Quantitative Aptitude PYQs

Shortcut Trick

Total Interest for 1 year = 4824 ÷ 6 = ₹804

Overall Average Rate = (804 ÷ 8800) × 100 = 201/22 %

Using Alligation: (11% − 201/22%) : (201/22% − 7%) = 41/22 : 47/22 = 47 : 41

Sum at 11% = (47 ÷ 88) × 8800 = ₹4700

∴ The correct answer is ₹4700.

Alternate Method

Given: Total Principal = ₹8800, Total Time = 6 years, Total SI = ₹4824

Formula: SI = (P × R × T) ÷ 100

⇒ Let sum at 11% be x and sum at 7% be (8800 − x)

⇒ [(x × 11 × 6) ÷ 100] + [((8800 − x) × 7 × 6) ÷ 100] = 4824

⇒ 66x + 42(8800 − x) = 482400

⇒ 66x + 369600 − 42x = 482400

⇒ 24x = 482400 − 369600

⇒ 24x = 112800

⇒ x = 112800 ÷ 24 = 4700

∴ The correct answer is ₹4700.

Additional Information

Simple Interest Formula

The basic formula is SI = P × R × T / 100, where P is principal, R is rate per annum, and T is time in years.

Alligation in Interest

When a sum is divided into two parts with different rates, the ratio of the parts is found by the difference between individual rates and the average rate.

Effective Rate of Interest

The effective rate for a period is R × T. For 11% over 6 years, the total interest is 66% of the principal part.

Shortcut Trick

Ratio of Profit Share = Ratio of (Investment × Time).

Ram profit : Shyam profit = 625 : (2500 − 625) = 625 : 1875 = 1 : 3.

Profit Ratio = (3000 × x) : (5000 × (x + 4)) = 3x : 5(x + 4).

1 ÷ 3 = 3x ÷ (5x + 20) ⇒ 5x + 20 = 9x ⇒ 4x = 20 ⇒ x = 5.

∴ The correct answer is 5.

Alternate Method

Given → Investment (Ram = 3000, Shyam = 5000), Time (Ram = x, Shyam = x + 4), Total Profit = 2500.
Formulas: Profit Share = (Investment × Time) | Total Profit = Ram Profit + Shyam Profit.

⇒ Ram profit share = 625.
⇒ Shyam profit share = 2500 − 625 = 1875.
⇒ (Ram Investment × Ram Time) ÷ (Shyam Investment × Shyam Time) = Ram Profit ÷ Shyam Profit.
⇒ (3000 × x) ÷ (5000 × (x + 4)) = 625 ÷ 1875.
⇒ 3x ÷ 5(x + 4) = 1 ÷ 3.
⇒ 9x = 5x + 20 ⇒ 4x = 20 ⇒ x = 5.

∴ The correct answer is 5.

Additional Information

Partnership Basics

In a partnership, the profit is distributed in the ratio of the product of capital invested and the time period of investment.

Ratio and Proportion

If P₁ and P₂ are profit shares, C₁ and C₂ are capitals, and T₁ and T₂ are time periods, then P₁/P₂ = (C₁ × T₁) / (C₂ × T₂).

Solution :

 Concept used :

 Follow the BODMAS rule to solve this question, as per the question is given below:

 Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

 Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

 Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

 Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated

 Given :

⇒ 119% of 89 + 161% of 61 - 79% of 131 = ?

Approximate the number to its nearest value 

⇒ 120% of 90 + 160% of 60 - 80% of 130 = ?

⇒ 120 × 90 / 100 + 160×60 / 100 - 80 × 130 / 100 = ?

⇒ 12× 9 + 16×6 - 8×13 = ?

⇒ 108 + 96 - 104 = ?

⇒ 204 - 104 = ?

⇒ ? = 100

Hence 100 is the correct answer.

Solution :

 Concept used :

 Follow the BODMAS rule to solve this question, as per the question is given below:

 Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

 Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

 Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

 Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated

 Given :

⇒ 3.99 × 1295.82 ÷ (35.71)² + 75.13% of 59.78 = ?

Approximate the number to its nearest value 

⇒ 4 × 1296 ÷ (36)² + 75% of 60 = ?

⇒ 4 × 1296 ÷ 1296 + 75 × 60 / 100 = ?

⇒ 4 × 1 + 45 = ?

⇒ ? = 49

Hence 49 is the correct answer.

Solution :

 Concept used :

 Follow the BODMAS rule to solve this question, as per the question is given below:

 Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

 Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

 Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

 Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated

 Given :

⇒ ( 32.97 + 43.12) ÷ 2.15 × 40.32 =  ? × 5.13 + 19.98

Approximate the number to its nearest value 

⇒ ( 33 + 43 ) ÷ 2 × 40 = ? × 5 + 20

⇒ 76 ÷ 2 × 40 = ? × 5 + 20

⇒ 38 × 40 = ? × 5 + 20

⇒  1520 - 20 = ? × 5

⇒ 1500 = ? × 5

⇒ ? = 1500 / 5

⇒ ? = 300

Hence 300 is the correct answer.

Given:

24.96% of 2015.76 - √(575.98) = 119.87 × (?)2

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

24.96% of 2015.76 - √(575.98) = 119.87 × (?)2

⇒ 25% of 2016 - √(576) = 120 × (?)²

⇒ (25/100) × 2016 - 24 = 120 × (?)²

⇒ (2016/4) - 24 = 120 × (?)²

⇒ 504 - 24 = 120 × (?)²

⇒ 480 = 120 × (?)²

⇒ (?)² = 480/120

⇒ (?)² = 4

⇒ ? = 2

∴ 2 will come in place of the question mark ('?').

Shortcut Trick

Given: B + D = 60 and C = D + 15, B = C − 5.

Substitute C in B: B = (D + 15) − 5 = D + 10.

Now, (D + 10) + D = 60 ⇒ 2D = 50 ⇒ D = 25.

Then B = 25 + 10 = 35 years.

Age of B after 10 years = 35 + 10 = 45 years.

∴ The correct answer is 45 years.

Alternate Method

Given:

1. A + C = 85

2. B + D = 60

3. C = D + 15

4. B = C − 5

Calculations:

⇒ From (4), substitute C: B = (D + 15) − 5

⇒ B = D + 10

⇒ Substitute B in (2): (D + 10) + D = 60

⇒ 2D + 10 = 60 ⇒ 2D = 50 ⇒ D = 25

⇒ Present age of B = D + 10 = 25 + 10 = 35 years

⇒ Age of B after 10 years = 35 + 10 = 45 years

∴ The correct answer is 45 years.

Additional Information

Ages involving Multiple Variables

When multiple relations are given, express all variables in terms of a single common variable to simplify calculations.

Future Age Calculation

If the present age is x, the age after n years is always x + n. Always identify if the question asks for present or future age.

Shortcut Trick

Using π = 3, Perimeter of Semi-circle A = r(π + 2) = r(3 + 2) = 5r = 25 ⇒ r_A = 5 cm.

Area of Circle B = πr_B² = 3 × r_B² = 147 ⇒ r_B² = 49 ⇒ r_B = 7 cm.

Difference x = r_B − r_A = 7 − 5 = 2 cm.

Radius of Circle C = x + 4 = 2 + 4 = 6 cm.

Total Perimeter = 2π(r_B + r_C) = 2 × 3 × (7 + 6) = 6 × 13 = 78 cm.

∴ The correct answer is 78 cm.

Alternate Method

Given: Perimeter of Semi-circle A = 25 cm, Area of Circle B = 147 cm², π = 3.

Formula: Perimeter of semi-circle = πr + 2r; Area of circle = πr²; Circumference = 2πr.

⇒ 3r_A + 2r_A = 25 ⇒ 5r_A = 25 ⇒ r_A = 5 cm.

⇒ 3 × (r_B)² = 147 ⇒ r_B² = 49 ⇒ r_B = 7 cm.

⇒ x = r_B − r_A = 7 − 5 = 2 cm.

⇒ r_C = x + 4 = 2 + 4 = 6 cm.

⇒ Perimeter of B = 2 × 3 × 7 = 42 cm.

⇒ Perimeter of C = 2 × 3 × 6 = 36 cm.

⇒ Total Sum = 42 + 36 = 78 cm.

∴ The correct answer is 78 cm.

Additional Information

Properties of Semi-circle

The perimeter includes the curved arc (πr) and the straight diameter (2r). Total Perimeter = r(π + 2).

Circle Mensuration

For a circle with radius r, the Area = πr² and the Circumference = 2πr. When π is given as a specific value like 3, always use that instead of 22/7.

Shortcut Trick

Change in total amount = New Total − Old Total

Total of 2 new students = 2 × 17 = ₹34

New average = 12 + 1 = ₹13

New total = (x + 2) × 13. Also, New total = 12x + 34.

⇒ 13x + 26 = 12x + 34 ⇒ x = 8. Total amount = 10 × 13 = ₹130.

∴ The correct answer is ₹130.

Alternate Method

Given:

⇒ Initial number of students = x

⇒ Initial average = ₹12

⇒ Average of 2 new students = ₹17

⇒ New average = 12 + 1 = ₹13

Calculations:

⇒ Initial total expenditure = 12 × x = 12x

⇒ Total spent by 2 new students = 2 × 17 = ₹34

⇒ New total expenditure = 12x + 34

⇒ Also, New total = (x + 2) × 13

⇒ 12x + 34 = 13x + 26

⇒ 13x − 12x = 34 − 26

⇒ x = 8

⇒ Total students = 8 + 2 = 10

⇒ Total amount spent = 10 × 13 = ₹130

∴ The correct answer is ₹130.

Additional Information

Average Expenditure Formula

Total Expenditure = Average × Number of Persons.

Net Deviation Method

The sum of deviations of all observations from the average is always zero.

Average of New Joiners

New Average = (Old Sum + New Sum) ÷ (Old Count + New Count).

General Solution:

 

Calculations:

Given:

Company A: Manufactured = 2500, Sold = 2150 ⇒ Unsold = 2500 − 2150 = 350

Company B: Sold = 2800

Formula used:

Percentage = (Part ÷ Whole) × 100

Calculation:

⇒ (350 ÷ 2800) × 100

⇒ 350/2800 = 1/8 = 0.125

⇒ 0.125 × 100 = 12.5%

Thus, the unsold pens from Company A are 12.5% of the sold pens of Company B.

General Solution:

 

Calculations:

Given Data:

Company B:

Pens Sold = 2800

Profit per pen = ₹25

Total Profit = 2800 × 25 = ₹70,000

Company D:

Pens Sold = 3600

Profit per pen = ₹90

Total Profit = 3600 × 90 = ₹3,24,000

Difference: 3,24,000 − 70,000 = 2,54,000

Thus, the total profit earned by Company D is ₹2,54,000 more than that of Company B.

General Solution:

 

Calculations:

Company A sold = 2150

Company C sold = 1650

Together = 2150 + 1650 = 3800

Company B manufactured = 3200, sold = 2800 → Unsold = 3200 - 2800 = 400

Company D manufactured = 4000, sold = 3600 → Unsold = 4000 - 3600 = 400

Together = 400 + 400 = 800

Ratio = 3800 / 800 = 19 / 4

Hence, the correct answer is 19:4.

General Solution:

 

Calculations:

Original profit per pen = ₹20

Increase 25% ⇒ new profit per pen = 20 × (1 + 25/100) = 20 × 1.25 = ₹25

Pens sold = 2150

Total profit = 25 × 2150 = ₹53,750

Hence, the correct answer is ₹53,750

General Solution:

 

Calculations:

Manufactured pens by each company:

A = 2500, B = 3200, C = 1800, D = 4000

Total Manufactured = 2500 + 3200 + 1800 + 4000 = 11,500

Average = Total ÷ Number of Companies = 11,500 ÷ 4 = 2875

Hence, the correct answer is 2875.

General Solution:

 

Calculations:

Given:

Company C: Manufactured = 1800, Sold = 1650

Unsold pens = 1800 − 1650 = 150

80% of unsold are sold next year

Original profit per pen = ₹85

Next year profit per pen = 20% less than ₹85

Formula used:

Unsold = Manufactured − Sold

Number sold next year = Unsold × 80%

Reduced profit per pen = Original profit × (1 − 20%)

Total profit from additional sales = (Number sold next year) × (Reduced profit per pen)

Calculations:

Unsold = 1800 − 1650 = 150

Number sold next year = 150 × 80% = 150 × 0.8 = 120

Reduced profit per pen = 85 × (1 − 0.20) = 85 × 0.8 = 68

Total profit = 120 × 68 = 8160

Hence, the correct answer is ₹8,160.

Shortcut Trick

Initial Ratio Milk : Water = x : y. Final Ratio = y :

x. Condition: Water added = Initial total mixture (x + y).

In dilution, Milk remains constant. x = y × (y / (x + (x + y))).

Using ratio logic: (x)/(y) → (y)/(x + added). Since milk is constant, equate parts: x = y × (x+y+y)/(x). Solving gives x = 8, y = 4.

Water added = Total Initial = 8 + 4 = 12 litres.

∴ The correct answer is 12 litres.

Alternate Method

Given: Initial Milk = 8 litres. Let initial Water = W litres.

Formula: Ratio = Milk / Water. Initial Ratio = 8/W. Final Ratio = W/8.

⇒ Total initial mixture = 8 + W

⇒ Water added = 8 + W

⇒ Final Water = Initial Water + Added Water = W + (8 + W) = 2W + 8

⇒ Final Ratio (Milk : Water) = 8 : (2W + 8)

⇒ According to question, Final Ratio = W : 8

⇒ 8 / (2W + 8) = W / 8

⇒ 64 = 2W² + 8W

⇒ W² + 4W − 32 = 0

⇒ (W + 8)(W − 4) = 0 ⇒ W = 4 (as quantity cannot be negative)

⇒ Water added = Initial mixture = 8 + 4 = 12 litres.

∴ The correct answer is 12 litres.

Additional Information

Constant Quantity Concept

When only one component (like water) is added to a mixture, the quantity of the other component (like milk) remains unchanged in absolute terms.

Ratio Reversal Rule

If initial ratio is a:b and final is b:a, the multiplier for the non-added substance must be adjusted to keep its units consistent across both ratios.

Replacement vs Addition

Replacement keeps the total volume constant, whereas addition increases the total volume of the mixture by the amount added.

Shortcut Trick

Using the speed-time relation for constant distance: S_normal = [2 × S₁ × S₂] ÷ [S₂ − S₁] where S₁ and S₂ are speed changes.

However, since time differences differ, we use: [S(S+8)/8] × 4 = [S(S−8)/8] × 6

⇒ 4(S + 8) = 6(S − 8) → 4S + 32 = 6S − 48 → 2S = 80 → S = 40 kmph.

Distance = [40 × (40+8) ÷ 8] × 4 = 5 × 48 × 4 = 960 km.

∴ The correct answer is 960 km.

Alternate Method

Given: Speed increase = 8 kmph, Time decrease = 4 hrs; Speed decrease = 8 kmph, Time increase = 6 hrs.

Formula Used: Distance (D) = [S × (S ± ΔS) ÷ ΔS] × ΔT

⇒ Case 1: D = [S(S + 8) ÷ 8] × 4

⇒ Case 2: D = [S(S − 8) ÷ 8] × 6

⇒ [S(S + 8) ÷ 8] × 4 = [S(S − 8) ÷ 8] × 6

⇒ 4S + 32 = 6S − 48

⇒ 2S = 80 ⇒ S = 40 kmph

⇒ D = [40 × (40 + 8) ÷ 8] × 4

⇒ D = 5 × 48 × 4 = 960 km

∴ The correct answer is 960 km.

Additional Information

Constant Distance Formula

When a person changes speed from S₁ to S₂ and the time difference is ΔT, Distance = (S₁ × S₂ × ΔT) ÷ |S₁ − S₂|.

Average Speed

For two equal distances covered at speeds x and y, Average Speed = (2xy) ÷ (x + y).

Speed-Time Relationship

If distance is constant, speed is inversely proportional to time (S ∝ 1/T). Ratio of speeds S₁:S₂ = T₂:T₁.

Given That,

Series 1:           

⇒ 9 + 22 = 31 

⇒ 31 + 33 = 64

⇒  64 + 46 = 110

⇒ 110 + 61 =171

⇒ 171 + 78 = 249 

22, 33, 46, 61, 78

Series 2:

⇒ 22 + 11 = 33

⇒ 33 + 13 = 46

⇒ 46 + 15 = 61

⇒ 61 + 17 = 78

Series 3:

Difference of 2

∴ 259 is the wrong number in the given pattern.

Given that,

Series 1

⇒ 11 + 78 = 89

⇒ 89 - 65 = 24

⇒ 24  + 52 = 76

⇒ 76 - 39 = 37 

⇒ 37 + 26 = 63

⇒ 63 - 13 = 50

Series 2

⇒ 13 × 6 = 78

⇒ 13 × 5 = 65

⇒ 13 × 4 = 52

⇒ 13 × 3 = 39

⇒ 13 × 2 = 26

⇒ 13 × 1 = 13

∴ The wrong number in the series is 36

Given that,

256, 260, 251, 267, 243, 278

⇒ 256 + 2² = 260

⇒ 260 - 32 = 251

⇒ 251 + 42 = 267

⇒ 267 - 52 = 242

⇒ 242 + 62 = 278

∴ 243 is the wrong number in the given pattern