Quantitative Aptitude PYQs

Given:

Expression: x²ⁿ - y²ⁿ + 1

Divisor: xⁿ + yⁿ

Statement I: n is odd

Statement II: n is even

Formula Used:

(a² - b²) = (a - b)(a + b)

Remainder when P is divided by Q = P - (Q × Quotient)

Calculation:

Case 1: Using Statement I (n is odd)

⇒ x²ⁿ - y²ⁿ can be factored as (xⁿ - yⁿ)(xⁿ + yⁿ)

⇒ When divided by (xⁿ + yⁿ), remainder is (xⁿ - yⁿ) + 1

∴ Remainder can be determined using Statement

I. Case 2: Using Statement II (n is even)

⇒ x²ⁿ - y²ⁿ can be factored as (xⁿ + yⁿ)(xⁿ - yⁿ)

⇒ When divided by (xⁿ + yⁿ), remainder is (xⁿ - yⁿ) + 1

∴ Remainder can be determined using Statement

II. ∴ The correct answer is Option 2.

Given:

Product of N and M = 252

N and M have the same digits in reverse order

Statement I: N + M = 33

Statement II: N > M

Formula Used:

Let N = 10a + b and M = 10b + a, where a and b are digits.

N × M = 252

N + M = 33

Calculation:

From N + M = 33:

⇒ (10a + b) + (10b + a) = 33

⇒ 11a + 11b = 33

⇒ a + b = 3

From N × M = 252:

⇒ (10a + b)(10b + a) = 252

⇒ 100ab + 10a² + 10b² + ab = 252

⇒ 101ab + 10a² + 10b² = 252

From a + b = 3, possible values are:

a = 1, b = 2 or a = 2, b = 1

Case 1: a = 1, b = 2:

⇒ N = 10(1) + 2 = 12

⇒ M = 10(2) + 1 = 21

⇒ N × M = 12 × 21 = 252 (valid)

Case 2: a = 2, b = 1:

⇒ N = 10(2) + 1 = 21

⇒ M = 10(1) + 2 = 12

⇒ N × M = 21 × 12 = 252 (valid)

From Statement II: N > M:

⇒ N = 21, M = 12 (valid)

∴ The correct answer is 21.

Given:

Number: (54D)¹⁰⁰

Last digit of expansion: 1

Statement I: D > 5

Statement II: D is a multiple of 3

Formula Used:

1. To find the last digit of a number raised to a power, consider the cyclicity of the last digit.
2. Cyclicity of a number is determined by the powers of its last digit.

Calculation:

⇒ The last digit of the number (54D) depends on D as the last digit of 54D is D.

⇒ For the expansion (54D)¹⁰⁰, the cyclicity of D determines the last digit.

⇒ To satisfy the condition that the last digit is 1, D must have a cyclicity where D⁴ gives 1 (as D¹⁰⁰ = (D⁴)²⁵).

⇒ D = 7 satisfies the condition as 7⁴ = 2401, and the last digit is 1.

Using Statement I:

⇒ D > 5 narrows down possibilities to D = 6, 7, 8, or 9.

⇒ From these, only D = 7 satisfies the cyclicity condition.

∴ Statement I alone is sufficient.

Using Statement II:

⇒ D is a multiple of 3 narrows down possibilities to D = 3, 6, or 9.

⇒ None of these satisfy the cyclicity condition for the last digit to be 1.

∴ Statement II alone is not sufficient.

∴ The correct answer is 1.

Given:

Statement-1: ABC is not an obtuse-angled triangle.

Statement-2: Angle C is acute.

Formula Used:

In a triangle, the sum of angles is 180°: ∠A + ∠B + ∠C = 180°.

Acute angle: An angle less than 90°.

Calculation:

We have,

∠A = ∠B - ∠C

⇒ ∠B = ∠A + ∠C

We know,

∠A + ∠B + ∠C = 180°

⇒ ∠A + (∠A + ∠C) + ∠C = 180°   (As  ∠B = ∠A + ∠C)

⇒ 2∠A + 2∠C = 180°

⇒ 2(∠A + ∠C) = 180° 

⇒ (∠A + ∠C) = 90°

Since  (∠A + ∠C) = 90°, it is mathematically certain that ∠A must be less than 90° (because ∠C must be greater than 0° for a triangle to exist).

So, we can answer this question even without using any of the Statements.

Hence, the correct answer is option (4).

Given:

Triangle ABC is right-angled at B.

AC = 20 cm.

Statement I: AB = 12 cm.

Statement-II: BC = 16 cm.

Formula Used:

Hypotenuse (AC)² = AB² + BC² (Pythagoras theorem)

Circum-radius = Hypotenuse / 2

Calculation:

Since Δ ABC is a right-angled triangle at B.

So, AC must be the Hypotenuse.

For the right-angled triangle:

Circum-radius = Hypotenuse / 2

So, Circum-radius of Δ ABC = AC/2 = 20/2 = 10 cm

Thus, the Question can be answered even without using any of the Statements 

∴ The correct answer is option (4).

Shortcut Trick

Area of parallelogram = side1 × side2 × sin(θ).

Given: Area = 7√3 and θ = 60° ⇒ AB × DA × sin(60°) = 7√3.

⇒ AB × DA × (√3/2) = 7√3 ⇒ AB × DA = 14.

For Statement-I: 14 = 2 × 7 (both prime). Perimeter = 2(2 + 7) = 18 units. Sufficient.

For Statement-II: 14 = 2 × 7 (both natural numbers > 1). Perimeter = 18 units. Sufficient.

∴ The correct answer is 18 units.

Alternate Method

Given: Parallelogram ABCD, ∠ABC = 60°, Area = 7√3 sq. units.

Formula Used: Area of parallelogram = a × b × sin(θ); Perimeter = 2(a + b).

⇒ Let sides AB = a and DA = b.

⇒ Area = a × b × sin(60°) = 7√3

⇒ a × b × (√3/2) = 7√3

⇒ a × b = 7 × 2 = 14

Case 1: Statement-I (a, b are prime numbers)

⇒ Factors of 14 are 1, 2, 7, 14.

⇒ Pairs (a, b) such that a × b = 14 are (1, 14) and (2, 7).

⇒ Since both must be prime, only (2, 7) is possible.

⇒ Perimeter = 2(2 + 7) = 18 units. (Sufficient)

Case 2: Statement-II (a, b are natural numbers > 1)

⇒ Factors of 14 are 1, 2, 7, 14.

⇒ Pairs (a, b) such that a × b = 14 and a, b > 1 is only (2, 7).

⇒ Perimeter = 2(2 + 7) = 18 units. (Sufficient)

∴ The correct answer is 18 units.

Additional Information

Area of Parallelogram

It can be calculated as base × height or as the product of adjacent sides and the sine of the included angle (ab sinθ).

Prime Numbers

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself (e.g., 2, 3, 5, 7).

Data Sufficiency

A question is answered by a statement if that statement leads to a unique value for the required quantity.

Given:

AP × PB = 48 square units

Statement-I: AP = 8 units

Statement-II: CP = 10 units

Concept Used:

In a circle, if two chords intersect (either inside or outside the circle), the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

Formula used:

If chords AB and CD intersect at P, then  AP × PB = CP × PD

Calculation:

So, to find CP × PD, we need AP × PB, which is already given in the question. 

So, we can answer this question without using any of the Statements.

Hence, the correct answer is option (3).

Given:

AB = 6 units

BC = 18 units

CD = 6 units

DA = 9 units

Formula Used:

1. For a quadrilateral, the diagonal can be determined using geometric properties and given lengths.
2. Additional conditions are provided in the statements to narrow down the possibilities.

Calculation:

Statement-I: The length of BD is an integer greater than 13.

⇒ Possible values for BD are 14, 15, 16, etc.

Statement-II: The length of BD is an even integer.

⇒ Possible values for BD are 14, 16, 18, etc.

Using both statements together:

⇒ From Statement-I, BD > 13 and integer. From Statement-II, BD is even.

⇒ The only value satisfying both is BD = 14.

∴ The correct answer is 3.

Shortcut Trick

For an isosceles triangle with sides 10, 10, and base x, area A = (x/4) × √(4×10² − x²) = 48.

By using Pythagorean triplets for height h and half-base (x/2): h² + (x/2)² = 10². Possible triplets (h, x/2) are (8, 6) or (6, 8).

Case 1: h = 8, x/2 = 6 ⇒ x = 12. Area = 1/2 × 12 × 8 = 48 units².

Case 2: h = 6, x/2 = 8 ⇒ x = 16. Area = 1/2 × 16 × 6 = 48 units².

Statement-I (x is even) applies to both 12 and 16. Statement-II (h > x/2) selects h=8, x=12. Thus, both are needed together.

∴ The correct answer is both the Statements together.

Alternate Method

Given: Isosceles triangle ABC, AB = AC = 10 units, Area = 48 square units.

Formula Used: Area = 1/2 × base × height; Pythagoras Theorem: h² + (b/2)² = a²

⇒ Let BC = b and height AD = h. In ΔABD, h² + (b/2)² = 10² = 100.

⇒ Area = 1/2 × b × h = 48 ⇒ b × h = 96.

⇒ Substitute h = 96/b into Pythagoras: (96/b)² + b²/4 = 100.

⇒ 9216/b² + b²/4 = 100 ⇒ 36864 + b⁴ = 400b² ⇒ b⁴ − 400b² + 36864 = 0.

⇒ Let y = b²: y² − 400y + 36864 = 0. Factors of 36864 that sum to 400 are 256 and 144.

⇒ b² = 256 or b² = 144 ⇒ b = 16 (then h = 6) or b = 12 (then h = 8).

⇒ Statement-I: b is an even integer. Both 12 and 16 are even. Not sufficient alone.

⇒ Statement-II: h > b/2. For b=16, h=6 (6 < 8 False). For b=12, h=8 (8 > 6 True). Not sufficient alone as we don't know if b must be an integer from this statement alone.

⇒ Using both: Statement-II narrows it to b = 12.

∴ The correct answer is using both the Statements together.

Additional Information

Area of Isosceles Triangle

Calculated as (b/4)√(4a² − b²) where a is the equal side and b is the base.

Data Sufficiency Logic

If multiple valid geometric configurations exist, supplementary statements are required to isolate a unique solution.

Altitude Property

In an isosceles triangle, the altitude to the base bisects the base and creates two congruent right-angled triangles.

Given:

Diagonals of the rhombus are in the ratio 5 : 12.

Sum of the diagonals = 34 cm (Statement-I).

Length of a side = 13 cm (Statement-II).

Formula Used:

Diagonal 1 = (5 / (5 + 12)) × Sum of diagonals.

Diagonal 2 = (12 / (5 + 12)) × Sum of diagonals.

Side of rhombus = √((Diagonal 1 / 2)² + (Diagonal 2 / 2)²).

Calculation:

Using Statement-I:

⇒ Sum of diagonals = 34.

⇒ Diagonal 1 = (5 / 17) × 34 = 10 cm.

⇒ Diagonal 2 = (12 / 17) × 34 = 24 cm.

Using formula for side:

⇒ Side = √((10 / 2)² + (24 / 2)²).

⇒ Side = √(5² + 12²).

⇒ Side = √(25 + 144) = √169 = 13 cm.

Using Statement-II:

Length of side is directly given as 13 cm.

From Statement-II, if side = 13 cm and diagonals are in ratio 5 : 12, we can calculate:

⇒ Side = √((Diagonal 1 / 2)² + (Diagonal 2 / 2)²).

⇒ Solving for diagonal lengths confirms the same result as from Statement-I.

∴ The correct answer is 2.

Given:

Option 1: Arithmetic mean

Option 2: Harmonic mean

Option 3: Geometric mean

Option 4: Median

Formula Used:

Median is the middle value of a dataset when arranged in ascending or descending order.

It is not influenced by extreme values, unlike mean-based measures.

Calculation:

⇒ Arithmetic mean, Harmonic mean, and Geometric mean are all influenced by extreme observations.

⇒ Median remains unaffected by extreme observations, as it solely depends on the middle value(s).

∴ The correct answer is Median.

Given:

To find the average ratio like price/unit, work done/hour, kilometre/hour under certain conditions.

Arithmetic mean

Geometric mean

Harmonic mean

Mode

Correct Option: 1

Formula Used:

Harmonic Mean (HM) = n / (Σ(1/x))

Arithmetic Mean (AM) = Σx / n

Geometric Mean (GM) = (Πx)^1/n

Calculation:

⇒ Average ratio calculations for price/unit, work done/hour, kilometre/hour use arithmetic mean for simplicity and direct averaging.

∴ The correct answer is arithmetic mean.

Given:

Marks more than 10: 100 candidates

Marks more than 20: 75 candidates

Marks more than 30: 60 candidates

Marks more than 40: 40 candidates

Formula Used:

Class Frequency = Number of candidates in a specific range

Average Marks = (Σ(Midpoint × Frequency)) / Total Frequency

Calculation:

Class intervals: 10-20, 20-30, 30-40, 40 and above

Frequencies: 100 - 75 = 25, 75 - 60 = 15, 60 - 40 = 20, 40 (last class)

Midpoints: (10 + 20)/2 = 15, (20 + 30)/2 = 25, (30 + 40)/2 = 35, 40 (open-ended class, assume midpoint = 45)

⇒ Σ(Midpoint × Frequency) = (15 × 25) + (25 × 15) + (35 × 20) + (45 × 40)

⇒ Σ(Midpoint × Frequency) = 375 + 375 + 700 + 1800

⇒ Σ(Midpoint × Frequency) = 3250

⇒ Total Frequency = 100

⇒ Average Marks = Σ(Midpoint × Frequency) / Total Frequency

⇒ Average Marks = 3250 / 100

⇒ Average Marks = 32.5

∴ The correct answer is 32.5.

Given:

Arithmetic mean of 200 observations = 60

Each observation is multiplied by 5

Formula Used:

New Arithmetic Mean = Old Arithmetic Mean × Multiplication Factor

Calculation:

⇒ New Arithmetic Mean = 60 × 5

⇒ New Arithmetic Mean = 300

∴ The correct answer is 300.

Given:

Frequencies = 45, 40, 55

Means = 2, 2.5, 2

Formula Used:

Combined Mean = (Σ(Frequency × Mean)) / Σ(Frequency)

Calculation:

⇒ Σ(Frequency × Mean) = (45 × 2) + (40 × 2.5) + (55 × 2)

⇒ Σ(Frequency × Mean) = 90 + 100 + 110

⇒ Σ(Frequency × Mean) = 300

⇒ Σ(Frequency) = 45 + 40 + 55

⇒ Σ(Frequency) = 140

⇒ Combined Mean = 300 / 140

⇒ Combined Mean = 2.14

∴ The correct answer is 2.14.

Given:

Option 1: Arithmetic mean

Option 2: Median

Option 3: Mode

Option 4: Geometric mean

Option 5: Not provided

Correct option: 2

Formula Used:

Positional average is determined by the position of data values in a sorted dataset, such as Median.

Calculation:

⇒ Median is the middle value of a sorted dataset, making it a positional average.

∴ The correct answer is Median.

Given:

Two frequency distributions are given.

Median of the first group = 20.8 m

Mean of the second group = 17.3 m

Unknown frequencies:

First group: u, v

Second group: 3u, 2v

Use median condition for the first group

Class intervals (width h = 5)

Median value 20.8 lies in the class 20–25

Lower limit L = 20

Frequency of median class f = 75

Cumulative frequency before median class:

cf = u + v + 11 + 52 = u + v + 63

Total frequency N = u + v + 11 + 52 + 75 + 22

⇒ N = u + v + 160

Median formula:

Median = L + [ (N/2 − cf) ÷ f ] × h

Substitute values:

⇒ 20.8 = 20 + [ ((u + v + 160)/2 − (u + v + 63)) ÷ 75 ] × 5

⇒ 0.8 = [ (u + v + 160 − 2u − 2v − 126) ÷ 150 ] × 5

⇒ 0.8 = [ (34 − (u + v)) ÷ 150 ] × 5

⇒ 0.8 = (34 − u − v) ÷ 30

⇒ 34 − u − v = 24

⇒ u + v = 10 ……… (1)

Use mean condition for the second group

Midpoints x: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5

Second group frequencies:

3u, 2v, 40, 50, 30, 28

Σf = 3u + 2v + 148

Σfx = (3u × 2.5) + (2v × 7.5) + (40 × 12.5) + (50 × 17.5) + (30 × 22.5) + (28 × 27.5)

⇒ Σfx = 7.5u + 15v + 500 + 875 + 675 + 770

⇒ Σfx = 7.5u + 15v + 2820

Mean formula:

Mean = Σfx ÷ Σf

Substitute mean = 17.3

⇒ (7.5u + 15v + 2820) ÷ (3u + 2v + 148) = 17.3

⇒ 7.5u + 15v + 2820 = 17.3(3u + 2v + 148)

⇒ 7.5u + 15v + 2820 = 51.9u + 34.6v + 2560.4

⇒ 44.4u + 19.6v = 259.6

⇒ Divide by 2

⇒ 22.2u + 9.8v = 129.8 ……… (2)

From (1): v = 10 − u

Substitute in (2):

⇒ 22.2u + 9.8(10 − u) = 129.8

⇒ 22.2u + 98 − 9.8u = 129.8

⇒ 12.4u = 31.8

⇒ u ≈ 2.56 ≈ 3 (nearest integer)

∴ The value of u is 3.

Given:

Two frequency distributions are given.

Median of the first group = 20.8 m

Mean of the second group = 17.3 m

Unknown frequencies:

First group: u, v

Second group: 3u, 2v

Use median condition for the first group

Class intervals (width h = 5)

Median value 20.8 lies in the class 20–25

Lower limit L = 20

Frequency of median class f = 75

Cumulative frequency before median class:

cf = u + v + 11 + 52 = u + v + 63

Total frequency N = u + v + 11 + 52 + 75 + 22

⇒ N = u + v + 160

Median formula:

Median = L + [ (N/2 − cf) ÷ f ] × h

Substitute values:

⇒ 20.8 = 20 + [ ((u + v + 160)/2 − (u + v + 63)) ÷ 75 ] × 5

⇒ 0.8 = [ (u + v + 160 − 2u − 2v − 126) ÷ 150 ] × 5

⇒ 0.8 = [ (34 − (u + v)) ÷ 150 ] × 5

⇒ 0.8 = (34 − u − v) ÷ 30

⇒ 34 − u − v = 24

⇒ u + v = 10 ……… (1)

Use mean condition for the second group

Midpoints x: 2.5, 7.5, 12.5, 17.5, 22.5, 27.5

Second group frequencies:

3u, 2v, 40, 50, 30, 28

Σf = 3u + 2v + 148

Σfx = (3u × 2.5) + (2v × 7.5) + (40 × 12.5) + (50 × 17.5) + (30 × 22.5) + (28 × 27.5)

⇒ Σfx = 7.5u + 15v + 500 + 875 + 675 + 770

⇒ Σfx = 7.5u + 15v + 2820

Mean formula:

Mean = Σfx ÷ Σf

Substitute mean = 17.3

⇒ (7.5u + 15v + 2820) ÷ (3u + 2v + 148) = 17.3

⇒ 7.5u + 15v + 2820 = 17.3(3u + 2v + 148)

⇒ 7.5u + 15v + 2820 = 51.9u + 34.6v + 2560.4

⇒ 44.4u + 19.6v = 259.6

⇒ Divide by 2

⇒ 22.2u + 9.8v = 129.8 ……… (2)

From (1): v = 10 − u

Substitute in (2):

⇒ 22.2u + 9.8(10 − u) = 129.8

⇒ 22.2u + 98 − 9.8u = 129.8

⇒ 12.4u = 31.8

⇒ u ≈ 2.56 ≈ 3 (nearest integer)

u + v = 10

⇒ v = 10 − u

⇒ v = 10 − 3

⇒ v = 7

∴ The value of v is 7.

Given:

Median = 24

Formula used:

Median = L + [( (N/2 − cf) / f ) × h]

Calculation:

Total number of students, N = 55 + k

⇒ N/2 = (55 + k)/2

Median value 24 lies in the class 20–30

Lower limit (L) = 20

Class width (h) = 10

Cumulative frequency before median class (cf) = 30

Frequency of median class (f) = (30 + k) − 30 = k

Substitute in median formula:

⇒ 24 = 20 + [ ( (55 + k)/2 − 30 ) ÷ k ] × 10

⇒ 4 = [ (55 + k − 60) ÷ (2k) ] × 10

⇒ 4 = ( (k − 5) ÷ (2k) ) × 10

⇒ 4 = (5(k − 5)) ÷ k

⇒ 4k = 5k − 25

⇒ k = 25

∴ The value of k is 25.

Calculation:

Median = L + [( (N/2 − cf) / f ) × h]

Calculation:

Total number of students, N = 55 + k

⇒ N/2 = (55 + k)/2

Median value 24 lies in the class 20–30

Lower limit (L) = 20

Class width (h) = 10

Cumulative frequency before median class (cf) = 30

Frequency of median class (f) = (30 + k) − 30 = k

Substitute in median formula:

⇒ 24 = 20 + [ ( (55 + k)/2 − 30 ) ÷ k ] × 10

⇒ 4 = [ (55 + k − 60) ÷ (2k) ] × 10

⇒ 4 = ( (k − 5) ÷ (2k) ) × 10

⇒ 4 = (5(k − 5)) ÷ k

⇒ 4k = 5k − 25

⇒ k = 25

Convert to class frequencies

Σf = 80

Σ(f × x) = 25 + 375 + 625 + 630 + 315 = 1970

Mean = Σ(f × x) ÷ Σf

⇒ Mean = 1970 ÷ 80

⇒ Mean = 24.625

∴ The mean of the distribution is 24.625.