after other workers. If you scan the rules, you’ll see that H fits that description, as does the FI pair from the first rule. Let’s therefore start by putting an “FI” and an “H” at the top of the sketch.
A good way to start is to find a worker or workers who might come at the top of the sketch; that is, anyone who doesn’t seem to be hired after other workers. If you scan the rules, you’ll see that H fits that description, as does the FI pair from the first rule. Let’s therefore start by putting an “FI” and an “H” at the top of the sketch.
1) F and I were the only two workers hired on their day, so whatever is true of one is true of the other. That means they’re an inseparable pair. We’ve already put our “FI” pair at the top of the sketch, so let’s move on. Where to now? We may as well find entities related to the ones we’ve already put on the page. Rule 6, containing I, links up nicely:
6) Break this information down into parts. First of all, D was hired after I. Draw a line down from the “FI” pair to a “D.” Secondly, D was hired before E, so draw a line down from the “D” and add an “E.”
5) We now continue to incorporate the rules that link up with the entities already on the page: “H was hired before D,” so we can draw another line up from D connecting to the H we drew earlier.
4) Now we can work down from E: This rule says that E was hired before B, so draw a line down from the “E,” and add a “B.”
8) This seems like the logical rule to incorporate next: B was hired before J, so draw a line down from the “B,” and add a “J.”
7) Now that J and B are placed, we can deal with Rule 7: G was hired after J and B. Add a “G” below the “J.”
2) Like F and I, C and G were hired on the same day (along with no one else that day), so add C to the G we’ve just placed below J.
3) We’re left with this one general rule: Except for the two pairs of workers in Rules 1 and 2 (FI and CG), on every other day exactly one employee was hired. Basically a loophole closer, no need to draw anything. (Note that this does confirm a total of seven hiring days, although they need not be consecutive.)
Much should be clear just by reviewing the master sketch: H and FI must have been hired on days 1 and 2, in either order. D must have been hired on day 3; E on day 4; B on day 5; J on day 6; and GC on the last day, day 7. All that’s ambiguous, in fact, is the H vs. FI relationship. Talk about concrete information! Small wonder that there are so many “non-if” questions.
The Final Visualization: And here is that neat, accessible sketch:
As long as you constructed your sketch carefully, this question is a gimme. The master sketch clearly shows that the last two workers hired were G and C, choice (D).
Again, if your sketch is complete and accurate, you’re rewarded with another ten-second question. The FI pair and H were hired on days 1 and 2 (although we don’t know in which order), D was hired on day 3, and E on day 4, choice (A).
Here you’re looking for the workers who must have been hired before J. Again, let your master sketch do the work for you. Just count the entities who must have come before J: F, I, H, D, E, and B, which makes a total of six, choice (A).
Thanks to the master sketch, you know nearly everything about the entities. It’s an easy (and quick) matter of checking each choice looking for the one that must be true.
(A) No, D must have been fourth, not first.
(B) and (C) H could have been hired first, but so could the FI pair. Each of these choices is possible only. One of ‘em is true, but we can’t figure out which.
(D) No—clearly the first two workers hired could be I and F, with H third. (D) therefore need not be true.
(E) is the only choice left, but for the record: The sketch clearly shows something we’ve known from the very beginning: that either H was hired first, or F and I were. (E) must be true and is the answer.
If E was hired on Monday, the earliest day on which B could have been hired is Tuesday. If B was hired on Tuesday, the earliest J could have been hired is Wednesday. And if J was hired on Wednesday, the earliest G (as well as C, of course) could have been hired is Thursday, answer choice (D).Online CAT LRDI Course @ INR 3999 only