Aptitude Solved PYQs Bank (36 Questions)
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04. Aptitude & Reasoning Solved PYQs & Comprehensive Practice Bank

Module Focus: High-Frequency PYQs & Speed Aptitude (Quant, Logical Reasoning, Verbal English)
Target Speed: 30 Questions in 30 Minutes
Key Coverage: Speed Math, Time & Work, Speed-Distance-Time, Boats & Streams, Permutations & Combinations, Probability, Profit & Loss, Alligations, Number Systems, Syllogisms, Circular & Linear Arrangements, Blood Relations, Direction Sense, Error Spotting, Conditionals, Para-Jumbles.

1. Quantitative Aptitude Solved PYQs

PYQ 1: Pipes & Cisterns with Leakage

Problem: Pipe A can fill a tank in 8 hours and Pipe B can fill it in 12 hours. Due to a leak at the bottom, it takes 2 hours longer to fill the tank when both pipes are open. How long will the leak alone take to empty the full tank?

Detailed Step-by-Step Solution & Proof: 1. Calculate Ideal Combined Rate (without leak): $$\text{Time}_{A+B} = \frac{8 \times 12}{8 + 12} = \frac{96}{20} = 4.8 \text{ hours}$$

  1. Calculate Actual Time (with leak): $$\text{Actual Time} = 4.8 + 2 = 6.8 \text{ hours} = \frac{34}{5} \text{ hours}$$

  2. LCM Capacity Method:

  3. Let Total Tank Capacity $= \text{LCM}(8, 12) = 24 \text{ units}$.
  4. Rate of Pipe A ($r_A$) $= \frac{24}{8} = +3 \text{ units/hour}$.
  5. Rate of Pipe B ($r_B$) $= \frac{24}{12} = +2 \text{ units/hour}$.

  6. Combined Inflow Rate ($r_A + r_B$) $= 3 + 2 = +5 \text{ units/hour}$.

  7. Determine Net Rate with Leak ($L$): $$\text{Net Rate} = \frac{\text{Capacity}}{\text{Actual Time}} = \frac{24}{34/5} = \frac{120}{34} = \frac{60}{17} \text{ units/hour}$$

  8. Solve for Leak Outflow Rate ($L$): $$\text{Net Rate} = (r_A + r_B) - L$$ $$\frac{60}{17} = 5 - L \implies L = 5 - \frac{60}{17} = \frac{85 - 60}{17} = \frac{25}{17} \text{ units/hour}$$

  9. Time for Leak Alone to Empty Full Tank (24 units): $$\text{Time}_{\text{Leak}} = \frac{24}{25/17} = \frac{24 \times 17}{25} = \frac{408}{25} = 16.32 \text{ hours} \quad (16 \text{ hours } 19.2 \text{ minutes})$$

Diagram / Rate Summary:

+-------------------------------------------------------------+
| Pipe A Inflow (+3 u/h)  \                                   |
|                          +---> Net Inflow = +60/17 u/h      |
| Pipe B Inflow (+2 u/h)  /      (Fills 24 units in 16.32 h)  |
|                                                             |
| Leak Outflow L (-25/17 u/h) ---> Empties 24 u in 16.32 h    |
+-------------------------------------------------------------+

Correct Answer: 16.32 hours (or $16\frac{8}{25}$ hours)

PYQ 2: Boats & Streams - Still Water Speed Derivation

Problem: A boat travels $30 \text{ km}$ downstream and returns upstream to the starting point in a total of $8 \text{ hours}$. If the speed of the stream is $2 \text{ km/h}$, find the speed of the boat in still water.

Detailed Step-by-Step Solution & Proof: 1. Let the speed of the boat in still water be $b \text{ km/h}$. - Downstream Speed ($U$) $= b + 2 \text{ km/h}$ - Upstream Speed ($V$) $= b - 2 \text{ km/h}$

  1. Formulate Total Time Equation: $$\text{Time}_{\text{downstream}} + \text{Time}_{\text{upstream}} = T_{\text{total}}$$ $$\frac{30}{b + 2} + \frac{30}{b - 2} = 8$$

  2. Algebraic Simplification: $$30 \left( \frac{(b - 2) + (b + 2)}{(b + 2)(b - 2)} \right) = 8$$ $$30 \left( \frac{2b}{b^2 - 4} \right) = 8 \implies \frac{60b}{b^2 - 4} = 8$$ $$60b = 8b^2 - 32 \implies 8b^2 - 60b - 32 = 0$$

  3. Solve Quadratic Equation: Divide entire equation by 4: $$2b^2 - 15b - 8 = 0$$ $$(2b + 1)(b - 8) = 0$$ Roots: $b = -\frac{1}{2}$ or $b = 8$. Since physical speed must be positive: $b = 8 \text{ km/h}$.

Correct Answer: 8 km/h

PYQ 3: Conditional Probability without Replacement

Problem: Two cards are drawn sequentially without replacement from a standard pack of 52 cards. What is the probability that the first card is a King and the second card is an Ace?

Detailed Step-by-Step Solution & Proof: 1. Event $E_1$ (First card is a King): - Total Cards $= 52$, Total Kings $= 4$. $$P(E_1) = \frac{4}{52} = \frac{1}{13}$$

  1. Event $E_2 \mid E_1$ (Second card is an Ace given first was a King):
  2. Remaining cards in deck $= 51$.

  3. Number of Aces remaining $= 4$. $$P(E_2 \mid E_1) = \frac{4}{51}$$

  4. Joint Probability using Multiplication Theorem: $$P(E_1 \cap E_2) = P(E_1) \times P(E_2 \mid E_1) = \frac{4}{52} \times \frac{4}{51} = \frac{1}{13} \times \frac{4}{51} = \frac{4}{663}$$

Correct Answer: 4 / 663

PYQ 4: Time & Work - Alternate Days Cycle

Problem: A can complete a work in 12 days, B in 16 days, and C in 24 days. If they work on alternate days in the order of A on Day 1, B on Day 2, C on Day 3, A on Day 4, and so on, in how many days will the work be completed?

Detailed Step-by-Step Solution & Proof: 1. Define Total Work using LCM: $$\text{Total Work} = \text{LCM}(12, 16, 24) = 48 \text{ units}$$

  1. Calculate Daily Efficiencies:
  2. Efficiency of A ($e_A$) $= \frac{48}{12} = 4 \text{ units/day}$
  3. Efficiency of B ($e_B$) $= \frac{48}{16} = 3 \text{ units/day}$

  4. Efficiency of C ($e_C$) $= \frac{48}{24} = 2 \text{ units/day}$

  5. Analyze 1 Complete Cycle (3 Days): $$\text{Work in 1 cycle (3 days)} = e_A + e_B + e_C = 4 + 3 + 2 = 9 \text{ units}$$

  6. Determine Completed Cycles: $$\text{Number of full cycles} = \left\lfloor \frac{48}{9} \right\rfloor = 5 \text{ cycles}$$

  7. Work completed in 5 cycles ($5 \times 3 = 15 \text{ days}$) $= 5 \times 9 = 45 \text{ units}$.

  8. Remaining Work $= 48 - 45 = 3 \text{ units}$.

  9. Evaluate Partial Day for Remaining Work:

  10. Day 16 is A's turn. A can do 4 units/day.

  11. Time required by A for 3 units $= \frac{\text{Remaining Work}}{e_A} = \frac{3}{4} \text{ day}$.

  12. Total Time: $$\text{Total Days} = 15 + \frac{3}{4} = 15.75 \text{ days} = 15\frac{3}{4} \text{ days}$$

Correct Answer: 15.75 days (or $15\frac{3}{4}$ days)

PYQ 5: Speed, Distance & Time - Relative Speed & Platform Crossing

Problem: Train X of length $200 \text{ m}$ is moving at $72 \text{ km/h}$. Train Y of length $300 \text{ m}$ is moving in the opposite direction at $54 \text{ km/h}$.
(a) How long will it take for the two trains to completely cross each other?
(b) How long will Train X take to cross a stationary platform of length $280 \text{ m}$?

Detailed Step-by-Step Solution & Proof: 1. Convert Speeds to m/s: - Speed of Train X ($v_X$) $= 72 \times \frac{5}{18} = 20 \text{ m/s}$ - Speed of Train Y ($v_Y$) $= 54 \times \frac{5}{18} = 15 \text{ m/s}$

  1. Part (a): Crossing Each Other (Opposite Directions):
  2. Relative Speed ($v_{\text{rel}}$) $= v_X + v_Y = 20 + 15 = 35 \text{ m/s}$

  3. Total Distance ($D_1$) $= L_X + L_Y = 200 + 300 = 500 \text{ m}$ $$\text{Time}_1 = \frac{D_1}{v_{\text{rel}}} = \frac{500}{35} = \frac{100}{7} \approx 14.29 \text{ seconds}$$

  4. Part (b): Train X Crossing Platform:

  5. Total Distance ($D_2$) $= L_X + L_{\text{platform}} = 200 + 280 = 480 \text{ m}$ $$\text{Time}_2 = \frac{D_2}{v_X} = \frac{480}{20} = 24 \text{ seconds}$$

Correct Answer: (a) 14.29 s, (b) 24 s

PYQ 6: Permutations & Combinations - Vowels Together Constraint

Problem: In how many different ways can the letters of the word 'CORPORATION' be arranged such that all the vowels always come together?

Detailed Step-by-Step Solution & Proof: 1. Analyze Letter Frequencies: - Total letters $= 11$ - Vowels: O, O, A, I, O $\implies$ 5 vowels (Count: O=3, A=1, I=1) - Consonants: C, R, P, R, T, N $\implies$ 6 consonants (Count: R=2, C=1, P=1, T=1, N=1)

  1. Bundle Vowels as One Block ($\mathbf{V}$):
  2. Items to arrange $= 6 \text{ consonants} + 1 \text{ Vowel-Block} = 7 \text{ items}$.

  3. Among consonants, 'R' repeats 2 times. $$\text{Arrangements of 7 items} = \frac{7!}{2!} = \frac{5040}{2} = 2520$$

  4. Internal Arrangements of the Vowel Block ($\mathbf{V}$):

  5. 5 vowels with 'O' repeating 3 times. $$\text{Internal Arrangements} = \frac{5!}{3!} = \frac{120}{6} = 20$$

  6. Apply Fundamental Principle of Counting: $$\text{Total Ways} = 2520 \times 20 = 50,400$$

Correct Answer: 50,400

PYQ 7: Profit, Loss & Successive Discounts

Problem: A retailer marks an item $50\%$ above its Cost Price (CP) and offers two successive discounts of $20\%$ and $10\%$. If the final Selling Price (SP) is ₹$1,080$, find the Cost Price and the net Profit Percentage.

Detailed Step-by-Step Solution & Proof: 1. Let Cost Price $= 100x$. 2. Marked Price (MP): $$\text{MP} = 100x \times (1 + 0.50) = 150x$$

  1. Apply 1st Discount ($20\%$): $$\text{Price}_1 = 150x \times (1 - 0.20) = 150x \times 0.80 = 120x$$

  2. Apply 2nd Discount ($10\%$): $$\text{SP} = 120x \times (1 - 0.10) = 120x \times 0.90 = 108x$$

  3. Solve for $x$ given $\text{SP} = ₹1,080$: $$108x = 1080 \implies x = 10$$ $$\text{Cost Price (CP)} = 100 \times 10 = ₹1,000$$

  4. Calculate Profit and Profit Percentage: $$\text{Profit} = \text{SP} - \text{CP} = 1080 - 1000 = ₹80$$ $$\text{Profit \%} = \left( \frac{80}{1000} \right) \times 100\% = 8\%$$

Correct Answer: Cost Price = ₹1,000, Profit % = 8%

PYQ 8: Mixtures & Alligations - Repeated Dilution Formula

Problem: A container holds $80 \text{ liters}$ of pure milk. $8 \text{ liters}$ of milk is extracted and replaced with $8 \text{ liters}$ of water. This operation is repeated 2 more times (total 3 operations). What is the final quantity of pure milk remaining in the container?

Detailed Step-by-Step Solution & Proof: 1. Repeated Replacement Formula: $$Q_{\text{final}} = Q_{\text{initial}} \times \left(1 - \frac{x}{V}\right)^n$$ Where: - $Q_{\text{initial}} = 80 \text{ L}$ (Initial pure volume) - $x = 8 \text{ L}$ (Volume replaced per cycle) - $V = 80 \text{ L}$ (Total container capacity) - $n = 3$ (Number of operations)

  1. Substitute Values: $$Q_{\text{final}} = 80 \times \left(1 - \frac{8}{80}\right)^3 = 80 \times \left(1 - \frac{1}{10}\right)^3 = 80 \times \left(\frac{9}{10}\right)^3$$

  2. Evaluate: $$Q_{\text{final}} = 80 \times \frac{729}{1000} = \frac{58320}{1000} = 58.32 \text{ liters}$$

Correct Answer: 58.32 liters

PYQ 9: Number System - Remainder Theorem & Modular Arithmetic

Problem: What is the remainder when $7^{84}$ is divided by $342$?

Detailed Step-by-Step Solution & Proof: 1. Express $7^{84}$ in terms of powers close to a multiple of $342$. - Note that $7^3 = 343$. 2. Express $343$ modulo $342$: $$343 \equiv 1 \pmod{342}$$

  1. Rewrite $7^{84}$ using power of 3: $$7^{84} = (7^3)^{28} = (343)^{28}$$

  2. Apply Modular Congruence Property ($(a)^n \equiv (b)^n \pmod m$): $$7^{84} \equiv (1)^{28} \pmod{342} \equiv 1 \pmod{342}$$

Correct Answer: 1

PYQ 10: Simple vs Compound Interest 3-Year Difference

Problem: The difference between Compound Interest (compounded annually) and Simple Interest on a sum of money for 3 years at $10\%$ per annum is ₹$155$. Find the principal sum.

Detailed Step-by-Step Solution & Proof: 1. Standard 3-Year Interest Difference Formula: $$D_3 = P \times \left(\frac{r}{100}\right)^2 \times \left(3 + \frac{r}{100}\right)$$

  1. Substitute Given Values ($D_3 = 155, r = 10$): $$155 = P \times \left(\frac{10}{100}\right)^2 \times \left(3 + \frac{10}{100}\right)$$ $$155 = P \times \left(\frac{1}{10}\right)^2 \times \left(\frac{31}{10}\right)$$ $$155 = P \times \frac{1}{100} \times \frac{31}{10} = P \times \frac{31}{1000}$$

  2. Solve for Principal $P$: $$P = \frac{155 \times 1000}{31} = 5 \times 1000 = ₹5,000$$

Correct Answer: ₹5,000

2. Logical Reasoning Solved PYQs

PYQ 11: Circular Seating Arrangement (Mixed Facing Directions)

Problem: 6 individuals ($A, B, C, D, E, F$) sit around a circular table. - $A$ and $B$ are facing outside the center. - $C, D, E, F$ are facing inside the center. - $C$ sits second to the right of $A$. - $B$ sits opposite to $D$. - $E$ is an immediate neighbor of both $C$ and $B$. Who sits to the immediate left of $F$?

Detailed Step-by-Step Logical Deduction: 1. Direction Rules: - Facing IN $\implies$ Right is Counter-Clockwise (CCW), Left is Clockwise (CW). - Facing OUT $\implies$ Right is Clockwise (CW), Left is Counter-Clockwise (CCW).

  1. Step 1: Place $A$ (Facing OUT):
  2. Assign Position 1 to $A$ (Facing OUT).

  3. Right of $A$ (Facing OUT) is Clockwise. Second to right places $C$ at Position 5 (or 2 steps CW).

  4. Step 2: Place $E$ and $B$:

  5. $E$ is adjacent to both $C$ and $B$.

  6. If $C$ is at Pos 5, $E$ must be at Pos 4, and $B$ at Pos 3.

  7. Step 3: Place $D$ and $F$:

  8. $B$ (at Pos 3) is opposite $D \implies D$ is at Pos 6.
  9. The remaining position (Pos 2) is occupied by $F$.

Circular Diagram Map:

                  Pos 1: A (OUT)
                 /             \
    Pos 6: D (IN)               Pos 2: F (IN)
        |                           |
    Pos 5: C (IN)               Pos 3: B (OUT)
                 \             /
                  Pos 4: E (IN)
  1. Step 4: Evaluate Immediate Left of $F$:
  2. $F$ is at Pos 2 and faces IN.
  3. For facing IN, Left means Clockwise $\implies$ Position 3.
  4. Position 3 is occupied by $B$.

Correct Answer: B

PYQ 12: Linear Seating Arrangement (8 Persons Facing North)

Problem: 8 friends $P, Q, R, S, T, U, V, W$ sit in a straight line facing North. 1. $R$ sits 3rd to the left of $W$. 2. $W$ is not at any extreme end. 3. Only two people sit between $W$ and $T$. 4. $P$ sits 2nd to the right of $T$. 5. $S$ sits to the immediate left of $U$. 6. $Q$ does not sit at an extreme end.

Who sits at the extreme left end of the row?

Detailed Step-by-Step Logical Mapping: 1. Create 8 slots (1 to 8 left to right). 2. Condition 1 & 2: $R \text{ _ _ } W$. Since $W$ is not extreme end, possible slots for $(R, W)$ are $(1, 4), (2, 5), (3, 6), (4, 7)$. 3. Condition 3 & 4: Two people between $W$ and $T$. $P$ is 2nd to right of $T$ ($T \text{ _ } P$). - If $W=4 \implies T=7 \implies P=9$ (Invalid, out of bounds). - If $W=5 \implies T=2 \implies P=4$. - Test $(R=2, W=5, T=2)$ collision! $R$ and $T$ both at 2 (Invalid). - If $W=6 \implies R=3$. Two between $W(6)$ and $T \implies T=3$ (Collision with $R$) or $T=9$ (Invalid). - If $W=4, R=1$: Two between $W(4)$ and $T \implies T=7$. $P$ at 2nd right of $T(7) \implies P=8$. - Slots filled so far: 1:R, 4:W, 7:T, 8:P.

  1. Condition 5 & 6: $S$ is immediate left of $U$ ($SU$ block). Remaining free contiguous slots are 5 & 6. So $S=5, U=6$.
  2. Remaining slots: 2 and 3. $Q$ cannot be at extreme end, so remaining $V$ and $Q$ fill slots 2 and 3. Since $Q$ at 3, $V$ at 2.

Final Linear Layout:

Slot:   [ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ] [ 7 ] [ 8 ]
Person:   R     V     Q     W     S     U     T     P

Correct Answer: R

PYQ 13: Syllogism - "Only a Few" & Venn Diagram Logic

Problem: Statements: 1. Only a few Pens are Pencils. 2. All Pencils are Erasers. 3. No Eraser is a Sharpener.

Conclusions: I. Some Pens are Erasers.
II. No Pencil is a Sharpener.
III. All Pens can never be Pencils.

Which of the conclusions logically follow?

Detailed Venn Diagram Logic & Proof: - "Only a few A are B" implies two simultaneous conditions: 1. Some A are B (Intersection exists). 2. Some A are NOT B (A cannot be completely inside B).

+-------------------------------------------------------+
|  [ PENS ]                                             |
|   +-----------------------+                           |
|   |  (Pen & Pencil & Eraser) |                         |
|   |   +----------------+  |     [ SHARPENERS ]        |
|   |   |   PENCILS      |  |      +--------------+     |
|   |   |  +----------+  |  |      |              |     |
|   |   |  |          |  |  |      |              |     |
|   +---+--+----------+--+--+      +--------------+     |
|       |    ERASERS     |               ^              |
|       +----------------+               | NO OVERLAP   |
|                                        v              |
+-------------------------------------------------------+
  1. Check Conclusion I (Some Pens are Erasers):

  2. Since Some Pens are Pencils and ALL Pencils are Erasers, the Pen-Pencil intersection MUST lie inside Erasers. $\implies$ Follows.

  3. Check Conclusion II (No Pencil is a Sharpener):

  4. All Pencils are inside Erasers, and No Eraser is a Sharpener. $\implies$ Pencils can never overlap with Sharpeners. $\implies$ Follows.

  5. Check Conclusion III (All Pens can never be Pencils):

  6. From Statement 1 ("Only a few Pens are Pencils"), by definition Some Pens are NOT Pencils. Thus, 100% of Pens can never enter Pencils. $\implies$ Follows.

Correct Answer: All I, II, and III follow

PYQ 14: Syllogism - Possibility and Negative Deduction

Problem: Statements: 1. Some Cats are Dogs. 2. All Dogs are Birds. 3. No Bird is a Fish.

Conclusions: I. Some Cats are not Fish.
II. All Birds being Cats is a possibility.

Detailed Step-by-Step Proof: 1. Conclusion I: - Cats that are Dogs are also Birds (since All Dogs are Birds). - No Bird is a Fish $\implies$ Those specific Cats which are Dogs can NEVER be Fish. - Therefore, at least those Cats are not Fish $\implies$ Follows.

  1. Conclusion II:
  2. The statement "All Dogs are Birds" allows Birds to expand and encompass all Cats without violating any negative boundary with Fish.
  3. A single valid Venn diagram exists where the circle of Cats is entirely inside Birds $\implies$ Follows.

Correct Answer: Both I and II follow

PYQ 15: Coded Blood Relations Tree Deduction

Problem: Symbols: - $A + B \implies A \text{ is the father of } B$ - $A - B \implies A \text{ is the sister of } B$ - $A \times B \implies A \text{ is the mother of } B$ - $A \div B \implies A \text{ is the brother of } B$

Given the expression: $\mathbf{P + Q \times R - S \div T}$
How is $P$ related to $T$?

Detailed Family Tree Mapping: 1. $P + Q \implies P$ (Male) is Father of $Q$. 2. $Q \times R \implies Q$ (Female) is Mother of $R$. 3. $R - S \implies R$ (Female) is Sister of $S$. 4. $S \div T \implies S$ (Male) is Brother of $T$.

Gen 1:            [ P ] (Male, Grandfather)
                    |
Gen 2:            ( Q ) (Female, Mother)
                    |
Gen 3:   +----------+----------+
         |          |          |
      ( R ) (F)   [ S ] (M)    [ T ] (Gender unstated)
  1. $R, S, T$ are siblings.
  2. $Q$ is the mother of $T$.
  3. $P$ is the father of $Q$.
  4. Therefore, $P$ is the Maternal Grandfather of $T$.

Correct Answer: Maternal Grandfather

PYQ 16: Direction Sense & Shortest Distance Vector

Problem: A person starts at point A and walks $10 \text{ m}$ North, turns right and walks $15 \text{ m}$ to point B, turns right and walks $18 \text{ m}$ to point C, then turns left and walks $9 \text{ m}$ to point D.
Find the shortest (direct) distance between point A and point D.

Detailed Coordinate System Proof: 1. Assign origin $(0,0)$ to starting Point A. 2. Step 1: North $10 \text{ m} \implies (0, +10)$. 3. Step 2: Turn Right (East) $15 \text{ m} \implies \text{Point B} = (0+15, 10) = (15, 10)$. 4. Step 3: Turn Right (South) $18 \text{ m} \implies \text{Point C} = (15, 10-18) = (15, -8)$. 5. Step 4: Turn Left (East) $9 \text{ m} \implies \text{Point D} = (15+9, -8) = (24, -8)$.

   North (+y)
       ^
       |       (0,10) -------------> B (15,10)
       |         |                     |
       |  10m N  |                     | 18m S
  A ---+---------+---------------------+----------> East (+x)
(0,0)  |                               |
       |                         C (15,-8) ---------> D (24,-8)
       |                                      9m E
       v
  1. Apply Pythagorean Theorem for Direct Distance $AD$: $$\Delta x = 24 - 0 = 24 \text{ m}, \quad \Delta y = -8 - 0 = -8 \text{ m}$$ $$AD = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{24^2 + (-8)^2} = \sqrt{576 + 64} = \sqrt{640} = 8\sqrt{10} \approx 25.30 \text{ meters}$$

Correct Answer: 8√10 meters (approx. $25.30 \text{ m}$)

PYQ 17: Data Sufficiency - Comparative Age Evaluation

Problem: Is person $X$ older than person $Y$? - Statement 1: $X$ was born 3 years before $Z$, and $Z$ is 2 years younger than $Y$. - Statement 2: The average age of $X$ and $Y$ is strictly greater than $Y$'s age.

Which statement(s) is/are sufficient to answer the question?

Detailed Mathematical Proof: 1. Evaluate Statement 1 Alone: - "$X$ born 3 years before $Z$" $\implies \text{Age}(X) = \text{Age}(Z) + 3$. - "$Z$ is 2 years younger than $Y$" $\implies \text{Age}(Y) = \text{Age}(Z) + 2$. - Compare $\text{Age}(X)$ and $\text{Age}(Y)$: $$\text{Age}(X) - \text{Age}(Y) = (\text{Age}(Z) + 3) - (\text{Age}(Z) + 2) = +1 \text{ year}$$ - Since $\text{Age}(X) > \text{Age}(Y)$, $X$ is definitively older than $Y$. - Statement 1 alone is SUFFICIENT.

  1. Evaluate Statement 2 Alone:
  2. $\frac{\text{Age}(X) + \text{Age}(Y)}{2} > \text{Age}(Y)$
  3. Multiply by 2: $\text{Age}(X) + \text{Age}(Y) > 2 \cdot \text{Age}(Y) \implies \text{Age}(X) > \text{Age}(Y)$.
  4. Statement 2 alone is also SUFFICIENT.

Correct Answer: Either Statement 1 alone or Statement 2 alone is sufficient

3. Verbal English Error Spotting & Grammar PYQs

PYQ 18: Subject-Verb Agreement with Proximity Rule

Problem: Identify the error in the following sentence:
"Neither the principal nor the teachers was present at the annual sports meet yesterday."

Grammatical Analysis & Proof: - Rule: When two subjects are joined by correlative conjunctions such as neither...nor, either...or, or not only...but also, the verb must agree with the subject closest to it (Subject Proximity Rule). - Here, the subjects are "the principal" (singular) and "the teachers" (plural). - The verb follows "the teachers" (plural), requiring the plural past tense auxiliary were, not the singular was.

Corrected Sentence:
"Neither the principal nor the teachers were present at the annual sports meet yesterday."

PYQ 19: Correlative Conjunctions & Inversion Structure

Problem: Identify the error in the following sentence:
"Scarcely had the train arrived at the platform than all the passengers rushed towards the doors."

Grammatical Analysis & Proof: - Rule 1 (Pairing): The adverbs Scarcely and Hardly must be paired with when or before (never than). (No sooner is paired with than). - Rule 2 (Inversion): When Scarcely begins a sentence, subject-auxiliary inversion occurs (Scarcely had + Subject + Past Participle), which is correctly implemented here.

Corrected Sentence:
"Scarcely had the train arrived at the platform when all the passengers rushed towards the doors."

PYQ 20: Third Conditional Clause Misconjunction

Problem: Correct the grammatical error in the conditional sentence:
"If he would have worked harder, he would have passed the examination."

Grammatical Analysis & Proof: - Rule (Third Conditional Structure): Unfulfilled past conditional situations require:
If + Subject + Past Perfect (had + V3), Subject + would have + V3 - The conditional (if) clause must NOT contain would have.

Corrected Sentence:
"If he had worked harder, he would have passed the examination."

PYQ 21: Dangling / Misplaced Modifier Correction

Problem: Spot the error in the sentence:
"Walking down the street, the trees looked vibrant and beautiful to John."

Grammatical Analysis & Proof: - Rule: A participial phrase at the beginning of a sentence ("Walking down the street") must modify the grammatical subject immediately following the comma. - As written, the sentence absurdly implies that the trees were walking down the street. - The modifier must explicitly attach to John.

Corrected Sentence:
"Walking down the street, John noticed that the trees looked vibrant and beautiful."

PYQ 22: Vocabulary in Context - Double Fillers

Problem: Select the pair of words that best completes the sentence logically and grammatically:
"Despite the board's initial __ towards the proposed merger, the CEO's compelling presentation eventually ____ their lingering doubts."

Options:
A) enthusiasm ... heightened
B) skepticism ... dispelled
C) indifference ... amplified
D) hostility ... fostered

Contextual & Etymological Proof: - The concession word "Despite" signals a contrast between the initial negative state of the board and the outcome achieved by the "compelling presentation". - Skepticism (doubt/reluctance) fits the first slot. - Dispelled (drove away / eliminated) fits the second slot, matching the removal of "lingering doubts".

Correct Answer: B) skepticism ... dispelled

PYQ 23: Para-Jumbles (Logical Paragraph Ordering)

Problem: Arrange the following 4 sentences in the most logical sequence to form a coherent paragraph:
- A: This rapid evolution of artificial intelligence has reshaped traditional corporate workflows.
- B: Consequently, organizational leaders must prioritize continuous upskilling to maintain competitive advantage.
- C: Over the past decade, automated algorithms have transitioned from basic task execution to complex predictive modeling.
- D: As a result of this shift, employees face unprecedented demands to adapt to hybrid human-AI environments.

Structural Analysis: 1. Sentence C introduces the overarching topic ("Over the past decade...") without pronouns or transition markers $\implies$ Ideal introductory sentence. 2. Sentence A builds on sentence C, referencing "This rapid evolution..." (pointing directly to C's transition from task execution to predictive modeling). Mandatory pair: C $\rightarrow$ A. 3. Sentence D explains the direct consequence on workforce demands ("As a result of this shift..."). 4. Sentence B offers the final leadership takeaway ("Consequently, organizational leaders must...").

Correct Order: C - A - D - B

PYQ 24: Reading Comprehension & Critical Inference

Passage:
"While automation inevitably eliminates routine manual tasks, empirical evidence suggests that technology shifts workforce demand toward higher-order analytical and interpersonal skills rather than reducing net employment overall."

Question: Which of the following is an underlying assumption of the passage?
A) Technological advancement eliminates all forms of physical labor completely.
B) Human workers are capable of acquiring higher-order analytical and interpersonal skills.
C) Net employment will decline sharply over the next decade.
D) Routine manual tasks require higher-order cognitive processing.

Critical Deduction & Proof: - The passage asserts that demand shifts toward higher-order skills rather than reducing net employment overall. - For net employment to remain stable as demand shifts, workers must be able to acquire these higher-order skills. If workers could not learn these skills, overall net employment would inevitably collapse, contradicting the author's premise. - Thus, Option B is a necessary assumption.

Correct Answer: B) Human workers are capable of acquiring higher-order analytical and interpersonal skills.

4. CoCubes Real Placement Solved PYQs Bank (PrepInsta, FACE Prep, IndiaBIX Archives)

Module Focus: Real CoCubes placement exam questions with option choices, speed shortcuts, and detailed solution proofs.
Exam Benchmark: 15–20 Questions per section in 15–20 Minutes (Moderate difficulty, 0 negative marking).

4.1 CoCubes Quantitative Aptitude Solved PYQs

PYQ 25: Time, Speed & Distance - Train Meeting Point

Problem: Two trains start from a station at 10:00 AM and 10:30 AM, traveling in the same direction at $60 \text{ km/h}$ and $75 \text{ km/h}$ respectively. At what distance from the station will the second train catch up with the first train?
Options:
A) $120 \text{ km}$
B) $150 \text{ km}$
C) $180 \text{ km}$
D) $200 \text{ km}$

Detailed Step-by-Step Solution & Speed Shortcut: 1. Initial Distance Lead: Train 1 travels alone for $30 \text{ minutes} = 0.5 \text{ hours}$. $$\text{Lead} = 60 \text{ km/h} \times 0.5 \text{ h} = 30 \text{ km}$$ 2. Relative Speed: Both trains move in same direction $\implies v_{\text{rel}} = 75 - 60 = 15 \text{ km/h}$. 3. Catch-up Time: $$\text{Time} = \frac{\text{Lead}}{v_{\text{rel}}} = \frac{30}{15} = 2 \text{ hours}$$ 4. Distance Covered: $$\text{Distance} = 75 \text{ km/h} \times 2 \text{ h} = 150 \text{ km}$$

Correct Answer: B) 150 km

PYQ 26: Time & Work - Man-Days Inverse Proportionality

Problem: 28 children can complete a piece of work in 50 days. How many children are required to finish the exact same work in 35 days?
Options:
A) 36 children
B) 40 children
C) 42 children
D) 45 children

Detailed Step-by-Step Solution & Speed Shortcut: 1. Use inverse proportionality formula $M_1 D_1 = M_2 D_2$: $$28 \times 50 = M_2 \times 35$$ 2. Solve for $M_2$: $$M_2 = \frac{28 \times 50}{35} = \frac{4 \times 50}{5} = 40 \text{ children}$$

Correct Answer: B) 40 children

PYQ 27: Ratio & Proportion - Transitive Ratio Multiplication

Problem: If $a : b = 3 : 4$ and $b : c = 5 : 7$, find the ratio $a : c$.
Options:
A) $15 : 28$
B) $3 : 7$
C) $5 : 4$
D) $12 : 35$

Detailed Step-by-Step Solution & Speed Shortcut: 1. Express as fractions: $\frac{a}{b} = \frac{3}{4}, \frac{b}{c} = \frac{5}{7}$. 2. Multiply ratios to eliminate $b$: $$\frac{a}{c} = \frac{a}{b} \times \frac{b}{c} = \frac{3}{4} \times \frac{5}{7} = \frac{15}{28}$$

Correct Answer: A) 15 : 28

PYQ 28: Algebraic Variable Substitution

Problem: If $a = \frac{2}{3}b$, $b = \frac{2}{3}c$, and $c = \frac{2}{3}d$, what fraction of $d$ is $b$?
Options:
A) $\frac{2}{3}$
B) $\frac{4}{9}$
C) $\frac{8}{27}$
D) $\frac{1}{3}$

Detailed Step-by-Step Solution & Speed Shortcut: 1. Substitute $c = \frac{2}{3}d$ into $b = \frac{2}{3}c$: $$b = \frac{2}{3} \left(\frac{2}{3}d\right) = \frac{4}{9}d$$

Correct Answer: B) 4/9

PYQ 29: Profit & Loss - Successive Discounts

Problem: A trader marks his goods $40\%$ above the cost price and allows two successive discounts of $15\%$ and $10\%$ on the marked price. Find his net profit percentage.
Options:
A) $7.1\%$
B) $8.0\%$
C) $10.5\%$
D) $12.0\%$

Detailed Step-by-Step Solution & Speed Shortcut: 1. Let Cost Price $(\text{CP}) = 100$. 2. Marked Price $(\text{MP}) = 100 \times 1.40 = 140$. 3. Selling Price $(\text{SP})$ after discounts: $$\text{SP} = 140 \times (1 - 0.15) \times (1 - 0.10) = 140 \times 0.85 \times 0.90 = 140 \times 0.765 = 107.1$$ 4. Net Profit $\% = \text{SP} - \text{CP} = 107.1 - 100 = 7.1\%$.

Correct Answer: A) 7.1%

4.2 CoCubes Logical Reasoning Solved PYQs (Syllogisms & Seating)

PYQ 30: Syllogism - The 100-50 Analytical Method

Problem:
Statements: 1. All books are pens. ($\text{Book}^{100} \text{ -- Pen}^{50}$) $[+]$ 2. Some pens are pencils. ($\text{Pen}^{50} \text{ -- Pencil}^{50}$) $[+]$

Conclusions: I. Some books are pencils.
II. Some pens are books.

Options:
A) Only conclusion I follows
B) Only conclusion II follows
C) Either I or II follows
D) Neither I nor II follows

Detailed Logical Proof (100-50 Rule): 1. Conclusion I: Middle term Pen has values $50$ in P1 and $50$ in P2 ($50+50=100$, condition fails). No connection between Book and Pencil $\implies$ Does NOT follow. 2. Conclusion II: Immediate conversion of P1 ("All $\text{Book}^{100}$ are $\text{Pen}^{50}$") $\to$ "Some $\text{Pen}^{50}$ are $\text{Book}^{50}$" $\implies$ Follows.

Correct Answer: B) Only conclusion II follows

PYQ 31: Syllogism - Negative Premise & Particular Conclusion

Problem:
Statements: 1. No cat is a dog. ($\text{Cat}^{100} \text{ -- Dog}^{100}$) $[-]$ 2. All dogs are elephants. ($\text{Dog}^{100} \text{ -- Elephant}^{50}$) $[+]$

Conclusions: I. Some elephants are not cats.
II. No cat is an elephant.

Options:
A) Only conclusion I follows
B) Only conclusion II follows
C) Both I and II follow
D) Neither I nor II follows

Detailed Logical Proof (100-50 Rule): 1. Middle term Dog has $100$ in P1 and $100$ in P2 (Valid). Sign $(-)+(+)=(-)$. 2. Conclusion I: "Some $\text{Elephant}^{50}$ are not $\text{Cat}^{100}$" (Type O, negative). Income of Elephant is $50$, Expense is $50$. Income of Cat is $100$, Expense is $100$. $\implies$ Follows. 3. Conclusion II: "No $\text{Cat}^{100}$ is an $\text{Elephant}^{100}$" (Type E). Elephant Expense $100 >$ Income $50$. $\implies$ Does NOT follow.

Correct Answer: A) Only conclusion I follows

PYQ 32: Circular Seating Arrangement (6 Persons Facing Center)

Problem:
6 friends ($A, B, C, D, E, F$) sit around a circular table facing the center. - $A$ sits 2nd to the left of $C$. - $B$ sits immediate right of $C$. - $D$ sits 2nd to the right of $B$. - $E$ sits between $A$ and $D$.

Who sits opposite to $C$?
Options:
A) $D$
B) $E$
C) $F$
D) $A$

Detailed Placement Logic & Map: 1. Clockwise = Left, Counter-Clockwise = Right (Facing Center). 2. Anchor $C$ at Slot 1. - 2nd left of $C \implies A$ at Slot 3. - Immediate right of $C \implies B$ at Slot 6. 3. 2nd right of $B \implies D$ at Slot 5. 4. $E$ between $A$ (Slot 3) and $D$ (Slot 5) $\implies E$ at Slot 4. 5. Remaining $F$ at Slot 2. 6. Opposite of Slot 1 ($C$) is Slot 4 ($E$).

Correct Answer: B) E

PYQ 33: Linear Seating Arrangement (5 Persons Facing North)

Problem:
5 students ($P, Q, R, S, T$) sit in a row facing North. - $S$ sits between $T$ and $Q$. - $Q$ sits to the immediate left of $R$. - $P$ sits at the extreme left end.

Who sits at the extreme right end of the row?
Options:
A) $Q$
B) $R$
C) $S$
D) $T$

Detailed Placement Logic: 1. Slots 1 to 5 (Left to Right). 2. Slot 1 = $P$ (Extreme left). 3. Immediate left of $R$ is $Q \implies QR$ block. 4. $S$ between $T$ and $Q \implies T - S - Q$. 5. Arrangement: $P(1) - T(2) - S(3) - Q(4) - R(5)$. 6. Extreme right end (Slot 5) = $R$.

Correct Answer: B) R

4.3 CoCubes Verbal English Solved PYQs (Error Spotting)

PYQ 34: Error Spotting - Subject-Verb Proximity Rule

Problem: Identify the part containing the grammatical error:
"Neither the manager (A) / nor his subordinates (B) / was present at (C) / the emergency meeting. (D)"
Options:
A) Part A
B) Part B
C) Part C
D) Part D

Grammatical Analysis & Proof: - Under neither...nor, the verb must agree with the subject closest to it ("his subordinates" - plural). - "was present" (singular) must be corrected to "were present" (plural). - Error is in Part C.

Correct Answer: C) Part C

PYQ 35: Error Spotting - Correlative Conjunction Pairing

Problem: Identify the part containing the grammatical error:
"No sooner had the alarm rung (A) / than the security guard (B) / locked all the entrance gates (C) / immediately. (D)"
Options:
A) Part A
B) Part B
C) Part C
D) No Error

Grammatical Analysis & Proof: - No sooner is correctly paired with than. - Subject-auxiliary inversion (No sooner had + subject + V3) is correctly applied. - The sentence is grammatically sound.

Correct Answer: D) No Error

PYQ 36: Error Spotting - Misplaced Participial Modifier

Problem: Spot the error in the sentence:
"While driving to work, (A) / a dog ran across the road (B) / and nearly caused (C) / a major accident. (D)"
Options:
A) Part A
B) Part B
C) Part C
D) Part D

Grammatical Analysis & Proof: - The opening modifier "While driving to work" must modify the subject of the main clause. - As written, "a dog" becomes the subject driving to work, creating a misplaced modifier error. - Correct phrasing: "While I was driving to work..." or "While driving to work, the driver saw a dog..." - Error is in Part A.

Correct Answer: A) Part A