Quantitative Aptitude PYQs

Calculation:

I. x2 – 16x + 55 = 0

⇒ x2 – 11x – 5x + 55 = 0

⇒ x(x – 11) – 5(x – 11) = 0 

⇒ (x – 11) (x – 5) = 0

⇒ x = 11, 5

II. y2 – 17y + 52 = 0

⇒ y2 – 13y – 4y + 52 = 0

⇒ y(y – 13) – 4(y – 13) = 0

⇒ (y – 13) (y – 4) = 0

⇒ y = 13, 4

Comparison between x and y (via tabulation):

 

∴ The relation between x and y cannot be determined.

Calculation:

I. 2x² – 7x – 15 = 0

⇒ 2x² – 10x + 3x – 15 = 0

⇒ 2x (x - 5) + 3 (x - 5) = 0

⇒ (2x + 3) (x - 5) = 0

⇒ x = -3/2, 5

 

II. 3y² – 6y – 9 = 0

⇒ 3y² – 9y + 3y – 9 = 0

⇒ 3y (y - 3) + 3 (y - 3) = 0

⇒ (3y + 3) (y - 3) = 0

⇒ y = -1, 3

Comparison between x and y (via Tabulation):

∴ x = y or relationship between x and y cannot be established

Calculation:

⇒ 5x² + 8x + 3 = 0

⇒ 5x² + 5x + 3x + 3 = 0

⇒ 5x(x + 1) + 3(x + 1) = 0

⇒ (5x + 3) (x + 1) = 0

⇒ x = -3/5, -1

⇒ 6y² + 7y – 3 = 0

⇒ 6y² + 9y – 2y – 3 = 0

⇒ 3y × (2y + 3) – 1(2y + 3) = 0

⇒ (3y - 1) (2y + 3) = 0

⇒ y = 1/3, -3/2

∴ The relation between x and y cannot determine.

I. 4x² – 10x + 4 = 0

⇒ 4x² – 8x – 2x + 4 = 0

⇒ 4x(x – 2) – 2(x - 2) = 0

⇒ (x – 2)(4x – 2) = 0

⇒ x = 2 or 1/2

II. 3y² – 11y + 6 = 0

⇒ 3y² – 9y – 2y + 6 = 0

⇒ 3y(y – 3) – 2(y – 3) = 0

⇒ (y – 3)(3y – 2) = 0

⇒ y = 3 or 2/3

 

∴ Relation can't be formed or x = y

I. 2x² – 11x + 15 = 0

⇒ 2x² – 6x – 5x + 15 = 0

⇒ 2x(x – 3) – 5(x – 3) = 0

⇒ (x – 3)(2x – 5) = 0

⇒ x = 3 or 5/2

II. 2y² – 9y + 10 = 0

⇒ 2y² – 4y – 5y + 10 = 0

⇒ 2y(y – 2) – 5(y – 2) = 0

⇒ (y – 2)(2y – 5) = 0

⇒ y = 2 or 5/2

∴ x ≥ y

Calculation:

 

The total number of students (male + female) in another school 'Z' = (100+30)% × 860

⇒ 130% × 860 = 1118

∵ The ratio of male to female in school Z = 6:7

∴ The number of female students in school 'Z' = (7/13) × 1118 = 602

The average total number of female students in school 'A' and school 'Z' together = (870 + 602)/2 = 736

Hence, the correct answer is 736.

Calculation:

 

Number of female students who are selected from school 'D' = (p - 15)% × 335

Number of female students who are selected from school 'E' = (p + 10)% × 220

∵ The average number of females selected for quizzes in schools D & E is 83,

⇒ [(p - 15)% × 335 + (p + 10)% × 220]/2 = 83

⇒ (p - 15)% × 335 + (p + 10)% × 220 = 83 × 2

⇒ (p - 15)% × 335 + (p + 10)% × 220 = 166

⇒ 335p - 15% × 335 + 220p + 10% × 220 = 166

⇒ 555p - 50.25 + 22 = 166

⇒ 555p - 28.25 = 166 

⇒ 555p = 166 + 28.25

⇒ 555p = 194.25

⇒ p = 194.25/555

⇒ p = 0.35

∵ The percentage value of p is 0.35

∴ The numerical value of p is 0.35 × 100 = 35

Hence, the value of p is 35.

Calculation:

 

The total number of female students in school C = 445

The total number of female students in school A = 870

The average total number of female students in school 'C' and school 'X' = 498

Total number of female students in school 'C' and school 'X' = 498 × 2 = 996

Number of female students in school 'X' = 996 - 445 = 551

Required percentage = (551/870) × 100 = 63.33%

Hence, the correct answer is 63.33%.

Calculation:

 

Difference between the number of male students in school 'A' to that in school 'E' = 420 - 210 = 210

Let the number of male students in school 'M' = m

ATQ, m - 630 = 210

⇒ m = 840

Required sum = 315 + 840 = 1155

Hence, the correct answer is 1155.

 

Let the total number of students in school 'G' = T

ATQ, [(T + 860)/(4300 + T)] × 360° = 120°

⇒ (T + 860)/(4300 + T) = 1/3

⇒ 3T + 2580 = 4300 + T

⇒ T = 860

Required difference = 1290 - 860 = 430

Hence, the correct answer is 430.

Shortcut Trick

Average of 5 even numbers is 60, so middle (3^rd) is 60 → highest is 64.

3^rd highest even number is the middle one = 60.

2^nd lowest odd number = (5/12 × 60) + 4 = 25 + 4 = 29.

If 2^nd lowest odd is 29, then 3^rd is 31, 4^th is 33, and highest (5^th) is 35.

Difference = 64 − 35 = 29.

∴ The correct answer is 29.

Alternate Method

Given: Average of 5 consecutive even numbers = 60.

⇒ Let numbers be x−4, x−2, x, x+2, x+4. Average x = 60.

⇒ Third highest even number (which is also the middle number) = 60.

⇒ Highest even number = 60 + 4 = 64.

⇒ Second lowest odd number = (5/12 × 60) + 4 = 25 + 4 = 29.

⇒ Let 5 consecutive odd numbers be y, y+2, y+4, y+6, y+8.

⇒ Second lowest is y+2 = 29 ⇒ y = 27.

⇒ Highest odd number = 27 + 8 = 35.

⇒ Difference = 64 − 35 = 29.

∴ The correct answer is 29.

Additional Information

Average of Consecutive Series

For any odd number of terms in an arithmetic progression, the average is always the middle term.

Consecutive Even/Odd Numbers

Both consecutive even and consecutive odd numbers have a common difference of 2. If the first term is a, the n^th term is a + 2(n−1).

Difference between Extremes

In a series of n consecutive even/odd numbers, the difference between the largest and smallest number is always 2 × (n − 1).

Shortcut Trick

Seeta marks = 456 − 24 = 432 marks.

54% of Maximum Marks = 432 ⇒ 1% = 8 marks.

Passing Marks (34%) = 34 × 8 = 272 marks.

Difference = 456 − 272 = 184 marks.

∴ The correct answer is 184 marks.

Alternate Method

Given:

Marks scored by Raman = 456

Marks scored by Seeta = 54% of Total Marks

Difference between Raman and Seeta = 24 marks

Minimum Passing Percentage = 34%

Calculations:

⇒ Marks of Seeta = Marks of Raman − 24

⇒ Marks of Seeta = 456 − 24 = 432

⇒ 54% of Maximum Marks = 432

⇒ Maximum Marks = (432 × 100) ÷ 54 = 800

⇒ Passing Marks = 34% of 800

⇒ Passing Marks = (34 × 800) ÷ 100 = 272

⇒ Required Difference = 456 − 272 = 184

∴ The correct answer is 184 marks.

Additional Information

Percentage to Value Conversion

To find the absolute value from a percentage: Value = (Percentage ÷ 100) × Total Value.

Passing Marks Concept

Passing marks is the threshold required to qualify. If a student gets 'x' marks and fails by 'y' marks, Passing Marks = x + y.

Unitary Method in Percentage

If p% = V, then 1% = V/p. This helps in calculating any other percentage (like 100% or passing %) quickly.

Shortcut Trick

Total amount = ₹50,000

Remaining after A = 1 − 2/3 = 1/3

Remaining after B = 1/3 × (1 − 1/5) = 1/3 × 4/5 = 4/15

Share of C = 4/15 × 2/5 = 8/75 of total

C's share = 50,000 × 8/75 = 2,000 × 8/3 = 16,000/3 ≈ ₹5,333.33

∴ The correct answer is ₹5,334.

Alternate Method

Given: Total = ₹50,000 | A's share = 2/3 | B's share = 1/5 of remaining | C : D = 2 : 3

⇒ Share of A = 50,000 × 2/3 = ₹33,333.33

⇒ Remaining amount = 50,000 − 33,333.33 = ₹16,666.67

⇒ Share of B = 16,666.67 × 1/5 = ₹3,333.33

⇒ Amount left for C and D = 16,666.67 − 3,333.33 = ₹13,333.34

⇒ Ratio C : D = 2 : 3 (Total parts = 5)

⇒ Share of C = (2/5) × 13,333.34

⇒ Share of C = ₹5,333.336 ≈ ₹5,334

∴ The correct answer is ₹5,334.

Additional Information

Ratio and Proportion

If an amount X is divided in ratio a:b, then share of first part = [a/(a+b)] × X.

Successive Fractions

When dealing with remainders, the final remaining fraction is (1 − f₁) × (1 − f₂) × ... × (1 − f_n).

Shortcut Trick

Ratio of Mother to Son: Present = 4 : 1 and After 8 years = 16 : 5.

Equating ratio difference: (4:1)×11 and (16:5)×3 → 44:11 and 48:15.

Change in units = 15 − 11 = 4 units → 8 years. So, 1 unit = 2 years.

Present age of son = 11 units × 2 = 22 years.

Value of x = 22² = 484. Then, x + 2 = 484 + 2 = 486.

∴ The correct answer is 486.

Alternate Method

Given: Present ratio M : S = 4 :

1. Let ages be 4s and s.

Formula: Ratio = (Initial Age + Time) ÷ (Initial Age + Time)

⇒ (4s + 8) ÷ (s + 8) = 16 ÷ 5

⇒ 5 × (4s + 8) = 16 × (s + 8)

⇒ 20s + 40 = 16s + 128

⇒ 4s = 88

⇒ s = 22

⇒ x = s² = 22² = 484

⇒ x + 2 = 484 + 2 = 486

∴ The correct answer is 486.

Additional Information

Ratio Difference Method

To solve age problems quickly, ensure the difference between ratio parts remains constant over time by multiplying with the cross-difference.

Square of Numbers

Memorizing squares up to 30 helps in fast calculations. For example, 21² = 441, 22² = 484, 23² = 529.

Ages Property

The difference between the ages of two people always remains constant, regardless of how many years pass.

Shortcut Trick

Total cards = 52. Red face cards (King, Queen, Jack of Hearts and Diamonds) = 2 × 3 = 6.

Remaining Total Outcome = 52 − 6 = 46.

Total face cards in a deck = 12. Remaining face cards = 12 − 6 = 6.

Probability = Favorable / Total = 6 / 46 = 3 / 23.

∴ The correct answer is 3 / 23.

Alternate Method

Given:

Total cards in a standard deck = 52.

Face cards include King, Queen, and Jack for each of the 4 suits.

⇒ Total face cards = 4 suits × 3 cards/suit = 12.

⇒ Red face cards = 2 suits (Hearts, Diamonds) × 3 cards/suit = 6.

Calculations:

⇒ Total cards after removing red face cards (n(S)) = 52 − 6 = 46.

⇒ Remaining face cards in the deck (n(E)) = 12 − 6 = 6 (only black face cards remain).

⇒ Probability P(E) = n(E) / n(S)

⇒ P(E) = 6 / 46

⇒ P(E) = 3 / 23

∴ The correct answer is 3 / 23.

Additional Information

Standard Deck Composition

A deck has 52 cards: 26 Red (Hearts, Diamonds) and 26 Black (Spades, Clubs). Each suit has 13 cards: Ace, 2-10, and 3 Face cards.

Probability Formula

P(Event) = (Number of favorable outcomes) ÷ (Total number of possible outcomes).

Face Cards vs Number Cards

Face cards (K, Q, J) total 12. Number cards (2-10) total 36. Aces are generally not considered face cards unless specified.

Solution:

The series follows the following pattern:

Hence, the correct answer is 74.

Solution:

The logic followed here is as follows:

⇒ 500 ×  0.1 = 50

⇒ 50 × 0.2 = 10

⇒ 10 × 0.3 = 3

⇒ 3 × 0.4 = 1.2

⇒ 1.2 × 0.5 = 0.6

Hence, 0.6 will come in the place of '?'.

Solution:

The logic followed here is as follows:

The series follows the difference of prime numbers start from 61.

Hence, 161 will come in the place of '?'.

Solution:

The series follows the following pattern :

 

Hence, 92 is the correct answer.

Solution:

The series follows the following pattern :

⇒ 40 + 5² + 1 = 66

⇒ 66 + 7² + 1 = 116

⇒ 116 + 92 + 1 = 198

⇒ 198 + 112 + 1 = 320

⇒ 320 + 132 + 1 = 490

⇒ The next term in the series will be 490.

Hence, 490 is the correct answer.