Quantitative Aptitude PYQs

Given that,

2, 9, 30, 93, 282, 846

⇒ 2 × 3 + 3 = 9

⇒ 9 × 3 + 3 = 30

⇒ 30 × 3 + 3 = 93

⇒ 93 × 3 + 3 = 282

⇒ 282 × 3 + 3 = 849

∴ 846 is the wrong number in the given pattern

8, 4, 4, 6, 12, 36

⇒ 8 × 1/2 = 4

⇒ 4 × 1 = 4

⇒ 4 × 3/2 = 6

⇒ 6 × 2 = 12

⇒ 12 × 5/2 = 30

∴ In the above series 36 is wrong number series

Shortcut Trick

Total CP = (150 × 220) + 7000 = 33000 + 7000 = 40,000.

Effective CP per unit = 40000 ÷ 150 = 800/3.

SP per unit = 400 × (100 − 5)% = 400 × 0.95 = 380.

Gain% = [(380 − 800/3) ÷ (800/3)] × 100.

Gain% = [(1140 − 800) ÷ 800] × 100 = 340 ÷ 8 = 42.5%.

∴ The correct answer is 42.5%.

Alternate Method

Given: Purchase Price = 220, Units = 150, Transportation = 7000, MP = 400, Discount = 5%.

Formulas: Total CP = Purchase Cost + Overhead Expenses | SP = MP × (100 − D%) ÷ 100 | Gain% = (Total SP − Total CP) ÷ Total CP × 100.

⇒ Total cost of 150 calculators = 150 × 220 = Rs. 33,000.

⇒ Total Cost Price (CP) including transportation = 33,000 + 7,000 = Rs. 40,000.

⇒ Total Marked Price (MP) for 150 calculators = 150 × 400 = Rs. 60,000.

⇒ Discount amount = 5% of 60,000 = (5 ÷ 100) × 60,000 = Rs. 3,000.

⇒ Total Selling Price (SP) = 60,000 − 3,000 = Rs. 57,000.

⇒ Profit = 57,000 − 40,000 = Rs. 17,000.

⇒ Gain Percentage = (17,000 ÷ 40,000) × 100 = 170 ÷ 4 = 42.5%.

∴ The correct answer is 42.5%.

Additional Information

Effective Cost Price

Always include overhead expenses like transportation, labor, or repairs into the original cost price to find the net CP before calculating profit.

Marked Price and Discount

Discount is always calculated on the Marked Price (MP). SP = MP − Discount.

Shortcut Trick

Ratio of Class A to Class B = 3 : 4. Let students be 300 and 400.

Vacation from A = 33.33% (1/3) = 100; Remaining = 200.

Vacation from B = 25% (1/4) = 100; Remaining = 300.

Total Remaining = 200 + 300 = 500 units → 200 students. Total Vacation = 100 + 100 = 200 units.

Vacation students = (200 ÷ 500) × 200 = 80.

∴ The correct answer is 80.

Alternate Method

Given: Ratio of students in Class A and Class B is 3 : 4.

Let students in A = 3x and B = 4x.

Remaining students in A:

⇒ 3x × (1 − 33.33%) = 3x × (1 − 1/3) = 3x × 2/3 = 2x

Remaining students in B:

⇒ 4x × (1 − 25%) = 4x × (3/4) = 3x

According to question:

⇒ 2x + 3x = 200

⇒ 5x = 200 ⇒ x = 40

Total vacation students:

⇒ (3x × 1/3) + (4x × 1/4) = x + x = 2x

⇒ 2 × 40 = 80

∴ The correct answer is 80.

Additional Information

Percentage to Fraction Conversion

Standard conversions: 33.33% = 1/3, 25% = 1/4, 20% = 1/5, 12.5% = 1/8. These simplify complex ratio calculations.

Base Change Property

If the remaining portion is given, always calculate the Multiplying Factor (1 − fraction spent) to find the absolute value directly.

Weighted Average in Ratios

The total percentage of change = (n₁p₁ + n₂p₂) ÷ (n₁ + n₂), where n is the ratio part and p is the percentage.

Shortcut Trick

Weighted Average = (Price₁ × Ratio₁ + Price₂ × Ratio₂) ÷ Total Ratio

Average Price = (10 × 2 + 15 × 3) ÷ (2 + 3)

Average Price = (20 + 45) ÷ 5 = 65 ÷ 5 = 13

∴ The correct answer is Rs. 13 per litre.

Alternate Method

Given:

Price of 1^st variety (P₁) = Rs. 10/litre

Price of 2^nd variety (P₂) = Rs. 15/litre

Ratio of mixing (n₁ : n₂) = 2 : 3

Formula Used:

Average Price = (Total Cost) ÷ (Total Quantity)

Calculations:

⇒ Let quantities be 2L and 3L respectively.

⇒ Cost of 1^st type = 10 × 2 = Rs. 20

⇒ Cost of 2^nd type = 15 × 3 = Rs. 45

⇒ Total Cost = 20 + 45 = Rs. 65

⇒ Total Quantity = 2 + 3 = 5 litres

⇒ Average Price = 65 ÷ 5 = Rs. 13 per litre

∴ The correct answer is Rs. 13 per litre.

Additional Information

Alligation Rule

It states that (Quantity of Cheaper) / (Quantity of Dearer) = (CP of Dearer − Mean Price) / (Mean Price − CP of Cheaper).

Weighted Average Formula

Average = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where n represents the weights/ratios and x represents the values/prices.

Shortcut Trick
Quantity A: Prime sums (2, 3, 5, 7, 11) occur (1, 2, 4, 6, 2) times = 15. P = 15/36 = 0.416.
Quantity B: CP of Mangoes = 11/1.1 = ₹10/kg; CP of Apples = 18/1.2 = ₹15/kg. Avg CP = 570/50 = ₹11.4/kg.
By Alligation: Ratio (Apples : Mangoes) = (11.4 − 10) : (15 − 11.4) = 1.4 : 3.6 = 7 : 18.
Quantity B = 7/18 ≈ 0.388. Comparing A (0.416) and B (0.388), A > B.
∴ The correct answer is Quantity A > Quantity B.
Alternate Method
Quantity A: Probability of prime sum
⇒ Total outcomes for two dice = 6 × 6 = 36.
⇒ Prime sums: 2, 3, 5, 7, 11.
⇒ Outcomes for Sum 2: (1,1) → 1 way.
⇒ Outcomes for Sum 3: (1,2), (2,1) → 2 ways.
⇒ Outcomes for Sum 5: (1,4), (2,3), (3,2), (4,1) → 4 ways.
⇒ Outcomes for Sum 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 ways.
⇒ Outcomes for Sum 11: (5,6), (6,5) → 2 ways.
⇒ Total prime outcomes = 1 + 2 + 4 + 6 + 2 = 15.
⇒ Quantity A = 15/36 = 5/12 ≈ 0.4167.

Quantity B: Ratio of weights
⇒ CP of Mangoes = SP / (1 + P%) = 11 / 1.1 = ₹10/kg.
⇒ CP of Apples = SP / (1 + P%) = 18 / 1.2 = ₹15/kg.
⇒ Overall Average CP = Total Cost / Total Quantity = 570 / 50 = ₹11.4/kg.
⇒ Using Alligation: Ratio (Apples : Mangoes) = (11.4 − 10) : (15 − 11.4) = 1.4 : 3.6.
⇒ Quantity B = 1.4 / 3.6 = 14/36 = 7/18 ≈ 0.3889.
⇒ Comparing Quantity A (0.4167) and Quantity B (0.3889), we find A > B.
∴ The correct answer is Quantity A > Quantity B.
Additional Information
Dice Sum Probabilities
For two dice, the number of ways to get a sum S is (S−1) for 2 ≤ S ≤ 7 and (13−S) for 8 ≤ S ≤ 12.
Alligation Rule
Quantity1 : Quantity2 = (Mean Price − CP2) : (CP1 − Mean Price).
Profit and Loss
Cost Price (CP) = Selling Price (SP) × 100 / (100 + Profit%).

Shortcut Trick
Quantity A: Round robin matches = n(n−1)⁄2 = (8 × 7) ÷ 2 = 28. Total = 28 + 1 (final) = 29.
Quantity B: Net change in average = 2 kg for 15 students ⇒ Total increase = 30 kg.
Difference = Shyam − Ram ⇒ 30 = 58 − Ram ⇒ Ram = 58 − 30 = 28 kg.
Comparing A (29) and B (28), we get A > B.
∴ The correct answer is Quantity A > Quantity B.
Alternate Method

Quantity A:
⇒ Matches in round robin = 8 × (8 − 1)⁄2 = 28 matches
⇒ Total matches = 28 (League) + 1 (Final) = 29 matches
Quantity B:
⇒ New average = 40 + 2 = 42 kg
⇒ Sum of weights (initial) = 15 × 40 = 600 kg
⇒ Sum of weights (new) = 15 × 42 = 630 kg
⇒ Ram + 30 = Shyam ⇒ Ram = 58 − 30 = 28 kg
∴ The correct answer is Quantity A > Quantity B.
Additional Information
Round Robin Tournament
In a single round robin with n teams, each team plays every other team once. Total matches = n(n − 1)⁄2.
Average Replacement Formula
Weight of New Person = Weight of Removed Person + (Number of members × Increase in Average).
Quantity Comparison
Always calculate the absolute numerical values for both quantities before comparing magnitudes (>, <, =).

Shortcut Trick
Quantity A: SP of 4 bananas at 75% = ₹18 ⇒ SP of 8 bananas at 75% = ₹36.
New SP for 15% profit = 36 × (115 ÷ 75) = 36 × (23 ÷ 15) = 12 × (23 ÷ 5) = ₹55.2.
Quantity B: Area = 122 − π × 52 = 144 − 25 × 3.14 = 144 − 78.5 = 65.5 cm2.
Comparing A (55.2) and B (65.5), we find A < B.
∴ The correct answer is Quantity A < Quantity B.
Alternate Method

Quantity A:
⇒ SP of 4 bananas = ₹18 (at 25% loss)
⇒ CP of 4 bananas = 18 ÷ 0.75 = ₹24
⇒ CP of 8 bananas = 24 × 2 = ₹48
⇒ Required SP (15% profit) = 48 × 1.15 = ₹55.2
Quantity B:
⇒ Area of Square = Side2 = 122 = 144 cm2
⇒ Area of 2 Semicircles = Area of 1 Circle = πr2
⇒ Area of Circle = 3.14 × 52 = 3.14 × 25 = 78.5 cm2
⇒ Remaining Area = 144 − 78.5 = 65.5 cm2
⇒ 55.2 < 65.5 ∴ Quantity A < Quantity B
∴ The correct answer is Quantity A < Quantity B.
Additional Information
Profit and Loss Relation
SP = CP × (100 + Profit%) / 100 and CP = SP × 100 / (100 − Loss%).
Mensuration of Circles
Area of a circle is πr2. Two semicircles of the same radius always form one complete circle.
Comparison of Quantities
In data sufficiency or quantity comparison, always calculate the numerical values to the same precision before concluding the relationship.

Shortcut Trick
For Quantity A: Total SI % = 10% + 12% + 13% = 35%.
For Quantity B: Effective CI % for 11% for 3 years is approximately (1.11)3 − 1 ≈ 36.76%.
Since Principal is same (₹12000), compare interest percentages: 35% < 36.76%.
Therefore, Amount A < Amount B.
∴ The correct answer is Quantity A < Quantity B.
Alternate Method

Quantity A (Scheme 1 - SI):
⇒ Total Simple Interest % = r1 + r2 + r3 = 10% + 12% + 13% = 35%
⇒ Amount A = Principal × (1 + 35/100) = 12000 × 1.35 = ₹16200
Quantity B (Scheme 2 - CI):
⇒ Amount B = Principal × (1 + r/100)n
⇒ Amount B = 12000 × (1 + 11/100)3 = 12000 × (1.11)3
⇒ Amount B = 12000 × 1.367631 = ₹16411.572
Comparison:
⇒ 16200 < 16411.572
∴ The correct answer is Quantity A < Quantity B.
Additional Information
Simple Interest with Variable Rates
When rates change annually, Total SI = P × (r1 + r2 + ... + rn)/100. The effective rate is the simple sum of individual rates.
Compound Interest Formula
Amount = P(1 + r/100)n. For 3 years, the effective rate is slightly more than 3 × r due to the interest-on-interest effect.
Comparison Logic
If the average rate of SI is less than or equal to the CI rate, then Amount from CI will always be greater than Amount from SI for n > 1.

Shortcut Trick

Quantity I: Ratio 7 : 8, difference is 1 unit. Given 1 unit = 3 years ⇒ Ankita (7 units) = 7 × 3 = 21 years.

Quantity II: Sum of ages = 25 × 2 = 50. Younger person (Ankita) = (Sum − Diff) ÷ 2 = (50 − 2) ÷ 2 = 24 years.

Comparing both: 21 years < 24 years.

∴ The correct answer is Quantity I < Quantity II.

Alternate Method

Quantity I: Age of Ankita

Given: Ratio of Ankita : Ankit = 7 : 8. Age difference = 3 years.

⇒ Let ages be 7x and 8x.

⇒ 8x − 7x = 3

⇒ x = 3

⇒ Ankita = 7 × 3 = 21 years.

Quantity II: Age of Ankita

Given: Average of Pankti and Ankita = 25 years. Pankti − Ankita = 2 years.

⇒ Total age = 25 × 2 = 50 years.

⇒ P + A = 50 and P − A = 2

⇒ 2A = 48

⇒ Ankita = 24 years.

Comparing Quantity I (21) and Quantity II (24): 21 < 24.

∴ The correct answer is Quantity I < Quantity II.

Additional Information

Ratio and Proportion in Ages

If the ratio of ages is a:b and the difference is D, then 1 unit = D ÷ |a − b|.

Average and Difference Rule

If sum (S) and difference (D) of two quantities are known, the smaller value = (S − D) ÷ 2 and larger value = (S + D) ÷ 2.

Shortcut Trick

Net weight loss = Total weight of excluded people − (Old average × 2)

Net weight loss = (72 × 2) − (48 × 2) = 144 − 96 = 48 kg

This loss causes the remaining (n − 2) people to lose 1 kg each: (n − 2) × 1 = 48 ⇒ n = 50

Value of (n + 15) = 50 + 15 = 65

∴ The correct answer is 65.

Alternate Method

Given: Initial average of n people = 48 kg. Average of 2 excluded people = 72 kg. New average = 47 kg.

Formula: Total Weight = Average × Number of people

⇒ Initial Total Weight = 48n

⇒ Total Weight of 2 excluded people = 72 × 2 = 144 kg

⇒ New Total Weight = (n − 2) × 47

⇒ 48n − 144 = 47(n − 2)

⇒ 48n − 144 = 47n − 94

⇒ 48n − 47n = 144 − 94

⇒ n = 50

⇒ n + 15 = 50 + 15 = 65

∴ The correct answer is 65.

Additional Information

Weighted Average

If two groups with averages A₁ and A₂ and sizes n₁ and n₂ are combined, the combined average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Average Deviation Method

Total change in average × final number of observations = total deviation from the original average due to added or removed items.

Shortcut Trick

For 2 years, the difference between Compound Interest (CI) and Simple Interest (SI) is given by: Difference = P × (r/100)²

Given: Difference = ₹232 and Principal (P) = ₹5800.

232 = 5800 × (r/100)² ⇒ 232/5800 = (r/100)²

4/100 = (r/100)² ⇒ (2/10)² = (r/100)² ⇒ r/100 = 0.2 ⇒ r = 20.

∴ The correct answer is 20%.

Alternate Method

Given: Principal (P) = ₹5800, Time (T) = 2 years, Difference (CI − SI) = ₹232.

Formula: Difference for 2 years = Pr² / 100²

⇒ 232 = (5800 × r²) / 10000

⇒ 232 = (58 × r²) / 100

⇒ 23200 = 58 × r²

⇒ r² = 23200 ÷ 58

⇒ r² = 400

⇒ r = √400 = 20%

∴ The correct answer is 20%.

Additional Information

Difference for 3 years

The difference between CI and SI for 3 years is given by: Difference = P(r/100)² × (3 + r/100).

Effective Rate of Interest

The effective annual rate for CI for 2 years is calculated using x + y + (xy/100), where x and y are the rates for each year.

Simple Interest Formula

SI is calculated on the original principal throughout the period: SI = (P × r × t) / 100.

Common Calculation

Formula Used:

Girls = 2 × Average − Boys

Calculation:

2001

⇒ Girls = 2 × 140 − 160

⇒ = 280 − 160 = 120

2002

⇒ Girls = 2 × 160 − 200

⇒ = 320 − 200 = 120

2003

⇒ Girls = 2 × 220 − 200

⇒ = 440 − 200 = 240

2004

⇒ Girls = 2 × 180 − 120

⇒ = 360 − 120 = 240

2005

⇒ Girls = 2 × 150 − 200

⇒ = 300 − 200 = 100

Given:

2003 → Boys = 200, Girls = 240

2005 → Boys = 200, Girls = 100

Avg age:

2003 → Boys = 18, Girls = 16

2005 → Boys = 20, Girls = 15

Formula Used:

Total Age = (Number × Avg Age)

Calculation:

2003 Total Age

⇒ Boys = 200 × 18 = 3600

⇒ Girls = 240 × 16 = 3840

⇒ Total = 3600 + 3840 = 7440

2005 Total Age

⇒ Boys = 200 × 20 = 4000

⇒ Girls = 100 × 15 = 1500

⇒ Total = 4000 + 1500 = 5500

⇒ Ratio = 7440 : 5500

⇒ Divide by 20

⇒ 372 : 275

Final Answer: 372 : 275

Given:

I: x² − 32x + 255 = 0

II: y² − 11y + 30 = 0

Formula used:

ax² + bx + c = 0

Calculations:

From I:

x² − 32x + 255 = 0

Or, x² − 15x − 17x + 255 = 0

Or, x(x − 15) − 17(x − 15) = 0

Or, (x − 15)(x − 17) = 0

So, x = 15 or x = 17

From II:

y² − 11y + 30 = 0

Or, y² − 5y − 6y + 30 = 0

Or, y(y − 5) − 6(y − 5) = 0

Or, (y − 5)(y − 6) = 0

So, y = 5 or y = 6

Comparison:

x values = 15, 17

y values = 5, 6

⇒ Both values of x are greater than both values of y

∴ Quantity-I > Quantity-II

Given:

I: x² − 10x + 24 = 0

II: y² − 27y + 182 = 0

Formula used:

ax² + bx + c = 0

Calculations:

From I:

x² − 10x + 24 = 0

Or, x² − 6x − 4x + 24 = 0

Or, x(x − 6) − 4(x − 6) = 0

Or, (x − 6)(x − 4) = 0

So, x = 6 or x = 4

From II:

y² − 27y + 182 = 0

Or, y² − 13y − 14y + 182 = 0

Or, y(y − 13) − 14(y − 13) = 0

Or, (y − 13)(y − 14) = 0

So, y = 13 or y = 14

Comparison:

x values = 4, 6

y values = 13, 14

⇒ Both values of y are greater than both values of x

∴ Quantity-II > Quantity-I

Given:

I: x² − 30x + 189 = 0

II: 2y² − 29y + 105 = 0

Formula used:

ax² + bx + c = 0

Calculations:

From I:

x² − 30x + 189 = 0

Or, x² − 21x − 9x + 189 = 0

Or, x(x − 21) − 9(x − 21) = 0

Or, (x − 21)(x − 9) = 0

So, x = 21 or x = 9

From II:

2y² − 29y + 105 = 0

Or, 2y² − 14y − 15y + 105 = 0

Or, 2y(y − 7) − 15(y − 7) = 0

Or, (2y − 15)(y − 7) = 0

So, y = 15/2 or y = 7

Comparison:

x values = 9, 21

y values = 7, 7.5

⇒ Both values of x are greater than both values of y

∴ Quantity-I > Quantity-II