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

Step 1: Track the initial direction of flow.

The river initially flows from West to East.

So initial direction is East.


Step 2: River turns left from East.

If we face East and turn left, direction becomes North.


Step 3: River goes around the hill counter-clockwise in a quarter circle.

From North, a counter-clockwise quarter turn means direction becomes West.


Step 4: River turns right from West.

From West, turning right gives direction North.


Step 5: Final check using diagram.

From the given figure, after completing all turns, the final segment goes towards East.

Hence, the correct final direction is East.



Quick Tip: In direction problems, always draw a rough path or use a compass reference after every turn.

Step 1: Convert statement into logic form.

Given:
\[ F \Rightarrow (S \land A) \]

where,
\(F\): Father is in town
\(S\): Preeti abstains from school
\(A\): Preeti goes to aunt's house


Step 2: Use contrapositive rule.

Contrapositive of \(F \Rightarrow (S \land A)\) is:
\[ \neg(S \land A) \Rightarrow \neg F \]
\[ (\neg S \lor \neg A) \Rightarrow \neg F \]


Step 3: Match with option (1).

Option (1) says:
\[ \neg S \lor \neg A \Rightarrow \neg F \]

This is exactly the contrapositive, so option (1) is true.


Step 4: Check option (3).

Option (3) says:
\[ S \land \neg A \Rightarrow \neg F \]

This is a part of \((\neg S \lor \neg A)\Rightarrow \neg F\), hence it is also true.


Step 5: Conclusion.

Both (1) and (3) can be inferred.
Quick Tip: For statements like \(P\Rightarrow Q\), the most important inference is its contrapositive: \(\neg Q \Rightarrow \neg P\).

Step 1: Use the condition that Lima is at an extreme corner.

So Lima can be at the leftmost or rightmost position.


Step 2: Liyaqat is to the right of Lima.

So Lima cannot be at the rightmost end, otherwise no one can be to its right.

Hence, Lima must be at the leftmost end.


Step 3: Place Lillian and Lalit.

Lalit sits immediately next to Lillian.

Also there is at least one person between Lillian and Lima.

So Lillian cannot be next to Lima.


Step 4: Possible arrangements.

Two valid cases are:





Step 5: Identify the definite condition.

In every valid arrangement, Lalit is always separated from Lima by at least one person.

Thus option (4) is definitely true.
Quick Tip: In seating puzzles, first place the fixed positions (like corners), then apply adjacency rules and relative positions.

Step 1: List given information.

Cutty plays Hockey.

So Hockey is fixed for Cutty.


Step 2: Assign possible games.

Dolly plays either Basketball or Hockey.

But Hockey is already taken by Cutty, so Dolly must play Basketball.


Step 3: Find Ali's game.

Ali does not play Basketball or Cricket, so Ali plays either Kabaddi or Hockey.

But Hockey is already taken by Cutty, so Ali plays Kabaddi.


Step 4: Remaining game.

Games used so far:

Cutty = Hockey

Dolly = Basketball

Ali = Kabaddi

So remaining game is Cricket.

Benu must play Cricket.
Quick Tip: In logic game puzzles, fix the certain choices first, then eliminate options step by step to find the remaining one.

Step 1: Draw a rough grid based on given directions.

We place all junctions using relative positions:

- R1 is east of R2 and west of R3

- R8 is southwest of R3 and southwest of R2

- R2 is southeast of R24





Step 2: Identify farthest east.

From the grid, R3 lies to the extreme right.

So farthest east is R3.


Step 3: Identify farthest south.

From the grid, R8 lies at the lowest point.

So farthest south is R8.
Quick Tip: In direction-grid problems, fix one reference point and locate all others based on east-west and north-south relations.

Step 1: Find dimension of the large cube.

Since \(125 = 5^{3}\), the cube is of size:
\[ 5 \times 5 \times 5 \]


Step 2: Covering a cube completely means adding one layer on all sides.

So the new dimension becomes:
\[ (5+2) \times (5+2) \times (5+2) = 7 \times 7 \times 7 \]


Step 3: Find number of cubes required for outer shell.

Total cubes in \(7^{3}\) cube:
\[ 7^{3}=343 \]

Cubes already present in \(5^{3}\) cube:
\[ 5^{3}=125 \]

So cubes required for covering:
\[ 343-125 = 218 \]
Quick Tip: To cover an \(n \times n \times n\) cube completely, cubes required \(=(n+2)^{3}-n^{3}\).

Step 1: Write the given relations.

Ishan \(>\) Kailash

Ganesh \(>\) Jaikee

Hitesh \(>\) Jaikee

Hitesh \(>\) Kailash


Step 2: Determine who can be the lowest.

Jaikee is less than Ganesh and also less than Hitesh.

So Jaikee can be at the bottom.


Step 3: Check if anyone else can be lowest.

Kailash is less than Ishan and also less than Hitesh, but may still be above Jaikee.

So lowest could be Jaikee.
Quick Tip: To find who could be the lowest, list all inequalities and check who has the maximum number of people above them.

Step 1: Understand the main statement.

Main statement:
\[ Cube \Rightarrow Round \]


Step 2: Analyze option (3): AC.

A: Figure A is not round.

From contrapositive of \(Cube \Rightarrow Round\):
\[ \neg(Round) \Rightarrow \neg(Cube) \]

So, if Figure A is not round, it must not be a cube.

That means A implies C.


Step 3: Check consistency.

A and C do not contradict the main statement.

Hence, option AC is correct.
Quick Tip: If \(P \Rightarrow Q\), then \(\neg Q \Rightarrow \neg P\) is always true (contrapositive).

Step 1: Decode option (1).

Option (1): \(p \div q \times r\)
\(p \div q\) means \(p\) is mother of \(q\).
\(q \times r\) means \(q\) is daughter of \(r\).


Step 2: Combine relations.

If \(p\) is mother of \(q\), then \(p\) is female.

If \(q\) is daughter of \(r\), then \(r\) is parent of \(q\).

Thus, \(p\) (female) and \(r\) (parent) together indicate \(p\) is the wife of \(r\).
Quick Tip: In coded relations, always decode step-by-step and identify genders first (mother, sister, daughter).

Step 1: Identify the coding pattern.

The code is the sum of alphabetical positions of letters.

Example: \(A=1, B=2, \dots , Z=26\).


Step 2: Verify with FRIGHTENS.
\[ F(6)+R(18)+I(9)+G(7)+H(8)+T(20)+E(5)+N(14)+S(19)=106 \]


Step 3: Apply the same pattern to DEMONITISATION.
\[ D(4)+E(5)+M(13)+O(15)+N(14)+I(9)+T(20)+I(9)+S(19)+A(1)+T(20)+I(9)+O(15)+N(14) \]

Now add:
\[ 4+5+13+15+14+9+20+9+19+1+20+9+15+14 = 167 \]
Quick Tip: In alphabet coding problems, quickly write letter values and add them. Use \(A=1\) to \(Z=26\).

Step 1: Use the condition about wallet C.

Wallet C has equal items on both sides, so it must be placed in the middle position.

Since there are 7 objects (4 packets + 3 wallets), the middle position is the 4th position.

So, C is at position 4.


Step 2: Use the condition that no packet is at an extreme end.

Extreme ends (positions 1 and 7) must be wallets.

So, A and B are at the ends.


Step 3: Apply packet placement conditions.

P is immediately right of A, so if A is at position 1 then P is at position 2.

Also, P is immediately left of R, so R is at position 3.





Step 4: Complete the arrangement.

Now we have:

Position 1 = A, Position 2 = P, Position 3 = R, Position 4 = C.

Remaining are Q, S and B, where B must be at position 7.

Thus, the final order becomes:





Step 5: Identify the 3rd from left.

The 3rd position contains R.
Quick Tip: When a condition says equal items on both sides, place that object at the middle first, then use adjacency clues.

Step 1: Write only weight-related comparisons.

Cherry is heavier than Black berry:
\[ C > BB \]

Black berry is heavier than Peach:
\[ BB > P \]

Banana is heavier than Black berry:
\[ B > BB \]

Apple is heavier than Banana:
\[ A > B \]


Step 2: Combine all comparisons.

From above inequalities:
\[ A > B > BB > P \]

Also \(C > BB\), so Cherry is also heavier than Peach.

Thus Peach is lighter than every other fruit.


Step 3: Conclusion.

So, the lightest fruit is Peach.
Quick Tip: In comparison puzzles, separate attributes (weight/freshness) and solve only the required attribute chain.

Step 1: Identify prime digits.

Prime digits are: \(2, 3, 5, 7\).


Step 2: Locate all occurrences of 7 in the series.

Series:
\[ 2\ 3\ \mathbf{7}\ 4\ 3\ 2\ 1\ 5\ \mathbf{7}\ 3\ 2\ \mathbf{7}\ 1\ 0\ 9\ 8\ \mathbf{7}\ 5\ 4\ \mathbf{7}\ 2\ 3 \]


Step 3: Check each 7 for the condition.

Condition: Followed by an even number AND not preceded by a prime number.


- First 7: preceded by 3 (prime) and followed by 4 (even) \(\Rightarrow\) rejected.

- Second 7: preceded by 5 (prime), followed by 3 (odd) \(\Rightarrow\) rejected.

- Third 7: preceded by 2 (prime), followed by 1 (odd) \(\Rightarrow\) rejected.

- Fourth 7: preceded by 8 (not prime), followed by 5 (odd) \(\Rightarrow\) rejected.

- Fifth 7: preceded by 4 (not prime), followed by 2 (even) \(\Rightarrow\) accepted.


Step 4: Count accepted cases.

Only one such 7 satisfies the condition.
Quick Tip: In series questions, mark required digit occurrences and verify both left (preceding) and right (following) conditions.

Step 1: Find total items of each type.

Total tiffin boxes = 6

Total color boxes = 6

Total chocolate boxes = 6

So total items =
\[ 6+6+6=18 \]


Step 2: Items per kid.

There are 3 kids, total items are distributed equally, so each kid gets:
\[ \frac{18}{3}=6 items \]


Step 3: Apply condition of different counts for 3 categories.

Each kid must receive different numbers of (tiffin, color, chocolate).

The only possible distinct positive integers that sum to 6 are:
\[ 1,\ 2,\ 3 \]


Step 4: Check options.

Option (4) matches exactly this condition: each kid gets \(1,2,3\) in some order.

Hence option (4) can be true.
Quick Tip: If three distinct positive integers sum to 6, the only set possible is \(1,2,3\).

Step 1: Identify genders using given conditions.

No lady can be a teacher or engineer.

A is a teacher \(\Rightarrow\) A is male.

F is an engineer \(\Rightarrow\) F is male.


Step 2: Determine female members.

C is married to A \(\Rightarrow\) C is female.

F is married to D \(\Rightarrow\) D is female (since F is male).

B is sister of G \(\Rightarrow\) B is female.

Thus females are: \(\{B, C, D\}\).


Step 3: Assign professions.

C is a travel agent (given).

D is neither travel agent nor doctor (given).

So D must be Architect (since only professions left for ladies are doctor/architect and no two ladies have same profession).

Then B must be Doctor.


Step 4: Use architect condition for G.

G is an architect (given).

So we already have two architects: D and G.


Step 5: Remaining professions for males E.

We have two travel agents in total. One is C.

So the second travel agent must be E.


Quick Tip: In family-profession puzzles, first identify genders using restricted professions, then distribute unique jobs systematically.

Step 1: Use family relations.

Qills is father of Tills \(\Rightarrow\) Qills is male.

Sills is grandmother of Tills \(\Rightarrow\) Sills is female.

Uills is grandfather of Rills \(\Rightarrow\) Uills is male.


Step 2: Identify married condition.

Telegu person is female and married.

Also, no grandchild is married, so only elders can be married.


Step 3: Find couples.

Since Qills is father of Tills, Qills must be married to a female elder.

The Telegu female must be Pills, and she is married.

So Pills is married to Qills.


Step 4: Second couple.

Uills and Sills are grandfather and grandmother of the family, hence they form the second married couple.




Quick Tip: When “grandchildren are not married”, couples must be among elders (parents/grandparents). Use gender + language clues together.

Step 1: Observe the pattern carefully.

Each number is repeated twice.
\[ 0,0 \quad 6,6 \quad 20,20 \quad \ldots \]


Step 2: Express each unique term.
\[ 0 = 1^{2}-1 \]
\[ 6 = 2^{3}+2 \]
\[ 6 = 3^{2}-3 \]
\[ 20 = 4^{2}+4 \]
\[ 20 = 5^{2}-5 \]


Step 3: Next term.

Next should follow:
\[ 6^{2}+6 = 36+6=42 \]
Quick Tip: In series problems, check if terms are formed by \(n^{2}\pm n\) or \(n^{3}\pm n\). Repetition also indicates grouping.

Step 1: Decode the statement inside-out.

"My daughter's father" means Rahul himself.

So, "father of my daughter's father" means Rahul's father.


Step 2: Apply the full statement.

"Your only brother is the father of my daughter's father" becomes:

Samir's only brother is Rahul's father.


Step 3: Relationship conclusion.

If Samir's brother is Rahul's father, then Samir is Rahul's uncle.
Quick Tip: In blood relation puzzles, simplify step-by-step starting from the innermost phrase like “my daughter’s father”.

Step 1: Fix the shortest building.

Green building is the shortest and Pradeep lives there.

So:
\[ Pradeep \rightarrow Green \rightarrow Shortest \]


Step 2: Use the clue about same initial letter and second tallest.

Only possible match is:

Rishi in Red (R and R match).

Hence Rishi lives in Red building and it is the second tallest.





Step 3: Determine Lalita's building colour.

Lalita's building is neither Blue nor White.

Also it is taller than the buildings of Meenal and Rishi.

Since Rishi is already in second tallest, Lalita must be in tallest building.

So Lalita's building must be Orange (the remaining suitable colour).


Quick Tip: In logic-grid puzzles, lock fixed extremes first (shortest/tallest), then use unique matching clues like same-letter conditions.

Step 1: Place the fixed movies.

Chanakya Slot I = Horror (given).

Chanakya Slot II = Religion (given in question).

DT Slot II = Thriller (given).

Maratha Mandir Slot II = History (given).

Regal has Documentary in one slot (given).

Drama is only in Slot II.





Step 2: Place Children and Cartoon together.

Children and Cartoon must be in the same theatre.

Only theatre with both slots free is PVR, so PVR gets Children and Cartoon.





Step 3: Decide Regal's other film.

Regal already has Documentary.

Drama must be in Slot II and since Regal still has one slot open, Drama fits there.

Thus Regal can screen Documentary and Drama.



Quick Tip: In slot puzzles, fill the fixed entries first, then place paired conditions (like Children+Cartoon) and finally assign remaining constraints (like Drama only in Slot II).

Step 1: Understand all fixed conditions.

There are 6 people sitting in a circle facing inwards.

Given:

- K and L are not adjacent.

- Z and P sit opposite each other.

- M sits immediately right of P.

- If K is not between Z and M, then N is not next to P.


Step 2: Check option (1) arrangement.

Option (1): N K Z L M P (clockwise).

So circle is:
\[ N \rightarrow K \rightarrow Z \rightarrow L \rightarrow M \rightarrow P \rightarrow N \]


Step 3: Verify Z opposite P.

In a 6-person circle, opposite positions differ by 3 seats.

From P, 3 seats ahead is Z. Hence, Z is opposite P. Condition satisfied.


Step 4: Verify M is immediate right of P.

Facing inward, immediate right means anticlockwise neighbor.

In the circle, anticlockwise of P is M. So condition satisfied.


Step 5: Check the conditional statement.

We must see whether K is between Z and M.

Between Z and M clockwise: Z \(\rightarrow\) L \(\rightarrow\) M (K not included).

Between Z and M anticlockwise: Z \(\rightarrow\) K \(\rightarrow\) N \(\rightarrow\) P \(\rightarrow\) M (K included).

So K is not between Z and M in one direction, hence condition triggers:
\[ If K is not between Z and M, then N is not next to P. \]

But in option (1), N is next to P (since P is adjacent to N).

So conditional is violated.


Step 6: Conclusion.

Option (1) does not satisfy all conditions.
Quick Tip: In circular seating with conditional statements, always verify fixed placements first (opposite/right), then check the "if-then" condition carefully.

Step 1: Write given totals.

Total candidates: 10

M.Tech = 2, MBA = 2, MBBS = 2, LLB = 4

Selected = 6


Step 2: Apply selection constraints.

Exactly one M.Tech is selected.

Two MBBS candidates are selected.

At least one MBA is selected.


So far selected:
\[ 1(M.Tech) + 2(MBBS) = 3 \]

Remaining seats = \(6-3=3\).


Step 3: Fill remaining seats.

Remaining must include at least 1 MBA.

Possible case:
\[ 1(MBA) + 2(LLB) = 3 \]

This fits all conditions.


Step 4: Conclusion.

Thus, the selection can be:
\[ 1 M.Tech,\ 2 MBBS,\ 1 MBA,\ 2 LLB \]

So option (3) can be TRUE.


Quick Tip: In selection problems, lock the fixed constraints first, then fill remaining slots using remaining counts.

Step 1: Convert statements into set relations.

All students are young:
\[ Students \subseteq Young \]

Some young are short:
\[ Some Young \cap Short \neq \emptyset \]

All short are stout:
\[ Short \subseteq Stout \]

Most stout are clever:
\[ Most Stout \subseteq Clever \]

All clever are courageous:
\[ Clever \subseteq Courageous \]


Step 2: Derive what is guaranteed.

Since most stout are clever and all clever are courageous, we can say:
\[ Most Stout \subseteq Courageous \]

So option (3) is supported, not false.


Step 3: Check option (4).

Option (4) says: All stout are courageous.

But we only know most stout are clever (hence courageous).

The remaining stout (not clever) may or may not be courageous.

So "all stout are courageous" is not guaranteed and is the most definitely false option.
Quick Tip: “Most” never means “all”. If most A are B and all B are C, we can say most A are C, but never all A are C.

Step 1: List restrictions.

Arvind and Romy are not MBA.

Saurabh and Romy are not LLB.

Romy and Denu are not B.Tech.

Denu and Arvind are not M.Tech.


Step 2: Given Saurabh is B.Tech.

So Saurabh = B.Tech.


Step 3: Assign possible degrees to each.

Romy: Not MBA, Not LLB, Not B.Tech \(\Rightarrow\) Romy must be M.Tech.

Arvind: Not MBA, Not M.Tech \(\Rightarrow\) Arvind must be LLB.


Step 4: Remaining degree.

Degrees used:

Saurabh = B.Tech, Romy = M.Tech, Arvind = LLB.

Remaining degree is MBA, so Denu must be MBA.


Quick Tip: In qualification puzzles, eliminate impossible choices first; usually one person is forced into the remaining qualification.

Step 1: Represent forwarding as directed connections.

Given:

P \(\rightarrow\) Q, S, T

T \(\rightarrow\) R

P \(\leftrightarrow\) R (exchange) means P \(\rightarrow\) R and R \(\rightarrow\) P

Q \(\rightarrow\) S

S \(\rightarrow\) T


Step 2: Count all distinct paths from P to R.

Possible ways:


Way 1: Direct forwarding.
\[ P \rightarrow R \]


Way 2: Through T.
\[ P \rightarrow T \rightarrow R \]


Way 3: Through S and T.
\[ P \rightarrow S \rightarrow T \rightarrow R \]


Way 4: Through Q, S and T.
\[ P \rightarrow Q \rightarrow S \rightarrow T \rightarrow R \]


Step 3: Total ways.

Thus, total number of ways = \(4\).
 

Quick Tip: For file-transfer network questions, convert the information into a directed graph and count all valid paths from source to destination.