Quantitative Aptitude PYQs

Common solution:

Total students in A and C = 400 – 160 = 240

Boy/Girls in B = 160/2 = 80

Let boys in A be x.

Or, x + x + 40 + 40 = 240

Or, 2x = 160

So, x = 80

 

In A,

Not appeared boys = 25% of 80 = 20

Appeared boys = 80 – 20 = 60

Not appeared girls = 20 – 5 = 15

Appeared girls = 40 – 15 = 25

In B,

Not appeared boys = 15

Appeared boys = 80 – 15 = 65

Not appeared girls = 81 – (20 + 15 + 15 + 12 + 12) = 7

Appeared girls = 80 – 7 = 73

In C,

Not appeared: Appeared = 1 : 4

Total students = 120

Not appeared students = 120 × (1/5) = 24

Appeared students = 120 – 24 = 96

Not appeared girls = 12

Appeared girls = 80 – 12 = 68

Not appeared boys = 24 – 12 = 12

Appeared boys = 40 – 12 = 28

Calculation:

Students not appeared from C = 12 + 12 = 24

Ratio (D: C not appeared) = 5: 3

Total students in D = 24 x (5/3) = 40

Girls in D = 20% less than boys not appeared from A = 20% less than 20 = 16

Therefore, Boys in D = 40 – 16 = 24

Common solution:

Total students in A and C = 400 – 160 = 240

Boy/Girls in B = 160/2 = 80

Let boys in A be x.

Or, x + x + 40 + 40 = 240

Or, 2x = 160

So, x = 80

 

In A,

Not appeared boys = 25% of 80 = 20

Appeared boys = 80 – 20 = 60

Not appeared girls = 20 – 5 = 15

Appeared girls = 40 – 15 = 25

In B,

Not appeared boys = 15

Appeared boys = 80 – 15 = 65

Not appeared girls = 81 – (20 + 15 + 15 + 12 + 12) = 7

Appeared girls = 80 – 7 = 73

In C,

Not appeared: Appeared = 1 : 4

Total students = 120

Not appeared students = 120 × (1/5) = 24

Appeared students = 120 – 24 = 96

Not appeared girls = 12

Appeared girls = 80 – 12 = 68

Not appeared boys = 24 – 12 = 12

Appeared boys = 40 – 12 = 28

Calculation:

Total boys in A and B together = 80 + 80 = 160

Total girls in C = 80

Required percentage = (160/80) × 100 = 200%

Common solution:

Total students in A and C = 400 – 160 = 240

Boy/Girls in B = 160/2 = 80

Let boys in A be x.

Or, x + x + 40 + 40 = 240

Or, 2x = 160

So, x = 80

 

In A,

Not appeared boys = 25% of 80 = 20

Appeared boys = 80 – 20 = 60

Not appeared girls = 20 – 5 = 15

Appeared girls = 40 – 15 = 25

In B,

Not appeared boys = 15

Appeared boys = 80 – 15 = 65

Not appeared girls = 81 – (20 + 15 + 15 + 12 + 12) = 7

Appeared girls = 80 – 7 = 73

In C,

Not appeared: Appeared = 1 : 4

Total students = 120

Not appeared students = 120 × (1/5) = 24

Appeared students = 120 – 24 = 96

Not appeared girls = 12

Appeared girls = 80 – 12 = 68

Not appeared boys = 24 – 12 = 12

Appeared boys = 40 – 12 = 28

Calculation:

Required difference = (60 + 25) – (15 + 7) = 85 – 22 = 63

Common solution:

Total students in A and C = 400 – 160 = 240

Boy/Girls in B = 160/2 = 80

Let boys in A be x.

Or, x + x + 40 + 40 = 240

Or, 2x = 160

So, x = 80

 

In A,

Not appeared boys = 25% of 80 = 20

Appeared boys = 80 – 20 = 60

Not appeared girls = 20 – 5 = 15

Appeared girls = 40 – 15 = 25

In B,

Not appeared boys = 15

Appeared boys = 80 – 15 = 65

Not appeared girls = 81 – (20 + 15 + 15 + 12 + 12) = 7

Appeared girls = 80 – 7 = 73

In C,

Not appeared: Appeared = 1 : 4

Total students = 120

Not appeared students = 120 × (1/5) = 24

Appeared students = 120 – 24 = 96

Not appeared girls = 12

Appeared girls = 80 – 12 = 68

Not appeared boys = 24 – 12 = 12

Appeared boys = 40 – 12 = 28

Calculation:

Required ratio = (20 + 15): 28 = 35: 28 = 5: 4

Shortcut Trick

Area = 4x × 3x = 12x² = 588 ⇒ x² = 49 ⇒ x = 7 cm.

Perimeter = 2(4x + 3x) = 14x = 14 × 7 = 98 cm.

Circumference = 98 + 34 = 132 cm.

Using 2πr = 132 ⇒ 2 × (22/7) × r = 132 ⇒ r = 21 cm.

Area = (22/7) × 21 × 21 = 1386 cm².

∴ The correct answer is 1386 sq. cm.

Alternate Method

Given: Ratio L:B = 4:3, Area = 588 cm², Circumference (C) = Perimeter (P) + 34 cm.

Formula: Area = L × B; Perimeter = 2(L+B); Circumference = 2πr; Circle Area = πr².

⇒ Let Length = 4x and Breadth = 3x.

⇒ 4x × 3x = 588 ⇒ 12x² = 588 ⇒ x² = 49 ⇒ x = 7.

⇒ Length = 28 cm, Breadth = 21 cm.

⇒ Perimeter = 2(28 + 21) = 2 × 49 = 98 cm.

⇒ Circumference = 98 + 34 = 132 cm.

⇒ 2 × (22/7) × r = 132 ⇒ r = (132 × 7) / 44 = 21 cm.

⇒ Area of Circle = (22/7) × 21 × 21 = 1386 cm².

∴ The correct answer is 1386 sq. cm.

Additional Information

Circle Constants

For a circle with radius 7, the circumference is 44 and area is 154. For radius 21 (3 × 7), the area is 154 × 3² = 1386.

Rectangle Properties

Perimeter is the total length of the boundary: 2(l + b). Area is the product of dimensions: l × b.

Standard π values

Use π ≈ 22/7 for calculations involving multiples of 7 to simplify fractions quickly.

Shortcut Trick

By using the rule of Alligation:

Ratio of Boys to Girls = (66 − 60) : (72 − 66) = 6 : 6 = 1 :

1. Since the number of boys is 18, the number of girls (X) must also be 18.

∴ The correct answer is 18.

Alternate Method

Given:

Average marks of 18 boys = 72

Average marks of X girls = 60

Combined average marks of all students = 66

Formula Used:

Weighted Average = (N₁ × A₁ + N₂ × A₂) ÷ (N₁ + N₂)

Calculations:

⇒ 66 = (18 × 72 + X × 60) ÷ (18 + X)

⇒ 66 × (18 + X) = 1296 + 60X

⇒ 1188 + 66X = 1296 + 60X

⇒ 66X − 60X = 1296 − 1188

⇒ 6X = 108

⇒ X = 18

∴ The correct answer is 18.

Additional Information

Weighted Average

Used when two or more groups with different averages and quantities are combined into a single group.

Alligation Method

An easier alternative to find the ratio of quantities of ingredients or groups mixed together to produce a desired average mixture value.

Shortcut Trick

Effective Profit% = m − d − (m × d) ÷ 100, where m = Markup% and d = Discount%.

Initial Profit% = 40 − 10 − (40 × 10) ÷ 100 = 30 − 4 = 26%.

Given, 26% of Cost Price (CP) = Rs. 234 ⇒ 1% of CP = 234 ÷ 26 = Rs. 9.

New Profit% (with 25% discount) = 40 − 25 − (40 × 25) ÷ 100 = 15 − 10 = 5%.

New Profit = 5% of CP = 5 × 9 = Rs. 45.

∴ The correct answer is Rs. 45.

Alternate Method

Given: Markup = 40%, Discount₁ = 10%, Profit₁ = Rs. 234, Discount₂ = 25%.

Formulas: MP = CP × (100 + Markup%) ÷ 100 | SP = MP × (100 − Discount%) ÷ 100 | Profit = SP − CP.

⇒ Let Cost Price (CP) = 100 units.

⇒ Marked Price (MP) = 100 + 40% of 100 = 140 units.

⇒ Case 1 Selling Price (SP₁) = 140 − 10% of 140 = 140 − 14 = 126 units.

⇒ Profit₁ = 126 − 100 = 26 units. Given 26 units = Rs. 234 ⇒ 1 unit = Rs. 9. ⇒ Case 2 Selling Price (SP₂) = 140 − 25% of 140 = 140 − 35 = 105 units.

⇒ Profit₂ = 105 − 100 = 5 units.

⇒ Profit in Rupees = 5 × 9 = Rs. 45. ∴ The correct answer is Rs. 45.

Additional Information

Relation between CP and MP

CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%). This ratio is very useful when both discount and profit percentages are known.

Marked Price & Discount

Marked Price is also known as the list price. Discount is always calculated on the MP, while Profit or Loss is always calculated on the CP.

The pattern of the series:

84 – 4² = 68,

68 + 5³ = 193,

193 - 6² = 157,

157 + 7³ = 500,

500 - 8² = 436,

The pattern of the series:

18 + 4³ = 82,

82 + 5³ = 207,

207 + 6³ = 423,

423 + 7³ = 766,

766 + 8³ = 1278

The pattern of the series:

7 + 17² = 296,

296 + 19² = 657,

657 + 21² = 1098,

1098 + 23² = 1627,

1627 + 25² = 2252

Total complaints filed in A = 72

Ratio of complaints filed by Females: Males = 7: 2

Total parts = 7 + 2 = 9

Complaints filed by females = 72 x (7/9) = 56

Given, out of the female complaints, 10 are unaddressed.

Therefore, addressed complaints filed by females = 56 – 10 = 46

Total complaints filed in C = 84

Required difference = 84 – 46 = 38

Total complaints filed in A = 72

Out of these, Robbery complaints = 45

Therefore, Theft complaints = 72 − 45 = 27

Given, 20% of robbery complaints are unaddressed.

Unaddressed robbery complaints = 20% of 45 = 9

Addressed robbery complaints = 45 – 9 = 36

Required difference = 36 – 27 = 9

Total complaints in F = (72 + 48)/2 = 60

Given, Unaddressed in F: Addressed in C = 5: 9

Addressed complaints in C = 54

Unaddressed complaints in F = (5/9) x 54 = 30

Therefore, addressed complaints in F = 60 – 30 = 30

Unaddressed complaints in C and D together = 30 + 18 = 48

Addressed complaints in A and E together = 48 + 36 = 84

Required Ratio = 48:84 = 4:7

Total complaints filed in D and E = 48 + 96 = 144

Total addressed complaints in A and B = 48 + 63 = 111

Required Percentage = (144/111) x 100 = 129.73%

Shortcut Trick

Speed of P = 90 km/hr = 25 m/s. Length of P = (25 × 24) − 300 = 300 m.

Length of Q = 300 ÷ 1.5 = 200 m. Speed of Q = (4/5) × 90 = 72 km/hr = 20 m/s.

Time for Q = (Length of Q + Platform) ÷ Speed of Q = (200 + 520) ÷ 20 = 36 seconds.

∴ The correct answer is 36 seconds.

Alternate Method

Given: Speed of P = 90 km/hr; Bridge = 300 m; Time = 24 s; Ratio Speed P:Q = 5:4; Length P = 1.5 × Length Q.

Formula Used: Distance = Speed × Time; 1 km/hr = 5/18 m/s.

⇒ Speed of Train P in m/s = 90 × (5/18) = 25 m/s

⇒ Total Distance = (Length of P + 300) = 25 × 24 = 600 m

⇒ Length of Train P = 600 − 300 = 300 m

⇒ Length of Train Q = 300 ÷ 1.5 = 200 m

⇒ Speed of Train Q = (4/5) × 90 = 72 km/hr = 72 × (5/18) = 20 m/s

⇒ Time taken by Q = (Length of Q + Platform) ÷ Speed Q = (200 + 520) ÷ 20

⇒ Time = 720 ÷ 20 = 36 seconds

∴ The correct answer is 36 seconds.

Additional Information

Relative Speed

When two trains move in the same direction, relative speed = S₁ − S₂. In the opposite direction, relative speed = S₁ + S₂.

Crossing Objects

When a train crosses a platform, bridge, or another train, total distance covered is the sum of their lengths. If it crosses a pole or person, distance is just the train's length.

Unit Conversion

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Shortcut Trick

Let Total Work be LCM(18, 9, 6) = 18 units.

Efficiencies: A = 1, B = 2, C = 3.

Work equation: (A + C) × D + B × D = Total Work

(1 + 3) × D + 2 × D = 18 ⇒ 6D = 18 ⇒ D = 3.

Value of 3D = 3 × 3 = 9 days.

∴ The correct answer is 9 days.

Alternate Method

Given: A and C work for D days, and then B works for D days to finish the work.

Calculation:

⇒ Combined efficiency of A and C = 1 + 3 = 4 units/day

⇒ Work done by A and C in D days = 4 × D = 4D units

⇒ Efficiency of B = 2 units/day

⇒ Work done by B in D days = 2 × D = 2D units

⇒ Total work = 4D + 2D = 18 units

⇒ 6D = 18

⇒ D = 3 days

⇒ 3D = 3 × 3 = 9 days

∴ The correct answer is 9 days.

Additional Information

Combined Efficiency

When multiple people work together, their efficiencies (work per day) are added to find the total work done per unit time.

Work-Efficiency Relation

Total Work = Efficiency × Time. Using the LCM method simplifies fractional work into integer units for faster calculation.

Shortcut Trick

Initial ratio of Milk : Water = 7 : 3.

After removing 20 liters and adding 20 liters of Water, new ratio = 7 : 4.

Since only water is added, the Milk units remain constant (7 units).

Increase in water = 4 − 3 = 1 unit → 20 liters.

Total quantity after replacement = (7 + 4) × 20 = 220 liters.

Initial quantity of water = (3/10) × 220 = 66 liters.

∴ The correct answer is 66 liters.

Alternate Method

Given: Initial Ratio = 7 : 3, Replaced Quantity = 20 liters, Final Ratio = 7 : 4.

Formula: Quantity of Milk = Initial Total × (7/10). Replacement keeps total volume constant.

⇒ Let initial total quantity be Q liters. Initial Milk = 0.7Q, Initial Water = 0.3Q.⇒ Milk removed = 20 × (7/10) = 14 liters. Water removed = 20 × (3/10) = 6 liters.⇒ New Milk = 0.7Q − 14. New Water = 0.3Q − 6 + 20 = 0.3Q + 14.⇒ (0.7Q − 14) / (0.3Q + 14) = 7 / 4⇒ 4 × (0.7Q − 14) = 7 × (0.3Q + 14)⇒ 2.8Q − 56 = 2.1Q + 98⇒ 0.7Q = 154⇒ Q = 154 / 0.7 = 220 liters.⇒ Initial Water = 0.3 × 220 = 66 liters.

∴ The correct answer is 66 liters.

Additional Information

Mixture Replacement Rule

When a portion of mixture is replaced by one of its components, the component not added remains constant in the ratio comparison.

Initial Quantity Formula

Final Quantity = Initial Quantity × (1 − replaced volume / total volume)ⁿ for repeated replacements of a single component.

Alligation in Replacement

Replacement problems can also be solved by treating the added pure component as a 100% concentration mixture.

Shortcut Trick

The age difference between two people remains constant at any point in time.

Difference after 6 years = 6 years ⇒ Difference 3 years ago = 6 years.

Ratio 3 years ago (P : Q) = 7 : 9 ⇒ Ratio difference = 2 units.

2 units = 6 years ⇒ 1 unit = 3 years.

Q age 3 years ago = 9 × 3 = 27 years → Present age = 27 + 3 = 30 years.

∴ The correct answer is 30 years.

Alternate Method

Given: Ratio of P : Q (3 years ago) = 7 : 9. Condition after 6 years: Q = P + 6.

Method Used: Linear equations and age difference property.

⇒ Let ages 3 years ago be P = 7x and Q = 9x.

⇒ Present ages: P = 7x + 3, Q = 9x + 3.

⇒ Ages after 6 years: P = 7x + 3 + 6 = 7x + 9, Q = 9x + 3 + 6 = 9x + 9.

⇒ According to question: (9x + 9) − (7x + 9) = 6.

⇒ 2x = 6 ⇒ x = 3.

⇒ Present age of Q = 9x + 3 = 9(3) + 3 = 27 + 3 = 30.

∴ The correct answer is 30 years.

Additional Information

Constant Age Difference

The difference between the ages of two persons remains the same throughout their lives. If Q is 6 years older than P today, Q will always be 6 years older.

Ratio Balancing Method

When the age difference is known, the difference between the ratio parts can be directly equated to the real-time difference (e.g., Difference in ratio units = Real age difference).

Shortcut Trick

Let Total Income be 100 units.

Household = 30 units; Remaining = 70 units.

Shopping = 20% of 70 = 14 units; Remaining after shopping = 56 units.

Fixed Deposit (FD) = 3/5 × 56 = 33.6 units.

Difference (FD − Shopping) = 33.6 − 14 = 19.6 units → ₹1960. ∴ 1 unit = ₹100.

∴ The correct answer is ₹10000.

Alternate Method

Given: Household expenses = 30%, Shopping = 20% of remaining, MF : FD = 2 : 3.

⇒ Let Income = I

⇒ Remaining income after household = I − 0.30I = 0.70I

⇒ Amount spent on shopping = 20% of 0.70I = 0.14I

⇒ Remaining income after shopping = 0.70I − 0.14I = 0.56I

⇒ Amount in Fixed Deposits (FD) = 3 / (2 + 3) × 0.56I = 0.6 × 0.56I = 0.336I

⇒ Difference (FD − Shopping) = 0.336I − 0.14I = 0.196I

⇒ 0.196I = 1960

⇒ I = 1960 ÷ 0.196 = 10000

∴ The correct answer is ₹10000.

Additional Information

Percentage of Remaining

When spending from the remaining balance, multiply the current balance by (100 − x) / 100 to find the new balance.

Ratio Distribution

To divide a quantity Q in ratio a : b, the parts are [a / (a + b)] × Q and [b / (a + b)] × Q.

Base Change Rule

Always identify the base of the percentage. If it says 'of the remaining', the base is the balance after previous subtractions.