Quantitative Aptitude PYQs

General Solution:

Let the runs scored by A in Test matches = x

Then:

ODIs = 55% of x = 0.55x

T20s = Total of A’s average = 230 × 3 = 690

So, A's T20 runs = 690 − (x + 0.55x) = 690 − 1.55x

C's T20 runs = 40 more than one-third of A’s Test runs:

C's T20 = (1/3) × x + 40

We're told:

C and D scored 180 and 95 runs respectively in T20

So:

C's T20 = 180 = (1/3)x + 40

⇒ (1/3)x = 140 ⇒ x = 420

Now that we know x = 420, substitute values for A:

A's Test = 420

A's ODI = 0.55 × 420 = 231

A's T20 = 690 − (420 + 231) = 39

ODI data:

C's ODI = c

D's ODI = c − 42

B's ODI = (1/3) × A ODI = (1/3) × 231 = 77

Total ODI by B, C, D = 471

⇒ 77 + c + (c − 42) = 471

⇒ 2c + 35 = 471 ⇒ 2c = 436 ⇒ c = 218

So:

C's ODI = 218

D's ODI = 176

B's ODI = 77

Test Match Data:

D's Test = 4 × D ODI = 4 × 176 = 704

C's Test = D's Test + 8 = 712

Also:

C's Test = 2 × B's Test ⇒ B's Test = 712 / 2 = 356

T20 Data:

C's T20 = 180 (already given)

D's T20 = 95

B's T20 = (1/7) × B ODI = (1/7) × 77 = 11

Thus,

 

Calculations:

A's Original Total:

Test: 420

ODI: 231

T20: 39

Total A = 420 + 231 + 39 = 690

B's Original Total:

Test: 356

ODI: 77

T20: 11

Total B = 356 + 77 + 11 = 444

Adjustments:

B’s Disqualified Runs:

25% of 444 =

→ 444 × (25/100) = 111

→ New B total = 444 − 111 = 333

A gets 20% of D’s T20 runs added:

→ D’s T20 = 95

→ 20% of 95 = 19

→ New A total = 690 + 19 = 709

New Total = 333 + 709 = 1042

Thus, the correct answer is 1042.

Shortcut Trick

Ratio of Time for Rahul and Priya = 2 : 1

Ratio of Efficiency (Inverse of time) = 1 : 2

Total work = Rahul efficiency × Rahul days = 1 × 20 = 20 units

Work done by Rahul in last 8 days = 8 × 1 = 8 units

Work done by both in k days = 20 − 8 = 12 units

k = 12 ÷ (1 + 2) = 4 days

∴ The correct answer is 4 days.

Alternate Method

Given: Time taken by Rahul = 20 days. Ratio of time (Rahul : Priya) = 2 :

1. Formula Used: Total Work = Efficiency × Time; Work Done = (Combined Efficiency) × Time

⇒ Time for Priya = 20 × (1/2) = 10 days.

⇒ Total work (LCM of 20, 10) = 20 units.

⇒ Rahul finished the remaining work in 8 days.

⇒ Work done by Rahul alone = 8 × 1 = 8 units.

⇒ Remaining work done by Priya and Rahul together = 20 − 8 = 12 units.

⇒ Combined efficiency = 1 + 2 = 3 units/day.

⇒ k = 12 ÷ 3 = 4 days.

∴ The correct answer is 4 days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a task. If time ratio is a:b, efficiency ratio is b:a.

Work and Wages

Wages are generally distributed in the ratio of the work done by each individual, which is (Efficiency × Time).

Alternate Day Work

In problems where people work on alternate days, find the work done in one complete cycle (e.g., 2 days) and then calculate total time.

Shortcut Trick

Ratio of Length (L) to Breadth (B) is 7 : 3.

Product of ratios = 7 × 3 = 21 units². Given Area = 2352 cm².

1 unit² = 2352 ÷ 21 = 112 ⇒ 1 unit = √112 = 4√7.

Perimeter = 2(7 + 3) units = 20 units = 20 × 4√7 = 80√7 cm.

Square side = Perimeter ÷ 4 = 20√7 cm. Square Area = (20√7)² = 400 × 7 = 2800 cm².

∴ The correct answer is 2800 sq. cm.

Alternate Method

Given: 3 × L = 7 × B ⇒ L = 7x, B = 3x. Area = 2352 sq. cm.

Formula: Area = L × B | Perimeter = 2(L + B) | Square Area = (Perimeter/4)²

⇒ 7x × 3x = 2352

⇒ 21x² = 2352

⇒ x² = 112 ⇒ x = √112 = 4√7

⇒ Perimeter of Rectangle = 2(7x + 3x) = 20x = 20 × 4√7 = 80√7 cm

⇒ Side of Square (a) = Perimeter ÷ 4 = 80√7 ÷ 4 = 20√7 cm

⇒ Area of Square = a² = (20√7)² = 400 × 7 = 2800 sq. cm

∴ The correct answer is 2800 sq. cm.

Additional Information

Rectangle Properties

Area = Length × Breadth. Perimeter = 2 × (Length + Breadth). Diagonal = √(L² + B²).

Square Properties

Area = side². Perimeter = 4 × side. For a fixed perimeter, a square has the maximum area among all rectangles.

Ratio Application in Geometry

When dimensions are in ratio m:n, Area is proportional to (m×n) and Perimeter is proportional to (m+n).

Equation I: x² − 22x + 105 = 0

Find two numbers whose product = 105 and sum = 22:

⇒ Factors: (x − 15)(x − 7) = 0

⇒ x = 15, 7

Equation II: y² − 20y + 96 = 0

Find two numbers whose product = 96 and sum = 20:

⇒ Factors: (y − 12)(y − 8) = 0

⇒ y = 12, 8

Compare roots:

x₁ = 15 vs y₁ = 12 ⇒ x > y

x₂ = 7 vs y₂ = 8 ⇒ x < y

Since one x > y and one x < y ⇒ Relationship cannot be established

Thus, the correct answer is x = y or Relationship cannot be established.

Equation I: 4x² − 25 = 0

⇒ 4x² = 25 ⇒ x² = 25⁄4 ⇒ x = ±5⁄2

Equation II: y² − 10y + 25 = 0

⇒ (y − 5)² = 0 ⇒ y = 5

Possible x values: −2.5 and 2.5

y = 5

Both −2.5 < 5 and 2.5 < 5 ⇒ in every case x < y

Thus, the correct answer is x < y.

Equation I: x² − 11x + 24 = 0

Factor: Find two numbers that multiply to 24 and add up to −11

⇒ (x − 3)(x − 8) = 0

⇒ x = 3, 8

Equation II: y² + 6y − 135 = 0

Factor: Find two numbers that multiply to −135 and add up to 6

⇒ (y + 15)(y − 9) = 0

⇒ y = −15, 9

Now compare:

x = 3, 8

y = −15, 9

Compare each x with both y values:

3 < 9 and 3 > −15

8 < 9 and 8 > −15

So: x is less than y in one case and greater in another → No definite relation

Thus, the correct answer is x = y or Relationship cannot be established.

Equation I: x² − 35x + 294 = 0

⇒ Discriminant = 35² − 4×294 = 1225 − 1176 = 49

⇒ Roots: x = [35 ± 7] ÷ 2 ⇒ x = 21 or 14

Equation II: y² − 38y + 345 = 0

⇒ Discriminant = 38² − 4×345 = 1444 − 1380 = 64

⇒ Roots: y = [38 ± 8] ÷ 2 ⇒ y = 23 or 15

Possible values:

x ∈ {14, 21}

y ∈ {15, 23}

Since sometimes x < y (e.g. 14 < 15) and sometimes x > y (e.g. 21 > 15), the relationship between x and y cannot be uniquely determined.

Thus, the correct answer is x = y or relationship cannot be established.

Equation I: 12x² + 7x − 12 = 0

⇒ Discriminant = 7² − 4 × 12 × (−12) = 49 + 576 = 625, √625 = 25

⇒ x = [−7 ± 25] ÷ (2 × 12) = [−7 ± 25] ÷ 24

x₁ = (18) ÷ 24 = 3/4 = 0.75

x₂ = (−32) ÷ 24 = −4/3 ≈ −1.333

Equation II: 8y² − 18y + 9 = 0

⇒ Discriminant = (−18)² − 4 × 8 × 9 = 324 − 288 = 36, √36 = 6

⇒ y = [18 ± 6] ÷ (2 × 8) = [18 ± 6] ÷ 16

y₁ = 24 ÷ 16 = 3/2 = 1.5

y₂ = 12 ÷ 16 = 3/4 = 0.75

Comparison of roots:

x₂ = −1.333 < y₂ = 0.75

x₁ = 0.75 = y₂ = 0.75

x₁ = 0.75 < y₁ = 1.5

x₂ = −1.333 < y₁ = 1.5

All x-values are ≤ all y-values.

Thus, the correct answer is x ≤ y.

General Solution:

Use the formula:

Cakes Sold = Total Baked × (Percentage Sold / 100)

Cakes Unsold = Total Baked − Cakes Sold

 

Calculations:

Find unsold cakes of Bakery C and D:

C (unsold) = 132

D (unsold) = 192

Total to distribute = 132 + 192 = 324

Divide equally between A and B:

Each receives = 324 ÷ 2 = 162

Find the original total cakes with Bakery A = 320

New total = 320 + 162 = 482

Thus, the correct answer is 482.

General Solution:

Use the formula:

Cakes Sold = Total Baked × (Percentage Sold / 100)

Cakes Unsold = Total Baked − Cakes Sold

 

Calculations:

Given Information:

Average number of cakes, cupcakes, and brownies in Bakery A = 320

So, total items baked in A = 320 × 3 = 960

Out of this, the number of cakes baked = 320 (already given)

Let the number of brownies = b

Given: Ratio of brownies to cakes = 3:4

⇒ b / 320 = 3 / 4 ⇒ b = (3/4) × 320 = 240

Find the number of cupcakes in A:

Total baked = cakes + brownies + cupcakes = 960

⇒ Cupcakes = 960 − (320 + 240) = 400

Number of cakes sold by Bakery D:

Total cakes baked by D = 480

Sold % = 60%

⇒ Cakes sold = 60% of 480 = (60/100) × 480 = 288

Find the required difference:

Difference = Cupcakes in A − Cakes sold by D = 400 − 288 = 112

Thus, the correct answer is 112.

General Solution:

Use the formula:

Cakes Sold = Total Baked × (Percentage Sold / 100)

Cakes Unsold = Total Baked − Cakes Sold

 

Calculations:

Original Data for Bakery D:

Baked = 480

Sold = 60% of 480 = 288

Unsold = 480 − 288 = 192

Changes Given:

Baked increases by 20%

→ New baked = 480 + (20% of 480)

→ 480 + 96 = 576

Cakes sold decrease by 6:

→ New sold = 288 − 6 = 282

→ New unsold = 576 − 282 = 294

Required Ratio: Sold : Unsold = 282 : 294

Thus, the correct answer is 47:49.

General Solution:

Use the formula:

Cakes Sold = Total Baked × (Percentage Sold / 100)

Cakes Unsold = Total Baked − Cakes Sold

 

Calculations:

Find cakes sold by A and C:

A Sold = 320 × 65 ÷ 100 = 208

C Sold = 240 × 45 ÷ 100 = 108

Find the difference:

⇒ 208 − 108 = 100

Find % of cakes baked by B (which is 520):

⇒ (100 ÷ 520) × 100 = 19.23%

Thus, the correct answer is 19.23%.

General Solution:

Use the formula:

Cakes Sold = Total Baked × (Percentage Sold / 100)

Cakes Unsold = Total Baked − Cakes Sold

 

Calculations:

Bakery D:

Unsold cakes = 192

Donated cakes = 50% of 192 = 192 × 0.5 = 96

Bakery B:

Sold cakes = 208

Returned cakes = 25% of 208 = 208 × 0.25 = 52

Required difference = 96 − 52 = 44

Thus, the correct answer is 44.

Shortcut Trick

Milk in X = 5/7 and Milk in Y = 3/7.

Since X and Y are mixed in ratio 7:5, let volumes be 7 and 5 units.

Total Milk = (7 × 5/7) + (5 × 3/7) = 5 + 15/7 = 50/7 units.

Total Mixture = 7 + 5 = 12 units.

Milk in 84 litres = (Total Milk / Total Mixture) × 84 = (50/7 / 12) × 84 = 50 litres.

∴ The correct answer is 50 litres.

Alternate Method

Given: Ratio in X = 5:2, Ratio in Y = 3:4, Mixing Ratio X:Y = 7:5, Total New Mixture = 84 litres.

⇒ Quantity of X in new mixture = (7 / 12) × 84 = 49 litres

⇒ Quantity of Y in new mixture = (5 / 12) × 84 = 35 litres

⇒ Milk from X = 49 × (5 / 7) = 35 litres

⇒ Milk from Y = 35 × (3 / 7) = 15 litres

⇒ Total Milk in new mixture = 35 + 15 = 50 litres

∴ The correct answer is 50 litres.

Additional Information

Alligation Method

Used to find the ratio in which two ingredients are mixed. Mean concentration = (Quantity₁ × C₁ + Quantity₂ × C₂) / (Quantity₁ + Quantity₂).

Ratio of Fractions

When mixing containers of different ratios, always convert the component (Milk or Water) into a fraction of the total volume before applying the mixing ratio.

Shortcut Trick

Initial average for 12 bags = 45 kg.

New average for 15 bags = 45 + 2 = 47 kg.

Sum of 3 new bags = (New average × Total count) − (Old average × Old count)

Sum = (47 × 15) − (45 × 12) = 705 − 540 = 165 kg.

3w + 30 = 165 ⇒ 3w = 135 ⇒ w = 45 kg.

∴ The correct answer is 45 kg.

Alternate Method

Given:

Number of initial rice bags = 12

Initial average weight = 45 kg

Weights of added bags = (w + 5), (w − 15), (w + 40)

Calculations:

⇒ Total weight of 12 bags = 12 × 45 = 540 kg

⇒ Sum of 3 new bags = (w + 5) + (w − 15) + (w + 40) = 3w + 30 kg

⇒ New total count of bags = 12 + 3 = 15

⇒ New average weight = 45 + 2 = 47 kg

⇒ Total weight of 15 bags = 15 × 47 = 705 kg

⇒ 540 + (3w + 30) = 705

⇒ 3w + 570 = 705

⇒ 3w = 705 − 570 = 135

⇒ w = 135 ÷ 3 = 45 kg

∴ The correct answer is 45 kg.

Additional Information

Average formula

Average = (Sum of all observations) ÷ (Number of observations). Sum = Average × Count.

Deviation method

Net change in average = (Total surplus/deficit) ÷ (Total number of items). This is useful for finding the value of new entrants.

Weighted Average

Combined Average = (n₁ × a₁ + n₂ × a₂) ÷ (n₁ + n₂), where n is count and a is average.

Shortcut Trick

Ratio of Grandfather : Shagun was 5 : 1 (10 years ago) and will be 3 : 1 (after 5 years).

Gap between ratios = 10 + 5 = 15 years.

Equating differences: (5−1) × 1 = 4; (3−1) × 2 = 4 ⇒ New ratios: 5 : 1 and 6 : 2.

Difference of 1 unit = 15 years. Current ages: (5×15)+10 = 85 and (1×15)+10 = 25.

5 years ago ages: 80 and 20 ⇒ Ratio = 20 : 80 = 1 : 4.

∴ The correct answer is 1 : 4.

Alternate Method

Let Shagun's age 10 years ago be x and Grandfather's age be 5x.

⇒ Present ages: Shagun = x + 10, Grandfather = 5x + 10.

⇒ After 5 years: Shagun = x + 15, Grandfather = 5x + 15.

⇒ According to question: 5x + 15 = 3 × (x + 15)

⇒ 5x + 15 = 3x + 45

⇒ 2x = 30 ⇒ x = 15.

⇒ 5 years ago, Shagun's age = (x + 10) − 5 = x + 5 = 15 + 5 = 20.

⇒ 5 years ago, Grandfather's age = (5x + 10) − 5 = 5x + 5 = 75 + 5 = 80.

⇒ Ratio = 20 : 80 = 1 : 4.

∴ The correct answer is 1 : 4.

Additional Information

Ratio Difference Method

When the time gap is the same for two people, the difference between their ages remains constant. Use this to balance ratios.

Age Linear Equations

Always define variables at a specific point in time (like 10 years ago) to simplify calculations.

Calculation

Aman’s total = 70 + 40 + 55 + 45 = 210

Raman’s total = 65 + 45 + 60 + 55 = 225

Aman’s average = 210 ÷ 4 = 52.5

Raman’s average = 225 ÷ 4 = 56.25

Percentage difference = [[56.25 − 52.5] /52.5 ] ×100

= [3.75/52.5] × 100 ≈ 7.14%

Calculation

Class IX English = 45

Class X Geography = 45 × 1.30 = 58.5

Percentage = (58.5 / 90) × 100 = 65%

Calculations:

Math = 70 + 20% of 70 = 70 + 14 = 84

English = 40 + 20% of 40 = 40 + 8 = 48

Science = 55 + 20% of 55 = 55 + 11 = 66

History = 45 + 20% of 45 = 45 + 9 = 54

Total = 84 + 48 + 66 + 54 = 252

Average = 252 ÷ 4 = 63

∴ Aman’s average marks in Class X = 63

Calculation

Let Raman’s marks = x

(66 + x) / 2 = 62 → 66 + x = 124 → x = 58

So, total 58 + 55 = 113