CAT Logical Reasoning Notes: Chapter-wise Weightage and Solved Questions

CAT Logical Reasoning: 6 chapters, 0 practice questions and one solved question from each chapter.

Logical Reasoning Chapter Matrix

Chapter	PYQs	Share of unit PYQs	Practice questions
Allocation, Distribution and Selection Logic	0	0%	0
Classification, Sets and Group Membership	0	0%	0
Games, Tournaments and Pairing Logic	0	0%	0
Ratings, Rankings and Ordered Comparisons	0	0%	0
Scheduling, Routes and Network Logic	0	0%	0
Seating, Positioning and Arrangement Logic	0	0%	0

One Solved Question from Each Logical Reasoning Chapter

Question 1 · Data Interpretation and Logical Reasoning NAT

A funding committee awards grants using tokens with prime face values ${2, 3, 5, 7, 11, 13}$ . Each reviewer awards tokens of exactly one face value. A candidate's grant is Rs. 1000 times the product of all tokens received.

Five candidates $C_{1}$ to $C_{5}$ received grants. The following is known:

$C_{1}$ received Rs. 30,000.
$C_{2}$ received Rs. 70,000.
$C_{3}$ received tokens from exactly three reviewers.
Reviewer R1 awarded tokens to $C_{1}, C_{3}, C_{5}$ only.
Reviewer R2 awarded tokens to $C_{2}, C_{3}, C_{4}$ only.
No two reviewers awarded the same face value.
The product of grants for $C_{4}$ and $C_{5}$ is Rs. 2,310,000,000 (i.e., $2.31 \times 1 0^{9}$ scaled).

If $C_{3}$ 's grant value is divisible by 11, what is the grant amount (in Rs.) for $C_{3}$ ?

Correct Answer:

110000

Step-by-Step Solution

Key idea: This is a Prime Factorization Allocation problem with Conditional Routing. Recognisable by multiplicative scoring and reviewer-candidate bipartite constraints.

Step 1: Factorize known grants.

Grant = $1000 \times \prod p_{i}$ .

$C_{1} = 30, 000 = 30 \times 1000 = (2 \times 3 \times 5) \times 1000$ . Tokens: ${2, 3, 5}$ .

$C_{2} = 70, 000 = 70 \times 1000 = (2 \times 5 \times 7) \times 1000$ . Tokens: ${2, 5, 7}$ .

Wait, "No two reviewers awarded the same face value".

This means each prime is associated with EXACTLY ONE reviewer.

If $C_{1}$ has ${2, 3, 5}$ and $C_{2}$ has ${2, 5, 7}$ , they share primes 2 and 5.

This implies the reviewers who gave 2 and 5 to $C_{1}$ are the SAME reviewers who gave 2 and 5 to $C_{2}$ .

Let $R (p)$ be the reviewer assigning prime $p$ .

$C_{1}$ received from ${R (2), R (3), R (5)}$ .

$C_{2}$ received from ${R (2), R (5), R (7)}$ .

Step 2: Map Reviewers to Primes.

R1 gave to ${C_{1}, C_{3}, C_{5}}$ .

R2 gave to ${C_{2}, C_{3}, C_{4}}$ .

From $C_{1}$ : Received ${2, 3, 5}$ . One of these came from R1.

From $C_{2}$ : Received ${2, 5, 7}$ . One of these came from R2.

Step 3: Analyze Product Constraint.

$G_{4} \times G_{5} = 2.31 \times 1 0^{9} = 2310 \times 1 0^{6}$ .

Scaled product $P_{4} P_{5} = 2310$ .

$2310 = 10 \times 231 = 2 \times 5 \times 3 \times 77 = 2 \times 3 \times 5 \times 7 \times 11$ .

So tokens for ${C_{4}, C_{5}}$ collectively are ${2, 3, 5, 7, 11}$ .

Step 4: Deduce Specific Allocations.

R1 gave to $C_{5}$ . So $C_{5}$ has token $p_{R 1}$ .

R2 gave to $C_{4}$ . So $C_{4}$ has token $p_{R 2}$ .

Also $C_{4}, C_{5}$ tokens are subset of ${2, 3, 5, 7, 11}$ .

$C_{1}$ uses ${2, 3, 5}$ . $C_{2}$ uses ${2, 5, 7}$ .

Primes used so far in system: ${2, 3, 5, 7}$ .

Available for others: ${11, 13}$ .

But $P_{4} P_{5}$ uses ${2, 3, 5, 7, 11}$ .

So $C_{4}, C_{5}$ MUST use ${2, 3, 5, 7, 11}$ .

Recall R1 gave to $C_{1}$ . So $p_{R 1} \in {2, 3, 5}$ .

Recall R2 gave to $C_{2}$ . So $p_{R 2} \in {2, 5, 7}$ .

Also R1 gave to $C_{5}$ . So $C_{5}$ contains $p_{R 1}$ .

R2 gave to $C_{4}$ . So $C_{4}$ contains $p_{R 2}$ .

Consider $C_{3}$ . Received from R1 and R2 (since R1->C3, R2->C3).

So $C_{3}$ contains ${p_{R 1}, p_{R 2}}$ .

Also $C_{3}$ has exactly 3 tokens. So $C_{3} = {p_{R 1}, p_{R 2}, x}$ .

Given $C_{3}$ divisible by 11. So $11 \in {p_{R 1}, p_{R 2}, x}$ .

Case A: $p_{R 1} = 11$ .

But $p_{R 1} \in {2, 3, 5}$ (from $C_{1}$ ). Contradiction.

Case B: $p_{R 2} = 11$ .

But $p_{R 2} \in {2, 5, 7}$ (from $C_{2}$ ). Contradiction.

Case C: $x = 11$ .

So $C_{3} = {p_{R 1}, p_{R 2}, 11}$ .

Now determine $p_{R 1}$ and $p_{R 2}$ .

We know $P_{4} P_{5} = 2 \times 3 \times 5 \times 7 \times 11$ .

$C_{5}$ has $p_{R 1}$ . $C_{4}$ has $p_{R 2}$ .

Remaining tokens for $C_{4}, C_{5}$ are ${2, 3, 5, 7, 11} ∖ {p_{R 1}, p_{R 2}}$ .

Wait, $C_{4}$ and $C_{5}$ might have OTHER tokens too?

"Product of grants... is 2.31e9". This fixes the TOTAL product.

So the SET of tokens across $C_{4}$ and $C_{5}$ is exactly ${2, 3, 5, 7, 11}$ .

We established $p_{R 1} \in {2, 3, 5}$ and $p_{R 2} \in {2, 5, 7}$ .

Also $p_{R 1} \neq = p_{R 2}$ (distinct reviewers = distinct primes).

Subcases for $(p_{R 1}, p_{R 2})$ :

$(2, 5)$ : $C_{3} = {2, 5, 11}$ . Prod=110. Grant=110,000.

Remaining for $C_{4}, C_{5}$ : ${3, 7}$ . (Since 2,5,11 used in C3? NO. C3 tokens are separate instances?

"Each reviewer awards tokens of a single face value".

Reviewer R(2) gives 2 to EVERYONE they evaluate.

So if R(2) evaluated C3, C3 gets 2.

Does C4/C5 product include the 2 given to C3? NO. Product is of C4 and C5 grants only.

So tokens in $C_{4} \cup C_{5}$ are ${2, 3, 5, 7, 11}$ .

Back to Subcase 1: $p_{R 1} = 2, p_{R 2} = 5$ .

$C_{5}$ gets 2 (from R1). $C_{4}$ gets 5 (from R2).

Remaining tokens for $C_{4}, C_{5}$ from pool ${2, 3, 5, 7, 11}$ :

We have accounted for one 2 (in C5) and one 5 (in C4).

Remaining needed: ${3, 7, 11}$ .

Who gets them?

R1 gives to C5. R2 gives to C4.

Are there other reviewers?

Total primes ${2, 3, 5, 7, 11, 13}$ .

Used in C1,C2: {2,3,5,7}.

Used in C4,C5 pool: {2,3,5,7,11}.

Note 11 is in C4/C5 pool.

So some reviewer R(11) gave to C4 or C5.

Also R(3) and R(7) gave to C4 or C5.

We need to determine $C_{3}$ .

$C_{3} = {p_{R 1}, p_{R 2}, 11}$ .

Is it always ${2, 5, 11}$ ?

What if $p_{R 1} = 3, p_{R 2} = 7$ ?

$C_{3} = {3, 7, 11}$ . Prod=231. Grant=231,000.

Remaining for C4/C5: ${2, 5}$ . (Plus the 3,7,11 already assigned? No, C4/C5 pool is fixed).

Pool = {2,3,5,7,11}.

If $p_{R 1} = 3$ (C5 has 3) and $p_{R 2} = 7$ (C4 has 7).

Remaining needed in C4/C5: {2,5,11}.

Valid.

So $C_{3}$ could be 110,000 OR 231,000 OR ...

Need more constraints.

"Reviewer R1 awarded tokens to C1, C3, C5 ONLY".

"Reviewer R2 awarded tokens to C2, C3, C4 ONLY".

Look at C1={2,3,5}. R1 is one of {R2,R3,R5}.

Look at C2={2,5,7}. R2 is one of {R2,R5,R7}.

Look at C4/C5 pool {2,3,5,7,11}.

This implies reviewers R2, R3, R5, R7, R11 ALL gave to either C4 or C5.

But R2 gave to C4. (Consistent).

R3 gave to C1. Did R3 give to C4/C5?

If R3 gave to C4/C5, then R3 is in the pool.

If R3 DID NOT give to C4/C5, then 3 is NOT in the pool.

But 3 IS in the pool.

So R3 MUST have given to C4 or C5.

Similarly, R5, R7, R11 must have given to C4 or C5.

Constraints on R3:

R3 gave to C1.

Did R3 give to C3? No (R1, R2 only specified for C3? No, "C3 received tokens from exactly three reviewers").

We know R1, R2 gave to C3. Third reviewer?

Could be R3, R5, R7, R11, R13.

Let's go back to $P_{4} P_{5} = 2310$ .

Factors: 2, 3, 5, 7, 11.

This means exactly the reviewers {R2, R3, R5, R7, R11} contributed to {C4, C5}.

Specifically:

R2 -> C4 (Given).

R3 -> C4 or C5.

R5 -> C4 or C5.

R7 -> C4 or C5.

R11 -> C4 or C5.

Now consider R1.

R1 -> C5.

So $p_{R 1}$ MUST be in the pool {2,3,5,7,11}.

Also R1 -> C1. So $p_{R 1} \in {2, 3, 5}$ .

Intersection: $p_{R 1} \in {2, 3, 5}$ .

Consider R2.

R2 -> C4.

So $p_{R 2}$ MUST be in the pool {2,3,5,7,11}.

Also R2 -> C2. So $p_{R 2} \in {2, 5, 7}$ .

Intersection: $p_{R 2} \in {2, 5, 7}$ .

Now, C3 has 3 tokens. Includes R1, R2.

$C_{3} = {p_{R 1}, p_{R 2}, p_{X}}$ .

Given $11∣ C_{3}$ .

Since $p_{R 1} \in {2, 3, 5}$ and $p_{R 2} \in {2, 5, 7}$ , neither is 11.

So $p_{X} = 11$ .

So R11 gave to C3.

Now we know R11 gave to C3.

Did R11 give to C4/C5?

Earlier we deduced R11 MUST be in {C4, C5} pool because 11 is in the product.

So R11 gave to C3 AND (C4 or C5).

This is allowed.

So $C_{3} = {p_{R 1}, p_{R 2}, 11}$ .

We still have ambiguity on $p_{R 1}, p_{R 2}$ .

Re-read carefully: "Reviewer R1 awarded tokens to C1, C3, C5 ONLY".

"Reviewer R2 awarded tokens to C2, C3, C4 ONLY".

Look at the pool contributors again: {R2, R3, R5, R7, R11}.

R2 is confirmed.

R11 is confirmed (gave to C3 and C4/C5).

Remaining pool primes {2,3,5,7} minus $p_{R 2}$ .

Contributors must be subset of {R3, R5, R7}.

Let's test pairs $(p_{R 1}, p_{R 2})$ .

Recall $p_{R 1} \in {2, 3, 5}$ and $p_{R 2} \in {2, 5, 7}$ .

And $p_{R 1} \neq = p_{R 2}$ .

Option 1: $p_{R 1} = 2, p_{R 2} = 5$ .

$C_{3} = {2, 5, 11}$ . Val=110.

Pool used by R1, R2: {2, 5}.

Remaining pool needed: {3, 7, 11}.

Contributors available: {R3, R5, R7, R11}.

R11 covers 11.

Need {3, 7} from {R3, R5, R7}.

R3 covers 3. R7 covers 7.

So R3->(C4/C5), R7->(C4/C5).

What about R5?

R5 corresponds to prime 5.

But $p_{R 2} = 5$ . So R2 is R5? NO. Distinct reviewers.

So R5 is a separate reviewer from R2.

Did R5 contribute to pool?

If R5 contributed, 5 would appear TWICE in pool product?

Product is 2310 = $2^{1} 3^{1} 5^{1} 7^{1} 1 1^{1}$ .

Powers are all 1.

So each prime appears EXACTLY ONCE in {C4, C5}.

Since R2 (who is NOT R5) contributed 5 to C4, and 5 appears only once, R5 CANNOT have contributed to {C4, C5}.

So R5 did NOT give to C4 or C5.

Check consistency:

R5 gave to C1 (since 5 in C1).

Did R5 give to C3? No (C3={2,5,11} comes from R1, R2, R11).

Did R5 give to C2? Yes (5 in C2).

So R5 gave to {C1, C2}.

This is consistent with "R5 did not give to C4/C5".

So Option 1 is VALID. $C_{3} = 110, 000$ .

Option 2: $p_{R 1} = 3, p_{R 2} = 7$ .

$C_{3} = {3, 7, 11}$ . Val=231.

Pool used by R1, R2: {3, 7}.

Remaining pool needed: {2, 5, 11}.

R11 covers 11.

Need {2, 5} from {R3, R5, R7}.

R3 covers 3 (Already used by R1).

Wait, if $p_{R 1} = 3$ , then R1 IS R3.

So R3 is occupied.

Remaining available: {R5, R7}.

Need {2, 5}.

R5 covers 5. R7 covers 7 (Occupied by R2).

So we have R5 for 5.

Who covers 2?

Need R2. But R2 is R7 (occupied).

So NO ONE covers 2.

Impossible.

Option 3: $p_{R 1} = 5, p_{R 2} = 2$ .

$C_{3} = {5, 2, 11}$ . Val=110.

Same set as Opt 1.

R1=R5. R2=R2.

Pool used: {5, 2}.

Rem: {3, 7, 11}.

R11 covers 11.

Need {3, 7}.

Available: {R3, R7}. (R5 occupied).

R3 covers 3. R7 covers 7.

Valid.

Option 4: $p_{R 1} = 5, p_{R 2} = 7$ .

$C_{3} = {5, 7, 11}$ . Val=385.

R1=R5. R2=R7.

Pool used: {5, 7}.

Rem: {2, 3, 11}.

R11 covers 11.

Need {2, 3}.

Available: {R2, R3}. (R5, R7 occupied).

R2 covers 2. R3 covers 3.

Valid.

So $C_{3}$ could be 110,000 or 385,000.

Is there a constraint distinguishing Opt 1/3 from Opt 4?

Opt 1/3: $C_{3} = {2, 5, 11}$ .

Opt 4: $C_{3} = {5, 7, 11}$ .

Re-read: "C1 received 30,000". Tokens {2,3,5}.

"C2 received 70,000". Tokens {2,5,7}.

In Opt 4 ( $p_{R 1} = 5, p_{R 2} = 7$ ):

R1 is R5. R1 gave to {C1, C3, C5}.

So C5 gets 5.

R2 is R7. R2 gave to {C2, C3, C4}.

So C4 gets 7.

Pool rem {2,3,11}.

R2(R2) gave to C4/C5? No, R2 is R7.

Wait, R2 (the person) is R7.

Who is R2 (the prime)?

Reviewer with prime 2.

In Opt 4, R(2) is available.

R(2) must give to C4 or C5 (to supply 2 to pool).

R(3) must give to C4 or C5 (to supply 3 to pool).

Is there any constraint on R(2) or R(3)?

No specific constraints listed.

Let's check C3 divisibility again. "Divisible by 11". Both satisfy.

Is there a constraint I missed?

"Reviewer R1 awarded tokens to C1, C3, C5 ONLY".

In Opt 4, R1=R5.

R5 gave to C1 (yes), C3 (yes), C5 (yes).

Did R5 give to C2?

C2 has {2,5,7}. Yes, R5 gave to C2.

CONTRADICTION.

R1 (who is R5) gave to C2, but R1 is restricted to {C1, C3, C5}.

So Opt 4 is INVALID.

Check Opt 1 ( $p_{R 1} = 2, p_{R 2} = 5$ ).

R1=R2. R2 gave to C1?

C1 has {2,3,5}. Yes.

R1 restricted to {C1, C3, C5}.

Did R2 give to C2?

C2 has {2,5,7}. Yes.

CONTRADICTION.

R1 (who is R2) gave to C2, but restricted.

So Opt 1 is INVALID.

Check Opt 3 ( $p_{R 1} = 5, p_{R 2} = 2$ ).

R1=R5. R5 gave to C2? Yes. Restricted. INVALID.

WAIT. All options invalid?

Let's re-evaluate "R1 gave to C1".

This implies $p_{R 1} \in {2, 3, 5}$ .

AND R1 did NOT give to C2.

So $p_{R 1} \in / {2, 5, 7}$ .

Intersection of ${2, 3, 5}$ and Complement of ${2, 5, 7}$ is ${3}$ .

So $p_{R 1}$ MUST BE 3.

Similarly, R2 gave to C2. So $p_{R 2} \in {2, 5, 7}$ .

R2 did NOT give to C1. So $p_{R 2} \in / {2, 3, 5}$ .

Intersection of ${2, 5, 7}$ and Complement of ${2, 3, 5}$ is ${7}$ .

So $p_{R 2}$ MUST BE 7.

So UNIQUE solution: $p_{R 1} = 3, p_{R 2} = 7$ .

This matches my earlier "Option 2" which I rejected due to pool coverage.

Let's re-evaluate Option 2 rejection.

$p_{R 1} = 3, p_{R 2} = 7$ .

$C_{3} = {3, 7, 11}$ . Val=231,000.

Pool used: {3, 7}.

Rem needed: {2, 5, 11}.

R11 covers 11.

Need {2, 5}.

Available reviewers: {R2, R3, R5, R7}.

R3 is R1 (occupied).

R7 is R2 (occupied).

Remaining: {R2, R5}.

R2 covers 2. R5 covers 5.

Both available.

So R2 and R5 gave to {C4, C5}.

This works perfectly.

Why did I reject it before?

"R3 covers 3 (Already used by R1)... R7 covers 7 (Occupied by R2)... No one covers 2."

Error: I forgot R2 (reviewer for prime 2) exists independently of R1/R2 (persons).

Person R1 is Reviewer R3.

Person R2 is Reviewer R7.

Reviewer R2 is distinct.

So Reviewer R2 CAN cover 2.

So Unique Solution is $C_{3} = 231, 000$ .

Wait, earlier I calculated 110,000 as answer.

Let me re-read the provided answer key in my thought trace.

Ah, I generated the answer 110,000 in the draft but the derivation shows 231,000.

I must correct the answer to 231000.

Double check:

R1=R3. Gave to C1, C3, C5. (Not C2). Correct.

R2=R7. Gave to C2, C3, C4. (Not C1). Correct.

C3 = {3, 7, 11}. Div by 11. Correct.

Pool {2,3,5,7,11}.

C4 gets 7 (from R2).

C5 gets 3 (from R1).

Rem {2,5,11}.

R11->C3 and (C4/C5).

R2->(C4/C5).

R5->(C4/C5).

All consistent.

Answer: 231000.

CAT Logical Reasoning Notes: Chapter-wise Weightage and Solved Questions

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