Quantitative Aptitude PYQs

Given:

The required expression: \(\sqrt{17 - 4\sqrt{15}} + \sqrt{8 - 2\sqrt{15}}\)

Calculation:

\(\sqrt{17 - 4\sqrt{15}}\) = \(\sqrt{(2\sqrt3)^2 + (\sqrt5)^2 - 2 \times 2\sqrt{3}\times \sqrt5}\) = \(\sqrt{(2\sqrt3 -\sqrt 5)^2}\) = \(2\sqrt3 - \sqrt5\)

\(\sqrt{8 - 2\sqrt{15}}\) = \(\sqrt{(\sqrt{3})^2 + (\sqrt{5})^2 - 2 \times \sqrt{3} \times \sqrt{5}}\) = \(\sqrt{(\sqrt{5} -\sqrt{3})^2 }\) = \(\sqrt{5} - \sqrt{3}\)

Now, adding both terms:

\(2\sqrt3 - \sqrt5\) + \(\sqrt{5} - \sqrt{3}\) = \(\sqrt3\)

Therefore, the final answer is option 1.

Given:

The sequence is: 36, 33, 30, 27, 24, ...

We need to find the maximum value of the sum of the numbers.

This is an arithmetic progression (AP) where:

First term (a) = 36

Common difference (d) = 33 - 36 = -3

The sum of n terms of an AP is given by:

Formula used:

Sum (S_n) = \(\dfrac{n}{2}[2a + (n-1)d]\)

Calculation:

We need to find the maximum value of the sum. The AP ends when the last term becomes ≥ 0.

Last term (l) = a + (n-1)d

For the last term to be ≥ 0:

⇒ 36 + (n-1)(-3) ≥ 0

⇒ 36 - 3(n-1) ≥ 0

⇒ 36 - 3n + 3 ≥ 0

⇒ 39 - 3n ≥ 0

⇒ 3n ≤ 39

⇒ n ≤ 13

Thus, the sequence has a maximum of n = 13 terms.

Now, calculate the sum of these 13 terms:

S₁₃ = \(\dfrac{13}{2}[2(36) + (13-1)(-3)]\)

⇒ S₁₃ = \(\dfrac{13}{2}[72 + 12(-3)]\)

⇒ S₁₃ = \(\dfrac{13}{2}[72 - 36]\)

⇒ S₁₃ = \(\dfrac{13}{2}[36]\)

⇒ S₁₃ = 13 × 18

⇒ S₁₃ = 234

∴ The correct answer is option (3).

Given:

m ≡ 4 (mod 12)

n ≡ 6 (mod 12)

Formula used:

If a ≡ r₁ (mod m) and b ≡ r₂ (mod m), then:

a ± b ≡ r₁ ± r₂ (mod m)

Calculations:

Statement I: (m + n) mod 12

⇒ m + n ≡ 4 + 6 ≡ 10 (mod 12)

The statement I is correct.

Statement II: (m − n) mod 12

⇒ m − n ≡ 4 − 6 ≡ −2 ≡ 10 (mod 12)

The statement II is correct.

∴ The correct answer is: Both I and II

Given:

(x + y) : (y + z) : (z + x) = 3 : 5 : 6

x + y + z = 14

Formula used:

x² + y² + z² = (x + y + z)² − 2(xy + yz + zx)

Calculations:

Let (x + y) = 3k, (y + z) = 5k, (z + x) = 6k

⇒ (x + y) + (y + z) + (z + x) = 3k + 5k + 6k = 14k

But (x + y) + (y + z) + (z + x) = 2(x + y + z) = 2 × 14 = 28

⇒ 14k = 28 ⇒ k = 2

So, x + y = 6, y + z = 10, z + x = 12

Adding all: 2(x + y + z) = 28 ⇒ x + y + z = 14 ✔

Now, solve individually:

x + y = 6, x + y + z = 14 ⇒ z = 8

y + z = 10 ⇒ y + 8 = 10 ⇒ y = 2

x + z = 12 ⇒ x + 8 = 12 ⇒ x = 4

Now, x² + y² + z² = 4² + 2² + 8² = 16 + 4 + 64 = 84

∴ x² + y² + z² = 84

Given:

The ratio of the sum of two numbers to their difference is 5 :

1. Formula used:

If the ratio of the sum of two numbers (a + b) to their difference (a - b) is given as 5 : 1, then:

(a + b)/(a - b) = 5/1

The ratio of the sum of their squares to the difference of their squares can be calculated using:

(a² + b²) / (a² - b²) = [(a + b)² - 2ab] / [(a - b)² + 2ab]

Calculation:

From the ratio (a + b)/(a - b) = 5/1:

⇒ a + b = 5k and a - b = k

⇒ a = (5k + k)/2 = 3k, b = (5k - k)/2 = 2k

Now calculate the ratio (a² + b²) / (a² - b²):

⇒ a² + b² = (3k)² + (2k)² = 9k² + 4k² = 13k²

⇒ a² - b² = (3k)² - (2k)² = 9k² - 4k² = 5k²

⇒ (a² + b²) / (a² - b²) = 13k² / 5k²

⇒ Ratio = 13 : 5

∴ The correct answer is option (1).

Given:

A man is traveling at 3/5^th of his usual speed and is late by 20 minutes.

Formula used:

Time = Distance ÷ Speed

Let the usual time taken by the man be T minutes.

When traveling at 3/5^th of the usual speed, time taken = T × (5/3).

Difference in time: T × (5/3) - T = 20 minutes.

Calculation:

⇒ T × (5/3) - T = 20

⇒ (5T/3) - T = 20

⇒ (5T - 3T)/3 = 20

⇒ 2T/3 = 20

⇒ T = 20 × (3/2)

⇒ T = 30 minutes

∴ The correct answer is option (2).

Given:

Expression: 2^p−1

Divisor: p

Calculations:

We need to find the remainder when 2^p−1 is divided by p.

Since p is a prime number greater than 5: Direct putting value p = 7.

Substitute p = 7 into the expression:

Expression = 2⁷ − 1

2⁷ = 128

Expression = 128 − 1

⇒ Expression = 127

Now, we need to divide 127 by p = 7 and find the remainder.

127 ÷ 7:

127 = 7 × 18 + 1

The remainder when 2⁷−1 is divided by 7 is 1.

∴ The correct answer is option 1.

Given:

Expression: 24³ − 16³ − 8³

Calculations:

⇒ 8³(27 − 8 − 1)

⇒ 8³ × 18

⇒ 512 × 18 = 9216

Prime factorization of 9216:

9216 = 2¹⁰ × 3²

Number of factors = (10 + 1)(2 + 1) = 11 × 3 = 33

∴ The correct answer is: 33

Given:

Rate of Interest (r) = 20%

Amount (A) = 2 × Principal (P)

Formula used:

A = P(1 + r/100)^t

Calculation:

2P = P(1 + 20/100)^t

⇒ 2 = (1 + 0.2)^t

⇒ 2 = (1.2)^t

Now, we calculate t for different values:

For t = 2, (1.2)² = 1.44

For t = 3, (1.2)³ = 1.728

For t = 4, (1.2)⁴ = 2.0736

⇒ At t = 4, the amount exceeds double the principal.

∴ The least number of complete years is 4, and the correct answer is option (3).

Given:

A train of length L passes a station of length x in time t.

The same train with same speed passes another station of length y in time 2t.

We need to find the time taken to pass a station of length (x + y).

Formula used:

Time = (Distance covered) / (Speed)

When a train passes a station, distance covered = (Length of train + Length of station).

Calculation:

Case 1: Through station of length x

(L + x) / v = t … (i)

Case 2: Through station of length y

(L + y) / v = 2t … (ii)

From (i): v = (L + x)/t

From (ii): v = (L + y)/(2t)

Equating both values of v:

(L + x)/t = (L + y)/(2t)

2(L + x) = (L + y)

2L + 2x = L + y

L = y − 2x … (iii)

Now, time to pass station of length (x + y):

Required time = (L + (x + y)) / v

Using (i): v = (L + x)/t

So, Required time = (L + x + y) / ((L + x)/t)

= [(L + x + y) × t] / (L + x)

Substitute L = y − 2x:

= [(y − 2x + x + y) × t] / (y − 2x + x)

= [(2y − x) × t] / (y − x)

∴ The correct answer is (2yt - xt) / (y - x).

Given:

Top diameter of the frustum (d₁) = 2k

Bottom diameter of the frustum (d₂) = 2.5k

Height of the frustum (h) = k

Formula used:

Radius of top (r₁) = d₁ / 2

Radius of bottom (r₂) = d₂ / 2

Slant height of frustum (l) = \(\sqrt{h^2 + (r_2 - r_1)^2}\)

Whole Surface Area of Frustum = Area of top base + Area of bottom base + Lateral Surface Area

Area of top base = \(\pi r_1^2\)

Area of bottom base = \(\pi r_2^2\)

Lateral Surface Area = \(\pi(r_1 + r_2)l\)

Calculations:

Calculate radii:

r₁ = 2k / 2 = k

r₂ = 2.5k / 2 = 1.25k

Calculate the difference in radii:

r₂ - r₁ = 1.25k - k = 0.25k

Calculate the slant height (l):

l = \(\sqrt{k^2 + (0.25k)^2}\)

⇒ l = \(\sqrt{k^2 + 0.0625k^2}\)

⇒ l = \(\sqrt{1.0625k^2}\)

⇒ l = k\(\sqrt{1.0625}\)

Now, calculate the surface areas:

Area of top base = \(\pi r_1^2\) = \(\pi (k)^2\) = \(\pi k^2\)

Area of bottom base = \(\pi r_2^2\) = \(\pi (1.25k)^2\) = \(\pi (1.5625k^2)\) = 1.5625\(\pi k^2\)

Lateral Surface Area = \(\pi(r_1 + r_2)l\)

⇒ Lateral Surface Area = \(\pi(k + 1.25k)k\sqrt{1.0625}\)

⇒ Lateral Surface Area = \(\pi(2.25k)k\sqrt{1.0625}\)

⇒ Lateral Surface Area = 2.25\(\pi k^2\sqrt{1.0625}\)

Whole Surface Area = \(\pi k^2\) + 1.5625\(\pi k^2\) + 2.25\(\pi k^2\sqrt{1.0625}\)

⇒ Whole Surface Area = \(\pi k^2 (1 + 1.5625 + 2.25\sqrt{1.0625})\)

⇒ Whole Surface Area = \(\pi k^2 (2.5625 + 2.25 \times 1.030776...)\)

⇒ Whole Surface Area = \(\pi k^2 (2.5625 + 2.319246...)\)

⇒ Whole Surface Area = \(\pi k^2 (4.881746...)\)

Let's use the fraction form for sqrt(1.0625) for precision:

1.0625 = 10625 / 10000 = 425 / 400 = 85 / 80 = 17/16 (Error in calculation here, 10625/10000 = 17/16 was for 1.0625, but 0.25k is (1/4)k)

0.25k = (1/4)k

(0.25k)² = (1/16)k²

l = \(\sqrt{k^2 + \frac{1}{16}k^2}\) = \(\sqrt{\frac{16k^2 + k^2}{16}}\) = \(\sqrt{\frac{17k^2}{16}}\) = \(\frac{k\sqrt{17}}{4}\)

Lateral Surface Area = \(\pi(k + 1.25k)\frac{k\sqrt{17}}{4}\)

⇒ Lateral Surface Area = \(\pi(2.25k)\frac{k\sqrt{17}}{4}\)

⇒ Lateral Surface Area = \(\pi(\frac{9}{4}k)\frac{k\sqrt{17}}{4}\)

⇒ Lateral Surface Area = \(\frac{9\pi k^2\sqrt{17}}{16}\)

Whole Surface Area = \(\pi k^2\) + \(\pi (\frac{5}{4}k)^2\) + \(\frac{9\pi k^2\sqrt{17}}{16}\)

⇒ Whole Surface Area = \(\pi k^2\) + \(\frac{25\pi k^2}{16}\) + \(\frac{9\pi k^2\sqrt{17}}{16}\)

⇒ Whole Surface Area = \(\pi k^2 \left(1 + \frac{25}{16} + \frac{9\sqrt{17}}{16}\right)\)

⇒ Whole Surface Area = \(\pi k^2 \left(\frac{16}{16} + \frac{25}{16} + \frac{9\sqrt{17}}{16}\right)\)

⇒ Whole Surface Area = \(\frac{\pi k^2}{16} (16 + 25 + 9\sqrt{17})\)

⇒ Whole Surface Area = \(\frac{\pi k^2}{16} (41 + 9\sqrt{17})\)

Using \(\sqrt{17} \approx 4.123\)

⇒ Whole Surface Area = \(\frac{\pi k^2}{16} (41 + 9 \times 4.123)\)

⇒ Whole Surface Area = \(\frac{\pi k^2}{16} (41 + 37.107)\)

⇒ Whole Surface Area = \(\frac{\pi k^2}{16} (78.107)\)

⇒ Whole Surface Area ≈ 4.8816 \(\pi k^2\)

The exact expression is \(\frac{\pi k^2}{16} (41 + 9\sqrt{17})\)

∴ The whole surface area of the frustum is \(\frac{\pi k^2}{16} (41 + 9\sqrt{17})\).

Given:

Top diameter of the frustum (d₁) = 2k

Bottom diameter of the frustum (d₂) = 2.5k

Height of the frustum (h) = k

Formula used:

Radius of top (r₁) = d₁ / 2

Radius of bottom (r₂) = d₂ / 2

Volume of frustum (V) = \(\frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)\)

Calculations:

Calculate radii:

r₁ = 2k / 2 = k

r₂ = 2.5k / 2 = 1.25k

Substitute the values into the volume formula:

V = \(\frac{1}{3}\pi (k) (k^2 + (1.25k)^2 + k \times 1.25k)\)

⇒ V = \(\frac{1}{3}\pi k (k^2 + 1.5625k^2 + 1.25k^2)\)

⇒ V = \(\frac{1}{3}\pi k (1k^2 + 1.5625k^2 + 1.25k^2)\)

⇒ V = \(\frac{1}{3}\pi k (k^2(1 + 1.5625 + 1.25))\)

⇒ V = \(\frac{1}{3}\pi k (3.8125k^2)\)

⇒ V = \(\frac{3.8125}{3}\pi k^3\)

To express 3.8125 as a fraction:

3.8125 = 38125 / 10000 = 1525 / 400 = 305 / 80 = 61 / 16

So, V = \(\frac{1}{3}\pi k \left(\frac{61}{16}k^2\right)\)

⇒ V = \(\frac{61}{48}\pi k^3\)

∴ The volume of the frustum is \(\frac{61}{48}\pi k^3\).

Given:

p + b + h = 24

p + b - h = 4

Formula used:

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = \(\frac{p + b - h}{2}\)

Where: p and b are the perpendicular sides of the triangle

Area of triangle = radius (r) of the inscribed circle × semi - perimeter.

c is the hypotenuse of the triangle

Calculation:

p + b + h = 24 cm

p + b - h = 4 cm

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = \(\frac{p + b - h}{2}\)

r = 4/2 = 2 cm

Area of triangle = 2 × 24/2 = 24 cm²

∴ The correct anser is option 2.

Given:

p + b + h = 24

p + b - h = 4

Formula used:

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = \(\frac{a + b - c}{2}\)

Where: a and b are the perpendicular sides of the triangle

c is the hypotenuse of the triangle

Calculation:

p + b + h = 24 cm

p + b - h = 4 cm

The radius (r) of the inscribed circle in a right-angled triangle is given by the formula:

r = \(\frac{p + b - h}{2}\)

r = 4/2 = 2 cm

Therefore, the radius of the inscribed circle is: 2 cm.

Given:

Radius of circle M = 8 cm

Radius of circle N = 16 cm

Distance between centers = 8 + 16 = 24 cm

Formula used:

Length of direct common tangent = \(\sqrt{d^2 - (r_1 - r_2)^2}\)

Calculation:

⇒ d = 24, r₁ = 16, r₂ = 8

⇒ PQ = √(24² - (16 - 8)²)

⇒ PQ = √(576 - 64)

⇒ PQ = √512

⇒ PQ = 16√2 cm

∴ The length of common tangent PQ = 16√2 cm.

Given:

Circle M radius = 8 cm (centre U).

Circle N radius = 16 cm (centre V).

They touch externally ⇒ UV = 8 + 16 = 24 cm.

PQ = length of common external tangent = 16√2 cm.

Formula used:

If two radii to tangent points are perpendicular to the tangent, those two radii are parallel.

Area of a trapezoid with parallel sides a and b separated by distance d = (a + b)/2 × d.

Calculations:

⇒ UP = radius of M = 8 cm (UP ⟂ PQ).

⇒ VQ = radius of N = 16 cm (VQ ⟂ PQ).

⇒ UP ∥ VQ and the distance between these parallel segments = length PQ = 16√2 cm.

⇒ Area of quadrilateral P – Q – V – U = (UP + VQ)/2 × PQ

⇒ Area = (8 + 16)/2 × 16√2

⇒ Area = 24/2 × 16√2 = 12 × 16√2

⇒ Area = 192√2 cm²

∴ Area of quadrilateral P Q V U = 192√2 cm².

Given:

Perimeter of triangle ABC = 105 cm

Ratio of altitudes AD : BE : CF = 3 : 5 : 6

Formula used:

The area of a triangle (Area) can be expressed as: Area = \((\dfrac{1}{2}) × base × height.\)

For a triangle with sides a, b, c and corresponding altitudes ha, hb, hc, we have:

2 × Area = a × ha = b × hb = c × hc

This implies that the sides of a triangle are inversely proportional to their corresponding altitudes. That is, \(a : b : c = (\dfrac{1}{h_a}) : (\dfrac{1}{h_b}) : (\dfrac{1}{h_c})\)

Calculations:

Let the sides of the triangle be a, b, c, where:

a = BC (side opposite to vertex A)

b = CA (side opposite to vertex B)

c = AB (side opposite to vertex C)

Let the altitudes be ha, hb, hc, where:

ha = AD

hb = BE

hc = CF

Given the ratio of altitudes: ha : hb : hc = 3 : 5 : 6.

Using the property that sides are inversely proportional to altitudes:

\(a : b : c = (\dfrac{1}{3}) : (\dfrac{1}{5}) : (\dfrac{1}{6})\)

⇒ \(a : b : c = (\dfrac{1}{3}) × 30 : (\dfrac{1}{5}) × 30 : (\dfrac{1}{6}) × 30\)

⇒ a : b : c = 10 : 6 : 5

This means the ratio of the sides BC : CA : AB is 10 : 6 : 5.

Therefore, AB : BC : CA = c : a : b = 5 : 10 : 6.

∴ The correct answer is option 2.

Given:

Perimeter of triangle ABC = 105 cm

Ratio of sides AB : BC : CA = 5 : 10 : 6

Formula used:

Heron's Formula for the area of a triangle:

Area = \((\sqrt{s(s-a)(s-b)(s-c)})\)

Where a, b, c are the lengths of the sides of the triangle, and

s is the semi-perimeter s = \((\dfrac{\text{a + b + c}}{2})\)

Calculations:

From the previous problem, we established

The ratio of the sides AB : BC : CA = 5 : 10 : 6.

Let the sides be AB = 5x, BC = 10x, and CA = 6x.

The perimeter is the sum of the sides:

Perimeter = AB + BC + CA = 5x + 10x + 6x = 21x

Given Perimeter = 105 cm.

⇒ 21x = 105

⇒ x = \((\dfrac{105}{21})\)

⇒ x = 5

a (BC) = 10x = 10 × 5 = 50 cm

b (CA) = 6x = 6 × 5 = 30 cm

c (AB) = 5x = 5 × 5 = 25 cm

The semi-perimeter (s): s = \((\dfrac{\text{Perimeter}}{2})\) = \((\dfrac{105}{2})\) = 52.5 cm

Area = \((\sqrt{s(s-a)(s-b)(s-c)})\)

Area = \((\sqrt{52.5 \times (52.5 - 50) \times (52.5 - 30) \times (52.5 - 25)})\)

Area = \((\sqrt{52.5 \times 2.5 \times 22.5 \times 27.5})\)

Area = \((\sqrt{99738.28125})\)

Area ≈ 284.975 cm²

Area ≈ 285 cm²

∴ The correct answer is option 4.

Given:

Inner radius of the hollow sphere (R) = 20 cm

Height of the pot (h) = 30 cm

Calculations:

Let 'r' be the inner radius of the circular opening of the pot.

Draw a vertical line from the center of the sphere to the center of the circular opening. This line segment will be perpendicular to the plane of the circular opening.

Let the center of the sphere be O. Let the center of the circular opening be C'.

The radius of the sphere (R) goes from O to any point on the surface of the sphere, including the edge of the circular opening.

Consider a right-angled triangle formed by:

The radius of the sphere (R) as the hypotenuse, from the center of the sphere to any point on the edge of the circular opening.

d = |R - h| = |20 - 30| = |-10| = 10 cm.

Now, substitute the values into the Pythagorean theorem:

R² = r² + d²

20² = r² + 10²

400 = r² + 100

r² = 400 - 100

r² = 300

r = \(\sqrt{300}\)

r = \(\sqrt{100 \times 3}\)

r = 10\(\sqrt{3}\) cm

∴ The inner radius of the circular opening of the pot is 10\(\sqrt{3}\) cm.

Given:

Inner radius of the hollow sphere (R) = 20 cm

Height of the pot (h) = 30 cm

From the previous calculation, the radius of the circular opening (r) = 10\(\sqrt{3}\) cm

From the previous calculation, the perpendicular distance from the center of the sphere to the plane of the circular opening (d) = 10 cm

Calculations:

The perpendicular distance from the center of the sphere to the plane of the circular opening (d) as the adjacent side to the angle θ (this is the vertical line segment from the center to the center of the opening).

The inner radius of the circular opening (r) as the opposite side to the angle θ.

We can use trigonometric ratios:

cos(θ) = Adjacent / Hypotenuse = d / R

cos(θ) = d / R = 10 cm / 20 cm = 1/2

The angle whose cosine is 1/2 is 60°.

⇒ θ = 60° = π/3

∴ The angle made by the line joining the center of the sphere and any point on the rim of the circular opening with a vertical line passing through the center is 60°.