Quantitative Aptitude PYQs

Explanation:

We are given:

\( p = \tan(2\alpha) - \tan(\alpha) \)

\( q = \cot(\alpha) - \cot(2\alpha) \)

We rewrite q  in terms of tangent since \(\cot(\theta) = \frac{1}{\tan(\theta)} \)

\( q = \frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)} \)

Now we compute \(\cot(\theta) = \frac{1}{\tan(\theta)} \)

\( \frac{p}{q} = \frac{\tan(2\alpha) - \tan(\alpha)}{\frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)}} \)

Next, we find a common denominator for q :

\( \frac{1}{\tan(\alpha)} - \frac{1}{\tan(2\alpha)} = \frac{\tan(2\alpha) - \tan(\alpha)}{\tan(\alpha) \cdot \tan(2\alpha)} \)

Substituting this back into the formula for \(\frac{p}{q} \)

\( \frac{p}{q} = \frac{\tan(2\alpha) - \tan(\alpha)}{\frac{\tan(2\alpha) - \tan(\alpha)}{\tan(\alpha) \cdot \tan(2\alpha)}} \)

Simplifying, we get:

\( \frac{p}{q} = \tan(\alpha) \cdot \tan(2\alpha) \)

∴ The correct answer is Option (c): \( \tan(\alpha) \cdot \tan(2\alpha) \)

Calculation: 

We are given:

\( p = \tan(2\alpha) - \tan(\alpha) \)

\( q = \cot(\alpha) - \cot(2\alpha) \)

We need to find p + q.

\( p + q = \left( \frac{\sin^2(2\alpha) - \cos^2(2\alpha)}{\sin(2\alpha) \cdot \cos(2\alpha)} \right) + \left( \frac{\cos^2(\alpha) - \sin^2(\alpha)}{\sin(\alpha) \cdot \cos(\alpha)} \right) \)

Simplifying both terms:

\( = \frac{-2\cos(4\alpha)}{\sin(4\alpha)} + \frac{2\cos(2\alpha)}{\sin(2\alpha)} \)

Now, factorizing and simplifying further:

\( = \frac{2(\sin(4\alpha) \cdot \cos(2\alpha) - \cos(4\alpha) \cdot \sin(2\alpha))}{\sin(4\alpha) \cdot \sin(2\alpha)} \)

Recognizing the sine identity, we get:

\( = \frac{2 \sin(4\alpha - 2\alpha)}{\sin(4\alpha) \cdot \sin(2\alpha)} \)

Finally, simplifying this:

\( = \frac{2 \sin(2\alpha)}{\sin(4\alpha) \cdot \sin(2\alpha)} \)

The final result is:

\( p + q = 2 cosec (4\alpha) \)

∴ The correct answer is Option (4)

Calculation: 

We are given:

\( p = \tan(2\alpha) - \tan(\alpha) \)

\( q = \cot(\alpha) - \cot(2\alpha) \)

We need to find \(\tan^2(\alpha) \) in terms of p  and q .

We start by simplifying the expression \(\frac{p}{p + 2q} \)

\( \frac{p}{p + 2q} = \frac{1}{1 + \frac{2q}{p}} \)

This simplifies further as:

\( \frac{1}{1 + \frac{2}{\tan(\alpha) \cdot \tan(2\alpha)}} \)

Now, simplify the fraction inside the denominator:

\( = \frac{1}{1 + \frac{2}{\tan(\alpha) \cdot \tan(2\alpha)}} = \frac{1}{1 + \frac{1}{2} \cdot \tan(\alpha) \cdot \tan(2\alpha)} \)

Now, expand both terms using trigonometric identities:

\( = \frac{\sin(\alpha) \cdot \sin(2\alpha)}{\sin(\alpha) \cdot \cos(\alpha) + \cos(\alpha) \cdot \cos(2\alpha)} \)

Simplifying this expression gives:

\( = \tan^2(\alpha) \)

∴ The correct answer is Option (c):

Calculation: 

We are given:

\( 2\sin(\alpha) + \cos(\alpha) = 2 \)

 

\( \cos(\alpha) = 2(1 - \sin(\alpha)) \)

\( \cos^2(\alpha) = 4(1 - \sin(\alpha))^2 \)

Use the identity \(\cos^2(\alpha) = 1 - \sin^2(\alpha) \)

\( 1 - \sin^2(\alpha) = 4(1 - 2\sin(\alpha) + \sin^2(\alpha)) \)

\( 1 - \sin^2(\alpha) = 4 - 8\sin(\alpha) + 4\sin^2(\alpha) \)

Rearrange the terms to form a quadratic equation:

\( 5\sin^2(\alpha) - 8\sin(\alpha) + 3 = 0 \)

Using the quadratic formula:

\( \sin(\alpha) = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(3)}}{2(5)} \)

\( \sin(\alpha) = \frac{8 \pm \sqrt{64 - 60}}{10} \)

\( \sin(\alpha) = \frac{8 \pm 2}{10} \)

\( \sin(\alpha) = 1 \) or \( \sin(\alpha) = \frac{3}{5} \)

Since \(0^\circ < \alpha < 90^\circ \), we select \(\sin(\alpha) = \frac{3}{5} \)

\( \cos^2(\alpha) = 1 - \sin^2(\alpha) = 1 - \left( \frac{3}{5} \right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \)

\( \cos(\alpha) = \frac{4}{5} \)

\( \tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \)

∴ The correct answer is Option (c):

Calculation: 

We know:

\( \sin(\alpha) = \frac{3}{5} \quad \text{and} \quad \cos(\alpha) = \frac{4}{5} \)

 

\( \sin(2\alpha) = 2\sin(\alpha)\cos(\alpha) \)

\( \cos(2\alpha) = 1 - 2\sin^2(\alpha) \)

\( \sin(2\alpha) = 2 \times \frac{3}{5} \times \frac{4}{5} = \frac{24}{25} \)

\( \cos(2\alpha) = 1 - 2 \times \left( \frac{3}{5} \right)^2 = 1 - 2 \times \frac{9}{25} = \frac{7}{25} \)

\( 2\sin(2\alpha) + \cos(2\alpha) = 2 \times \frac{24}{25} + \frac{7}{25} \)

Simplifying:

\( 2\sin(2\alpha) + \cos(2\alpha) = \frac{48}{25} + \frac{7}{25} = \frac{55}{25} = \frac{11}{5} \)

∴ The correct answer is Option (2):

Calculation:

 

We are given the equation for the ratio of sides using the Sine Rule:

\( \frac{a}{\sin(3x)} = \frac{b}{\sin(x)} \)

 

\( \frac{a}{b} = \frac{\sin(3x)}{\sin(x)} \)

Step 3: Use the identity for sin(3x), which is \(\sin(3x) = 3\sin(x) - 4\sin^3(x) \), and substitute it into the equation:

\( 2 = \frac{3\sin(x) - 4\sin^3(x)}{\sin(x)} \)

\( 2\sin(x) = 3\sin(x) - 4\sin^3(x) \)

\( -\sin(x) + 4\sin^3(x) = 0 \)

\( \sin(x)(4\sin^2(x) - 1) = 0 \)

We have two possible solutions for this equation:

\( \sin(x) = 0 \), which gives x = \(0^\circ \) (not valid in this case).

\( 4\sin^2(x) - 1 = 0 \), which simplifies to:

\( \sin^2(x) = \frac{1}{4} \quad \Rightarrow \quad \sin(x) = \frac{1}{2} \)

\( x = 30^\circ \)

∴ The correct answer is Option (2)

Explanation:

We are given the triangle with angles:

\( \angle B = x = 30^\circ\)

\( \angle A = 3x = 90^\circ \)

\(\angle C = 180^\circ - 120^\circ = 60^\circ \)

Step 1: Check if the sum of the angles is 180°:

\( \angle A + \angle B + \angle C = 90^\circ + 30^\circ + 60^\circ = 180^\circ \)

This confirms that the angles satisfy the angle sum property of a triangle.

**Statement

I. The triangle is right-angled.**

Since\( \angle A = 90^\circ \), the triangle is right-angled.

**Statement III:

III. The angles A, C and B of the triangle are in AP.**

The angles \(30^\circ 60^\circ , \text and 90^\circ \) are in Arithmetic Progression because:

\( 60^\circ - 30^\circ = 30^\circ \quad \text{and} \quad 90^\circ - 60^\circ = 30^\circ \)

This confirms that the angles are in AP.

Statement II is not correct because there is no mention of a side being 3 times the other.

∴ The correct answer is Option (I) and (III) are correct.

Hence, the correct answer is Option 3.

Given:

• The height of the chimney (PR) is given as 50 meters.
• The man is 100 meters away from the base of the chimney (point P).
• The angle of elevation from the man (at point M) to the highest point of the smoke (point Q) is 45^∘.
• There is a straight line from point P through point R to point Q, perpendicular to PM.

Calculation:

For the right-angled triangle △ MPQ the tangent of the angle of elevation at point M is

⇒ tan 45^∘ = PQ/PM = \(\frac{PQ}{100}\) 

Since tan 45^∘ = 1, we have PQ = 100 m.

Finding the angle ∠RMQ:

∠RMQ = ∠PMQ - ∠PMR

In △ PMR, tan ∠PMR = PR/PM = 50/100 = 1/2.

Using the formula tan(A - B) = (tan A - tan B) / (1 + tan A ⋅ tan B):

⇒ tan ∠RMQ = (tan 45^∘ - tan ∠PMR) / (1 + tan 45^∘ ⋅ tan ∠PMR)

⇒ tan ∠RMQ = (1 - 1/2) / (1 + 1 ⋅ 1/2) = (1/2) / (3/2) = 1/3

⇒ ∠RMQ = tan^-1\(\frac{1}{3}\)

Hence, the Correct answer is Option 2.

Calculation:

We are given that k is a root of x2−4x+1=0

Solving the quadratic equation:

\(x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(1)}}{2(1)} = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} \)

⇒ \(x = 2 \pm \sqrt{3} \)

Thus, the two possible values of k are

⇒ \(k = 2 + \sqrt{3} \quad \text{or} \quad k = 2 - \sqrt{3} \)

We need to find  \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) \)

Using the identity for the sum of inverse tangents

\(\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \)

⇒ \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - k \cdot \frac{1}{k}}\right) \)

\(= \tan^{-1}\left(\frac{k + \frac{1}{k}}{1 - 1}\right) = \tan^{-1} \left(\frac{k + \frac{1}{k}}{0}\right) \)

This expression results in an undefined value, but we know from properties of the inverse tangent that

⇒ \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{k}\right) = \frac{\pi}{2}\)

Hence, the correct answer is Option 4.

Calculation:

Given,

\( \tan^{-1}(k) + \tan^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{4} \)

Use the addition formula for inverse tangents:

\( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \)

Using this for \(\tan^{-1}(k) + \tan^{-1}\left(\frac{1}{2}\right) \), we get:

\( \tan^{-1}\left(\frac{k + \frac{1}{2}}{1 - k \cdot \frac{1}{2}}\right) = \frac{\pi}{4} \)

Since \(\tan\left(\frac{\pi}{4}\right) = 1 \), we have:

\( \frac{k + \frac{1}{2}}{1 - \frac{k}{2}} = 1 \)

\( k + \frac{1}{2} = 1 - \frac{k}{2} \)

\( 2k + 1 = 2 - k \)

\( 2k + k = 2 - 1 \)

\( 3k = 1 \)

\( k = \frac{1}{3} \)

Hence, the correct answer is Option 3.

Calculation:

Given,

Line 1: \(m^2x + ny - 1 = 0 \)

Line 2: \(n^2x - my + 2 = 0 \)

Slope of Line 1, \(m_1 = -\dfrac{m^2}{n}\)

Slope of Line 2, \(m_2 = \dfrac{n^2}{m}\)

For perpendicular lines, \(m_1 \times m_2 = -1\).

\(\frac{-m^2}{n} \times \frac{-n^2}{m} = -1\)

mn -1 = 0

Hence, the correct answer is Option 1.

Calculation:

Given points A(0,0), B(p,1), and C(1,q), which form an equilateral triangle. We are told (0 < p,q < 1).

Compute the squared lengths of the sides:

\(AB^{2} = (p - 0)^{2} + (1 - 0)^{2} = p^{2} + 1\)

\(AC^{2} = (1 - 0)^{2} + (q - 0)^{2} = 1 + q^{2}\)

\(BC^{2} = (1 - p)^{2} + (,q - 1,)^{2} = (1 - p)^{2} + (q - 1)^{2} = 2,(1 - p)^{2}\)

Because the triangle is equilateral, all three squared lengths are equal:

⇒ \(AB^{2} = AC^{2} \)

⇒ \(p^{2} + 1 ;=; 1 + q^{2} ;\Longrightarrow; p^{2} = q^{2} ;\Longrightarrow; p = q \)

both p and q are positive in (0,1)

Let p = q = t Then

⇒ \(AB^{2} = t^{2} + 1 \)

⇒ \(BC^{2} = 2,(1 - t)^{2} \)

Equate AB² and BC²:

⇒ \(t^{2} + 1 ;=; 2,(1 - t)^{2} ;=; 2\bigl(1 - 2t + t^{2}\bigr) ;=; 2 - 4t + 2t^{2}. \)

Simplify:

⇒ \(t^{2} + 1 = 2 - 4t + 2t^{2};\Longrightarrow;0 = 2 - 4t + 2t^{2} - (t^{2} + 1) = t^{2} - 4t + 1. \)

So t satisfies:

⇒ \(t^{2} - 4t + 1 = 0 \quad\Longrightarrow\quad t = \frac{4 \pm √{16 - 4}}{2} = \frac{4 \pm 2√{3}}{2} = 2 \pm √{3}. \)

Since 0 < t < 1, we take t = 2 - √3 (note 2 + √3> 1) is not allowed. Hence,

⇒ \(p = q = 2 - √{3}.\)

Therefore:

 \(p + q ;=; (2 - √{3}) + (2 - √{3}) ;=; 4 - 2√{3}.\)

Hence, the correct answer is Option 4.

Calculation:

Given points A(1,1), B(0,0), and C(2,0). The angle bisectors meet at P (the incenter).

Compute the side lengths:

\(a = BC = \sqrt{(2 - 0)^{2} + (0 - 0)^{2}} = 2\)

\(b = CA = \sqrt{(1 - 2)^{2} + (1 - 0)^{2}} = \sqrt{2}\)

\(c = AB = \sqrt{(1 - 0)^{2} + (1 - 0)^{2}} = \sqrt{2}\)

Therefore, \(a + b + c = 2 + \sqrt{2} + \sqrt{2} = 2 + 2\sqrt{2}.\)

By the incenter formula,

\(X_{P} ;=; \frac{a,x_{A} + b,x_{B} + c,x_{C}}{a + b + c} ;=; \frac{,2\cdot 1 + \sqrt{2}\cdot 0 + \sqrt{2}\cdot 2,}{,2 + 2\sqrt{2},} ;=; \frac{2 + 2\sqrt{2}}{2 + 2\sqrt{2}} ;=; 1.\)

\(Y_{P} ;=; \frac{a,y_{A} + b,y_{B} + c,y_{C}}{a + b + c} ;=; \frac{,2\cdot 1 + \sqrt{2}\cdot 0 + \sqrt{2}\cdot 0,}{,2 + 2\sqrt{2},} ;=; \frac{2}{2 + 2\sqrt{2}} ;=; \frac{1}{,1 + \sqrt{2},} ;=; \sqrt{2} ;-; 1. \)

∴ The incenter is \(P = \bigl(1,;\sqrt{2} - 1\bigr) \).

Hence, the correct answer is Option 1.

Calculation:

Given points A(3, −1) and B(1, 1). Let P be the midpoint of AB, and Q be a point on the perpendicular bisector of AB that lies √2 units from P.

Compute the midpoint P of AB:

\(P = \Bigl(\tfrac{3 + 1}{2},,\tfrac{-1 + 1}{2}\Bigr) = (2,,0) \)

Compute the slope of AB:

\(m_{AB} = \frac{,1 - (-1),}{,1 - 3,} = \frac{2}{-2} = -1\)

Therefore, the equation of line AB is:

\(y - 1 = -1,(x - 1);\Longrightarrow;x + y - 2 = 0\)

The perpendicular bisector of AB must pass through P(2, 0) and have slope perpendicular to −1 (i.e. slope +1):

\(y - 0 = 1,(x - 2);\Longrightarrow;y = x - 2\)

Any point Q on this bisector satisfies y = x - 2. Write Q = (x, x − 2).

We require the distance PQ = √2. Since P(2, 0),

\(\text{Distance}^2 = (x - 2)^2 + \bigl((,x - 2,) - 0\bigr)^2 = 2\)

⇒ \((x - 2)^2 + (x - 2)^2 = 2;\Longrightarrow;2,(x - 2)^2 = 2;\Longrightarrow;(x - 2)^2 = 1\)

Thus,

⇒ \(x - 2 = \pm 1;\Longrightarrow;x = 3\text{ or }x = 1\)

If x = 3, then y = 3 - 2 = 1. So one solution is Q(3, 1).

If x = 1, then y = 1 - 2 = -1. So the other solution is Q(1, −1).

Hence, the correct answer is Option 2.

Calculation:

Given vertices A(1, 2) and D(−2, 6), where AD is the altitude from A onto BC in an equilateral triangle ABC.

Compute the slope of AD:

\(m_{AD} = \frac{,y_D - y_A,}{,x_D - x_A,} = \frac{,6 - 2,}{,-2 - 1,} = \frac{4}{-3} = -\tfrac{4}{3}.\)

Because AD ⟂ BC, the slope of BC, \(m_{BC}\), satisfies

\(m_{AD} \cdot m_{BC} = -1 ;\Longrightarrow; \Bigl(-\tfrac{4}{3}\Bigr) ,m_{BC} = -1 ;\Longrightarrow; m_{BC} = \frac{-1}{, -\tfrac{4}{3},} = \tfrac{3}{4}.\)

Equation of the line BC 

\(y - 6 = \tfrac{3}{4},(x + 2).\)

\(4(y - 6) = 3(x + 2);\Longrightarrow;4y - 24 = 3x + 6.\)

\(4y - 24 - 3x - 6 = 0 \)

\(;\Longrightarrow; -3x + 4y - 30 = 0 ;\Longrightarrow; 3x - 4y + 30 = 0.\)

Hence, the correct answer is Option 4.

Calculation:

Given two diameters of the circle are along the lines x + y = 0 and x - y = 0. The center of the circle is at their intersection.

We have, 

\(x - y = 0 ;\Longrightarrow; x = y.\)

Substitute into x + y = 0:

\(x + x = 0 ;\Longrightarrow; 2x = 0 ;\Longrightarrow; x = 0,; y = 0.\)

Thus, the center is (0, 0).

Also radius,

\(r = \frac{10}{2} = 5.\)

the equation of the circle with center (0,0) and radius 5 is

\(x^{2} + y^{2} = 5^{2} = 25.\)

Hence, the correct answer is Option 2.

Calculation:

Given,

The circle’s equation is

\(x^{2} + y^{2} + 2x + 2y + 1 = 0\)

Rewrite by completing squares:

\((x + 1)^{2} + (y + 1)^{2} = 1\)

∴ center = (-1, -1), radius r = 1

For a square inscribed in this circle with sides parallel to the axes, its diagonal equals the circle’s diameter = 2. If the square has side length a, then

\(a\sqrt{2} = 2 ;\Longrightarrow; a = \sqrt{2}.\)

Each vertex lies a half‐side \(= \tfrac{a}{2} = \tfrac{1}{\sqrt{2}} \) from the center along both axes. Since the center is ( -1, -1), the four vertices are

\(\displaystyle \bigl(-1 \pm \tfrac{1}{\sqrt{2}},; -1 \pm \tfrac{1}{\sqrt{2}}\bigr).\)

One of these vertices is

\(\bigl(-1 + \tfrac{1}{\sqrt{2}},; -1 - \tfrac{1}{\sqrt{2}}\bigr).\)

Hence, the correct answer is Option 3

Calculation:

Given the parabola

y2 = 4x 

and a tangent to this parabola that is inclined at an angle of 45°with the positive x-axis, Hence, its slope is 

\(m ;=; \tan(45^\circ) ;=; 1.\)

A standard parametric form for y2 = 4x is 

\(\bigl(x(t),,y(t)\bigr) ;=;\bigl(t^{2},,2t\bigr), \)

since  \(y^{2} = 4x \implies (2t)^{2} = 4,t^{2} \)

The slope of the tangent at \(\bigl(t^{2},,2t\bigr) \)

Differentiate y2 = 4x 

\(2y,\frac{dy}{dx} ;=; 4 \)

\(;\Longrightarrow; \frac{dy}{dx} ;=; \frac{4}{,2y,} ;=; \frac{2}{,y,}. \)

At the point \(\bigl(x,y\bigr) = \bigl(t^{2},,2t\bigr) \) one has \(y = 2t \)

\(\left.\frac{dy}{dx}\right|_{,(t^{2},,2t)} \) \(= \frac{2}{,2t,} = \frac{1}{t}.\)

We require this slope to equal 1.

\(\frac{1}{t} ;=; 1 ;\Longrightarrow; t = 1. \)

Now point of contact

Substitute t = 1 in \(\bigl(x(t),,y(t)\bigr) = \bigl(t^{2},,2t\bigr)\). 

\(x(1) = 1^{2} = 1, \)

\(y(1) = 2 \cdot 1 = 2. \)

Thus the point of contact of the tangent of slope 1 is (1, 2)

Hence, the correct answer is Option 4.

Calculation:

Given,

Hyperbola equation: \(25x^{2} - 75y^{2} = 225\)

Divide both sides by 225 to obtain standard form:

\(\frac{25x^{2}}{225} - \frac{75y^{2}}{225} = 1 ;\Longrightarrow; \frac{x^{2}}{9} - \frac{y^{2}}{3} = 1\)

Thus, \(a^{2} = 9\) and \(b^{2} = 3\).

Compute \(c\) from \(c^{2} = a^{2} + b^{2}\):

\(c^{2} = 9 + 3 = 12 ;\Longrightarrow; c = \sqrt{12} = 2\sqrt{3}.\)

The foci are at \((\pm c,,0)\), so the distance between them is \(2c\):

\(2c = 2 \times 2\sqrt{3} = 4\sqrt{3}.\)

∴ The distance between the two foci is \(4\sqrt{3}\) units.

Hence, the correct answer is Option 2.

Calculation:

Given any point on the ellipse is 3sinα, 5cosα. In the standard parametric form of an ellipse,

\(x = a,\sinα,\quad y = b,\cosα\)

we identify

\(a = 3,\quad b = 5 \)

Since (b > a), the semi-major axis is (b = 5) and the semi-minor axis is (a = 3). The eccentricity e of an ellipse is

\(e = \sqrt{,1 - \frac{(\text{semi-minor})^{2}}{(\text{semi-major})^{2}},} = \sqrt{,1 - \frac{a^{2}}{b^{2}},} \)

Substitute (a = 3) and (b = 5):

\(e = \sqrt{,1 - \frac{3^{2}}{5^{2}},} = \sqrt{,1 - \frac{9}{25},} = \sqrt{\frac{16}{25}} = \frac{4}{5} \)

Hence, the correct answer is Option 2.