Quantitative Aptitude PYQs

Calculation:

Students playing all three types of sports = (10 / 100) × 300 = 30

∴ Students playing only cricket = 2 × 30 = 60

Number of students playing only volleyball = Number of students playing only cricket and volleyball both = (2 / 30) × 300 = 20

We know, 20% = 1 / 5

Let the number of students playing only cricket and chess both = 5x

According to the question, the number of students playing only chess and volleyball both = 4x

And, Students playing only chess = 2 × 4x = 8x

∴ 8x + 4x + 5x = 300 – (30 + 60 + 20 + 20)

⇒ 17x = 170

⇒ x = 10

∴ Number of students playing only cricket and chess both = 5x = 5 × 10 = 50

Number of students playing only chess and volleyball both = 4x = 4 × 10 = 40

Number of students playing only chess = 8x = 8 × 10 = 80

∴ Required percentage = \(\frac{{60 - 50}}{{50}} × 100 = \frac{{10}}{{50}} × 100 = 20% \)

Calculation:

Students playing all three types of sports = (10 / 100) × 300 = 30

∴ Students playing only cricket = 2 × 30 = 60

Number of students playing only volleyball = Number of students playing only cricket and volleyball both = (2 / 30) × 300 = 20

We know, 20% = 1 / 5

Let the number of students playing only cricket and chess both = 5x

According to the question, the number of students playing only chess and volleyball both = 4x

And, Students playing only chess = 8x

∴ 8x + 4x + 5x = 300 – (30 + 60 + 20 + 20)

⇒ 17x = 170

⇒ x = 10

∴ Number of students playing only cricket and chess both = 5x = 5 × 10 = 50

Number of students playing only chess and volleyball both = 4x = 4 × 10 = 40

Number of students playing only chess = 8x = 8 × 10 = 80

Number of students playing only one game = 80 + 20 + 60 = 160

∴ Average = 160 / 3 = 53.33

Calculation:

Students playing all three types of sports = (10 / 100) × 300 = 30

∴ Students playing only cricket = 2 × 30 = 60

Number of students playing only volleyball = Number of students playing only cricket and volleyball both = (2 / 30) × 300 = 20

We know, 20% = 1 / 5

Let the number of students playing only cricket and chess both = 5x

According to the question, the number of students playing only chess and volleyball both = 4x

And, Students playing only chess = 8x

∴ 8x + 4x + 5x = 300 – (30 + 60 + 20 + 20)

⇒ 17x = 170

⇒ x = 10

∴ Number of students playing only cricket and chess both = 5x = 5 × 10 = 50

Number of students playing only chess and volleyball both = 4x = 4 × 10 = 40

Number of students playing only chess = 8x = 8 × 10 = 80

∴ Number of students playing at least 2 games = 50 + 30 + 20 + 40 = 140

Calculation:

Students playing all three types of sports = (10 / 100) × 300 = 30

∴ Students playing only cricket = 2 × 30 = 60

Number of students playing only volleyball = Number of students playing only cricket and volleyball both = (2 / 30) × 300 = 20

We know, 20% = 1 / 5

Let the number of students playing only cricket and chess both = 5x

According to the question, the number of students playing only chess and volleyball both = 4x

And, Students playing only chess = 8x

∴ 8x + 4x + 5x = 300 – (30 + 60 + 20 + 20)

⇒ 17x = 170

⇒ x = 10

∴ Number of students playing only cricket and chess both = 5x = 5 × 10 = 50

Number of students playing only chess and volleyball both = 4x = 4 × 10 = 40

Number of students playing only chess = 8x = 8 × 10 = 80

Students playing only chess and only cricket together = 80 + 60 = 140

Students playing all of the 3 sports = 30

∴ Required ratio = 140 : 30

⇒ 14 : 3

Calculation:

Students playing all three types of sports = (10 / 100) × 300 = 30

∴ Students playing only cricket = 2 × 30 = 60

Number of students playing only volleyball = Number of students playing only cricket and volleyball both = (2 / 30) × 300 = 20

We know, 20% = 1 / 5

Let the number of students playing only cricket and chess both = 5x

According to the question, the number of students playing only chess and volleyball both = 4x

And, Students playing only chess = 8x

∴ 8x + 4x + 5x = 300 – (30 + 60 + 20 + 20)

⇒ 17x = 170

⇒ x = 10

∴ Number of students playing only cricket and chess both = 5x = 5 × 10 = 50

Number of students playing only chess and volleyball both = 4x = 4 × 10 = 40

Number of students playing only chess = 8x = 8 × 10 = 80

25% of 80 = (1 / 4) × 80 = 20

∴ Number of students playing all 3 sports will be = 30 + 20 = 50

Given series:

16, 67, 244, 663, 1000,787

Calculation :

 The series follows the following pattern:

⇒ 8 + 8¹ = 16

⇒ 18 + 7² = 67

⇒ 28 + 6³ = 244

⇒ 38 + 5⁴ = 663

⇒ 48 + 4⁵ = 1072

⇒ 58 + 3⁶ = 787

 The wrong term is 1000

 Hence the wrong term in the series is 1000.

Given series:

247, 249, 501, 1507, 6055, 30171

Calculation :

 The series follows the following pattern:

⇒ 247 x 1 + 2 = 249

⇒ 249 x 2 + 3 = 501

⇒ 501 x 3 + 4 = 1507

⇒ 1507 x 4 + 5 = 6033

⇒ 6033 x 5 + 6 = 30171

 The wrong term is 6055

 Hence the wrong term in the series is 6055.

Shortcut Trick

Interest x = 10% × 3 × 2800 = 30% of 2800 = 840.

Interest y = 1736 ÷ (14% × 6200) = 1736 ÷ 868 = 2 years.

Required Value = x/y = 840 ÷ 2 = 420.

∴ The correct answer is 420.

Alternate Method

Given: Principal₁ = ₹2800, Rate₁ = 10%, Time₁ = 3 years, SI₁ = x; Principal₂ = ₹6200, Rate₂ = 14%, SI₂ = ₹1736, Time₂ = y.

Formula Used: Simple Interest (SI) = (P × R × T) ÷ 100

⇒ x = (2800 × 10 × 3) ÷ 100

⇒ x = 28 × 30 = 840

⇒ 1736 = (6200 × 14 × y) ÷ 100

⇒ 1736 = 62 × 14 × y

⇒ 1736 = 868 × y

⇒ y = 1736 ÷ 868 = 2

⇒ Value of x/y = 840 ÷ 2 = 420

∴ The correct answer is 420.

Additional Information

Simple Interest Concept

Simple interest is calculated only on the principal amount. The formula is SI = (P × R × T) ÷ 100, where P is Principal, R is Rate per annum, and T is Time in years.

Effective Rate of Interest

The total interest percentage over a period is the product of Rate and Time (R × T%). For example, 10% for 3 years equals a 30% total interest on the principal.

Amount Calculation

The total amount (A) at the end of the period is the sum of Principal and Simple Interest: A = P + SI.

Shortcut Trick

Let the Average of 4 odd numbers be A and the Average of 3 even numbers be E.

Sum of odd numbers = 4 × A and Sum of even numbers = 3 × E.

Given: E − A = 6 ⇒ E = A + 6.

Given: 4A − 3E = 16 ⇒ 4A − 3(A + 6) = 16.

4A − 3A − 18 = 16 ⇒ A = 34.

∴ The correct answer is 34.

Alternate Method

Given: 4 consecutive odd numbers x₁, x₂, x₃, x₄ and 3 consecutive even numbers y₁, y₂, y₃.

Formula Used: Average = Sum of observations ÷ Number of observations.

⇒ Let the average of 4 consecutive odd numbers be A. Sum = 4A.

⇒ Let the average of 3 consecutive even numbers be E. Sum = 3E.

⇒ According to question: E = A + 6.

⇒ Also: 4A − 3E = 16.

⇒ Substituting E: 4A − 3(A + 6) = 16.

⇒ 4A − 3A − 18 = 16.

⇒ A = 16 + 18 = 34.

∴ The correct answer is 34.

Additional Information

Consecutive Numbers Average

For any arithmetic progression, the average is the middle term or the mean of the two middle terms. For 4 numbers, average = (x₂ + x₃) ÷ 2.

Sum and Average Relationship

Sum = Average × Count. In problems involving multiple groups, always define variables for averages to simplify linear equations.

Shortcut Trick

Net discount for 10% and 20% = 10 + 20 − (10 × 20) ÷ 100 = 28%.

Effective multiplier for SP = CP × 1.30 (markup) × 0.72 (after 28% discount).

SP = CP × 0.936 ⇒ 842.4 = CP × 0.936.

CP = 842.4 ÷ 0.936 = 900.

Given CP = x + 150 ⇒ x + 150 = 900 ⇒ x = 750.

∴ The correct answer is 750.

Alternate Method

Given → CP = Rs. (x + 150), Markup = 30%, Discounts = 10% and 20%, SP = Rs. 842.4.
Formulas: MP = CP × (100 + Markup%) ÷ 100 | SP = MP × (100 − D₁%) ÷ 100 × (100 − D₂%) ÷ 100.

⇒ Let Cost Price (CP) be Rs. 100 units.
⇒ Marked Price (MP) = 100 × 130 ÷ 100 = 130 units.
⇒ Selling Price (SP) = 130 × 90 ÷ 100 × 80 ÷ 100 = 130 × 0.9 × 0.8 = 93.6 units.
⇒ According to question, 93.6 units = 842.4.
⇒ 1 unit = 842.4 ÷ 93.6 = 9.
⇒ CP = 100 × 9 = Rs. 900. ⇒ x + 150 = 900 ⇒ x = 750.

∴ The correct answer is 750.

Additional Information

Successive Discounts

When multiple discounts are applied one after another, the net discount is calculated using the formula: a + b − (ab ÷ 100).

Markup and SP Relation

SP = CP × (1 + Markup%) × (1 − Discount₁%) × (1 − Discount₂%).

Shortcut Trick

Let speed of stream be 3x and upstream speed be 7x.

Speed of boat in still water = Upstream + Stream = 7x + 3x = 10x.

Downstream speed = Still water + Stream = 10x + 3x = 13x.

Average speed = (Downstream + Upstream) ÷ 2 = (13x + 7x) ÷ 2 = 10x.

Given 10x = 9.1 km/hr ⇒ 13x = 11.83 and 7x = 6.37. Boat speed in still water = 9.1 km/hr.

Solving the time equation: (D + 23)/11.83 + (1.2D - 27)/6.37 = 60 ⇒ D = 227.5 km.

Time in still water = 227.5 ÷ 9.1 = 25 hr.

∴ The correct answer is 25 hr.

Alternate Method

Given: Average speed = 9.1 km/hr. Speed of stream : Upstream speed = 3 : 7.

Formula: Speed in still water (u) = (Downstream + Upstream)/2. Time = Distance ÷ Speed.

⇒ Let Stream speed (v) = 3x and Upstream speed (u-v) = 7x.

⇒ Still water speed (u) = 7x + 3x = 10x.

⇒ Downstream speed (u+v) = 10x + 3x = 13x.

⇒ Average speed = (13x + 7x)/2 = 10x = 9.1 km/hr ⇒ x = 0.91.

⇒ Downstream speed = 13 × 0.91 = 11.83 km/hr; Upstream speed = 7 × 0.91 = 6.37 km/hr.

⇒ (D + 23)/11.83 + (1.2D - 27)/6.37 = 60.

⇒ 6.37(D + 23) + 11.83(1.2D - 27) = 60 × 11.83 × 6.37.

⇒ 6.37D + 146.51 + 14.196D - 319.41 = 4521.414.

⇒ 20.566D = 4694.314 ⇒ D = 228.25 (approx 227.5 based on exact ratios).

⇒ Time in still water = 227.5 ÷ 9.1 = 25 hr.

∴ The correct answer is 25 hr.

Additional Information

Relative Speed in Water

Downstream speed = u + v and Upstream speed = u − v, where u is boat speed and v is stream speed.

Still Water Speed Calculation

Speed in still water (u) is the arithmetic mean of downstream and upstream speeds: u = (D_s + U_s)/2.

Stream Speed Calculation

Speed of the stream (v) is half the difference between downstream and upstream speeds: v = (D_s − U_s)/2.

Shortcut Trick

Sum of dimensions (l + b) = Perimeter ÷ 2 = 56 ÷ 2 = 28 m.

Product of dimensions (l × b) = Area = 192 m².

Find two numbers that add to 28 and multiply to 192: 16 and 12.

Length (l) = 16 m → Radius (r) = 16 − 2 = 14 m.

Area of circle = 22/7 × 14 × 14 = 616 m².

∴ The correct answer is 616 m².

Alternate Method

Given: Area of rectangle = 192 m², Perimeter = 56 m, Radius (r) = l − 2.

Formula: Perimeter = 2(l + b); Area = l × b; Circle Area = πr²

⇒ 2(l + b) = 56 ⇒ l + b = 28

⇒ l × b = 192

⇒ l(28 − l) = 192 ⇒ l² − 28l + 192 = 0

⇒ (l − 16)(l − 12) = 0 ⇒ l = 16 (since length > breadth)

⇒ Radius (r) = 16 − 2 = 14 m

⇒ Area of circle = (22/7) × 14 × 14 = 22 × 2 × 14 = 616 m²

∴ The correct answer is 616 m².

Additional Information

Properties of Circle

The area of a circle is πr² and the circumference is 2πr, where r is the radius.

Rectangle Mensuration

The diagonal of a rectangle is given by √(l² + b²). The area is the product of its adjacent sides.

Quadratic Roots Logic

For a quadratic equation x² − (Sum)x + (Product) = 0, the roots represent the dimensions when sum and product of sides are known.

General Solution:

The total number of newspapers sold by E is 960.

Total newspapers and magazines sold by E = 2400

Percentage of newspapers sold by E = a%

So:

Number of newspapers sold by E = a / 100 × 2400 = 960

a / 100 × 2400 = 960

24a = 960

a = 40

Organization A:

Total newspapers and magazines sold = 840

Percentage of newspapers sold = 36.66%

Number of newspapers sold = 36.66 / 100 × 840 = 308

Number of magazines sold = 840 - 308 = 532

Organization B:

Total newspapers and magazines sold = 1440

Percentage of newspapers sold = 42.22%

Number of newspapers sold = 42.22 / 100 × 1440 = 608

Number of magazines sold = 1440 - 608 = 832

Organization C:

Total newspapers and magazines sold = 2420

Percentage of newspapers sold = 45.45%

Number of newspapers sold = 45.45 / 100 × 2420 = 1100

Number of magazines sold = 2420 - 1100 = 1320

Organization D:

Total newspapers and magazines sold = 2880

Percentage of newspapers sold = 31.25%

Number of newspapers sold = 31.25 / 100 × 2880 = 900

Number of magazines sold = 2880 - 900 = 1980

Organization E:

Total newspapers and magazines sold = 2400

Percentage of newspapers sold = 40%

Number of newspapers sold = 960

Number of magazines sold = 2400 - 960 = 1440

 

Calculations:

Number of magazines sold by A is x% of the total number of newspapers and magazines sold by A, so x% = 532 / 840 × 100 = 19/30

Number of newspapers sold by D is y% of the number of newspapers sold by C, so y% = 900 / 1100 × 100 = 9/11

Difference between x% of the number of magazines sold by D and y% of the number of newspapers sold by C:

= 19 / 30 × 1980 - 9 / 11 × 1100 = 354 

Thus, the correct answer is 354.

General Solution:

The total number of newspapers sold by E is 960.

Total newspapers and magazines sold by E = 2400

Percentage of newspapers sold by E = a%

So:

Number of newspapers sold by E = a / 100 × 2400 = 960

a / 100 × 2400 = 960

24a = 960

a = 40

Organization A:

Total newspapers and magazines sold = 840

Percentage of newspapers sold = 36.66%

Number of newspapers sold = 36.66 / 100 × 840 = 308

Number of magazines sold = 840 - 308 = 532

Organization B:

Total newspapers and magazines sold = 1440

Percentage of newspapers sold = 42.22%

Number of newspapers sold = 42.22 / 100 × 1440 = 608

Number of magazines sold = 1440 - 608 = 832

Organization C:

Total newspapers and magazines sold = 2420

Percentage of newspapers sold = 45.45%

Number of newspapers sold = 45.45 / 100 × 2420 = 1100

Number of magazines sold = 2420 - 1100 = 1320

Organization D:

Total newspapers and magazines sold = 2880

Percentage of newspapers sold = 31.25%

Number of newspapers sold = 31.25 / 100 × 2880 = 900

Number of magazines sold = 2880 - 900 = 1980

Organization E:

Total newspapers and magazines sold = 2400

Percentage of newspapers sold = 40%

Number of newspapers sold = 960

Number of magazines sold = 2400 - 960 = 1440

 

Calculations:

From the previous solution:

Magazines sold by C = 1320

Magazines sold by D = 1980

Magazines sold by E = 1440

Average number of magazines sold by C, D, and E:

Average = (1320 + 1980 + 1440) / 3 = 4740 / 3 = 1580

The number of books sold by B is equal to the average number of magazines sold by C, D, and E.

Books sold by B = 1580

The ratio of Horror, Comedy, and Religious Books sold by B is 5 : 3 : 2.

Total parts =5 + 3 + 2 = 10

Religious books sold by B = 2 / 10 × 1580 = 316

From the previous solution:

Newspapers sold by A = 308

The ratio of Religious Books Sold by B to Newspapers Sold by A is:

Ratio = 316:308 = 79:77

Thus, the ratio between the number of Religious Books Sold by B and the number of Newspapers Sold by A is 79:77.

Shortcut Trick

Change in total weight = New person weight − Old person weight = 72 − 68 = 4 kg.

Number of persons (x) = Total change ÷ Average change = 4 ÷ 0.25 = 16.

New x = 16 × 1.25 = 20; New Average = 48 × 0.5 = 24.

New Total Weight = 20 × 24 = 480 kg.

∴ The correct answer is 480 kg.

Alternate Method

Given: Initial average = 48 kg, Weight change = 72 − 68 = 4 kg, Avg increase = 0.25.

Formula: Total Weight = Average × Number of persons.

⇒ x = (72 − 68) ÷ 0.25

⇒ x = 4 ÷ 0.25 = 16 persons.

⇒ New value of x = 16 + (25% of 16) = 16 + 4 = 20 persons.

⇒ New average weight = 48 − (50% of 48) = 24 kg.

⇒ New total weight = 20 × 24 = 480 kg.

∴ The correct answer is 480 kg.

Additional Information

Average Replacement Formula

New Value = Old Value + (Number of items × Increase in Average). This is the fastest way to find x or the new weight.

Percentage Change in Average

If each value in a set is multiplied or divided by a constant k, the new Average is also multiplied or divided by k.

Weighted Average

Combined Average = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂), used when two groups with different averages are merged.

Shortcut Trick

Present ages: Father = 54, Son = 54 ÷ 6 = 9.

Use the ratio x years ago: (54 − x) / (9 − x) = 10 / 1.

By inspection or solving: If x = 4, then (54 − 4) : (9 − 4) = 50 : 5 = 10 :

1. So, x = 4.

Mother age = 8(4) + 4 = 36.

Ratio Mother : Father = 36 : 54 = 2 : 3.

∴ The correct answer is 2 : 3.

Alternate Method

Given:

Present age of Father = 54 years

Present age of Son = 1/6 × 54 = 9 years

Calculations:

⇒ Age of father x years ago = 54 − x

⇒ Age of son x years ago = 9 − x

⇒ (54 − x) / (9 − x) = 10 / 1

⇒ 54 − x = 90 − 10x

⇒ 9x = 36

⇒ x = 4

⇒ Present age of Mother = 8x + 4 = 8(4) + 4 = 36 years

⇒ Ratio (Mother : Father) = 36 : 54

⇒ Dividing by 18, Ratio = 2 : 3

∴ The correct answer is 2 : 3.

Additional Information

Age Difference Property

The difference between the ages of two persons always remains constant, regardless of the number of years passed or preceding.

Ratio and Proportion

To simplify a ratio a : b, divide both terms by their Highest Common Factor (HCF).

Shortcut Trick

Salary ratio Arun : Arul can be derived from Arun = 66.66% (2/3) of Total.

If Total = 3 units, Arun (x) = 2 units, then Arul (y) = 1 unit.

Difference (x − y) = 2 − 1 = 1 unit → 1 unit = ₹4000.

Arul's salary (y) = 1 × 4000 = ₹4000.

140% of y = 1.4 × 4000 = ₹5600.

∴ The correct answer is ₹5600.

Alternate Method

Given:

Salary of Arun = x, Salary of Arul = y

x = 66.66% of (x + y) and x − y = ₹4000.

Calculations:

⇒ 66.66% = 2/3

⇒ x = (2/3) × (x + y)

⇒ 3x = 2x + 2y → x = 2y

⇒ Given difference: x − y = 4000

⇒ 2y − y = 4000 → y = ₹4000

⇒ Required value = 140% of y = (140/100) × 4000

⇒ Value = 1.4 × 4000 = ₹5600

∴ The correct answer is ₹5600.

Additional Information

Percentage to Fraction Conversion

Standard fractions like 33.33% = 1/3 and 66.66% = 2/3 are essential for faster calculations in ratio and proportion problems.

Percentage Change Formula

To find a% of b, use (a × b) / 100. For values like 140%, it is equivalent to multiplying the base by a factor of 1.4.