Practice Question - 27 (Conditional) | 100 DILR Questions for CAT Preparation PDF Download

The Action: This set is a straightforward grouping set of selection. There are five kinds of fish to choose from—G, H, J, K and L—and four kinds of plants—w, x, y, and z (we’ll use lowercase for the plants; why not?). Our job is to select exactly five—three fish and two plants—of the nine entities. The Key Issues will, predictably, be:

1) What fish and plants are selected?

2) What fish can, must, or cannot be selected with what other fish?

3) What fish can, must, or cannot be selected with what plants?

4) What plants can, must, or cannot be selected with what other plants?

5) What plants can, must, or cannot be selected with what fish?

The Initial Setup: Make a list of the entities, and in each question, circle the ones selected and X out the ones rejected. This is the classic way to work out selection tasks. As it turns out, this particular set is so basic that most of the questions can be answered simply by looking over the rules with virtually no scratchwork. Perhaps you found that to be the case; if you did, bravo!

The Rules:
1) means that selecting G eliminates two other entities. We can write “If G, then no H and no y,” and remember to think through and jot down the contrapositive: “If H or y, then no G.”

2) Did you take the necessary time to translate this rule correctly? Many students did not, and, of course, they were therefore in the soup. Read it carefully: Choosing K is necessary for choosing H, hence “If H, then K” and, of course, the contrapositive “If no K, then no H.”

3) Same logical structure as Rule 2, same need to translate correctly: “If J, then w” and “If no w, then no J.”

4) Different wording from Rules 2 and 3, but same logic: “If K, then x” and “If no x, then no K.”

Key Deductions: You could combine rules 2 and 4 to realize that “If H, then x” and “If no x, then no H.” More importantly, give yourself a mental reminder of the numbers in the set before you move ahead: You will always need exactly three of the five fish and two of the four plants.

The Final Visualization: Be sure you’ve kept all of your notations neat and easily accessible:

Ans. 1 (B)
Check each rule against each choice. Rule 1 axes choice (A). Rule 2 kills choice (D). Choice (C) can be axed because it doesn’t obey Rule 3. And (E) violates Rule 4. That leaves (B).

Ans. 2 (B)
Jot down your roster, circle H as chosen, and then check your rules. Recognize that H leads to K, and K leads to x (Rules 2 and 4)—meaning answer choice (B). On Test Day, you wouldn’t waste any time on the other choices. For our purposes though, notice that a selection of Fish: H, K, L and plants: x, y eliminates answer choices (A), (C), (D), and (E). None of them need be true.

Ans. 3 (C)
If x and z are chosen, then the plant quota is filled; w and y (the other plants) can’t be selected. According to Rule 3, selecting J requires selecting w, so J can’t be one of the chosen fish. That eliminates answer choices (B), (D), and (E) right off the bat, leaving only answer choices (A) and (C). Without J, we’re left to choose three fish from the following group of four: G, H, K, and L. G and H can’t be selected together (Rule 1), which eliminates answer choice (A). So it must be the case that K and L are selected along with either G or H. The only choice that fits this description, and the only choice that remains, is answer choice (C): G, K, L.

Ans. 4 (C)
A mini-acceptability question, so just check the rules against the choices. The only difference is that ordinarily the four wrong choices violate rules, but here only one—the correct one—is a violator. As it happens, choice (C) violates Rule 2 by selecting an H without a K.
(A) We saw this group in the answer of Q. 14, so we know it works.
(B) and (E) Each is acceptable with w and x as the plants.
(D) is acceptable with x and z as the plants.

Ans. 5 (D)
Selecting y requires crossing G off our fish list (Rule 1). That leaves four fish—H, J, K, and L—from which we need to choose three. J would require the presence of plant w (Rule 3), while K requires the presence of plant x (Rule 4). But the hobbyist can’t select both plant w and plant x, because she’s already taken plant y, and there are only two plants to be selected. Since it’s impossible to select both w and x, it’s also impossible to select both J and K. The hobbyist can only take one of them, which means that the other two fish, H and L, must both be chosen. And now there’s really no ambiguity about the J/K selection—K must get the nod (Rule 2; we need K for H). So the complete group of three fish selected by the hobbyist must be H, K, and L, choice (D).

Ans. 6 (B)
Consider the situation. Four of the five choices are pairs of plants that can be chosen together. No extra conditions or limitations are specified. That means that any pair we have already seen chosen, in a previous question, can be tossed. In Q. 16, we had plants x and y, so cross out (D). The question stem to Q. 14 had the hobbyist selecting x and z, so (E) is out; by 14’s stem alone, we can see that (E)’s pair can be picked! Finally, in Question 12, the correct answer included w and x; that’s choice (A).
We’re left with only two to check. If you start with (B), you ask yourself: Can the hobbyist pick both w and y? Well, having selected y, she can’t select G (Rule 1), and having not selected x, she cannot select K (Rule 4). That means she can’t select H either, since Rule 2 states that H requires K. G, K, and H are gone; only J and L remain; we’ve dropped below the 3-fish requirement. So it’s choice (B) that contains an impossible plant combination.
For the record, plants w and z can be selected along with fish G, J, and L. (C) is acceptable.