Quantitative Aptitude PYQs

Shortcut Trick

Total Work = LCM(10, 25, 50) = 50 units.

Efficiency of (A+B) = 5, (B+C) = 2, (C+A) = 1.

Sum of efficiencies 2(A+B+C) = 5 + 2 + 1 = 8 → (A+B+C) = 4.

Time taken = Total Work ÷ Efficiency = 50 ÷ 4 = 12.5 days.

∴ The correct answer is 25/2 days.

Alternate Method

Given:

Time taken by (A + B) = 10 days

Time taken by (B + C) = 25 days

Time taken by (C + A) = 50 days

Formula Used:

Efficiency = Total Work ÷ Time

Calculations:

⇒ Total efficiency of 2(A + B + C) = 5 + 2 + 1 = 8

⇒ Efficiency of (A + B + C) = 8 ÷ 2 = 4

⇒ Days taken by (A + B + C) together = 50 ÷ 4

⇒ Days = 25/2 days

∴ The correct answer is 25/2 days.

Additional Information

Combined Work Formula

When pairs (A+B, B+C, C+A) are given, the sum of their individual efficiencies equals 2 × (A+B+C) efficiency.

Work and Wages

Wages are always distributed in the ratio of the efficiency of the workers or the amount of work done.

Pipes and Cisterns

This concept is identical to Time & Work, where an inlet pipe has positive efficiency and an outlet pipe has negative efficiency.

General Solution:

Given Data:

 

Find History marks (H) using ratio S : H = a : b ⇒ H = S × (b ÷ a):

Student A: S = 72, ratio 4 : 3 ⇒ H = 72 × (3 ÷ 4) = 72 × 0.75 = 54

Student B: S = 84, ratio 7 : 6 ⇒ H = 84 × (6 ÷ 7) = 84 × (6/7) = 72

Student C: S = 60, ratio 5 : 4 ⇒ H = 60 × (4 ÷ 5) = 60 × 0.8 = 48

Find Polity marks (P) from average (P + S) ÷ 2 = given ⇒ P = 2 × (given average) − S:

Student A: Average = 66, S = 72 ⇒ P = 2 × 66 − 72 = 132 − 72 = 60

Student B: Average = 78, S = 84 ⇒ P = 2 × 78 − 84 = 156 − 84 = 72

Student C: Average = 54, S = 60 ⇒ P = 2 × 54 − 60 = 108 − 60 = 48

Thus,

 

Calculations:

Total marks (B) = 84 + 72 + 72 = 228 → % = (228 / 300) × 100 = 76%

Total marks (A) = 72 + 54 + 60 = 186 → % = 62%

Total marks (C) = 60 + 48 + 48 = 156 → % = 52%

Average % (A & C) = (62 + 52) / 2 = 57%

Difference = 76 − 57 = 19%

Hence, the correct answer is 19%.

General Solution:

Given Data:

 

Find History marks (H) using ratio S : H = a : b ⇒ H = S × (b ÷ a):

Student A: S = 72, ratio 4 : 3 ⇒ H = 72 × (3 ÷ 4) = 72 × 0.75 = 54

Student B: S = 84, ratio 7 : 6 ⇒ H = 84 × (6 ÷ 7) = 84 × (6/7) = 72

Student C: S = 60, ratio 5 : 4 ⇒ H = 60 × (4 ÷ 5) = 60 × 0.8 = 48

Find Polity marks (P) from average (P + S) ÷ 2 = given ⇒ P = 2 × (given average) − S:

Student A: Average = 66, S = 72 ⇒ P = 2 × 66 − 72 = 132 − 72 = 60

Student B: Average = 78, S = 84 ⇒ P = 2 × 78 − 84 = 156 − 84 = 72

Student C: Average = 54, S = 60 ⇒ P = 2 × 54 − 60 = 108 − 60 = 48

Thus,

 

Calculations:

History total = 54 + 72 + 48 = 174

Polity total = 60 + 72 + 48 = 180

Ratio = 174 : 180 = 29 : 30

Hence, the correct answer is 29:30.

General Solution:

Given Data:

 

Find History marks (H) using ratio S : H = a : b ⇒ H = S × (b ÷ a):

Student A: S = 72, ratio 4 : 3 ⇒ H = 72 × (3 ÷ 4) = 72 × 0.75 = 54

Student B: S = 84, ratio 7 : 6 ⇒ H = 84 × (6 ÷ 7) = 84 × (6/7) = 72

Student C: S = 60, ratio 5 : 4 ⇒ H = 60 × (4 ÷ 5) = 60 × 0.8 = 48

Find Polity marks (P) from average (P + S) ÷ 2 = given ⇒ P = 2 × (given average) − S:

Student A: Average = 66, S = 72 ⇒ P = 2 × 66 − 72 = 132 − 72 = 60

Student B: Average = 78, S = 84 ⇒ P = 2 × 78 − 84 = 156 − 84 = 72

Student C: Average = 54, S = 60 ⇒ P = 2 × 54 − 60 = 108 − 60 = 48

Thus,

 

Calculations:

Given Changes:

Student A: Polity marks ↑ 20%

Student B: Sociology marks ↓ 25%

Student A:

Original P = 60 ⇒ New P = 60 + (20% of 60) = 60 + 12 = 72

S (unchanged) = 72

H (unchanged) = 54

Student B:

Original S = 84 ⇒ New S = 84 − (25% of 84) = 84 − 21 = 63

P (unchanged) = 72

H (unchanged) = 72

Find new totals:

Student A total = S + H + P = 72 + 54 + 72 = 198

Student B total = S + H + P = 63 + 72 + 72 = 207

Combined total:

198 + 207 = 405

Final Answer:

 

Hence, the correct answer is 405.

General Solution:

Given Data:

 

Find History marks (H) using ratio S : H = a : b ⇒ H = S × (b ÷ a):

Student A: S = 72, ratio 4 : 3 ⇒ H = 72 × (3 ÷ 4) = 72 × 0.75 = 54

Student B: S = 84, ratio 7 : 6 ⇒ H = 84 × (6 ÷ 7) = 84 × (6/7) = 72

Student C: S = 60, ratio 5 : 4 ⇒ H = 60 × (4 ÷ 5) = 60 × 0.8 = 48

Find Polity marks (P) from average (P + S) ÷ 2 = given ⇒ P = 2 × (given average) − S:

Student A: Average = 66, S = 72 ⇒ P = 2 × 66 − 72 = 132 − 72 = 60

Student B: Average = 78, S = 84 ⇒ P = 2 × 78 − 84 = 156 − 84 = 72

Student C: Average = 54, S = 60 ⇒ P = 2 × 54 − 60 = 108 − 60 = 48

Thus,

 

Calculations:

Given (from previous results for Student C):

Sociology (C) = 60

History (C) = 48

Polity (C) = 48

Find D's marks (25% more than C):

Multiplier for 25% more = 1.25

Sociology (D) = 60 × 1.25 = 75

History (D) = 48 × 1.25 = 60

Polity (D) = 48 × 1.25 = 60

Economics marks for D (135% of History):

Economics (D) = 150% of 60 = 1.5 × 60 = 90

Overall average across four subjects:

Total = 75 + 60 + 60 + 90 = 285

Average = 285 ÷ 4 = 71.25

Hence, the correct answer is 71.25.

General Solution:

Given Data:

 

Find History marks (H) using ratio S : H = a : b ⇒ H = S × (b ÷ a):

Student A: S = 72, ratio 4 : 3 ⇒ H = 72 × (3 ÷ 4) = 72 × 0.75 = 54

Student B: S = 84, ratio 7 : 6 ⇒ H = 84 × (6 ÷ 7) = 84 × (6/7) = 72

Student C: S = 60, ratio 5 : 4 ⇒ H = 60 × (4 ÷ 5) = 60 × 0.8 = 48

Find Polity marks (P) from average (P + S) ÷ 2 = given ⇒ P = 2 × (given average) − S:

Student A: Average = 66, S = 72 ⇒ P = 2 × 66 − 72 = 132 − 72 = 60

Student B: Average = 78, S = 84 ⇒ P = 2 × 78 − 84 = 156 − 84 = 72

Student C: Average = 54, S = 60 ⇒ P = 2 × 54 − 60 = 108 − 60 = 48

Thus,

 

Calculations:

Given:

Sociology : History ratio for student D = same as student B = 7 : 6

Polity : Sociology ratio for student D = same as student C = 48 : 60 = 4 : 5

Assume Sociology marks for student D = 35x

From S : H = 7 : 6 ⇒ History = 30x

From P : S = 4 : 5 ⇒ Polity = (4/5) × Sociology = (4/5) × 35x = 28x

Find required ratio H : P

H : P = 30x : 28x = 30 : 28 = 15 : 14

Hence, the correct answer is 15:14.

General Solution:

Given Data:

 

Find History marks (H) using ratio S : H = a : b ⇒ H = S × (b ÷ a):

Student A: S = 72, ratio 4 : 3 ⇒ H = 72 × (3 ÷ 4) = 72 × 0.75 = 54

Student B: S = 84, ratio 7 : 6 ⇒ H = 84 × (6 ÷ 7) = 84 × (6/7) = 72

Student C: S = 60, ratio 5 : 4 ⇒ H = 60 × (4 ÷ 5) = 60 × 0.8 = 48

Find Polity marks (P) from average (P + S) ÷ 2 = given ⇒ P = 2 × (given average) − S:

Student A: Average = 66, S = 72 ⇒ P = 2 × 66 − 72 = 132 − 72 = 60

Student B: Average = 78, S = 84 ⇒ P = 2 × 78 − 84 = 156 − 84 = 72

Student C: Average = 54, S = 60 ⇒ P = 2 × 54 − 60 = 108 − 60 = 48

Thus,

 

Calculations:

From previous results:

History (A) = 54

Polity: A = 60, B = 72, C = 48

Total Polity marks = 60 + 72 + 48 = 180

Percentage = (History(A) ÷ Total Polity) × 100 = (54 ÷ 180) × 100 = 0.3 × 100 = 30%

Hence, the correct answer is 30%.

Shortcut Trick

Ratio of (A/2 + B) : B = 5 : 4 → 4(A/2 + B) = 5B → 2A + 4B = 5B → 2A = B.

C after 15 years is 125% of present C → 15 years = 25% of C → C = 60 years.

Total age (A+B+C) = 40 × 3 = 120 years.

A + B = 120 − 60 = 60 years. Substitute A = B/2 → B/2 + B = 60 → 3B/2 = 60 → B = 40 years.

∴ The correct answer is 40 years.

Alternate Method

Given: Ratio ((A/2) + B) / B = 5/4, Average(A,B,C) = 40, C + 15 = 1.25 × C.

Step 1: Calculate present age of C

⇒ C + 15 = 1.25C

⇒ 0.25C = 15

⇒ C = 15 / 0.25 = 60 years.

Step 2: Calculate Sum of A and B

⇒ A + B + C = 40 × 3 = 120

⇒ A + B = 120 − 60 = 60 years.

Step 3: Relate A and B using the ratio

⇒ (A/2 + B) / B = 5/4

⇒ 4(A/2 + B) = 5B

⇒ 2A + 4B = 5B

⇒ B = 2A or A = B/2.

Step 4: Solve for B

⇒ B/2 + B = 60

⇒ 3B/2 = 60

⇒ B = (60 × 2) / 3 = 40 years.

∴ The correct answer is 40 years.

Additional Information

Average Age Concept

The sum of ages is always equal to the average multiplied by the number of individuals (Sum = Avg × n).

Percentage and Age

If an age increases by x% over 't' years, then (x/100) × Present Age = 't'. Here 25% of C = 15.

Ratio Proportions

Cross-multiplication is the most efficient way to simplify linear ratios involving two variables.

Shortcut Trick

Effective Rate of Interest for 20% CI (2 years) = 20 + 20 + (20 × 20)/100 = 44%.

Principal = 7776 ÷ 1.44 = ₹5400.

CI (10% for 2 years) = 21% of 5400 = 1134.

SI (7% for 5 years on 3P) = 35% of (3 × 5400) = 5670.

Total Interest = 1134 + 5670 = ₹6804.

∴ The correct answer is ₹6804.

Alternate Method

Given: Amount₁ = ₹7776, R₁ = 20%, T₁ = 2 years (CI); R₂ = 10%, T₂ = 2 years (CI); R₃ = 7%, T₃ = 5 years (SI on 3P).

Formula: Amount = P(1 + r/100)ⁿ | SI = (P × R × T)/100

⇒ 7776 = P(1 + 20/100)²

⇒ 7776 = P(6/5)² ⇒ 7776 = P × 36/25

⇒ P = (7776 × 25) ÷ 36 = 216 × 25 = ₹5400.

⇒ Interest₁ (CI at 10% for 2 yrs) = 5400 × [(1.1)² − 1] = 5400 × 0.21 = ₹1134.

⇒ Interest₂ (SI at 7% for 5 yrs on 3P) = (3 × 5400 × 7 × 5) ÷ 100 = 16200 × 0.35 = ₹5670.

⇒ Total Interest = 1134 + 5670 = ₹6804.

∴ The correct answer is ₹6804.

Additional Information

Effective Rate of Compound Interest

For 2 years, effective rate = x + y + (xy/100). For 10%, it is 10 + 10 + 1 = 21%.

Simple Interest Calculation

SI is always calculated on the Principal amount. SI = (P × R × T) ÷ 100.

Compound Interest vs Simple Interest

Difference for 2 years = P(R/100)². CI is always greater than or equal to SI for the same period and rate.

Shortcut Trick

Using Pythagoras triplet logic for Diagonal (x + 49), Length (x + 41), and Breadth (x):

(x + 49)² − (x + 41)² = x² ⇒ (2x + 90)(8) = x² ⇒ x² − 16x − 720 = 0.

Solving the quadratic, factors of 720 with difference 16 are 36 and 20. ∴ Breadth (x) = 36 cm.

Length = 36 + 41 = 77 cm. Square side = 36 − 6 = 30 cm.

Difference = (77 × 36) − (30 × 30) = 2772 − 900 = 1872 cm².

∴ The correct answer is 1872 cm².

Alternate Method

Given: Length (L) = x + 41, Breadth (B) = x, Diagonal (D) = x + 49. Square side (a) = x − 6.

Formula Used: L² + B² = D²; Area of Rectangle = L × B; Area of Square = a².

⇒ (x + 41)² + x² = (x + 49)²

⇒ x² + 82x + 1681 + x² = x² + 98x + 2401

⇒ x² − 16x − 720 = 0

⇒ (x − 36)(x + 20) = 0 ⇒ x = 36 cm (as length cannot be negative)

⇒ Area of Rectangle = (36 + 41) × 36 = 77 × 36 = 2772 cm²

⇒ Area of Square = (36 − 6)² = 30² = 900 cm²

⇒ Difference = 2772 − 900 = 1872 cm²

∴ The correct answer is 1872 cm².

Additional Information

Pythagoras Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: a² + b² = c².

Rectangle Properties

The diagonal of a rectangle of length 'L' and breadth 'B' is given by √(L² + B²). The area is L × B.

Quadratic Roots

For x² + bx + c = 0, identify two numbers that multiply to 'c' and add up to 'b' for quick factorisation.

Shortcut Trick

Let Income of P be I and Expenditure be E. Original Saving = I − E.

New Income = 2I. New Saving = 2I − E.

Increase in Saving = (2I − E) − (I − E) = I. Given I = ₹40000.

Income of Q = 115% of ₹40000 = ₹46000.

Saving of Q = Income − Expenditure = 46000 − 20000 = ₹26000.

∴ The correct answer is ₹26000.

Alternate Method

Given: Original Income of P = I, Expenditure = E.

Formula Used: Saving = Income − Expenditure

⇒ Original Saving = I − E

⇒ New Income = 2I

⇒ New Saving = 2I − E

⇒ (2I − E) − (I − E) = 40000

⇒ I = 40000

⇒ Income of Q = 1.15 × 40000 = 46000

⇒ Expenditure of Q = 20000

⇒ Saving of Q = 46000 − 20000 = 26000

∴ The correct answer is ₹26000.

Additional Information

Income-Expenditure Relation

The fundamental relation is Income = Expenditure + Savings. If expenditure is constant, any change in income reflects directly in savings.

Percentage Change

To calculate a value increased by x%, use Original Value × (1 + x/100).

Shortcut Trick

From Statement I: Amount = P + (P × R × T)/100 ⇒ 3120 = P(1 + 30/100) ⇒ Principal (P) = ₹2400.

From Statement II: CI − SI = PR²/100² ⇒ 24 = P × (R/100)². (Provides relation between P and R).

Using P from I in II: 24 = 2400 × (R/100)² ⇒ (R/100)² = 1/100 ⇒ Rate (R) = 10%.

Both P and R are needed to find Rajesh's CI amount: A = 2400 × (1.1)² = ₹2904.

∴ The correct answer is ₹2904.

Alternate Method

Given: Time (T) = 2 years, Rahul's Amount = ₹3120, Time for Rahul = 3 years, Rahul's Rate = 10%, Difference CI−SI = ₹24.

Formula Used: Amount (SI) = P(1 + RT/100); Difference (2yr) = PR²/100²; Amount (CI) = P(1 + R/100)^T

⇒ From Statement I: 3120 = P × (1 + (10 × 3)/100) ⇒ 3120 = 1.3P ⇒ P = ₹2400.

⇒ From Statement II: Difference = P(R/100)² ⇒ 24 = 2400 × (R/100)².

⇒ (R/100)² = 24/2400 = 1/100 ⇒ R/100 = 1/10 ⇒ R = 10%.

⇒ Rajesh's Amount = P(1 + R/100)² = 2400(1 + 10/100)² = 2400 × 1.21.

⇒ Total Amount = ₹2904.

∴ The correct answer is ₹2904.

Additional Information

CI and SI Difference (2 Years)

The difference between Compound Interest and Simple Interest for 2 years is given by D = P(R/100)².

CI and SI Difference (3 Years)

The difference for 3 years is given by D = [P(R/100)²] × [(300 + R)/100].

Effective Rate for CI

For 2 years at R%, the effective CI rate is (2R + R²/100)%. For 10%, it is 21%.

Shortcut Trick

From Statement I: Increase (2018 − 2017) = Increase (2019 − 2017).

This implies Profit 2018 = Profit 2019.

Therefore, the percentage change from 2018 to 2019 is 0%.

Statement II provides 2017 and 2019 values but lacks 2018 data, making it insufficient.

∴ The correct answer is 0%.

Alternate Method

Given: Profits for years 2017, 2018, and 2019. Required to find % change from 2018 to 2019.

Evaluating Statement I:

⇒ Let Profits be P₂₀₁₇, P₂₀₁₈, and P₂₀₁₉.

⇒ Increase (2017 to 2018) = P₂₀₁₈ − P₂₀₁₇

⇒ Increase (2017 to 2019) = P₂₀₁₉ − P₂₀₁₇

⇒ P₂₀₁₈ − P₂₀₁₇ = P₂₀₁₉ − P₂₀₁₇

⇒ P₂₀₁₈ = P₂₀₁₉

⇒ Percentage change = [(P₂₀₁₉ − P₂₀₁₈) ÷ P₂₀₁₈] × 100 = 0%.

⇒ Statement I is sufficient.

Evaluating Statement II:

⇒ P₂₀₁₇ = ₹20,000 crore, P₂₀₁₉ = ₹21,000 crore.

⇒ P₂₀₁₈ is unknown. We cannot calculate the change from 2018 to 2019.

⇒ Statement II is not sufficient.

∴ The correct answer is 0%.

Additional Information

Percentage Change

Percentage change is calculated as [(New Value − Old Value) ÷ Old Value] × 100. If Old = New, % change is always 0.

Data Sufficiency Logic

In data sufficiency, if a statement allows you to find a unique numerical value or a definite relationship (like equality), it is considered sufficient.

Shortcut Trick

The distance just before meeting depends only on the Relative Speed and the Remaining Time.

Relative Speed (Opposite Direction) = 29 + 36 = 65 m/s.

Distance = Speed × Time ⇒ 65 m/s × (69 × 60) seconds.

Since only the speeds (Statement II) and time (69 min) are required, Statement II alone is sufficient.

∴ The correct answer is 269.1 km.

Alternate Method

Given:

Time (t) = 69 minutes

Speed₁ (S₁) = 29 m/s, Speed₂ (S₂) = 36 m/s

Formula Used:

Relative Speed (Opposite) = S₁ + S₂

Distance = Relative Speed × Time

Calculations:

⇒ Relative Speed = 29 + 36 = 65 m/s

⇒ Relative Speed in km/h = 65 × (18/5) = 13 × 18 = 234 km/h

⇒ Distance before 69 mins = Speed × Time = 234 × (69/60) km

⇒ Distance = 234 × 1.15 = 269.1 km

∴ The correct answer is 269.1 km.

Additional Information

Relative Speed

When two objects move in opposite directions, their Relative Speed is the sum of their individual speeds (S₁ + S₂).

Distance Before Meeting

The distance between two objects 't' time before they meet is always equal to (Relative Speed × t), regardless of the total initial distance.

Unit Conversion

To convert m/s to km/h, multiply by 18/5. To convert minutes to hours, divide by 60.

Shortcut Trick

Ratio of Remaining Solution to Total Mixture = (1 − x/V)ⁿ

Given Solution : Water = 81 : 19 ⇒ Solution : Total Mixture = 81 : (81 + 19) = 81 : 100

√(81/100) = 9/10 (since process is done twice)

1 unit (10 − 9) = 10 litres → Total Capacity = 10 × 10 = 100 litres

Water in resultant mixture = (19/100) × 100 = 19 litres.

∴ The correct answer is 19 litres.

Alternate Method

Given: Volume replaced (x) = 10 litres, Number of times (n) = 2, Final Ratio (Sol:Wat) = 81:19

Formula: Final Solution = Initial Solution × (1 − x/V)ⁿ

⇒ 81/100 = (1 − 10/V)²

⇒ √(81/100) = 1 − 10/V

⇒ 9/10 = 1 − 10/V

⇒ 10/V = 1 − 9/10 = 1/10

⇒ V = 100 litres (Total Capacity)

⇒ Water in Resultant Mixture = (19 / (81+19)) × 100

⇒ Water = (19/100) × 100 = 19 litres

∴ The correct answer is 19 litres.

Additional Information

Replacement Formula

Quantity of pure liquid left = Total × (1 − replaced quantity/Total)ⁿ, where n is the number of operations.

Alligation in Replacement

Replacement problems can be simplified by tracking the ratio of the remaining original component to the total volume.