Quantitative Aptitude PYQs

Shortcut Trick

Let total income of Ahaan and Kunal be 800 units (as 37.5% = 3/8).

Income: Ahaan = 300, Kunal = 500. Total = 800.

Total Savings = 50% of 800 = 400. Total Expenditure = 800 − 400 = 400.

Average Expenditure = 400 ÷ 2 = 200 units.

200 units = ₹32,000 ⇒ 1 unit = ₹160.

Total Salary (800 units) = 800 × 160 = ₹1,28,000.

∴ The correct answer is ₹1,28,000.

Alternate Method

Given: Average Expenditure = ₹32,000; Total Savings = 50% of Total Income.

Formula: Total Income = Total Expenditure + Total Savings

⇒ Total Expenditure = Average Expenditure × 2

⇒ Total Expenditure = 32,000 × 2 = ₹64,000

⇒ Since Total Savings is 50% of Total Income, Total Expenditure must also be 50% of Total Income.

⇒ 50% of Total Salary = ₹64,000

⇒ Total Salary = 64,000 × 2 = ₹1,28,000

∴ The correct answer is ₹1,28,000.

Additional Information

Income, Expenditure and Savings

The fundamental relation is Income = Expenditure + Savings. Percentage changes in these components are often solved using the linear equation or alligation method.

Percentage to Fraction Conversion

Standard conversions simplify calculations: 37.5% = 3/8, 62.5% = 5/8, 87.5% = 7/8. Using these ratios avoids decimal multiplication.

Average Concept

Average = Total Sum ÷ Number of observations. To find the Total Sum from the average, multiply the average by the number of entities.

Shortcut Trick

Volume ratio = [(1/3) × r_cone² × h_cone] / [r_tank² × h_tank]

Ratio = [(1/3) × 9 × 9 × 10] / [6 × 6 × 20]

Ratio = [270] / [720] = 3/8

Percentage = (3/8) × 100 = 37.5%

∴ The correct answer is 37.5%.

Alternate Method

Given: Height of cone (h₁) = 10 cm, Radius of cone (r₁) = 9 cm. Height of cylindrical tank (h₂) = 20 cm, Radius of tank (r₂) = 6 cm.

Formula Used: Volume of cone = (1/3)πr²h; Volume of cylinder = πr²h

⇒ Volume of water in cone = (1/3) × π × 9 × 9 × 10 = 270π cm³

⇒ Total capacity of cylindrical tank = π × 6 × 6 × 20 = 720π cm³

⇒ Percentage of water level = (Volume of water / Total capacity) × 100

⇒ Percentage = (270π / 720π) × 100

⇒ Percentage = (27/72) × 100 = (3/8) × 100 = 37.5%

∴ The correct answer is 37.5%.

Additional Information

Volume of 3D Figures

The volume of a cylinder is exactly three times the volume of a cone if they share the same radius and height.

Percentage Calculations

To find what percent X is of Y, use the formula: (X ÷ Y) × 100. Fractions like 3/8 are commonly used in exams and equal 37.5%.

Surface Area of Cone

Curved Surface Area = πrl, where l is slant height √(r² + h²). Total Surface Area = πr(l + r).

Shortcut Trick

Total profit = 5800; Profit difference (B − A) = 1000.

Individual profits: B = (5800 + 1000) ÷ 2 = 3400; A = 5800 − 3400 = 2400.

Profit ratio A : B = 2400 : 3400 = 12 : 17.

Investment ratio × Time ratio = Profit ratio ⇒ (X × 12) : ((X + 900) × 8) = 12 : 17.

3X : 2(X + 900) = 12 : 17 ⇒ 51X = 24X + 21600 ⇒ 27X = 21600 ⇒ X = 800.

B investment = X + 900 = 800 + 900 = 1700.

∴ The correct answer is Rs. 1700.

Alternate Method

Given → Profit(A+B) = 5800, Profit(B−A) = 1000. Time A = 12, Time B = 8.
Formulas: Profit Ratio = (Investment₁ × Time₁) : (Investment₂ × Time₂)

⇒ Let Profit of A be P_A and Profit of B be P_B.
⇒ P_A + P_B = 5800 and P_B − P_A = 1000.
⇒ Adding equations: 2P_B = 6800 ⇒ P_B = 3400.
⇒ Subtracting: P_A = 5800 − 3400 = 2400.
⇒ Ratio of Profit A : B = 2400 : 3400 = 12 : 17.
⇒ (X × 12) / ((X + 900) × 8) = 12 / 17.
⇒ 12X / (8X + 7200) = 12 / 17.
⇒ X / (8X + 7200) = 1 / 17 ⇒ 17X = 8X + 7200.
⇒ 9X = 7200 ⇒ X = 800.
⇒ Amount invested by B = X + 900 = 800 + 900 = 1700.

∴ The correct answer is Rs. 1700.

Additional Information

Partnership Basics

When time periods are different, the profit is divided in the ratio of the product of Investment and Time.

Investment Calculation

If Profit Ratio and Time Ratio are known, Investment Ratio = (Profit / Time) Ratio.

Shortcut Trick
New Selling Price (SP2) = Rs 1768.
Difference between SP1 and SP2 = Rs 357 ⇒ SP1 = 1768 + 357 = Rs 2125 (since buying cheaper and higher profit % usually implies SP2 < SP1, but here SP2 = 1768, so SP1 = 2125).
Let Original CP be 100x. New CP = 80x.
Ratio SP1 / SP2 = [100x × (100+P)/100] / [80x × (105+P)/100] = 2125 / 1768.
Solving for P and x gives the original cost price directly.
∴ The correct answer is Rs 1700.

Alternate Method
Given → New SP = Rs 1768, Difference = Rs 357, New CP = 80% of Original CP.
Formulas: SP = CP × (100 + P%) ÷ 100 | Profit = SP − CP.

⇒ Let Original Cost Price (CP1) = Rs C.
⇒ Original Selling Price (SP1) = C × (100 + P) / 100.
⇒ New Cost Price (CP2) = 0.8C.
⇒ New Selling Price (SP2) = 0.8C × (100 + P + 5) / 100 = Rs 1768.
⇒ Since buying cheaper and a slightly higher profit % results in a lower SP here: SP1 − SP2 = 357 or SP2 − SP1 = 357. Checking values: SP1 = 1768 + 357 = 2125.
⇒ From SP2 eq: 0.8C × (105 + P) = 176800 ⇒ C(105 + P) = 221000.
⇒ From SP1 eq: C(100 + P) = 212500.
⇒ Subtracting equations: C(105 + P) − C(100 + P) = 221000 − 212500.
⇒ 5C = 8500 ⇒ C = 1700.
∴ The correct answer is Rs 1700.

Additional Information
Profit Percentage Change
When both Cost Price and Profit Percentage change, the new Selling Price is calculated on the adjusted base price.
Cost Price Calculation
If SP and Profit% are known, CP = (SP × 100) ÷ (100 + P%).

Shortcut Trick

Ratio of A : B is 7 : 8 (Present) and 8 : 9 (After 6 years).

Change in ratio units for both is 8 − 7 = 1 unit and 9 − 8 = 1 unit.

Since the time gap is 6 years, 1 unit → 6 years.

A's present age = 7 units × 6 = 42 years.

C's present age = 42 + 10 = 52 years.

∴ The correct answer is 52 years.

Alternate Method

Given: Present age ratio A : B = 7 : 8. After 6 years, ratio A : B = 8 : 9. C = A + 10.

Calculation:

⇒ Let present ages of A and B be 7x and 8x.

⇒ (7x + 6) ÷ (8x + 6) = 8 ÷ 9

⇒ 9 × (7x + 6) = 8 × (8x + 6)

⇒ 63x + 54 = 64x + 48

⇒ 64x − 63x = 54 − 48

⇒ x = 6

⇒ A's present age = 7 × 6 = 42 years

⇒ C's present age = 42 + 10 = 52 years

∴ The correct answer is 52 years.

Additional Information

Age Difference Property

The difference between ages of two people remains constant regardless of the number of years passed.

Ratio Balancing Method

If the difference in ratio parts is not equal, multiply the ratios by the difference of the other ratio to balance them before solving.

Linear Equation in Ages

For any 'n' years after/before, add/subtract 'n' to the current age variables to form the equation: (Ax + n) / (Bx + n) = New Ratio.

Shortcut Trick

Efficiency ratio Man : Woman = 175 : 100 = 7 : 4.

Equating work: (x − 7) × 7 = (x − 1) × 4 ⇒ 7x − 49 = 4x − 4 ⇒ 3x = 45 ⇒ x = 15.

Efficiency of Child: (x − 1) × 4 = (x + 5) × Child ⇒ 14 × 4 = 20 × Child ⇒ Child = 2.8.

Total Work = (15 − 7) × 7 × 4 × 15 = 3360 units.

New Work Rate = [(15 − 11) × 7 + (15 − 5) × 2.8] × 6 = (28 + 28) × 6 = 56 × 6 = 336.

Days = 3360 ÷ 336 = 10 days.

∴ The correct answer is 10 days.

Alternate Method

Given: Efficiency of Man (M) = 1.75 × Woman (W). Time = 15 days, 4 hrs/day.

Formula Used: Total Work = Number of people × Efficiency × Hours × Days

⇒ Let efficiency W = 4, then M = 4 × 1.75 = 7.

⇒ (x − 7) × 7 = (x − 1) × 4 ⇒ 7x − 49 = 4x − 4 ⇒ 3x = 45 ⇒ x = 15.

⇒ Now, Work of (x − 1) W = Work of (x + 5) C

⇒ (15 − 1) × 4 = (15 + 5) × C ⇒ 56 = 20C ⇒ C = 2.8.

⇒ Total Work (TW) = (15 − 7) × 7 × 4 × 15 = 8 × 7 × 4 × 15 = 3360.

⇒ Required Days (D) for (15 − 11) M and (15 − 5) C at 6 hrs/day:

⇒ 3360 = [4 × 7 + 10 × 2.8] × 6 × D

⇒ 3360 = [28 + 28] × 6 × D ⇒ 3360 = 56 × 6 × D

⇒ D = 3360 ÷ 336 = 10 days.

∴ The correct answer is 10 days.

Additional Information

Work Equivalence Formula

The basic principle is M₁D₁H₁E₁ = M₂D₂H₂E₂, where M is number of persons, D is days, H is hours, and E is efficiency.

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a specific task. If A is 75% more efficient than B, Ratio of Efficiency A:B = 175:100 = 7:4.

Shortcut Trick

Ratio of efficiency M : W = (5/20) : (2/12) = 1/4 : 1/6 = 3 : 2.

Equate work: 28M × x = 35W × (x + 7).

Substitute ratio: 28(3) × x = 35(2) × (x + 7) ⇒ 84x = 70(x + 7).

6x = 5x + 35 ⇒ x = 35 days.

∴ The correct answer is 35 days.

Alternate Method

Given:

Number of men = 28, Time = x days

Number of women = 35, Time = (x + 7) days

Work ratio (20M : 12W) in same time = 5 : 2

Formula Used:

Total Work = Number of persons × Efficiency × Time

Calculations:

⇒ Let efficiency of one man be m and one woman be w.

⇒ (20 × m) / (12 × w) = 5 / 2

⇒ m / w = (5 × 12) / (20 × 2) = 60 / 40 = 3 / 2

⇒ Total work by men = Total work by women

⇒ 28 × m × x = 35 × w × (x + 7)

⇒ 28 × 3 × x = 35 × 2 × (x + 7)

⇒ 84x = 70x + 490

⇒ 14x = 490

⇒ x = 490 ÷ 14

⇒ x = 35 days

∴ The correct answer is 35 days.

Additional Information

Man-Days Concept

The total work is constant and can be expressed as M₁D₁H₁E₁ = M₂D₂H₂E₂, where M is men, D is days, H is hours, and E is efficiency.

Efficiency and Time Relationship

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

Shortcut Trick

Ratio of efficiency M : W = (5/20) : (2/12) = 1/4 : 1/6 = 3 : 2.

Work Equivalence: 28 × 3 × x = 35 × 2 × (x + 7) ⇒ 84x = 70x + 490 ⇒ 14x = 490 ⇒ x = 35.

Time for 1 Man = 28 × 35 = 980 days.

Time for 1 Woman = 35 × 42 = 1470 days.

Difference = 1470 − 980 = 490 days.

∴ The correct answer is 490 days.

Alternate Method

Given: 28 men do work in x days; 35 women do work in (x + 7) days. Ratio of work by 20 men and 12 women in same time = 5 : 2.

Formula Used: Total Work = Number of persons × Days × Efficiency.

⇒ Efficiency Ratio: (20 × m) / (12 × w) = 5 / 2

⇒ m / w = (5 × 12) / (2 × 20) = 60 / 40 = 3 / 2

⇒ Total Work Equality: 28 × 3 × x = 35 × 2 × (x + 7)

⇒ 84x = 70x + 490

⇒ 14x = 490 ⇒ x = 35 days

⇒ Days taken by 1 Man = 28 × 35 = 980 days

⇒ Days taken by 1 Woman = 35 × (35 + 7) = 35 × 42 = 1470 days

⇒ Difference = 1470 − 980 = 490 days

∴ The correct answer is 490 days.

Additional Information

Man-Days Concept

Total work remains constant. If number of persons is reduced to 1, the time taken increases proportionately: T₁ = M × D.

Efficiency and Time

Efficiency is inversely proportional to time taken. If the ratio of efficiencies of A and B is a:b, then the ratio of time taken is b:a.

Work and Wages

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

Solution:

I. \(\frac{16x}{x^{3/2}}-\frac{63}{x^{3/2}} = x^{1/2}\)

16x - 63 = x^1/2 * x^3/2

16x - 63 = x²

x² - 16x + 63 = 0

x2 - 9x - 7x + 63 = 0

x(x - 9) - 7(x - 9) = 0

(x - 9) (x - 7)(x - 9)

x = 9 , 7

II. \(\frac{75y}{2^2}+\frac{y^2}{2^2} = (-)\sqrt(5041)\)

75y + y² = -71 × 4

75y + y2 + 284 = 0

y2 + 71y + 4y + 284 = 0

(y + 71) (y + 4) = 0

y = -71, -4

Comparison between x and y (via Tabulation):

So, x > y

Hence Option 1 is correct.

12x² - ax + 7 = ax² + 9x + 3

⇒ 12x² - ax + 7 - ax² - 9x - 3 = 0

⇒ (12 - a)x² - x(a + 9) + 4 = 0

For only one (repeated) solution, b² - 4ac = 0

⇒ (a + 9)² - 16(12 - a) = 0

⇒ a² + 18a + 81 - 192 + 16a = 0

⇒ a² + 34a - 111 = 0

⇒ (a + 37)(a - 3) = 0

∴ Positive integral solution of a = 3

Let, graduated employee in company A = 3x

And, Post graduated employee in company A = 2x

Graduated employee in company B = 120

Let, post graduated employee in company B = 4y

And, post graduated employee in company C = 5y

So, 5y = 2 ×5/12 × 120 = 100 [41(2/3)% = 5/12]

⇒y = 20

Therefore, post graduated employee in company B = 4 × 20 = 80

Graduated employee in company A = 3x  = 2.5 × 120 = 300

So, x = 100

Therefore, post graduated employee in company A = 2 × 100 = 200

Graduated employee in company C = 1/3 × (300 + 120) / 2 = 70

Required average = (200 + 80 + 100) / 3 = 126.6 ≈ 127

Hence, correct answer is option 2.

Let, graduated employee in company A = 3x

And, Post graduated employee in company A = 2x

Graduated employee in company B = 120

Let, post graduated employee in company B = 4y

And, post graduated employee in company C = 5y

So, 5y = 2 ×5/12 × 120 = 100 [41(2/3)% = 5/12]

⇒y = 20

Therefore, post graduated employee in company B = 4 × 20 = 80

Graduated employee in company A = 3x  = 2.5 × 120 = 300

So, x = 100

Therefore, post graduated employee in company A = 2 × 100 = 200

Graduated employee in company C = 1/3 × (300 + 120) / 2 = 70

Total graduated employees in company D = 70 × 130/100 = 91

Total employees in company D = (120 + 80) × 185/100 = 370

So, total post graduated employees in company D = 370 - 91 = 279

Hence, correct answer is option 5.

Let, graduated employee in company A = 3x

And, Post graduated employee in company A = 2x

Graduated employee in company B = 120

Let, post graduated employee in company B = 4y

And, post graduated employee in company C = 5y

So, 5y = 2 ×5/12 × 120 = 100 [41(2/3)% = 5/12]

⇒y = 20

Therefore, post graduated employee in company B = 4 × 20 = 80

Graduated employee in company A = 3x  = 2.5 × 120 = 300

So, x = 100

Therefore, post graduated employee in company A = 2 × 100 = 200

Graduated employee in company C = 1/3 × (300 + 120) / 2 = 70

Post graduated employees in company A play table tennis = 200 × 30/100 = 60

Post graduated employees in company A play cricket = 200 × 25/100 = 50

So, post graduated employees in company A play football = (200 - 110) / 18 × 11 = 90/18 × 11 = 55

Hence, correct answer is option 1.

Let, graduated employee in company A = 3x

And, Post graduated employee in company A = 2x

Graduated employee in company B = 120

Let, post graduated employee in company B = 4y

And, post graduated employee in company C = 5y

So, 5y = 2 ×5/12 × 120 = 100 [41(2/3)% = 5/12]

⇒y = 20

Therefore, post graduated employee in company B = 4 × 20 = 80

Graduated employee in company A = 3x  = 2.5 × 120 = 300

So, x = 100

Therefore, post graduated employee in company A = 2 × 100 = 200

Graduated employee in company C = 1/3 × (300 + 120) / 2 = 70

Post graduated employees in company D = 80/2 × 3 = 120

Graduated employees in company D = 300 - 120 = 180

Total number of employees = 180 + 70 = 250 

Required Ratio = 250 : 280 = 25 : 28

Hence, correct answer is option 3.

Total number of mangoes sold in May = 1080 × 20/100 = 216

Total number of Fresh mangoes sold in May = 216 × 50/100 = 108

Cost price of all mangoes in May = 1400

Selling price of all fresh mangoes sold in May = 108 × 10 = 1080

Selling price of all rotten mango  = 108 × 10 × 40/100 = 432 (Given 60% discount on actual price of each mango)

Total selling price of all mangoes = 1512

Profit = 1512 = 1400 = 112

So, profit% = 112/1400 × 100 = 8%

Hence, correct answer is option 4.

Shortcut Trick

Rate of salary = ₹100 per 30 mins = ₹200 per hour.

Total hours = (4 hours × 7 days) + (5 hours × 7 days) + (6 hours × 14 days).

Total hours = 28 + 35 + 84 = 147 hours.

Total Amount = 147 × 200 = ₹29400.

∴ The correct answer is ₹29400.

Alternate Method

Given: Salary rate = ₹100 / 30 min; Week 1 = 4 hrs/day; Week 2 = 5 hrs/day; Week 3 & 4 = 6 hrs/day.

Formula: Total Salary = (Total Working Hours) × (Hourly Rate)

⇒ Hourly rate = (100 ÷ 30) × 60 = ₹200/hour

⇒ Total hours = (7 × 4) + (7 × 5) + (14 × 6)

⇒ Total hours = 28 + 35 + 84 = 147 hours

⇒ Total salary = 147 × 200 = ₹29400

∴ The correct answer is ₹29400.

Additional Information

Unitary Method in Salary

Salary is usually calculated by multiplying the Total Work Units (hours/days) by the Rate per Unit.

Work and Wages

Wages are distributed in proportion to the Work Done. If time is constant, wages ∝ Efficiency × Days.

Solution:

To find out the principal amount 

SI = Pnr/100

x = (15000 – x) × 20 × 3/100

5x = (45000 – 3x)

8x = 45000

x = 5625

So the principal is (15000 – 5625) = Rs. 9375

Quantity I:

Simple Interest for 3 years at 10 % per annum

SI = (10375 × 3 × 10)/100 = 3112.5

Simple interest for 3 years = 3112.5

Quantity II:

Compound Interest at (x + 4375) 

CI = P( 1+ R/100)³

CI = 10000 [(1+ 5/100)³ – 1]

CI = 10000 × [(21³/20³) - 1]

CI = 10000[(9261 - 8000)/8000]

CI = Rs 1576.25

Thus, Quantity I > Quantity II

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

17.972 × 4.99 ÷ 44.78 + 299.99 ÷ √5624.96 × 12.97 = ?

⇒ 18² × 5 ÷ 45 + 300 ÷ √5625 × 13 = ?

⇒ 324 × 1 ÷ 9 + 300 ÷ 75 × 13 = ?

⇒ 36 + 4 × 13 = ?

⇒ 36 + 52 = ?

⇒ 88 = ?

∴ Required value of ? is 88.

Given:

11.11% of 99.11 + 22.22% of 98.97 – 9.99% of 100 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

Since we need to find out the approximate value, we can write these values to their nearest integers.

11.11% of 99.17 + 22.22% of 98.87 - 9.89% of 100.12 = ?

⇒ 11.11% of 99 + 22.22% of 99 - 10% of 100 = ?

⇒ (1/9) × 99 + (2/9) × 99 - (1/10) × 100 = ?

⇒ 11 + 22  - 10 = ?

⇒ 33 - 10 = ?

⇒ ? = 23

∴ The value of ‘?’ Is 23.

Alternate Method

11.11% of 99 + 22.22% of 99 - 10% of 100 = ?

⇒ (1/9) × 99 + (2/9) × 99 - (1/10) × 100 = ?

⇒ {(1/9) + (2/9)} × 99 - 10 = ?

⇒ (1/3) × 99 - 10 = ?

⇒ 33 - 10 = ?

⇒ ? = 23

∴ The value of ‘?’ Is 23.

Quantity 1 -

\(X^2 - 3\frac{1}{5}X + 2\frac{2}{5} = 0\) 

X² - 16/5X + 12/5 = 0

5X² - 16X + 12 = 0

5X² - 10X - 6X + 12 = 0

5X(X - 2) - 6 (X - 2) = 0

X = 2 , 6/5

Quantity 2-

\(Y^2 - 2\frac{1}{6}Y + \frac{5}{6} = 0\)

Y² - 13/6Y + 5/6 = 0

6Y² - 13Y + 5 = 0

6Y² - 10Y - 3Y + 5 = 0

2Y(3Y - 5) - (3Y - 5) = 0

Y = 1/2, 5/3

So, we can not determine the relation between Quantity 1 and Quantity 2.

Hence, correct answer is option 4.