Quantitative Aptitude PYQs

Formula used:

Herbivores = Total animals − (Carnivores + Omnivores)

Ratio of Herbivores : Omnivores = H : O

Omnivores = Herbivores × O-part ÷ H-part

Carnivores = (Carnivores + Omnivores) − Omnivores

Calculations:

Given:

H(A) = 70, H(B) = 76

O(D) = 60, O(E) = 77

Formula Used:

Ratio = (Sum₁ : Sum₂)

Calculation:

⇒ H(A + B) = 70 + 76 = 146

⇒ O(D + E) = 60 + 77 = 137

⇒ Ratio = 146 : 137

∴ Required ratio = 146 : 137

Shortcut Trick

Ratio of Marked Price (MP) to Cost Price (CP) is given by (100 + Profit%) ÷ (100 − Discount%).

MP ÷ CP = (100 + 20) ÷ (100 − 10) = 120 ÷ 90 = 4 ÷ 3.

Given CP = 3 units = &inr;495 ⇒ 1 unit = 495 ÷ 3 = &inr;165.

MP = 4 units = 4 × 165 = &inr;660.

Since MP = x + 200, then x + 200 = 660 ⇒ x = 460.

∴ The correct answer is 460.

Alternate Method

Given: CP = &inr;495, Profit = 20%, Discount = 10%, MP = &inr;(x + 200).
Formulas: SP = CP × (100 + Profit%) ÷ 100 | MP = SP × 100 ÷ (100 − Discount%).

⇒ Selling Price (SP) = 495 × (120 ÷ 100) = 495 × 1.2 = &inr;594.

⇒ Marked Price (MP) = 594 × (100 ÷ 90) = 594 × (10 ÷ 9) = 66 × 10 = &inr;660.

⇒ x + 200 = 660.

⇒ x = 660 − 200 = 460.

∴ The correct answer is 460.

Additional Information

Relation between CP and MP

The most efficient way to link Cost Price and Marked Price is the ratio formula: CP ÷ MP = (100 − D%) ÷ (100 + P%).

Markup Percentage

Markup% = (MP − CP) ÷ CP × 100. It represents how much more the shopkeeper marks the price above the cost price.

Shortcut Trick

Word SUCCESS has 7 letters: S, U, C, C, E, S, S. Letters other than S are U, C, C, E (4 letters).

Arrange U, C, C, E first: 4! ÷ 2! = 12 ways. These create 5 gaps for S: _ U _ C _ C _ E _.

Choose 3 gaps out of 5 for three S's: ⁵C₃ = 10 ways.

Total ways = 12 × 10 = 120 ways.

∴ The correct answer is 120 ways.

Alternate Method

Given: Word = SUCCESS. Total letters = 7 (S: 3, U: 1, C: 2, E: 1).

Formula Used: Permutations with repetition = n! ÷ (p₁! × p₂! ...)

⇒ Step 1: Arrange letters U, C, C, E. Total = 4, Repeated C = 2 times.

⇒ Ways = 4! ÷ 2! = (4 × 3 × 2 × 1) ÷ 2 = 12.

⇒ Step 2: There are 4 letters, so there are (4 + 1) = 5 possible gaps to place S.

⇒ Step 3: Choose 3 gaps out of 5 for the three S's = ⁵C₃.

⇒ ⁵C₃ = (5 × 4 × 3) ÷ (3 × 2 × 1) = 10.

⇒ Total ways = 12 × 10 = 120.

∴ The correct answer is 120 ways.

Additional Information

Gap Method

Used when certain items must not be together. Arrange other items first (n), then place restricted items in (n+1) gaps.

String Method (Bundling)

Used when certain items must ALWAYS be together. Treat the group as a single unit and arrange with other letters.

Permutation with Repetition

Total arrangements of n objects where p, q, r are identical = n! ÷ (p! × q! × r!).

Calculation:

19.98% of √(249.98 + ?) = 76.09 ÷ 18.78

⇒ 20% √(250 + ?) = 76 ÷ 19

⇒ (1/5)√(250 + ?) = 4

⇒ √(250 + ?) = 20

⇒ 250 + ? = 400

⇒ ? = 150

Approximation value is 150

Calculation:

136.10 ÷ 7.908 + 522.121 ÷ 17.91 − √1090 = ?

⇒ 136 ÷ 8 + 522 ÷ 18 − √1089 = ?

⇒ 17 + 29 − 33 = 13

Approximation value is 13

Calculation:

√1850 + ? = (35.901 ÷ 2.95) × 88.03

⇒ √1849 + ? = (36 ÷ 3) × 88

⇒ 43 + ? = 12 × 88

⇒ 43 + ? = 1056

⇒ ? = 1013

Approximation value is 1013

Calculation:

⇒ (482.56 × 19.87) ÷ 9.92 − 125.44 = ?

⇒ (480 × 20) ÷ 10 − 125

⇒ 960 − 125

⇒ 835

Approximation value is 835

Calculation:

54.867 − 35.975 − 11.1491 − 1.743 = ?

⇒ 55 − 36 − 11 − 2

⇒ 6

Approximation value is 6

Shortcut Trick

Ratio of Profit Share A : B = 560 : (840 − 560) = 560 : 280 = 2 :

1. Ratio of Time A : B = 12 : (12 − 3) = 12 : 9 = 4 : 3.

Ratio of Investment = (Profit ÷ Time) ⇒ A : B = (2 ÷ 4) : (1 ÷ 3) = 1/2 : 1/3 = 3 : 2.

Since 3 units = ₹15000, then 1 unit = ₹5000.

Investment of B (x) = 2 units = 2 × 5000 = ₹10000.

∴ The correct answer is ₹10000.

Alternate Method

Given: A's Investment = ₹15000, A's Time = 12 months, B's Time = 9 months, Total Profit = ₹840, A's Profit = ₹560.

Formulas: Profit Ratio = (Investment₁ × Time₁) : (Investment₂ × Time₂)

⇒ B's Profit Share = Total Profit − A's Profit Share = 840 − 560 = ₹280.

⇒ Ratio of Profits (A : B) = 560 ÷ 280 = 2 ÷ 1.

⇒ (A's Investment × A's Time) ÷ (B's Investment × B's Time) = 2 ÷ 1.

⇒ (15000 × 12) ÷ (x × 9) = 2 ÷ 1.

⇒ 180000 ÷ 9x = 2 ÷ 1.

⇒ 20000 ÷ x = 2 ⇒ 2x = 20000.

⇒ x = 10000.

∴ The correct answer is ₹10000.

Additional Information

Partnership Basics

When investments are for different time periods, profit is distributed in the ratio of the product of capital and time.

Investment Calculation

Capital = Profit ÷ Time. This is useful when profit ratios and time durations are known but investment is missing.

Shortcut Trick

Ratio of Father and Son (Present) = 4 : 1 (Difference = 3)

Ratio of Father and Son (5 years ago) = 5 : 1 (Difference = 4)

Equating differences: Present (16 : 4) and 5 years ago (15 : 3)

Change in units = 1 unit → 5 years. Father's present age = 16 × 5 = 80 years.

Mother's age = 80 − 2 = 78 years.

∴ The correct answer is 78 years.

Alternate Method

Given: Present Age of Father (F) = 4 × Son (S)

5 years ago: (F − 5) = 5 × (S − 5)

⇒ (4S − 5) = 5S − 25

⇒ 5S − 4S = 25 − 5

⇒ S = 20 years

⇒ Father's present age = 4 × 20 = 80 years

⇒ Mother's age = Father − 2 = 80 − 2 = 78 years

∴ The correct answer is 78 years.

Additional Information

Ratio Difference Method

When the time gap is same for both people, the difference between their ages in the ratio must be made equal by multiplying with the cross-difference.

Linear Equation in Ages

Age 'n' years ago is (x − n) and age 'n' years hence is (x + n), where x is the present age.

Age Difference Property

The numerical difference between the ages of two persons remains constant throughout their lives.

Shortcut Trick

Ratio of Speed of Boat (u) : Speed of Stream (v) = 5 : 2.

Given u = 25 km/hr ⇒ 5 units = 25, so 1 unit = 5. Thus, v = 10 km/hr.

Downstream Speed = 25 + 10 = 35 km/hr; Upstream Speed = 25 − 10 = 15 km/hr.

Time Upstream = 90 ÷ 15 = 6 hours. Total time = 11 hours ⇒ Time Downstream = 11 − 6 = 5 hours.

Downstream Distance = 35 × 5 = 175 km. Therefore, x + 75 = 175 ⇒ x = 100.

∴ The correct answer is 100.

Alternate Method

Given: Ratio u:v = 5:2. Still water speed (u) = 25 km/hr. Upstream dist = 90 km. Total time = 11 hrs.

Formula: Downstream speed (D) = u + v | Upstream speed (U) = u − v | Time = Distance ÷ Speed.

⇒ 5 units = 25 ⇒ 1 unit = 5. So, Stream speed (v) = 2 × 5 = 10 km/hr.

⇒ Downstream speed (D) = 25 + 10 = 35 km/hr.

⇒ Upstream speed (U) = 25 − 10 = 15 km/hr.

⇒ Total Time = (Downstream Dist / D) + (Upstream Dist / U)

⇒ 11 = [(x + 75) / 35] + [90 / 15]

⇒ 11 = [(x + 75) / 35] + 6

⇒ 5 = (x + 75) / 35

⇒ 175 = x + 75

⇒ x = 100.

∴ The correct answer is 100.

Additional Information

Relative Speed in Water

Downstream speed is the sum (u + v) and Upstream speed is the difference (u − v).

Speed of Boat and Stream from D & U

Still water speed u = (D + U) ÷ 2 and Stream speed v = (D − U) ÷ 2.

Average Speed for Round Trip

If distance is same, Average Speed = (Downstream × Upstream) ÷ Still water speed.

Shortcut Trick

Efficiency of Pipe B is given as 12 litres/minute.

Time taken by Pipe B to empty the full tank alone = 45 minutes.

Total Capacity = Rate of flow × Time taken.

Total Capacity = 12 litres/min × 45 minutes = 540 litres.

∴ The correct answer is 540 litres.

Alternate Method

Given: Filling time of Pipe A = 30 min, Emptying time of Pipe B = 45 min, Emptying rate of Pipe B = 12 litres/min.

⇒ Net efficiency of (A + B) = 3 − 2 = +1 unit/min.

⇒ Combined time to fill = 90 ÷ 1 = 90 minutes (matches given data).

⇒ Actual rate of Pipe B = 12 litres/min.

⇒ Capacity = Pipe B rate × Pipe B time.

⇒ Capacity = 12 × 45 = 540 litres.

∴ The correct answer is 540 litres.

Additional Information

Pipes and Cisterns Logic

Filling pipes have positive efficiency, while emptying pipes (leaks) have negative efficiency.

Capacity Calculation

Total Capacity = Rate of flow (litres/hr or min) × Total time required to fill or empty the tank.

Net Work Done

If 1/x is the filling rate and 1/y is the emptying rate, then net rate = (1/x − 1/y).

Given:

Primary = (Secondary + Tertiary) / 2

 

Formula used:

Primary Sector = (Secondary + Tertiary) ÷ 2

Tertiary Sector = 2 × Primary − Secondary

Calculations:

For A:

⇒ Tertiary = 2 × 72 − 92

⇒ 144 − 92 = 52

For B:

⇒ Tertiary = 2 × 74 − 78

⇒ 148 − 78 = 70

For C:

⇒ Tertiary = 2 × 76 − 72

⇒ 152 − 72 = 80

For D:

⇒ Tertiary = 2 × 84 − 96

⇒ 168 − 96 = 72

For E:

⇒ Tertiary = 2 × 88 − 64

⇒ 176 − 64 = 112

 

Calculations:

⇒ Primary employees in F = (72 + 74 + 76) ÷ 3

⇒ = 222 ÷ 3 = 74

Total employees in D = 84 + 96 + 72 = 252

Total employees in E = 88 + 64 + 112 = 264

⇒ Total employees in D and E together = 252 + 264 = 516

⇒ 516 : Total employees in F = 6 : 5

⇒ Total employees in F = 516 × 5 ÷ 6

⇒ 430

Let Secondary and Tertiary employees in F be 3x and x

⇒ 74 + 3x + x = 430

⇒ 74 + 4x = 430

⇒ 4x = 356

⇒ x = 89

⇒ Secondary employees in F = 3 × 89

⇒ 267

∴ Employees in secondary sector of State F = 267.

Shortcut Trick

Article 1: 25% gain ⇒ CP₁ : SP₁ = 4 : 5. Article 2: 20% loss ⇒ CP₂ : SP₂ = 5 : 4.

Since Selling Price (SP) is same, equate SP to LCM(5, 4) = 20.

Ratio 1 × 4 ⇒ CP₁ : SP = 16 : 20. Ratio 2 × 5 ⇒ CP₂ : SP = 25 : 20.

Total CP = 16 + 25 = 41 units. Total SP = 20 + 20 = 40 units.

Net Loss = 1 unit. Given gain is ₹60 (likely a typo in source, net change is 60). If 1 unit = 60.

∴ The correct answer is 960 and 1500.

Alternate Method

Formulas: SP = CP × (100 ± P/L)% | Profit = Total SP − Total CP.

⇒ Let SP of each article be S. ⇒ CP₁ = S ÷ 1.25 = 0.8S. CP₂ = S ÷ 0.80 = 1.25S. ⇒ Total CP = 0.8S + 1.25S = 2.05S. Total SP = 2S. ⇒ Net change = Total SP − Total CP = 2S − 2.05S = −0.05S. ⇒ Given magnitude of change is ₹60 ⇒ 0.05S = 60 ⇒ S = 60 ÷ 0.05 = 1200. ⇒ CP₁ = 0.8 × 1200 = 960. ⇒ CP₂ = 1.25 × 1200 = 1500. ∴ The correct answer is 960 and 1500.

Additional Information

Case of Same Selling Price

When two items are sold at the same price, one at x% gain and another at y% loss, the overall gain or loss is determined by calculating individual cost prices based on the common SP.

Note on Mathematical Discrepancy

In this problem, the calculation shows a net loss of ₹60 (4000 − 4100 units), but the question mentions a gain. Following the magnitude of 60 leads to the values in Option 3.

Shortcut Trick

Total Present Students = 80% of Boys + 70% of Girls = 237

0.8 × (3x + 20) + 0.7 × (2x − 10) = 237

2.4x + 16 + 1.4x − 7 = 237

3.8x + 9 = 237 ⇒ 3.8x = 228 ⇒ x = 228 ÷ 3.8 = 60

∴ The correct answer is 60.

Alternate Method

Given:

Number of Boys = 3x + 20

Number of Girls = 2x − 10

Formula Used:

Present students = Total students × (100% − Absent %)

Calculations:

⇒ Present Boys = (3x + 20) × (100% − 20%) = (3x + 20) × 0.8

⇒ Present Girls = (2x − 10) × (100% − 30%) = (2x − 10) × 0.7

⇒ Total Present = 0.8(3x + 20) + 0.7(2x − 10) = 237

⇒ 2.4x + 16 + 1.4x − 7 = 237

⇒ 3.8x + 9 = 237

⇒ 3.8x = 228

⇒ x = 228 ÷ 3.8 = 60

∴ The correct answer is 60.

Additional Information

Percentage Base Change

If a value decreases by p%, the remaining value is (100 − p)% of the original base.

Linear Equations

To find the value of an unknown variable x, group all terms with x on one side and constant terms on the other side of the equation.