Quantitative Aptitude PYQs

Concept:

Sum of angles of a triangle is 180^∘

Explanation:

ABC is a triangle and C = 60^∘.

So, A + B + 60^∘ = 180^∘

i.e., A + B = 120^∘ ...(i)

 \(\frac{\cos A+\cos B}{\cos \left(\frac{A-B}{2}\right)}\)

= \(\frac{2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)}{\cos \left(\frac{A-B}{2}\right)}\)

= \(2\cos \left(\frac{A+B}{2}\right)\)

= 2 cos(60^∘) (using (i))

= 2 × \(\frac12\) = 1

Option (3) is true.

Concept:

Formula:

(i) cot^-1x = tan^-1(1/x)

(ii) 2tan^-1x = tan^-1\(2x\over 1-x^2\)

(iii) tan-1 x - tan-1 y = tan-1 \(x-y\over 1+xy\)

(iv) cot(cot^-1x) = x

Explanation:

2cot^-13 = 2tan^-1(1/3) = tan-1\(\frac23\over 1-{\frac19}\) = tan^-1 \(\frac23\times\frac98\) = tan-1 \(\frac{3}{4}\)

So, \(\sqrt{15+\cot ^2\left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)} \)

= \(\sqrt{15+\cot ^2\left(\tan^{-1}1-\tan ^{-1} \frac34\right)} \)

= \(\sqrt{15+\cot ^2\left(\tan^{-1}\frac{1-\frac34}{1+1.\frac34}\right)} \)

= \(\sqrt{15+\cot ^2\left(\tan^{-1}\frac17\right)} \)

= \(\sqrt{15+\cot ^2\left(\cot^{-1}7\right)} \)

= \(\sqrt{15+7^2} \)

= \(\sqrt{64} \) = 8

Option (3) is true.

Concept:

Formula: 2sin A sin B = cos(A - B) - cos(A + B)

cos c + cod d = 2cos\(c+d\over2\)cos\(c-d\over2\)

Explanation:

sin 10°·sin 50° + sin 50°·sin 250° + sin 250°·sin 10° 

= \(\frac12\) (2sin 10°·sin 50° + 2sin 50°·sin 250° + 2sin 250°·sin 10°)

= \(\frac12\)(cos 40° - cos 60° + cos 200° - cos 300° + cos 240° - cos 260°)

= \(\frac12\){cos 40° - \(\frac12\) + cos 200° - cos (360° - 60°) + cos (180° + 60°) - cos 260°} 

= \(\frac12\){cos 40° - \(\frac12\) + cos 200° - \(\frac12\) - \(\frac12\) - cos 260°} 

= \(\frac12\){- \(\frac32\) + cos 200° + cos 40° - cos 260°}

= \(\frac12\){- \(\frac32\) + 2cos120° cos80° - cos260°}

= \(\frac12\){- \(\frac32\) + 2cos(180° - 60°) cos80° - cos260°}

= \(\frac12\){- \(\frac32\) - 2cos 60° cos80° - cos260°}

= \(\frac12\){- \(\frac32\) - cos80° - cos260°}

= \(\frac12\){- \(\frac32\) - (cos80° + cos260°)}

= \(\frac12\){- \(\frac32\) - (cos80° + cos(180° + 80°))}

= \(\frac12\){- \(\frac32\) - (cos80° - cos80°)}

= \(-\frac{3}{4}\)

Option (2) is true.

Concept:

Formula: tan^-1x - tan^-1y = tan^-1 \(x-y\over 1+xy\)

Explanation:

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

= \(\tan ^{-1}\left({\frac{a}{b}-{a-b\over a+b}\over1+\frac ab.{a-b\over a+b}}\right)\)

= \(\tan ^{-1}\left({\frac{a^2+ab-ab+b^2}{b(a+b)}\over{ab+b^2+a^2-ab\over b(a+b)}}\right)\)

= tan^-1 \(a^2+b^2\over a^2+b^2\) = tan^-11 = \(\frac{\pi}{4}\)

Option (2) is true.

Concept:

Then quadratic equation ax² + bx + c = 0 has real roots if 

b² - 4ac ≥ 0

Explanation:

(cos β - 1)x2 + (cos β)x + sin β = 0  will have real roots if

(cos β)2 - 4(cos β - 1)sin β ≥ 0...(i)

Now, -1 ≤ cos β ≤ 1 and sin β ≥ 0 for β ∈ [0, π]  

So, (i) is possible only when 

cos β - 1 ≥ 0

i.e., 1 - cos β ≤ 0

Hence option (2) is true.

Concept:

Pythagoras formula: In a right angle triangle,

base² + altitude² = hypotenuse²

Explanation:

In a triangle ABC, AB = 16 cm, BC = 63 cm and AC = 65 cm. 

Since 16² + 63² = 65²

So, ABC is a right angle triangle

then sin A = 63/65, sin B = 1, sin C = 16/65

Now, cos 2A + cos 2B + cos 2C

  = 1 - 2sin²A + 1 - 2sin2B + 1 - 2sin2C

  = 3 - 2(sin2A + sin2B + sin2C)

= 3 - 2(\(({63\over 65})^2+({16\over 65})^2+1\))

= 3 - 2(1 + 1) = -1 

Option (1) is true.

Explanation:

\(α+β=\frac{5 π}{4}\)

⇒ tan(α + β) = tan(5π/4)

⇒ \({\tanα+\tanβ\over1-\tanα\tan β}\) = tan(π + \(π\over4\))

⇒ \({\tanα+\tanβ\over1-\tanα\tan β}\) = tan\(π\over4\)

⇒ \({\tanα+\tanβ\over1-\tanα\tan β}\) = 1

⇒ tan α + tan β = 1 - tan α tan β

tan α + tan β + tan α tan β = 1....(i)

Now, \(f(\theta)=\frac{1}{1+\tan \theta}\)

⇒ f(α) = \(\frac{1}{1+\tan α}\), f(β) = \(\frac{1}{1+\tan β}\)

So, f(α) f(β) = \(\frac{1}{1+\tan α}\).\(\frac{1}{1+\tan β}\) = \(\frac{1}{1+\tanα+\tan β+\tanα.\tan β}\) = \(1\over 1+1\) = \(\frac{1}{2}\)

(using (i))

Option (2) is true.

Alternate Method \(α+β=\frac{5 π}{4}\)

Take α = π and β = π/4

then  f(α) f(β) = \(\frac{1}{1+\tan α}\).\(\frac{1}{1+\tan β}\) = 1.\(\frac{1}{2}\) = \(\frac{1}{2}\)

Concept:

cos2x = \(1-\tan^2x\over1+\tan^2x\)

Explanation:

tan α and tan β are the roots of the equation x2 - 6x + 8 = 0

So, tan α + tan β = 6 and tan α.tan β = 8

∴ \(\tan α+\tan β\over1- \tan α.\tan β\) = \(6\over 1-8\)

⇒ tan(α + β) = -6/7

Now, cos(2α + 2β) = cos2(α + β) = \(1-\tan^2(α + β)\over1+\tan^2(α + β)\) = \(1-\frac{36}{49}\over1+\frac{36}{49}\) = \(\frac{13}{85}\)

Option (2) is true.

Given:

Evaluate the expression: tan 65º + 2tan 45º - 2tan 40º - tan 25º

Concept Used:

We will use the values of the tangent function at specific angles.

Recall the following values:

• tan 45º = 1
• tan 65º ≈ 2.14
• tan 40º ≈ 0.84
• tan 25º ≈ 0.47

Calculation:

Substitute the known values of tangent into the expression:

tan 65º + 2tan 45º - 2tan 40º - tan 25º

≈ 2.14 + 2(1) - 2(0.84) - 0.47

≈ 2.14 + 2 - 1.68 - 0.47

≈ 4.14 - 2.15

≈ 1.99

≈ 2 (rounding to the nearest whole number)

∴ The value of the expression tan 65º + 2tan 45º - 2tan 40º - tan 25º is approximately 2.

Calculation:

Statement 1:

In any triangle, the sum of the angles is π, and the angles A, B, C are all positive and less than π.

For a triangle to be acute-angled, each angle must be less than π/2.

⇒ 0 < A, B, C < π/2.

Now, cot x = \(\frac{1}{\tan x}\), is positive in the range 0 < x < π/2.

⇒ If  0 < A, B, C < π/2, then cot A > 0, cot B > 0, and cot C > 0.

⇒ cot A cot B cot C > 0 implies that all the angles A, B, and C are acute, making the triangle an acute-angled triangle.

Hence, Statement 1 is correct.

Statement 2:

We know that, tan x is positive in the interval (0, π/2) and negative in π/2, π.

Now, in a triangle, if one angle is obtuse (e.g., A > π/2), then the remaining two angles are acute, making their tangents positive.

⇒ tan A tan B tan C will be negative because one of the tangents (corresponding to the obtuse angle) is negative while the others are positive.

⇒ For tan A tan B tan C to be positive, either all the angles must be acute, or exactly two angles must be obtuse (which is not possible in a triangle as the sum of angles must be π).

Hence, Statement 2 is incorrect.

∴ Statement 1 is correct only.

The correct answer is Option 1.

Concept:

equation of a circle of centre (a, b) and radius r is

(x - a)² + (y - b)² = r² 

Explanation:

Equation of circle is

x2 + y2 + 2x + 6y + 1 = 0

⇒ x2 + 2x + 1 + y2 + 6y + 9 = 9

⇒ (x + 1)² + (y + 3)² = 3²

So, centre of the circle is (a, b) = (-1, -3) and radius is c = 3

Therefore

a2 + b2 + c2 = 1 + 9 + 9 = 19

Option (1) is true.

Explanation:

(1, -1, 2) and (2, 1, -1) are the end points of a diameter of sphere.

So, centre of the sphere is the midpoint i.e., \(({1+2\over 2}, {-1+1\over 2}, {2-1\over 2})\) = \(({3\over 2}, 0, {1\over 2})\)

Now, centre of the sphere 

x2 + y2 + z2 + 2ux + 2vy + 2wz - 1 = 0 

is (-u, -v, -w)

Then u = -3/2, v = 0, w = -1/2

So, u + v + w = \(-{3\over 2}+0-{1\over 2}\) = -2.

Option (1) is true.

Explanation:

Then graph of the equation x = 5 is

So, we can see that the number of points represented by the equation x = 5 on the xy-plane is infinitely many.

Option (4) is true.

Concept:

For a plane ax + by + cz + d = 0:

• Direction ratios = <a, b, c>
• Direction cosines = <\(\frac{a}{\sqrt{a^2+b^2+c^2}}\), \(\frac{b}{\sqrt{a^2+b^2+c^2}}\), \(\frac{c}{\sqrt{a^2+b^2+c^2}}\)> = <l, m, n>

l² + m² + n² = 1

Calculation:

Given, 2x - 3y + 6z + 4 = 0

⇒ Direction ratios are 2, - 3, 6

⇒ Direction cosines are \(\frac{2}{\sqrt{2^2+(-3)^2+6^2}}\), \(\frac{-3}{\sqrt{2^2+(-3)^2+6^2}}\), \(\frac{6}{\sqrt{2^2+(-3)^2+6^2}}\)

⇒ Direction cosines are \(\frac{2}{7}\), \(\frac{-3}{7}\), \(\frac{6}{7}\)

⇒ l = \(\frac{2}{7}\), m = \(\frac{-3}{7}\), n = \(\frac{6}{7}\)

∴ 49(7l2 + m2 - n2)

= 49(7×\(\frac{4}{49}\) + \(\frac{9}{49}\) - \(\frac{36}{49}\))

= 28 + 9 - 36

= 1

∴ The value of  49(7l2 + m2 - n2) is 1.

The correct answer is Option 2.

Concept:

The line passing through (x₀, y₀, z₀) and having direction ratios <a, b, c> is given by:

\(\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}\)

Calculation:

Given, a line passing through (1, -1, 2) has direction ratios < 3, 2, 2 > 

⇒ The line is represented as \(\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{2}\) = t (let)

⇒ x = 3t + 1, y = 2t - 1, z = 2t + 2

Let P(x, y, z) be the point of intersection.

⇒ P = (3t + 1, 2t - 1, 2t + 2)

Now, P lies on the plane x + 2y + 3z = 18

⇒ 3t + 1 + 2(2t - 1) + 3(2t + 2) = 18

⇒ 3t + 1 + 4t - 2 + 6t + 6 = 18

⇒ 13t + 5 = 18

⇒ 13t = 13

⇒ t = 1

∴ P = (3t + 1, 2t - 1, 2t + 2) = (3 + 1, 2 - 1, 2 + 2) = (4, 1, 4)

∴ The point of intersection of line and plane is (4, 1, 4).

The correct answer is Option 3.

Explanation:

Equation of the plane passing through (1, 0, 0), (0, 1, 0) and (0, 0, 1) is

\(\frac x1+\frac y1+\frac z1\) = 1

⇒ x + y + z - 1 = 0....(i)

Distance of the plane (i) from origin 

p = \(\left|{0+0+0-1\over \sqrt{1^2+1^2+1^2}}\right|\)

⇒ p = \(1\over\sqrt3\)

⇒ 3p2 = 1

Option (4) is true.

Explanation:

l + 2m + n = 0

2l - 2m + 3n = 0

Solving by cross multiplication method

\({l\over 6+2}={m\over 2-3}={n\over-2-4}\)

⇒ \({l\over 8}={m\over -1}={n\over-6}\) = k (say

then l = 8k, m = -k, n = -6k

Since l, m, n are direction cosines then

l² + m² + n² = 1

⇒ 64k² + k² + 36k² = 1

⇒ 101k²  = 1

⇒ k² = 1/101

Then 

l2 + m2 - n2 

= l2 + m2 + n2 - 2n²

= 1 - 2n2 

= 1 -  72k² 

= 1 - \(72\over 101\) = \(\frac{29}{101}\)

Option (2) is true.

Explanation:

x + 2y - 1 = 0...(i)

2x - y - 1 = 0...(ii)

(ii) × 2 ⇒ 4x - 2y - 2 = 0...(iii)

(i) + (iii) implies

5x - 3 = 0 ⇒ x = 3/5

Putting in (i) we get

\(3\over5\) + 2y - 1 = 0

⇒ 2y = 2/5 ⇒ y = 1/5

So, the intersection point is (3/5, 1/5)

Let the mid point of AB be (a, b)

Then equation of AB is

\(\frac xa+\frac yb=2\)....(iv)

(3/5, 1/5) passing through (iv) so

\(\frac 3{5a}+\frac 1{5b}=2\)

⇒ 3b + a = 10ab

⇒ a + 3b = 10ab

So, the locus of (a, b) is

x + 3y = 10xy

Option (2) is true.

Explanation:

Let equation of the line perpendicular to x cos θ + y sin θ = 9 be

x sin θ - y cos θ = c....(i)

(i) passing through (-sin θ, cos θ)

• sin²θ - cos²θ = c ⇒ c = -1

Therefore equation of the required line is

x sin θ - y cos θ = -1

x sin θ - y cos θ + 1 = 0

Option (2) is true.

Explanation:

P, Q both lies on the line y = 2x + 3

So, P, Q are of the form (a, 2a + 3)

\(\left(1+\frac{2}{\sqrt{5}}, 5+\frac{4}{\sqrt{5}}\right),\left(1-\frac{2}{\sqrt{5}}, 5-\frac{4}{\sqrt{5}}\right)\) satisfying (a, 2a + 3)

So, option (1) is true.

Other coordinates are satisfying (a, 2a + 3).

Other options are false.