Quantitative Aptitude PYQs

Given:

If x² - 5x + 4 is a factor of x⁴ - px² + q.

Formula used:

If (x² - 5x + 4) is a factor of (x⁴ - px² + q), then the roots of x² - 5x + 4 (i.e., 1 and 4) must satisfy x⁴ - px² + q = 0.

Calculation:

Roots of x² - 5x + 4 are x = 1 and x = 4.

Substitute x = 1 in x⁴ - px² + q = 0

⇒ 1⁴ - p(1²) + q = 0

⇒ 1 - p + q = 0

⇒ q = p - 1 ----- (1)

Substitute x = 4 in x⁴ - px² + q = 0

⇒ 4⁴ - p(4²) + q = 0

⇒ 256 - 16p + q = 0

⇒ q = 16p - 256 ------ (2)

Equate q values from 1 and 2

⇒ p - 1 = 16p - 256

⇒ 255 = 15p

⇒ p = 17

Substitute p = 17 into q = p - 1

⇒ q = 17 - 1

⇒ q = 16

∴ The values of p and q are 17 and 16 respectively.

The correct answer is option (1).

Given:

Two quadratic equations: \(px^2 + px + 4 = 0\) and \(x^2 + qx + q = 0\)

Common root = 2

Formula used:

For a quadratic equation \(ax^2 + bx + c = 0\), if a root is given, substitute it into the equation to find the unknowns.

Calculation:

Substituting the root 2 into the first equation:

\(p(2)^2 + p(2) + 4 = 0\)

⇒ \(4p + 2p + 4 = 0\)

⇒ \(6p + 4 = 0\)

⇒ \(6p = -4\)

⇒ \(p = -\frac{2}{3}\)

Substituting the root 2 into the second equation:

\( (2)^2 + q(2) + q = 0 \)

⇒ \(4 + 2q + q = 0\)

⇒ \(4 + 3q = 0\)

⇒ \(3q = -4\)

⇒ \(q = -\frac{4}{3}\)

Now, calculate \(p + q\):

\(p + q = -\frac{2}{3} - \frac{4}{3}\)

⇒ \(p + q = -\frac{6}{3}\)

⇒ \(p + q = -2\)

∴ The correct answer is option (2).

Given:

Find the HCF of: f(x) = x⁸ + x⁴ + 1 and g(x) = x⁴ + x² + 1

Formula used:

a² + ab + b² = (a + b)² − ab

f(x) = x⁸ + x⁴ + 1 = (x⁴)² + x⁴ × 1 + 1²

So f(x) = x⁸ + x⁴ + 1 = a² + ab + b² where a = x⁴, b = 1

⇒ f(x) = (x⁴)² + x⁴ × 1 + 1² = x⁸ + x⁴ + 1

Similarly, g(x) = x⁴ + x² + 1 = a² + ab + b² where a = x², b = 1

So: f(x) = a² + ab + b² where a = x⁴, b = 1

g(x) = a² + ab + b² where a = x², b = 1

Now, we know that: a² + ab + b² = (a + b)² − ab

So: f(x) = (x⁴ + 1)² − x⁴

f(x) = (x⁴ + x² + 1)(x⁴ − x² + 1)

f(x) = x⁸ + x⁴ + 1 = (x⁴ + x² + 1)(x⁴ − x² + 1)

g(x) = x⁴ + x² + 1

Since x4 + x2 + 1 is common in both terms.

∴ HCF is: x⁴ + x² + 1

Given:

An arc AB of a circle subtends an angle x radian at the center O.

The area of sector AOB = Half of the square of the length of arc AB.

Formula used:

Area of sector = (1/2) × r² × x

Length of arc = r × x

Given condition: (1/2) × r² × x = (1/2) × (r × x)²

Calculations:

(1/2) × r² × x = (1/2) × r² × x²

⇒ r² × x = r² × x²

⇒ x = x²

⇒ x² - x = 0

⇒ x(x - 1) = 0

⇒ x = 0 or x = 1

Since x cannot be 0 (as it represents an angle),

∴ x = 1 radian.

Correct answer is option (3).

Given:

p and q are prime numbers

Statement I: Their LCM is always an odd number

Formula used:

LCM of two numbers is odd only if both numbers are odd

Among prime numbers, 2 is the only even prime

Calculation:

If p = 2, q = 3 ⇒ LCM = 6 (even)

So, LCM is not always odd

Statement I is incorrect

Statement II: Sum of their LCM and HCF is always even

HCF of two distinct primes = 1

LCM of p and q = p × q

Sum = p × q + 1

Check:

p = 2, q = 3 ⇒ LCM = 6, HCF = 1 ⇒ Sum = 6 + 1 = 7  odd

p = 3, q = 5 ⇒ LCM = 15, HCF = 1 ⇒ Sum = 15 + 1 = 16  even

p = 3, q = 7 ⇒ 21 + 1 = 22  even

But counter example: p = 2, q = 3 ⇒ 6 + 1 = 7 

Statement II is incorrect

∴ The correct answer is: Neither I nor II.

Given:

Frequency distribution of X with total observations = 205

x = 3, 5, 6, 7

Frequencies = f, f + 2, f - 3, f + 6

Formula used:

Sum of all frequencies = Total observations

Calculations:

f + (f + 2) + (f - 3) + (f + 6) = 205

⇒ 4f + 5 = 205

⇒ 4f = 200

⇒ f = 50

∴ The value of f is 50.

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56 (from previous calculation f = 50)

Total observations = 205

Formula used:

Median for grouped data = L + [(N/2 - CF)/f] × h

Where, L = lower boundary of median class, N = total observations, CF = cumulative frequency before median class, f = frequency of median class, h = class width

Calculations:

Cumulative frequencies:

3 → 50

5 → 50 + 52 = 102

6 → 102 + 47 = 149

7 → 149 + 56 = 205

N/2 = 205/2 = 102.5 → lies in class x = 6 (since CF before = 102)

Assuming class width h = 1, L = 6

Median = 6 + [(102.5 - 102)/47] × 1

⇒ Median = 6 + [0.5/47]

⇒ Median ≈ 6 + 0.0106 ≈ 6

∴ The median of the frequency distribution is approximately 6.

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56

Total observations = 205

Formula used:

Mode for grouped data = L + [(f1 - f0) / (2f1 - f0 - f2)] × h

Where, L = lower boundary of modal class, f1 = frequency of modal class, f0 = frequency of previous class, f2 = frequency of next class, h = class width

Calculations:

Highest frequency = 56 → modal class x = 7

f1 = 56, f0 = 47 (previous), f2 = 0 (no class after 7), h = 1, L = 7

Mode = 7 + [(56 - 47)/(2×56 - 47 - 0)] × 1

⇒ Mode = 7 + [9 / (112 - 47)]

⇒ Mode = 7 + 9 / 65 ≈ 7 + 0.138 ≈ 7

∴ The mode of the frequency distribution is approximately 7.

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56

Formula used:

For discrete data, suitable graphs are bar graph or histogram (for continuous intervals); here X is discrete.

Calculations:

Plot x-values on x-axis (3, 5, 6, 7)

Plot frequencies on y-axis (50, 52, 47, 56)

Draw bars corresponding to each x-value with height equal to its frequency

∴ The most appropriate graphical representation is a bar graph.

Given:

Frequency distribution of X:

x = 3, 5, 6, 7

Frequencies = 50, 52, 47, 56

Total observations = 205

Formula used:

Mean = Σ(f × x) ÷ Σf

Calculations:

f × x:

3 × 50 = 150

5 × 52 = 260

6 × 47 = 282

7 × 56 = 392

Σ(f × x) = 150 + 260 + 282 + 392 = 1084

Mean = 1084 ÷ 205 ≈ 5.29

∴ The mean of the frequency distribution is approximately 5.29.

Given:

Cumulative frequency distribution of marks:

Below 40 = 50

Below 50 = 125

Below 60 = 210

Below 70 = 315

Below 80 = 350

Formula used:

Number of students in a class = Cumulative frequency of upper limit - Cumulative frequency of lower limit

Calculations:

Students between 60 and 70 = 315 - 210 = 105

∴ The number of students who scored between 60 and 70 marks is 105.

Given:

Cumulative frequency distribution of marks:

Below 40 = 50

Below 50 = 125

Below 60 = 210

Below 70 = 315

Below 80 = 350

Formula used:

Number of students scoring more than a mark = Total students - Cumulative frequency of that mark

Calculations:

Total students = 350 (Below 80)

Students scoring more than 50 = 350 - 125 = 225

∴ The number of students who scored more than 50 marks is 225.

Calculations:

Sum of D = 8 + 12 + 14 + 15 + 16 = 65 (thousands)

Average of D = 65 ÷ 5 = 13 (thousands)

Differences from average:

2019: |8 − 13| = 5

2020: |12 − 13| = 1

2021: |14 − 13| = 1

2022: |15 − 13| = 2

2023: |16 − 13| = 3

∴ The circulation of D was closest to its average (13 thousands) in the years 2020 and 2021 (difference = 1 thousand).

Given:

Circulation figures (in thousands) for newspapers A, B, C, D, E from 2019 to 2022

Formula used:

Average circulation = (A + B + C + D + E) ÷ 5

Calculations:

2019: Total = 20 + 10 + 15 + 8 + 20 = 73

Average = 73 ÷ 5 = 14.6; D = 8 → Difference = 6.6

2020: Total = 12 + 12 + 18 + 12 + 12 = 66

Average = 66 ÷ 5 = 13.2; D = 12 → Difference = 1.2

2021: Total = 24 + 14 + 17 + 14 + 15 = 84

Average = 84 ÷ 5 = 16.8; D = 14 → Difference = 2.8

2022: Total = 26 + 10 + 16 + 15 + 9 = 76

Average = 76 ÷ 5 = 15.2; D = 15 → Difference = 0.2

∴ The circulation of newspaper D was closest to the average in 2022.

Given:

Circulation figures (in thousands) of newspapers A, B, C, D, E for years 2019–2023

Formula used:

Average for individual newspaper = Sum of its circulation ÷ 5

Average of all newspapers for all years = Total circulation of all newspapers ÷ (5 × 5)

Calculations:

Total circulation for each newspaper:

A = 20 + 12 + 24 + 26 + 22 = 104 → Average = 104 ÷ 5 = 20.8

B = 10 + 12 + 14 + 10 + 16 = 62 → Average = 62 ÷ 5 = 12.4

C = 15 + 18 + 17 + 16 + 14 = 80 → Average = 80 ÷ 5 = 16

D = 8 + 12 + 14 + 15 + 16 = 65 → Average = 65 ÷ 5 = 13

E = 20 + 12 + 15 + 9 + 11 = 67 → Average = 67 ÷ 5 = 13.4

Total circulation of all newspapers = 104 + 62 + 80 + 65 + 67 = 378

Average circulation of all newspapers = 378 ÷ 25 = 15.12

Compare individual averages with 15.12:

A = 20.8 > 15.12 ✅

B = 12.4 < 15.12 ❌

C = 16 > 15.12 ✅

D = 13 < 15.12 ❌

E = 13.4 < 15.12 ❌

∴ There are 2 newspapers (A and C) whose average circulation is more than the average of all newspapers.

Given:

From Chart-I: Number of houses = 1,20,000

Angle for Pucca houses = 110°

From Chart-II: Population = 21,00,000

Angle for population living in Pucca houses = 60°

Formula used:

Part of total = (Angle / 360) × Total Value

Average = Total Value / Number of units

Calculations:

Number of Pucca Houses = (110 / 360) × 1,20,000

⇒ Number of Pucca Houses = (11 / 36) × 1,20,000

⇒ Number of Pucca Houses = 11 × (120000 / 36)

⇒ Number of Pucca Houses = 110000 / 3

Population Living in Pucca Houses = (60 / 360) × 21,00,000

⇒ Population Living in Pucca Houses = (1 / 6) × 21,00,000

⇒ Population Living in Pucca Houses = 3,50,000

Average Persons per Pucca House = Population Living in Pucca Houses / Number of Pucca Houses

⇒ Average Persons per Pucca House = 350000 / (110000 / 3)

⇒ Average Persons per Pucca House = (350000 × 3) / 110000

⇒ Average Persons per Pucca House = (35 × 3) / 11

⇒ Average Persons per Pucca House = 105 / 11

⇒ Average Persons per Pucca House ≈ 9.5454...

∴ The correct answer is option 3.

Given:

From Chart-I: Initial total number of houses = 1,20,000

Initial angle for Kutcha houses = 40°

Increase in Kutcha houses = 5000

Formula used:

Number of houses for a category = (Angle for category / 360°) × Total number of houses

Angle for a category = (Number of houses for category / Total number of houses) × 360°

Calculations:

Initial number of Kutcha houses = (40 / 360) × 1,20,000

⇒ Initial number of Kutcha houses = (1 / 9) × 1,20,000

⇒ Initial number of Kutcha houses = 120000 / 9

New total number of houses = Initial total number of houses + 5000

⇒ New total number of houses = 1,20,000 + 5000

⇒ New total number of houses = 1,25,000

New number of Kutcha houses = Initial number of Kutcha houses + 5000

⇒ New number of Kutcha houses = (120000 / 9) + 5000

⇒ New number of Kutcha houses = (120000 + 45000) / 9

⇒ New number of Kutcha houses = 165000 / 9

New angle for Kutcha houses = (New number of Kutcha houses / New total number of houses) × 360°

⇒ New angle for Kutcha houses = ((165000 / 9) / 125000) × 360°

⇒ (165000 / (9 × 125000)) × 360°

⇒ (165 / (9 × 125)) × 360°

⇒ (11 / 75) × 360° = 11 × 4.8 = 52.8°

Change in angle = New angle for Kutcha houses - Initial angle for Kutcha houses

⇒ Change in angle = 52.8° - 40°

⇒ Change in angle = 12.8° ≈ 13° 

∴ The approximate change in angle for Kutcha houses in Pie Chart-I will be 12.8°.

Given:

From Chart-III: Total number of families = 3,60,000

Angle for families in "Houseless" category = 20°

Angle for families "Living in Kutcha houses" = 185°

Number of families from "Houseless" category shifting to Kutcha houses = 300

Formula used:

Number of families for a category = (Angle for category / 360°) × Total number of families

Average families per house = Total families in houses / Total number of houses of that type

Calculations:

Initial Families in Kutcha Houses = (185 / 360) × 3,60,000

⇒ Initial Families in Kutcha Houses = 185 × (3,60,000 / 360) = 1,85,000

Initial Houseless Families = (20 / 360) × 3,60,000

⇒ Initial Houseless Families = 20 × (3,60,000 / 360) = 20,000

New Families in Kutcha Houses = Initial Families in Kutcha Houses + 300

⇒ New Families in Kutcha Houses = 1,85,000 + 300 = 1,85,300

From previous calculations or Chart-I:

Number of Kutcha Houses = (40 / 360) × 1,20,000

⇒ Number of Kutcha Houses = 120000 / 9

Average families per Kutcha house = New Families in Kutcha Houses / Number of Kutcha Houses

⇒ Average families per Kutcha house = 1,85,300 / (120000 / 9)

⇒ Average families per Kutcha house = (1,85,300 × 9) / 120000

⇒ Average families per Kutcha house = 1667700 / 120000

⇒ Average families per Kutcha house = 16677 / 1200

⇒ Average families per Kutcha house ≈ 13.8975 ≈ 13.9

∴ The average number of families in every Kutcha house will be 13.90.

Given:

Total students = 60

Marks range considered = ≥ 60 and < 80

Calculations:

Number of students in the range = 16

Percentage = (16 ÷ 60) × 100

⇒ Percentage = 26.67% ≈ 26.7%

∴ Approximately 26.7% of students scored between 60 and 79 marks.

Given:

Initial arithmetic mean of 100 observations = 50

Misread observations = 82 and 6

Correct observations = 182 and 56

Formula used:

Mean = (Sum of all observations) ÷ (Number of observations)

Calculations:

Initial sum of observations = Mean × Number of observations = 50 × 100 = 5000

Difference in sum due to misread observations = (182 + 56) - (82 + 6) = 238 - 88 = 150

Corrected sum of observations = 5000 + 150 = 5150

Corrected mean = Corrected sum ÷ Number of observations = 5150 ÷ 100 = 51.5

∴ The corrected mean is 51.5