IPMAT Indore 2023 · Quantitative Ability · Set 1: Five drugs and three organisms

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 R. The grid gives

R=\{A, B, E\}

Why the others fail:

(a) Only A and E — leaves out B, but condition 3 forces B onto every organism A treats.

(c) A, B and C — C works on P and Q only; putting C on R would give it three organisms.

(d) Only A and B — leaves out E, which has nowhere else to go once E\ne Q and E\ne P.

Answer: Option 2 — A, B and E

Shortcut. The A–B link does most of the work. The table already puts A on R, and “wherever A works, B works” immediately puts B on R too. So the answer must contain both A and B, which kills option (a) at once.

Then ask where E can live: not Q (it would sit with D), and not P (that forces B onto a third organism). So E\to R, and only option (b) has all three. Two deductions, no full grid needed.

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 E. The grid gives E\to R and nothing else.

Why the others fail:

(b) Q and R — D works on Q, and D and E may never share an organism.

(c) Only P — P is filled by C and D; adding E would sit it beside D again.

(d) P and R — same clash on P, and it would also give P three drugs with E among them.

Answer: Option 1 — Only R

Shortcut. Eliminate on the D–E rule alone. C\to Q forces D\to Q, so E is barred from Q — that removes option (b) instantly.

P ends up as \{C,D\}, so E cannot be there either, removing (c) and (d). Only R remains, and you never have to complete the grid. The “D and E do not work on the same organism” condition is the single most powerful line in this set — use it first.

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.

Count each column.

Both Q and R reach the maximum of three.

Why the others fail: any option containing P is wrong, since P has exactly two drugs; and “Only Q” misses R.

Sanity check on the totals. The drugs supply 1+2+2+2+1=8 drug–organism pairs, and the organisms consume 2+3+3=8 ✔

Answer: Option 4 — Q and R

Shortcut. Count the pairs before counting the columns. The five drugs cover 1+2+2+2+1=8 pairs in total, spread over three organisms each needing 2 or 3. The only split of 8 into three parts from \{2,3\} is

8=2+3+3

So exactly one organism has two drugs and two have three — which immediately rules out “Only Q” (a) and “Only P” (b), since both name a single organism. Between (c) and (d), P is the small one, so the answer is Q and R.

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.

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 P. The grid gives P=\{C, D\} — exactly two drugs, the minimum allowed.

Why the others fail:

(a) B, C and D — B already works on Q and R; a third organism breaks the two-organism cap.

(c) Only B and D — without C, there is nothing to force D onto P, and B again overflows.

(d) A, B, and D — A works only on R; putting it on P would drag B along and overfill B.

Answer: Option 2 — Only C and D

Shortcut. P is the organism with only two drugs (from the 8=2+3+3 split in Q43), so the answer must name exactly two — which leaves only options (b) and (c).

Both contain D; the difference is C versus B. B is already committed to Q and R, so a third organism is impossible. Only C and D.

Note how the five questions interlock: solving Q43’s counting argument first makes Q45 a one-line elimination. In a set like this, do the counting question before the identification questions.