Quantitative Aptitude PYQs

Given:

Total players = 15

Time increase = 4 years

Players replaced = 5

Average age of removed players = 28

Formula used:

Change in total age = Number × years

Calculations:

Increase in total age in 4 years = 15 × 4 = 60

Since average remains same ⇒ total age must remain same

⇒ Replacement must reduce total age by 60

Removed players total age = 5 × 28 = 140

Let average of new players = x

New players total = 5x

⇒ 140 − 5x = 60

⇒ 5x = 140 − 60 = 80

⇒ x = 80 ÷ 5 = 16

∴ Average age of newcomers = 16 years.

Given:

Average of 15 numbers = 28

Each number is doubled and 5 is added to each.

Formula Used:

New average = (Old average × multiplier) + increment

Multiplier = 2 (each number is doubled)

Increment = 5 (added to each number)

Calculation:

⇒ New average = (28 × 2) + 5

⇒ New average = 56 + 5

⇒ New average = 61

∴ The correct answer is 61.

Shortcut Trick

When distance is constant, speed is inversely proportional to time.

Ratio of time (T₁ : T₂) = 2 : 3

Ratio of speed (S₁ : S₂) = 3 : 2

Decrease in speed = (3 − 2) / 3 × 100 = 1/3 × 100 = 33.33%

∴ The correct answer is 33.33%.

Alternate Method

Given: Total Distance = 1,200 km, Initial Time = 2 hr, Target Time = 3 hr.

Formula Used: Speed = Distance / Time

⇒ Initial Speed (S₁) = 1,200 ÷ 2 = 600 kmph

⇒ Final Speed (S₂) = 1,200 ÷ 3 = 400 kmph

⇒ Decrease in speed = 600 − 400 = 200 kmph

⇒ Percentage decrease = (Decrease / Initial Speed) × 100

⇒ Percentage decrease = (200 / 600) × 100 = 1/3 × 100 = 33.33%

∴ The correct answer is 33.33%.

Additional Information

Inversely Proportional Relation

When the distance is constant, if the time increases by x%, the speed must decrease to maintain the same distance. Formula: S₁ × T₁ = S₂ × T₂.

Percentage Decrease Formula

Percentage decrease in speed = [(Initial Speed − Final Speed) / Initial Speed] × 100.

Average Speed for Equal Distances

If an object travels equal distances at speeds x and y, the average speed is [2 × x × y] / (x + y).

Given:

Marked price = 2500

First discount = 20%

Second discount = 10%

Formula Used:

Selling price after discount = Marked price × (1 - discount)

Calculation:

⇒ Price after first discount = 2500 × (1 - 0.20)

⇒ Price after first discount = 2500 × 0.80 = 2000

⇒ Final selling price = 2000 × (1 - 0.10)

⇒ Final selling price = 2000 × 0.90 = 1800

∴ The correct answer is 1800.

Given:

Initial price = ₹40,000

First discount = 10%

Second discount = 20%

Formula Used:

Price after first discount = Initial price × (1 - First discount/100)

Price after second discount = Price after first discount × (1 - Second discount/100)

Calculation:

⇒ Price after first discount = ₹40,000 × (1 - 10/100)

⇒ Price after first discount = ₹40,000 × 0.90

⇒ Price after first discount = ₹36,000

⇒ Price after second discount = ₹36,000 × (1 - 20/100)

⇒ Price after second discount = ₹36,000 × 0.80

⇒ Price after second discount = ₹28,800

∴ The correct answer is ₹28,800.

Given:

Total solution = 70

Wine : Water = 5 : 2

Removed solution = 14

Added water = 14

Formula Used:

Quantity of component = Total × (ratio part ÷ total ratio)

Remaining quantity after removal = Initial × (1 - removed ÷ total)

Calculation:

⇒ Initial wine = 70 × 5 ÷ 7 = 50

⇒ Initial water = 70 × 2 ÷ 7 = 20

⇒ Remaining wine = 50 × (1 - 14 ÷ 70)

⇒ Remaining wine = 50 × (1 - 0.20) = 50 × 0.80 = 40

⇒ Remaining water = 20 × (1 - 14 ÷ 70)

⇒ Remaining water = 20 × 0.80 = 16

⇒ New water after adding = 16 + 14 = 30

⇒ Final ratio = 40 : 30

⇒ Final ratio = 4 : 3

∴ The correct answer is 4 : 3.

Given:

First number = \( \sqrt{3} \)

Second number = \( \sqrt[3]{5} \)

Third number = \( \sqrt[4]{6} \)

Formula Used:

If a = b^x, then compare by converting to decimal values

Calculation:

⇒ \( \sqrt{3} \) ≈ 1.732

⇒ \( \sqrt[3]{5} \) ≈ 1.710

⇒ \( \sqrt[4]{6} \) ≈ 1.565

⇒ Ascending order: \( \sqrt[4]{6} \) < \( \sqrt[3]{5} \) < \( \sqrt{3} \)

∴ The correct answer is \( \sqrt[4]{6} \) < \( \sqrt[3]{5} \) < \( \sqrt{3} \).

Given:

Slope = -2

Point = (3, 5)

Formula Used:

Point-slope form: y - y₁ = m × (x - x₁)

Calculation:

⇒ y - 5 = -2 × (x - 3)

⇒ y - 5 = -2x + 6

⇒ y = -2x + 6 + 5

⇒ y = -2x + 11

∴ The correct answer is y = -2x + 11.

Given:

Area of smaller triangle = 16

Area of larger triangle = 25

Perimeter of smaller triangle = 20

Formula Used:

Ratio of areas = (Ratio of sides)²

Perimeter ∝ side

Calculation:

⇒ Ratio of areas = 25 ÷ 16

⇒ Ratio of sides = √(25 ÷ 16)

⇒ Ratio of sides = 5 ÷ 4

⇒ Perimeter of larger triangle = 20 × 5 ÷ 4

⇒ Perimeter of larger triangle = 25

∴ The correct answer is 25.

Given:

AD = 4

DB = 12

AE = 3

EC = 9

Formula Used:

Area ratio of similar triangles = (Ratio of corresponding sides)²

Calculation:

⇒ AB = 4 + 12 = 16

⇒ AC = 3 + 9 = 12

⇒ AD ÷ AB = 4 ÷ 16 = 1 ÷ 4

⇒ AE ÷ AC = 3 ÷ 12 = 1 ÷ 4

⇒ Ratio of sides = 1 ÷ 4

⇒ Area ratio = (1 ÷ 4)²

⇒ Area ratio = 1 ÷ 16

∴ The correct answer is 1 ÷ 16.

Given:

Cost price = 3000

Discount₁ = 12%

Discount₂ = 5%

Difference in profit = 504

Formula Used:

Selling price = Marked price × (100 − discount%) ÷ 100

Profit = Selling price − Cost price

Calculation:

⇒ Let Marked price = M

⇒ SP₁ = M × (88 ÷ 100)

⇒ SP₂ = M × (95 ÷ 100)

⇒ Profit difference = SP₂ − SP₁

⇒ 504 = M × (95 ÷ 100) − M × (88 ÷ 100)

⇒ 504 = M × (7 ÷ 100)

⇒ M = 504 × (100 ÷ 7)

⇒ M = 7200

∴ The correct answer is 7200.

Given:

Ratio of marked prices (pen : pencil) = 5 : 6

Discount on pen = 18%

Total discount = 12%

Formula Used:

Discount amount = Marked price × discount% ÷ 100

Total discount % = (Total discount ÷ Total marked price) × 100

Calculation:

⇒ Let marked prices = 5x and 6x

⇒ Discount on pen = 5x × (18 ÷ 100) = 0.9x

⇒ Discount on pencil = 6x × (d ÷ 100) = (6d ÷ 100)x

⇒ Total marked price = 11x

⇒ Total discount = 0.9x + (6d ÷ 100)x

⇒ 12% of 11x = 0.9x + (6d ÷ 100)x

⇒ (12 ÷ 100) × 11x = 0.9x + (6d ÷ 100)x

⇒ 1.32x = 0.9x + (6d ÷ 100)x

⇒ 1.32 − 0.9 = (6d ÷ 100)

⇒ 0.42 = (6d ÷ 100)

⇒ d = (0.42 × 100) ÷ 6 = 7

∴ The correct answer is 7%.

Shortcut Trick

Ratio of number of bricks in Stack A : B : C = 12 : 8 : 16 = 3 : 2 : 4.

Assume base average = 110 kg.

Net Deviation = [3 × (112.5 − 110)] + [2 × (118.75 − 110)] + [4 × (107.25 − 110)]

Net Deviation = [3 × 2.5] + [2 × 8.75] + [4 × (−2.75)] = 7.5 + 17.5 − 11 = 14.

Overall Average = 110 + (14 ÷ 9) = 110 + 1.555... ≈ 111.55 kg.

∴ The correct answer is 111.55 kg.

Alternate Method

Given:

Stack A: 12 bricks at 112.5 kg | Stack B: 8 bricks at 118.75 kg | Stack C: 16 bricks at 107.25 kg

Formula Used:

Overall Average = (Total Weight) ÷ (Total Number of Bricks)

⇒ Total weight of Stack A = 12 × 112.5 = 1350 kg

⇒ Total weight of Stack B = 8 × 118.75 = 950 kg

⇒ Total weight of Stack C = 16 × 107.25 = 1716 kg

⇒ Total Weight of all bricks = 1350 + 950 + 1716 = 4016 kg

⇒ Total number of bricks = 12 + 8 + 16 = 36

⇒ Overall Average = 4016 ÷ 36

⇒ Overall Average = 1004 ÷ 9 = 111.555...

∴ The correct answer is 111.55 kg.

Additional Information

Weighted Average Formula

The weighted average is calculated as (w₁x₁ + w₂x₂ + ... + w_nx_n) ÷ (w₁ + w₂ + ... + w_n), where w is the weight (quantity) and x is the value.

Deviation Method

New Average = Assumed Average + (Net Deviation ÷ Total Count). This simplifies calculations by using smaller numbers.

Average Property

If every value in a set is increased or decreased by a constant k, the overall average also increases or decreases by k.

Formula Used:

Perfect square: A number that can be expressed as the square of an integer.

Largest three-digit number: 999.

Calculation:

⇒ Check if the square root of each option is an integer:

⇒ √951 ≈ 30.81 (not an integer)

⇒ √961 = 31 (an integer)

⇒ √971 ≈ 31.17 (not an integer)

⇒ √981 ≈ 31.34 (not an integer)

∴ The correct answer is 961.

Given:

A's investment = ₹30,000

B's investment = ₹40,000

B joined after 6 months

Total time = 1 year

Formula Used:

Profit-sharing ratio = (Investment × Time)

Calculation:

⇒ A's share = ₹30,000 × 12 months

⇒ A's share = ₹360,000

⇒ B's share = ₹40,000 × 6 months

⇒ B's share = ₹240,000

⇒ Profit-sharing ratio = A's share : B's share

⇒ Profit-sharing ratio = ₹360,000 : ₹240,000

⇒ Profit-sharing ratio = 3 : 2

∴ The correct answer is 3:2.

Given:

sin(A + B) = sinAcosB + cosAsinB

75° = 45° + 30°

Formula Used:

sin(A + B) = sinAcosB + cosAsinB

Calculation:

⇒ sin75° = sin(45 + 30)

⇒ sin75° = sin45 × cos30 + cos45 × sin30

⇒ sin45 = 1 ÷ √2, cos30 = √3 ÷ 2

⇒ cos45 = 1 ÷ √2, sin30 = 1 ÷ 2

⇒ sin75° = (1 ÷ √2 × √3 ÷ 2) + (1 ÷ √2 × 1 ÷ 2)

⇒ sin75° = (√3 ÷ 2√2) + (1 ÷ 2√2)

⇒ sin75° = (√3 + 1) ÷ (2√2)

∴ The correct answer is (√3 + 1) ÷ (2√2).

Given:

Ratio = 2 : 5 : 3 : 4

Increase = 20%, 30%, 40%, 10%

Formula Used:

New value = Original × (100 + increase%) ÷ 100

Calculation:

⇒ New values = 2 × (120 ÷ 100), 5 × (130 ÷ 100), 3 × (140 ÷ 100), 4 × (110 ÷ 100)

⇒ = 2.4, 6.5, 4.2, 4.4

⇒ Multiply by 10 ⇒ 24 : 65 : 42 : 44

∴ The correct answer is 24 : 65 : 42 : 44.

Given:

Total mixture = 560 litres

Ratio (milk : water) = 4 : 3

Mixture removed = 140 litres

Formula Used:

Quantity = Total × (part ÷ sum of ratio)

Remaining = Initial − Removed proportion

Calculation:

⇒ Milk initially = 560 × (4 ÷ 7) = 320

⇒ Milk removed = 140 × (4 ÷ 7) = 80

⇒ Milk remaining = 320 − 80 = 240

∴ The correct answer is 240 litres.

Given:

Total mixture = 60 litres

Initial ratio (wine : water) = 1 : 2

New ratio = 1 : 3

Formula Used:

Quantity = Total × (part ÷ sum of ratio)

Calculation:

⇒ Wine = 60 × (1 ÷ 3) = 20

⇒ Water = 60 × (2 ÷ 3) = 40

⇒ Let water added = x

⇒ New ratio = 20 : (40 + x) = 1 : 3

⇒ 20 ÷ (40 + x) = 1 ÷ 3

⇒ 60 = 40 + x

⇒ x = 20

∴ The correct answer is 20 litres.

Given:

secθ + tanθ = 5

θ is acute

Formula Used:

(secθ + tanθ)(secθ − tanθ) = 1

Calculation:

⇒ secθ − tanθ = 1 ÷ 5

⇒ Add equations:

⇒ (secθ + tanθ) + (secθ − tanθ) = 5 + 1 ÷ 5

⇒ 2secθ = 26 ÷ 5

⇒ secθ = 13 ÷ 5

∴ The correct answer is 13 ÷ 5.