CMAT 2018 Logical Reasoning Question Paper with Answer Key PDFs (January 20 - Afternoon Session)

Step 1: Identify the elder brother of Raj's father.

Raj's father's elder brother is Raj's uncle.


Step 2: Identify Sweety's relation with Raj's uncle.

Sweety is the grand-daughter of Raj's uncle.


Step 3: Convert that into relation with Raj.

Grand-daughter of Raj's uncle would be the daughter of Raj's cousin (or close family line), which is considered as Raj's niece.


Step 4: Conclusion.

Hence, Sweety is Raj's niece.
Quick Tip: In blood relation questions, first convert the statement into direct family terms like uncle, aunt, cousin etc., then map it to the required person.

Step 1: List the fixed conditions.

- At least 2 experts must be before school.

- At least 3 experts must be after school.

- \(M\) cannot be after school.

- \(J\) is only after school.

- \(W\) is always during lunch.

- If \(G\) is before school, then \(N\) must also be before school.


Step 2: Check minimum allocations.

Since \(J\) is compulsory after school and at least 3 experts must be after school, minimum after-school sessions = 3.

Also, before-school sessions must be at least 2.


Step 3: Reason why option (4) is not possible.

The given constraints force after school to have a minimum of 3 experts.

But lunch sessions have \(W\) fixed and may not reach the same count as after-school under valid distributions because at least 2 must be before school and \(M\) cannot be after school.

This makes it impossible for lunch and after-school counts to be equal.


Step 4: Conclusion.

Hence, statement (4) cannot be true and is the correct answer.
Quick Tip: In scheduling problems, first satisfy minimum constraints, then check which option violates the possibility of distribution.

Step 1: Use given fixed pairs.
\(W\) is wife of \(E\).
\(V\) is wife of \(B\).


Step 2: Use the statement about sisters.

R and S are A's sisters.

So A cannot be married to R or S.


Step 3: Apply restriction on C.

Neither R nor T is wife of C.

So C cannot be married to R or T.


Step 4: Apply restriction on D.

D is not married to R, S or T.

So D cannot be married to T also.


Step 5: Final elimination.

Since A cannot marry R or S, and the given pairing eliminates others, the only valid choice left for A's wife is T.


Step 6: Conclusion.

Therefore, A's wife is T.
Quick Tip: In marriage-arrangement puzzles, fix confirmed pairs first, then eliminate impossible options systematically.

Step 1: Understand the direction change.

Given: Southeast becomes East.

That means the whole compass has shifted 45° clockwise (to the right).


Step 2: Verify with second condition.

Northwest becomes West.

Northwest shifting 45° clockwise becomes West, which matches.


Step 3: Apply the same shift to North.

North shifted 45° clockwise becomes Northeast.


Step 4: Conclusion.

So the new direction for North will be Northeast.
Quick Tip: If one direction changes to another, check the angle shift (clockwise/anticlockwise) and apply the same shift to all directions.

Step 1: Assume Raza is the thief.

Then Kalia's statement: ``Raza is the thief'' becomes true, but ``I am not thief'' is also true.

So Kalia would be speaking truth twice, hence Kalia could be the truth-teller.


Step 2: Check Raza's statements.

If Raza is thief, then statement ``I am not thief'' is false.

Second statement ``Either Kalia or Shera is thief'' is also false.

So Raza can be the liar.


Step 3: Check Shera's statements.

If Raza is thief:

Shera says ``I am not thief'' (true) and ``Raza is not thief'' (false).

So Shera alternates between truth and lie, which matches the condition.


Step 4: Conclusion.

All three roles fit correctly only when Raza is the thief.

Hence, the thief is Raza.
Quick Tip: In truth-lie puzzles, test each person as thief and verify who becomes truth-teller, liar and alternator consistently.

Step 1: Use the given condition for A.

A visits Delhi and Lucknow.


Step 2: Apply restrictions.

- B cannot visit Lucknow, Patna.

- C cannot visit Mumbai, Delhi.

- D cannot visit Kolkata, Hyderabad.

- A cannot visit Bangalore, Chennai (already fixed).

Also Patna and Bangalore are visited neither by B nor by C.


Step 3: Check which city B can visit.

Since A already visited Delhi and Lucknow, B cannot visit Delhi or Lucknow.

Also B cannot visit Bangalore, Patna.

Thus B can only visit among remaining cities like Chennai, Kolkata, Hyderabad, Mumbai.


Step 4: Conclusion.

Among options given, only Mumbai is possible for B.
Quick Tip: In city-allocation puzzles, first fix forced visits, then remove restricted cities for each person and match possible options.

Step 1: Convert statements into inequalities.

W is greater than C but less than M:
\[ M > W > C \]

Y is greater than D but not less than M:
\[ Y > D \quad and \quad Y \ge M \]


Step 2: Combine comparisons.

From above:
\[ Y \ge M > W > C \]
and also
\[ Y > D \]


Step 3: Identify the greatest.

Since \(Y \ge M\) and \(Y\) can be equal to or greater than \(M\), either \(Y\) can be greatest or \(M\) can be greatest.


Step 4: Conclusion.

Thus, Y or M can be the greatest.
Quick Tip: Convert word statements into inequalities and then chain them together to find possible greatest/smallest values.

Step 1: Determine group sizes.

There are 10 students and 4 groups.

No two groups can have the same size and max size is 4.

So the only possible sizes are: 1, 2, 3, 4.


Step 2: Place I alone.

Since I must be alone:

Group 1 = {I}.


Step 3: Given A, D, F, J form a group.

Group 4 = {A, D, F, J}.


Step 4: Remaining students.

Remaining = {B, C, E, G, H}.

These must be divided into groups of size 2 and 3.


Step 5: Apply conditions.

- B and E cannot be together.

- F and E must be in different groups (already satisfied since F is in group 4).

- C and G must be in same group.


So group of size 3 can be: {B, C, G}.

Group of size 2 will be: {E, H}.


Step 6: Conclusion.

Thus the two groups can be:

{E, H} and {B, C, G}.
Quick Tip: When group sizes are restricted uniquely, first assign the fixed groups, then distribute remaining students according to constraints.

Step 1: Write the conditions.

- If B is inspected, C cannot be inspected immediately after B.

So pattern B,C is not allowed.

- If A is inspected, E cannot come after A.

So pattern A,E is not allowed.


Step 2: Check each option.


Option (1): A, B, C, D, E

Here B is immediately followed by C, violates condition.


Option (2): D, B, C, E, A

Here B is immediately followed by C, violates condition.


Option (3): D, C, B, A, E

Here A is followed by E, violates condition.


Option (4): D, C, B, E, A

Here B is followed by E, not C. Also A is last so E is not after A. Valid.


Step 3: Conclusion.

Thus, option (4) is the correct order.
Quick Tip: In ordering problems, first identify forbidden adjacent pairs and forbidden sequences, then verify options one-by-one quickly.

Step 1: Understand option (4) DEA.

(D) All honest persons are obese.

(E) All obese persons are good natured.

(A) All honest persons are good natured.


Step 2: Deduce logically.

From (D): Honest \(\Rightarrow\) Obese.

From (E): Obese \(\Rightarrow\) Good natured.

So by transitivity:
\[ Honest \Rightarrow Good natured \]

This is exactly statement (A).


Step 3: Check other options (brief).

Other combinations contain contradictory or non-deducible statements, so they cannot logically lead to the third statement.


Step 4: Conclusion.

Hence, DEA is the correct logically related set.
Quick Tip: If two universal statements form a chain like A \(\Rightarrow\) B and B \(\Rightarrow\) C, then A \(\Rightarrow\) C is a valid conclusion.

Step 1: List all direct connections from R1.

From R1, we can go directly to R2 and R6.


Step 2: Check if R6 connects to R7 directly.

Given roads include R6 and R7.

So, R1 \(\rightarrow\) R6 \(\rightarrow\) R7 is possible.


Step 3: Count intermediate places.

Route R1 - R6 - R7 has only one intermediate place (R6).

All other routes have more intermediate places.


Step 4: Conclusion.

Hence, the shortest route is R1 - R6 - R7.
Quick Tip: For shortest route with least intermediate places, always check for direct links first and then 2-step connections.

Step 1: Write the weight relations clearly.

A is lighter than C, and C is lighter than D:
\[ D > C > A \]

B is heavier than D:
\[ B > D \]

E is heavier than D:
\[ E > D \]


Step 2: Combine all relations.

Since B is the heaviest, we have:
\[ B > E > D > C > A \]


Step 3: Identify the lightest.

From the chain, A is at the bottom.

So A is the lightest.


Step 4: Conclusion.

Hence, the lightest rod can be A only.
Quick Tip: For comparison problems, convert statements into inequality chains and then identify top (heaviest) and bottom (lightest).

Step 1: Write the given fixed placements.

DT screens film B in Slot 1.

PVR screens film C in Slot 3.


Step 2: Use rule: no two theatres can show same film in same slot.

So in Slot 1, neither PVR nor Regal can show B.

In Slot 3, neither DT nor Regal can show C.


Step 3: Identify the only remaining placement for Regal.

Since Regal cannot show B in Slot 1 and cannot show C in Slot 3, Regal must show A in Slot 2.


Step 4: Conclusion.

Thus, Regal exhibits A in the second slot must be true.
Quick Tip: In slot-based arrangement questions, place the fixed items first, then eliminate conflicts slot-by-slot to find forced placements.

Step 1: We get the following:



Step 2: Check each option.

(A, E): Not directly connected. Only A connects to C and C connects to E.

(E, D): Not directly connected. E connects to C, and C connects to D.

(B, C): Not directly connected. B connects to A and E, but not C directly.


Step 3: Conclusion.

Hence, none of the given pairs are directly connected.
Quick Tip: Always write all direct links first. If a pair requires going through another node, then it is not directly connected.

Step 1: Observe the pattern.

The middle number is the sum of alphabetical positions of all letters in the two groups.

(A=1, B=2, ..., Z=26)


Step 2: Verify with first row.

EJO = 5 + 10 + 15 = 30

TYE = 20 + 25 + 5 = 50

Total = 30 + 50 = 80


Step 3: Verify with second row.

DHL = 4 + 8 + 12 = 24

PTX = 16 + 20 + 24 = 60

Total = 24 + 60 = 84


Step 4: Apply to third row.

CFI = 3 + 6 + 9 = 18

LOR = 12 + 15 + 18 = 45

Total = 18 + 45 = 63


Step 5: Conclusion.

So the missing number is 63.
Quick Tip: Whenever letters are involved with numbers, try converting letters into alphabetical positions and check sum/difference patterns.

Step 1: Use given position relations among corners.

Chair A is on the side joining P and R.

S is immediate right of P and R is to the left of P.

Also, R is between P and T.


Step 2: Place chair B and neighbours.

Chair B is along the side of Q and T.

Chairs D and E are next to B on either side.


Step 3: Determine placement of chair C.

After placing all conditions and arranging chairs on edges, chair C comes on the side whose endpoints are P and S.



Step 4: Conclusion.

Hence, chair C is placed on the side joining P and S.
Quick Tip: In polygon seating/edge problems, draw a rough figure and place fixed relations first, then locate remaining chairs by elimination.

Step 1: Use fixed condition of Neetu and Lakshaya.

Neetu sits immediately left of Lakshaya in the same row.

Also Neetu is opposite to Om.


Step 2: Use condition about Jai and Neetu.

Jai and Neetu have exactly one person between them.


Step 3: Use condition about Punita and Kabir.

Punita and Kabir have two persons between them, meaning they sit at extreme ends.


Step 4: Determine diagonal opposites from valid arrangement.

From the valid seating arrangement shown in the explanation, the diagonal opposite pairs possible are:

Quick Tip: For two-row seating, diagonal opposite means one person is not directly opposite but shifted one seat left/right. Always test using a row diagram.

Step 1: Use fixed conditions.

G goes only to Ram.

H goes only to Deepak.

C is not seen by any manager.

Each manager can meet only 3 employees.


Step 2: Use condition E, F and G go together.

Since G goes only to Ram, the group E, F, G must go to Ram.

So Ram meets: E, F, G.


Step 3: Place priority employee D.

D must be assigned to either Ram or Deepak.

Ram is already full (E, F, G), so D must go to Deepak.


Step 4: Place H.

H goes only to Deepak, so Deepak already has D and H.


Step 5: Decide third person for Deepak.

Now, remaining people include A, B, I.

But restrictions:

A does not go with F (F is with Ram, so A can go to Deepak).

I does not go with E (E is with Ram, so I can go to Deepak).

B and F cannot be with same manager (F is with Ram, so B can go to Deepak).

But B and I cannot go together.


So Deepak's third employee can be either B or I.


Step 6: Conclusion.

Thus Deepak meets: D, H, I or D, H, B.
Quick Tip: In manager-allocation puzzles, fill forced-only members first (like G only Ram, H only Deepak), then use exclusions to finalize remaining slots.

Step 1: Arrange toys as per color restrictions.

Each box color must not match any toy color inside it.

So, Green box cannot contain Green toys.

Blue box cannot contain Blue toys.

Red box cannot contain Red toys.


Thus, one valid toy placement is:

Green box: Blue + Red toys

Blue box: Green + Red toys

Red box: Green + Blue toys


Step 2: Distribute chocolates with constraints.

Total chocolates = 10.

Each box has at least 1 chocolate.

No two boxes have same chocolates.

Green box must have maximum possible chocolates.

Red box must have minimum possible chocolates.


So the only possible distribution is:

Green box = 7 chocolates

Blue box = 2 chocolates

Red box = 1 chocolate


Step 3: Identify which box contains Green and Red toys.

Green and Red toys are in the Blue box.

And Blue box has 2 chocolates.


Step 4: Conclusion.

Hence, the box having toys of Green and Red colors has 2 chocolates in it.
Quick Tip: When maximum and minimum distribution is required with all distinct counts, assign minimum = 1 to least box and then distribute remaining to maximize the required box.

Step 1: Apply condition when A buys R.

If A buys R, then B cannot buy P or S.


Step 2: Given A buys R and T.

So B cannot buy P and S.

So B must choose from remaining items: Q and U.


Step 3: Check second rule for validity.

If B buys Q, then C cannot buy U or T.

Since A already bought T, C cannot buy T anyway.

So this does not affect B's choice.


Step 4: Conclusion.

Hence, B buys Q and U.
Quick Tip: In selection constraint problems, first eliminate disallowed items, then allocate remaining items logically.

Step 1: Convert the main statement into logical form.

Main statement: ``You cannot catch the bus unless it is morning.''

This means:
\[ If you catch the bus, then it must be morning. \]

Symbolically:
\[ B \Rightarrow M \]


Step 2: Understand what the ordered pair must satisfy.

We need two statements such that:

- The first implies the second.

- Both should be consistent with the main statement.


Step 3: Check option (4) CD.

(C): This is not morning.

So \(M\) is false.

If it is not morning, then catching the bus is not possible (because bus can be caught only when it is morning).

So (C) implies (D).
\[ \neg M \Rightarrow \neg B \]


Step 4: Verify consistency.

This does not contradict the main statement and follows logically from it.


Step 5: Conclusion.

Hence, the correct ordered pair is CD.
Quick Tip: Statement “X unless Y” means “If X happens, Y must be true.” Convert to implication form to solve quickly.

Step 1: Decode each relation one by one.

Expression: \(a+b-c+d-e\times f\)

\[ a+b \Rightarrow a is sister of b \Rightarrow a is female \]

\[ b-c \Rightarrow b is brother of c \Rightarrow b is male \]

\[ c+d \Rightarrow c is sister of d \Rightarrow c is female \]

\[ d-e \Rightarrow d is brother of e \Rightarrow d is male \]

\[ e\times f \Rightarrow e is daughter of f \Rightarrow e is female \]


Step 2: Identify the gender of f.

Since \(e\) is daughter of \(f\), \(f\) can be either male or female.

So gender of \(f\) is not fixed.


Step 3: Count confirmed females.

Confirmed females: \(a, c, e\) = 3 females.

But \(f\) may also be female, which would make total females = 4.


Step 4: Conclusion.

Since \(f\) can be male or female, the exact number of females cannot be determined.
Quick Tip: If any person’s gender is not fixed (like parent in “daughter of”), then total females cannot be uniquely determined.

Step 1: Understand independence of coin tosses.

The result of the first toss does not affect the next toss.

So we treat the next toss as a fresh toss of 3 coins.


Step 2: Total outcomes for 3 coins.

Each coin has 2 outcomes (H or T).
\[ Total outcomes = 2^3 = 8 \]


Step 3: Count outcomes with at least 2 tails.

At least 2 tails means either 2 tails or 3 tails.

Outcomes with exactly 2 tails:
\[ TTH,\ THT,\ HTT \Rightarrow 3 outcomes \]

Outcome with 3 tails:
\[ TTT \Rightarrow 1 outcome \]

Total favorable outcomes = \(3+1=4\).


Step 4: Find probability.
\[ P = \frac{4}{8} = \frac{1}{2} = 50% \]
Quick Tip: Previous toss results do not change probabilities of the next toss. Always treat coin tosses as independent events.

Step 1: Identify family roles from statements.

B travels with his father E.

So E is male and is father of B.


Step 2: Identify grandparent role.

F travels with her grand-daughter in Audi.

So F is a grandmother (female).


Step 3: Identify the married couple in first generation.

Since F is a grandmother and E is a father, the most logical first generation married couple is E and F.


Step 4: Validate with remaining condition.

A travels with her daughter in Honda, confirming A is female and belongs to second generation.

Thus EF being the older generation married pair fits correctly.


Step 5: Conclusion.

Hence, one of the married couples is EF.
Quick Tip: In family puzzles, first fix direct relations like “father of” and “grandmother of”. Married couples often come from same generation and are inferred from parent-grandparent roles.

Step 1: Identify gender of R.

R is not the mother of Q, but Q is the son of R.

So R must be the father of Q.


Step 2: Use marriage relation.

P and R are a married couple.

So P is Q's mother.


Step 3: Use sibling relations.

T is brother of R.

So T is the paternal uncle of R's children.

U is brother of Q (so U is also child of P and R).


Step 4: Use relation of S.

S is daughter of P, so S is also child of P and R.


Step 5: Determine relation of T to S.

T is brother of R and S is daughter of R.

So T is S's uncle.
Quick Tip: If X is brother of someone’s father, then X is the uncle of that person. Always map relations step-by-step.