LRDI Practice Set 8

The Action:

After reading the introductory paragraph and rules during your set overview, you probably thought, “Whew! There’s a lot going on here.” And so there is. Six dogs are each either a greyhound or a labrador and each either male or female—the matching aspect of this complex hybrid set. The six dogs are awarded four ribbons, first, second, third, and fourth based on their performances in the dog show, while two of the dogs don’t get a ribbon at all. The ribbons make up the sequencing part of this hybrid set.

The Key Issues are:

1) What is each dog’s gender and breed?

2) What ribbon, if any, does each dog receive?

The Initial Setup: One of the big challenges inherent in this set is how to keep track of all the information. The ultimate decision about this will come when you deal with the rules. For now, list the dogs and the possibilities for each dog’s gender, type, and ribbon.

Note: it’s a good idea to include “Xs” to represent the two dogs who won’t get a ribbon:

Now we’re going to go through the rules step by step, just as you might have done when working on the set yourself. If you had trouble with this set, please take your time and follow this discussion carefully. The effort will be worth it. We promise.

The Rules:

1) simply sets the possible type of each dog. Each of the six dogs is either all greyhound or all labrador; no mixed breeds are possible. We’re not told how many of each breed are present, so there’s not much more to do with this rule right now.

2) Here’s a helpful numbers rule. Exactly two of the six dogs are female, and exactly four are male. Make a note of that: You need to distribute two f’s and four m’s. (Or you can use the universal male/female symbols for that purpose.)

3) This rule has a lot of information, so as always with two-part rules, break it up and take it one step at a time. First of all, both female dogs—there are only two females, remember—are awarded ribbons. It’s a very small step to realize that both of the dogs who don’t receive ribbons must therefore be male. (Make a note of that: Those “X” dogs, the losers, are male.) By the same token, the other two ribbon winners must be male too. So make a note next to the 1st through 4th place slots: Two are female, two are male. Secondly, we’re told that exactly one of the two female dogs is a labrador. Again, it’s not a big jump to recognize that the other female dog must be a greyhound. To sum up (always a good idea in a complex set): We have two females finishing in the top four and winning prizes; one is a greyhound, the other a lab. We also have two winning males, and two loser males, breeds as yet undetermined.

4) states that exactly one labrador receives a ribbon. This should sound very familiar, because just one rule ago we saw a labrador winning a ribbon; it was a  female. Well, Rules 3 and 4 must be talking about the same ribbon-winning labrador. So since we have a single winning labrador, therefore the other three winners must be greyhounds. Our winners, then, are two male greyhounds, one female greyhound, and the selfsame female labrador. Those are the dogs that will take 1st through 4th place, though to be sure we don’t yet know their “names.” (It’s worth noting another implication of this rule: Since there is one and only one winning labrador, and it’s a female, any male labrador that happens to wander by in a question will have to take one of the two “loser” slots.)

5) This is the rule that most explicitly deals with the sequencing aspect of this set. It’s also the rule around which the master sketch will truly be built. Sketch it out rather than trying to work with it on a sheer mental level: Since P and R place higher than S who places higher than Q and T, draw it that way. Create a P and R with arrows down to an S, then more arrows down from that S to a Q and T. The big question that ought to come to mind, and bravo to you if you thought to ask it, is: What about U, the only dog unnamed by Rule 5—where does U place? The answer, for the time being, is: Don’t know. But we will.

6) P and R are greyhounds. Fine; let’s just jot that down for the moment. Next to the P and R that you’ve just drawn, note each as a “grey” to remind yourself of these dogs’ types.

7) Same kind of thing: S and U are labradors. Next to the S in your master sketch add a “lab” for labrador. As we noted before, we haven’t placed “U” in the master sketch yet, so jot down “U=lab” or something; but as they say, hold that thought . . .

Key Deductions:

Now it’s time to put it all together, and as usual, we should look at our most significant “bloc” of information. Here, it’s got to be the sequencing information from Rule 5. Use it to pursue the question we asked a while ago regarding Rule 4: What are the winners’ names?

Can P and R not win ribbons? No way. Each of them places higher than three other dogs, according to Rule 5. So P and R win ribbons.

Can S not win a ribbon? No way. S places higher than Q and T, so at the very worst, S can finish no lower than 4th place. Now we have three winners named out of four: P, R, and S.

Who’s the fourth winner? Well, it could be either Q or T; whichever one wins a ribbon, the other is a “loser.” But what about U? Can U be the fourth winner?

No way—and that’s the whole ball of wax, the Big Deduction available here. Remember what Rule 3 told us (and your sketch should remind you of it, if nothing else): There’s only one ribbon-winning labrador. S, a labrador, is that winner—we just deduced that from Rule 5. Therefore labrador U must take one of the “loser” slots. Go back over these last few steps carefully, to make sure you understand ‘em. Let’s sum it all up:

• Our winners are P, R, S, and one of the Q/T pair. U, and either Q or T, are the losers. Who gets which ribbon? Read on:
• Given the demands of Rule 5, P and R must take the 1st and 2nd place ribbons, in either order. S will have to take 3rd place, leaving either Q or T to take the 4th place ribbon. Once again, of Q and T, the one who doesn’t win shares “loser” status with U.
• We also know that S, as the winning labrador, must be female. (Rules 3, 4, and 5 demand it.) Thus the other winning female, a greyhound, is either P, R, or the 4th place dog. That’s another bit of uncertainty. However, we do know that since there’s exactly one winning labrador (Rule 4), and that S is it, then the other ribbon winners must be greyhounds. No big deal for P and R—we already knew they were greyhounds from Rule 6. But this does allow us to deduce that the fourth place dog, be it Q or T, must be a greyhound.
• What about the losers’ breeds? One loser, U, is a labrador, we know that. The other loser, male, could be labrador or greyhound—no way to tell.
• Both “loser” dogs, of course, are male—we deduced that way back at Rule 3, but it’s been a while since then. . . .

The Final Visualization: Here’s your master visualization of the whole thing:

The Questions:

1. (E)

Which dogs can be greyhounds? Well, why not start with the dogs that we definitely know must be labradors? S and U are the only dogs that are explicitly identified (by Rule 7) as labradors, so any of the other dogs can be greyhounds. That’s P, Q, R, and T, answer choice (E). (Of course P and R are greyhounds for sure, but they still merit a place in the right answer.)

2. (B)

Not much to do with this “CANNOT be true” question but to try out the choices. Remember that since the right answer is impossible, the four wrong choices all could (or must) be true.

(A) Either P or R (both greyhounds) wins 2nd place, and either can be female. (A) need not be false.

(B) This is it. Only P and R may win the top two ribbons, and Rule 6 identified P and R as greyhounds, so there’s no way that a labrador of either gender could win the second place ribbon. Since (B) can’t be true, it’s the answer. For the record, here’s the skinny on the remaining choices:

(C) S, a female lab, must win the third place ribbon.

(D) and (E) Either Q or T could be a male greyhound winning 4th place. By the same token, either could be female. Remember, we still have one female ribbon winner unaccounted for.

3. (E)

This “must be true” question with no new information is testing for one of our deductions. We deduced many things, but the only dog that we deduced must be male is U, choice (E).

4. (A)

You’re looking for the choice that can be false, so the four wrong choices must be true. There’s no new information, so not much to do but to hit the choices.

(A) P and R win the first and second ribbons, but P doesn’t necessarily have to come in ahead of R. R could win first and P second.

Since (A) could be false, that’s all she wrote; the remaining choices all must be true as the master sketch clearly shows.

5. (E)

If Q is female, then Rule 3 says that Q must receive a ribbon. S is the one labrador that wins a ribbon, so, thanks to Rule 4, Q must be a greyhound. There are only two females, S and Q, so the remaining dogs, P, R, T, and U, must all be male. From here, it’s a simple matter of scanning the choices and looking for the one choice that doesn’t match up, and it’s (E): T, a losing male in this case, could be a greyhound, but he could also be a lab. (E) could be false and is the answer.

6. (B)

This time it’s T who wins the fourth place ribbon. Rule 4 is still in effect, so T must be a greyhound. Q and U are the two dogs who don’t win ribbons, and we deduced from Rule 3 that any dog who doesn’t win a ribbon must be male. Q must be male, choice (B).

(A) Not necessarily true: P could be the second ribbon-winning female.

(C) Ditto. T could be the second ribbon-winning female.

(D) Not necessarily true. As a loser, Q could just as easily be a greyhound.

(E) False. We’ve already deduced that T is a greyhound in this question.

7. (D)

Another question requiring a browse through the choices, once you’ve worked out that since the right answer could be true, the four wrong choices must be false. Actually, (A), (B), and (C) have been worked out long ago. Rule 5 made it clear that P, R, and S are definite ribbon winners, so none of those statements can be true; and since we deduced some time ago that U is a ribbon loser, (E) is impossible as well. (D) is left, and of course it’s possible: T might win a ribbon, or might not.