Quantitative Aptitude PYQs

Formula used:

sin θ = √(1 – cos2 θ)

tan θ = sin θ/cos θ

Calculation:

Since, sin θ = √(1 – cos² θ)

⇒ sin θ = √(1 – 1/5) = √(4/5) = 2/√5

⇒ tan θ = sin θ/cos θ = (2/√5) / (1/√5)

⇒ tan θ = 2

⇒ 2 tan θ/(1 – tan² θ)

= (2 × 2) / (1 – 4)       [∴ tan θ = 2]

= 4/(– 3)

∴ \(\frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\) = (– 4/3)

When the value of θ ranges as 0 < θ < 90°, the value of cos θ ranges as 0 > cos θ > 1

Similarly,

When the value of φ ranges as 0 < φ < 90°, the value of cos φ ranges as 0 > cos φ > 1

Now given,

⇒ cos θ < cos φ

⇒ θ > φ

In the figure given below,

Considering ∆PQR,

⇒ tan 45° = Perpendicular/Base = PQ/QR

⇒ 1 = (h + r)/QR

⇒ QR = (h + r)

∵ The flag is just visible from the point R, PR is a tangent to the hemispherical dome

⇒ QT = Radius of dome = r

⇒ ∠QTR = ∠QTP = 90°

Considering ∆QTR,

⇒ ∠TQR = ∠QRT = 45°

This means that ∆QTR is an isosceles triangle

⇒ RT = QT = r

Similarly, considering ∆QTP,

⇒ ∠TQP = ∠QPT = 45°

This means that ∆QTP is an isosceles triangle

⇒ PT = QT = r

⇒ PR = PT + RT = 2r

Now, applying Pythagoras theorem in ∆PQR,

⇒ PR² = PQ² + QR²

⇒ (2r)² = (h + r)² + (h + r)²

⇒ 4r² = 2(h + r)²

⇒ 2r² = (h + r)²

⇒ r√2 = h + r

⇒ h = r(√2 – 1)

⇒ h/r = (√2 – 1)

Given, sin (A + B) = √3/2

⇒ sin (A + B) = sin 60°

⇒ A + B = 60°      ----(1)

Also, cos B = √3/2

⇒ cos B = cos 30°

⇒ B = 30°

Substituting in (1),

⇒ A = 60° – 30° = 30°

∴ tan (2A – B) = tan (60° – 30°) = tan 30° = 1/√3

Considering statement 1,

⇒ [cos θ/(1 – sin θ)] + [cos θ/(1 + sin θ)] = 4

⇒ [cos θ (1 + sin θ) + cos θ (1 – sin θ)]/(1 – sin² θ) = 4

⇒ 2 cos θ/cos² θ = 4                     [∵ sin² θ + cos² θ = 1]

⇒ cos θ = 1/2

⇒ θ = 60° [∵ 0 < θ < 90°]

⇒ Statement 1 is correct

Considering statement 2,

⇒ 3 tan θ + cot θ = 5 cosec θ

⇒ 3 (sin θ/cos θ) + (cos θ/sin θ) = 5/sin θ

⇒ 3 sin² θ + cos² θ = 5 cos θ

⇒ 3 – 3 cos² θ + cos² θ = 5 cos θ [∵ sin² θ + cos² θ = 1]

⇒ 2 cos² θ + 5 cos θ – 3 = 0

⇒ (2 cos θ – 1)(cos θ + 3) = 0

⇒ cos θ = 1/2, – 3

⇒ θ = 60° [∵ 0 < θ < 90°]

⇒ Statement 2 is correct

∴ Both statements 1 and 2 are correct

Considering statement 1,

The value of cos² θ lies in the range [0, 1]

The value of 1 – cos² θ also lies in the range [0, 1]

⇒ 0 ≤ (1 – cos² θ) ≤ 1

⇒ 0 ≤ (p² + q²)/2pq ≤ 1

⇒ 0 ≤ (p² + q²)/2pq

⇒ 0 ≤ (p² + q²), is always true as square terms are always greater than zero

Similarly,

⇒ (p² + q²)/2pq ≤ 1

⇒ (p² + q²) ≤ 2pq

⇒ (p² + q²) – 2pq ≤ 0

⇒ (p – q)² ≤ 0

∵ A square term can never be less than zero,

⇒ (p – q)² = 0

⇒ p = q

⇒ Statement 1 is correct

Considering statement 2,

Given, tan² θ = 4pq/(p + q)² – 1

⇒ sec² θ = 4pq/(p + q)² [∵ sec² θ = 1 + tan² θ]

⇒ cos² θ = (p + q)²/4pq

The value of cos² θ lies in the range [0, 1]

⇒ 0 ≤ (p + q)²/4pq ≤ 1

⇒ 0 ≤ (p + q)²/4pq

⇒ 0 ≤ (p + q)², is always true as square terms are always greater than zero

Similarly,

⇒ (p + q)²/4pq ≤ 1

⇒ p² + q² + 2pq ≤ 4pq

⇒ p² + q² – 2pq ≤ 0

⇒ (p – q)² ≤ 0

∵ A square term can never be less than zero,

⇒ (p – q)² = 0

⇒ p = q

⇒ Statement 2 is correct

∴ Both statements 1 and 2 are correct

Considering statement 1

Using AM-GM inequality

\(\frac{{cos\theta + sec\theta }}{2} \ge \sqrt {cos\theta .sec\theta }\)

\(\Rightarrow \frac{{cos\theta + sec\theta }}{2} \ge \sqrt 1\)

⇒ cos θ + sec θ ≥ 2

Hence statement 1 is correct.

Considering statement 2

sec² θ + cosec² θ = 1 + tan² θ + 1 + cot² θ = 2 + tan² θ + cot² θ

Using AM-GM inequality

\(\frac{{ta{n^2}\theta + co{t^2}\theta }}{2} \ge \sqrt {{{\tan }^2}\theta .{{\cot }^2}\theta }\)

⇒ tan² θ + cot² θ ≥ 2

∴ sec² θ + cosec² θ = 2 + tan² θ + cot² θ ≥ 4

Hence, statement 2 is also correct.

Alternate solution∶

Considering statement 1,

Suppose cos θ + sec θ = 1.5

⇒ cos θ + 1/cos θ = 3/2

⇒ 2(cos² θ + 1) = 3 cos θ

⇒ 2 cos² θ – 3 cos θ + 3 = 0

For a quadratic equation of form ax² + bx + c, x = \(x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)

⇒ cos θ = [3 ± √(9 – 24)]/4

⇒ cos θ = (3 ± √– 15)/4, which is not possible as the value of cos θ is always a real number

⇒ Statement 1 is correct

Considering statement 2,

Suppose sec² θ + cosec² θ < 4

⇒ 1/cos² θ + 1/sin² θ < 4

⇒ sin² θ + cos² θ < 4 sin² θ cos² θ

⇒ 1 < sin² 2θ, which is not possible as the value of sin² x always lie between [0, 1]

⇒ Statement 2 is correct

∴ Both statements 1 and 2 are correct.

Given, sin² x + sin x = 1

⇒ sin x = 1 – sin² x

⇒ sin x = cos² x       ----(1)

As we know, (a + b)³ = a³ + 3a²b + 3ab² + b³

⇒ cos¹² x + 3 cos¹⁰ x + 3 cos⁸ x + cos⁶ x

= (cos⁴ x + cos² x)³ [Substituting from eq. (1)]

= (sin² x + cos² x)³                   [∵ sin² θ + cos² θ = 1]

= 1³

= 1

Given, (3sin θ + 5cos θ) = 4

Squaring both sides,

⇒ (3sin θ + 5cos θ)² = 4²

⇒ 9sin² θ + 25cos² θ + 30sin θ cos θ = 16 [∵ (a + b)² = a² + b² + 2ab]

⇒ 30sin θ cos θ = 16 – 9sin² θ – 25cos² θ      ----(1)

Now,

(3cos θ – 5sin θ)²

= 9cos² θ + 25sin² θ – 30sin θ cos θ

= 9cos² θ + 25sin² θ – 16 + 9sin² θ + 25cos² θ [Substituting from eq. (1)]

= 34(sin² θ + cos² θ) – 16

= 34 – 16 [∵ sin² θ + cos² θ = 1]

= 18

Given, cot θ (1 + sin θ) = 4m

⇒ (cos θ/sin θ) (1 + sin θ) = 4m     ----(1)

Also, cot θ (1 – sin θ) = 4n

⇒ (cos θ/sin θ) (1 – sin θ) = 4n     ----(2)

Multiplying (1) and (2), we get,

⇒ (cos² θ/sin² θ) (1 – sin² θ) = 16mn [∵ cos² θ = 1 – sin² θ]

⇒ cos⁴ θ/16sin² θ = mn      ----(3)

Now,

Using equation (1) & (2), we can write,

⇒ 16(m² + n²) = (cos² θ/sin² θ) (1 + sin² θ + 2 sin θ + 1 + sin² θ – 2 sin θ)

⇒ (m² + n²) = (1/8) (cos² θ/sin² θ) (1 + sin² θ)

⇒ (m² + n²)² = (1/64) (cos⁴ θ/sin⁴ θ) (1 + sin² θ)²

Comparing with equation (3),

⇒ (m² + n²)² ≠ mn

Also, (m² + n²)² ≠ m²n²

Again using equation (1) & (2), we can write,

⇒ 16(m² – n²) = (cos² θ/sin² θ) (1 + sin² θ + 2 sin θ – 1 – sin² θ + 2 sin θ)

⇒ (m² – n²) = (1/4) (cos² θ/sin² θ) (sin θ)

⇒ (m² – n²)² = (1/16) (cos⁴ θ/sin⁴ θ) (sin² θ)

⇒ (m² – n²)² = (cos⁴ θ/16sin² θ)

Comparing with equation (3),

⇒ (m² – n²)² = mn

Let the given triangle be ∆ABC, as shown below,

⇒ BC = base = (u² – v²)

⇒ AC = hypotenuse = (u² + v²)

According to Pythagoras theorem,

⇒ (AC)² = (AB)² + (BC)²

⇒ (u² + v²)² = (AB)² + (u² – v²)²

⇒ u⁴ + v⁴ + 2u²v² = (AB)² + u⁴ + v⁴ – 2u²v²

⇒ (AB)² = 4u²v²

⇒ AB = 2uv

Now,

Area of triangle = 1/2 × base × height

⇒ 1/2 × BC × AB = 2016

⇒ (u² – v²) × uv = 2016

⇒ uv(u + v)(u – v) = 32 × 7 × 9

We can write,

⇒ uv(u + v)(u – v) = 16 × 9√2 × 7√2

Thus, we get, u = 8√2 and v = √2,

∴ Perimeter of triangle = AB + BC + AC = 2uv + u² – v² + u² + v²

= 2uv + 2u² = 32 + 256 = 288 units

Let the radius of the circle be ‘r’ and the side of square be ‘a’ in the figure given below,

Area of equilateral triangle = A = (√3/4) × (side)² = (√3/4)l²

Semi-perimeter of triangle = s = 3l/2

∵ Radius of incircle of triangle = Area/semi-perimeter

⇒ r = (√3/4)l² / (3l/2)

⇒ r = l/2√3

Now,

Diagonal of square = Diameter of circle

⇒ (side of square) × √2 = 2 × radius of circle

⇒ a√2 = l√3

⇒ a = l√6

∴ Area of square = (side)² = a² = l²/6

Given, length of room = l = 6 m

Breadth of room = b = 4 m

Height of room = h = 2.5 m

⇒ Total area of walls of 1 room = 2lh + 2bh = 30 + 20 = 50 m²

⇒ Total area of walls in 5 identical rooms = 5(50) = 250 m²

Now, two rooms have one square window each of side 2.5 m,

⇒ Area of the two windows = 2 × (2.5)² = 12.5 m²

⇒ Total area to be painted = 250 – 12.5 = 237.5 m²

∵ Area of 20 m² can be painted in 1 can,

∴ Total no. of cans required = 237.5/20 = 11.875 ≅ 12

Let ABCD be the parallelogram T and WXYZ be the parallelogram S as shown below,

Considering statement 1,

As we know, the line segments joining the mid-points of the sides of a triangle divides the triangle into four equal triangles

∵ In ∆ABD, AW = WB & AZ = ZD

⇒ Area (∆AWZ) = 1/4 × Area (∆ABD)       ----(1)

Also, in ∆BCD, BX = XC & CY = YD

⇒ Area (∆CXY) = 1/4 × Area (∆BCD)      ----(2)

Adding (1) and (2), we get,

⇒ Area (∆AWZ) + Area (∆CXY) = 1/4 × [Area (∆ABD) + Area (∆BCD)]

⇒ Area (∆AWZ) + Area (∆CXY) = 1/4 × [Area (ABCD)]      ----(3)

Similarly,

⇒ Area (∆DYZ) = 1/4 × Area (∆ADC)      ----(4)

⇒ Area (∆BWX) = 1/4 × Area (∆ABC)      ----(5)

Adding (4) and (5), we get,

⇒ Area (∆DYZ) + Area (∆BWX) = 1/4 × [Area (ABCD)] ----(6)

Now, adding (3) and (6), we get,

⇒ Area (∆AWZ) + Area (∆CXY) + Area (∆DYZ) + Area (∆BWX) = 1/2 × [Area (ABCD)]

⇒ Area (ABCD) – Area (WXYZ) = 1/2 × [Area (ABCD)]

⇒ 1/2 × Area (ABCD) = Area (WXYZ)

⇒ Area (ABCD)/Area (WXYZ) = 2

⇒ (Area of parallelogram T)∶ (Area of parallelogram S) = 2∶ 1

⇒ Statement 1 is correct

Considering statement 2,

According to the mid-point theorem, the length of the line segment joining the mid-points of two sides of a triangle is half the length of the third side

In ∆ABD, WZ = 1/2 × BD

In ∆ABC, WX = 1/2 × AC

In ∆BCD, XY = 1/2 × BD

In ∆ADC, YZ = 1/2 × AC

Adding the four equations, we get,

⇒ WZ + WX + XY + YZ = AC + BD

⇒ Perimeter (WXYZ) = Sum of diagonals of ABCD

⇒ Perimeter of parallelogram S is equal to the sum of diagonals of parallelogram T

⇒ Statement 2 is not correct

∴ Only statement 1 is correct

Formula∶

For a triangle with sides of lengths a, b, and c units,

Semi-perimeter = s = (a + b + c)/2

Area of triangle = √[s(s – a)(s – b)(s – c)]

Calculation∶

Hence, for the given triangle,

⇒ s = (5 + 6 + 7)/2 = 18/2 = 9 cm

∴ Area of given triangle = √(9 × 4 × 3 × 2) = √36 × √3 × √2

= 6 × 1.732 × 1.414 = 14.69 ≅ 14.7 cm²

As we know, the area of circle = π × (Radius)²

Given, area of circular plot = 144π m²

⇒ Radius of circular plot = √144 = 12 m

Now,

Width of path = 5 m

⇒ Radius of circular plot including the path surrounding it = 12 + 5 = 17 cm

∴ Required area = π × (17)² = 289π m²

Given, the lateral surface area of cone = 462 cm²

Also, the slant height of cone = 35 cm

∵ Lateral surface area of cone = π × radius × slant height

⇒ 462 = (22/7) × radius × 35

∴ Radius of base of cone = 21/5 = 4.2 cm

Let the angle between the generator and the axis of the cone be ‘θ’, shown below,

⇒ Radius of cone = OB = r (let)

When the semi-circle is rolled to form a cone, slant height of cone = Radius of semi-circle

⇒ AB = R (let)

Also, the circumference of the base of cone = Circumference of semi-circle

⇒ 2πr = πR

⇒ r/R = 1/2

Now, considering ∆OAB,

⇒ sin θ = OB/AB = r/R

⇒ sin θ = 1/2

⇒ sin θ = sin 30°

∴ θ = 30°

Let the radius of the base of the cylinder be ‘r’ cm and its height be ‘h’ cm

Given,

Lateral surface area of cylinder = (Total surface area of cylinder) /2

⇒ 2πrh = (2πrh + 2πr²) /2

⇒ 2πrh – πrh = πr²

⇒ πrh = πr²

⇒ h = r

∵ Height = h = π cm

∴ Radius of the base of cylinder = π cm

∵ Volume of cuboid = Length × Breadth × Height

⇒ Volume of rectangular block = 20 × 15 × 10 = 3000 cm³

Now, the least possible number of cubes will be cut from the rectangular block, when a cube cut is of the maximum possible size

⇒ Side of maximum possible cube = HCF of (20, 15, 10) = 5 cm

⇒ Volume of one cube cut = (5)³ = 125 cm³

∴ Least possible number of cubes = 3000/125 = 24