IPMAT Indore 2023 · Quantitative Ability · Question 44

The whole set rests on one completed grid. Build it once and all five questions are read off it.

Step 1 — write down what the table gives.

A\to R, B\to Q, C\to Q

Step 2 — use “wherever C works, D works”. Since C\to Q, we get D\to Q.

Step 3 — use “D and E never share an organism”. D is already on Q, so E is not on Q.

Step 4 — count organism P. P needs at least two drugs. Which drugs can reach P?

A cannot: if A\to P then B\to P too (condition 3). We will see this overfills the grid — hold that thought.

Each drug covers at most two organisms.

Work from the drug side instead. The five drugs supply at most 5\times 2=10 drug–organism pairs, and the three organisms need between 2+2+2=6 and 3+3+3=9 of them.

Step 5 — pin E. E\ne Q, so E works on P or R (or both). Suppose E\to P. P then needs a second drug. C\to P would force D\to P, and since D and E must stay apart, D\to P contradicts E\to P. So P’s second drug would have to be A or B — and A\to P drags B\to P as well, giving P three drugs \{A,B,E\} with A now on both P and R, B on P, Q and R — three organisms for B, breaking “at most two”. Every branch collapses, so

E\to R\text{only}

Step 6 — fill P. P still needs two drugs, and they cannot include E. If A\to P then B\to P, but B already works on Q, so B would cover P,Q and (via step 7) possibly R — and A would cover P and R. Checking the organism counts then breaks the “at most three drugs per organism” rule for R. The only consistent choice is

C\to P \text{and, by condition 3,}D\to P

So P=\{C,D\} — exactly two drugs ✔

Step 7 — fill R. A\to R and E\to R are already fixed. R needs at least two drugs and has them, but condition 3 says wherever A works B works, so B\to R as well:

R=\{A,B,E\}

Step 8 — read off Q. B\to Q, C\to Q, D\to Q:

Q=\{B,C,D\}

The completed grid.

Every condition checks out: each drug covers one or two organisms ✔; each organism is treated by two or three drugs ✔; A\subseteq B and C\subseteq D ✔; D=\{P,Q\} and E=\{R\} are disjoint ✔

An exhaustive search over all possible assignments confirms this grid is the only one consistent with the table and the four conditions.

Reading off D. The grid gives D\to P and D\to Q.

D\to Q comes straight from C\to Q and “wherever C works, D works”.

D\to P comes from the same rule once C\to P is forced in step 6.

Why the others fail:

(a) Only Q — then P would have to find two drugs among A, B, E, which breaks either the A–B rule or the two-organism cap on B.

(c) Q and R — E works on R, and D cannot share an organism with E.

(d) Only P — misses the forced D\to Q.

Answer: Option 2 — P and Q

Shortcut. Chain the two “if C then D” uses. C appears on Q in the table, so D\to Q is free. That alone removes option (d).

E lives on R, so D is barred from R — that removes (c). Between (a) and (b), note that P needs two drugs and the only pair that does not break another rule is \{C,D\}. Hence P and Q.

The habit worth taking away: in a conditional-grid set, apply every “if X then Y” rule to the table’s given entries before doing any case work. Here that single pass gives you three of the five answers.