Quantitative Aptitude PYQs

Concept:

In the binomial expansion of (a + b)ⁿ there are n + 1 terms.

Calculation:

⇒ {(2x – 3y)² (2x + 3y)²}² = {(2x -3y)× (2x + 3y) × (2x – 3y) × (2x + 3y)}²

As we know that, (a - b) × (a + b) = a² – b²

⇒ {(2x – 3y)² (2x + 3y)²}² = {(4x² – 9y²)²}² = (4x² – 9y²)⁴

As we know that, in the binomial expansion of (a + b)ⁿ there are n + 1 terms.

Here, a = 4x², b = - 9y² and n = 4.

Hence, there are 5 terms in the expansion of {(2x – 3y)² (2x + 3y)²}².

Concept:

The general term in the binomial expansion of (a + b)ⁿ is given by: \({T_{r + 1}} = {;^n}{C_r} \times {a^{n - r}} \times {b^r}\)

Calculation:

Given: In the expansion of (1+ ax)ⁿ, the first three terms are respectively 1, 12x and 64x².

2^st term in expansion of (1+ ax)ⁿ: \(^n{C_1} \times {\left( 1 \right)^{n - 1}} \times {\left( {ax} \right)^1} = n \times a \times x = 12;x\)

⇒ n × a = 12 ⇒ a = 12/n

3^rd term in expansion of (1+ ax)ⁿ: \(^n{C_2} \times {\left( 1 \right)^{n - 2}} \times {\left( {ax} \right)^2} = \frac{{n \times \left( {n - 1} \right)}}{2} \times {a^2} \times {x^2} = 64;{x^2}\)

\(\Rightarrow ;\frac{{n \times \left( {n - 1} \right)}}{2} \times {a^2} = 64\)     ...1)

By substituting the value of a in equation (1), we get

\( \Rightarrow ;\frac{{n \times \left( {n - 1} \right)}}{2} \times \frac{{144}}{{{n^2}}} = 64\)

⇒ 8n = 72 ⇒ n = 9.

Concept:

I. nth term of a GP is given by: T_n = ar^n-1,

Where a = first term and r = common ratio.

II. nth term of an AP is given by: T_n = a + (n - 1) × d,

Where a = first term and d = common difference.

Calculation:

Let us suppose that the numbers 1, 5 and 25 be the p^th, q^th and r^th term respectively of an AP with a = first term and d = common difference.

⇒ a_p = a + (p - 1) × d = 1    ----(1)

⇒ a_q = a + (q - 1) × d = 5    ----(2)

⇒ a_r = a + (r - 1) × d = 25    ----(3)

By subtracting (1) from (2), we get

⇒ (q - p) × d = 4    ----(4)

Similarly, by subtracting (2) from (3), we get

⇒ (r - q) × d = 20    ----(5)

By, dividing (4) and (5), we get

⇒ (r - p)/(q - p) = 5

∵ (r - p)/(q - p) is a rational number and 5 is also a rational number.

⇒ 1, 5 and 25 can be 3 terms of infinite number of AP’s.

So, The option 1^st and 2^nd are wrong 

Now, Let 1, 5 and 25 are the three terms of a G.P

So, Let P^th, Q^th and R^th term be 1, 5 and 25 resp.

So, a r^P-1 = 1    ----(6)

 a rQ-1 = 5         ----(7) 

a r^R-1 = 25        ----(8)

On doing (7) ÷ (6), we get 

rQ-P = 5        ----(9)

And on doing (8) ÷ (7), we get 

rR-Q = 5       ----(10)

Now, From (9) and(10), we get 

Q - P = R - Q

Q = (P + R)/2

So, there are several or infinite value of P, Q, R which can satisfy the above relation. 

So, There can be infinite number of G.P's.

∴ The numbers 1, 5 and 25 can be three terms of infinite terms of G.P.

Concept:

The nth term of an AP is given by: T_n = a + (n - 1) × d, where a = first term and d = common difference.

Calculation:

As we know that, the nth term of an AP is given by: T_n = a + (n - 1) × d, where a = first term and d = common difference.

Let a be the first term and d is the common difference.

\(\Rightarrow ;{a_{p + q}} = a + \left( {p + q - 1} \right) \times d\)     ...1)

\(\Rightarrow ;{a_{p - q}} = a + \left( {p - q - 1} \right) \times d\)     ...2)

By adding (1) and (2), we get

\(\Rightarrow ;{a_{p + q}} + {a_{p - q}} = 2;a + 2;\left( {p - 1} \right)d = 2 \times \left[ {a + \left( {p - 1} \right)d} \right] = 2 \times {a_p}\)

Concept:

Properties of Determinants

1. A × adj (A) = adj (A) × A = |A| × I, where A is a square matrix of order n and I is the identity matrix of order n.
2. adj (adj.......adj (A))| = \({\left| A \right|^{{{\left( {m - 1} \right)}^n}}}\), where A is a square matrix of order m and n represents the multiplicity of adj.
3. |A × B| = |A| × |B|, where A and B are square matrices of order n.
4. |A| = |A'|, where A' is transpose of matrix A of order n.
5. |- A| = (-1)n × |A|, where A is a square matrix of order n.

Concept:

Minimum value of \(m{\sec ^2}\theta + n;cose{c^2}\theta = ;{\left( {\sqrt m + \sqrt n } \right)^2}\)

Calculation:

Given: 25 cosec² x + 36 sec² x

As we know that, minimum value of \(m{\sec ^2}\theta + n;cose{c^2}\theta = ;{\left( {\sqrt m + \sqrt n } \right)^2}\)

Here, m = 36 and n = 25.

∴ The minimum value of 25 cosec² x + 36 sec² x : \({\left( {\sqrt {36} + \sqrt {25} } \right)^2} = 121\)

Concept:

If A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

Calculation:

Given: A and B be (3 × 3) matrices with det A = 4 and det B = 3

As we know that, if A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

⇒ det (2AB) = 23 × det (A) × det (B) = 8 × 4 × 3 = 96.

Concept:

If A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

If A is a non – singular square matrix of order n, then det (A-1­) = 1/det (A)

Calculation:

Given: Given: A and B be (3 × 3) matrices with det A = 4 and det B = 3

As we know that, if A and B are square matrices of order n, then det (m × AB) = mn × det (A) × det (B) where m ∈ R is a scalar.

⇒ det (3AB-1) = 33 × det (A) × det (B- 1)

As we know that, If A is a non – singular square matrix of order n, then det (A-1­) = 1/det (A)

\(\Rightarrow {\rm{;det}}\left( {3A{B^{ - 1}}} \right) = 27 \times 4 \times \frac{1}{3} = 36\)

Concept:

For any complex number, Z = a + ib, modulus of z is given by \(r = \sqrt {{a^2} + {b^2}}\)

Calculation:

Given: \(z = \frac{{1 + 2i}}{{1 - {{\left( {1 - i} \right)}^2}}}\)

\(\Rightarrow ;z = \frac{{1 + 2i}}{{1 - {{\left( {1 - i} \right)}^2}}} = \frac{{1 + 2i}}{{1 + 2i}} = 1\)

Now, we can write z = 1 = 1 + i × 0

On comparing z = 1 + i × 0 with z = a + ib, we get 

⇒ a = 1 and b = 0

As we know that \(r = \sqrt {{a^2} + {b^2}}\), where r = |z|

Here, a = 1 and b = 0.

\(\therefore ;r = \sqrt {{1^2} + {0^2}} = 1\)

Concept:

For any complex number, Z = a + ib, argument of Z is given by: \(\arg z = \theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right)\) and modulus \(r = \sqrt {{a^2} + {b^2}}\)

The principal argument of Z is given by: Arg (Z)

Case-1: If the point Z, lies in the first quadrant then, Arg(Z) = θ.

Case-2: If the point Z, lies in the second quadrant then, Arg(Z) = π – θ

Case-3: If the point Z, lies in the third quadrant then, Arg(Z) = θ - π

Case-4: If the point Z, lies in the fourth quadrant then, Arg(Z) = - θ

Note:

• Principal value of any complex number Z is unique and - π ≤ Arg (Z) < π.

Calculation:

Given: \(z = \frac{{1 + 2i}}{{1 - {{\left( {1 - i} \right)}^2}}}\)

\(⇒ ;z = \frac{{1 + 2i}}{{1 - {{\left( {1 - i} \right)}^2}}} = \frac{{1 + 2i}}{{1 + 2i}} = 1\)

We can write z = 1 as z = 1 + i × 0

On comparing z = 1 + i × 0 with z = a + ib, we get 

⇒ a = 1 and b = 0

As we know that, \(\arg z = \theta = {\tan ^{ - 1}}\left( {\frac{b}{a}} \right)\)

\(⇒ ;\arg z = \theta = {\tan ^{ - 1}}\left( {\frac{0}{1}} \right) = 0\)

⇒ z lies on the real axis of the argand plane.

∴  Arg (z) = 0.

Concept:

sin (A + B) = sin A × cos B + sin B × cos A

cos (A - B) = cos A × cos B + sin A × sin B

Calculation:

cos 236° = cos (180° + 56°) = - cos 56°

sin 124° = sin (90° + 34°) = cos 34°

cos 178° = cos (90° + 88°) =- sin 88°

sin 208° = sin (180° + 28°) = - sin 28°

By substituting these values in \(\frac{{\sin 34^\circ \cos 236^\circ - \sin 56^\circ \sin 124^\circ }}{{\cos 28^\circ \cos 88^\circ + \cos 178^\circ \sin 208^\circ }}\) we get

\(\Rightarrow ;\frac{{\sin 34^\circ \cos 236^\circ - \sin 56^\circ \sin 124^\circ }}{{\cos 28^\circ \cos 88^\circ + \cos 178^\circ \sin 208^\circ }} = \frac{{ - \sin 34^\circ \cos 56^\circ - \sin 56^\circ \cos 34^\circ ;}}{{\cos 28^\circ \cos 88^\circ + ;\sin 88^\circ \sin 28^\circ }}\)

As we know that, sin (A + B) = sin A × cos B + sin B × cos A and cos (A - B) = cos A × cos B + sin A × sin B.

\(\Rightarrow ;\frac{{\sin 34^\circ \cos 236^\circ - \sin 56^\circ \sin 124^\circ }}{{\cos 28^\circ \cos 88^\circ + \cos 178^\circ \sin 208^\circ }} = - \frac{{\sin \left( {34^\circ + 56^\circ } \right)}}{{\cos \left( {88^\circ - 28^\circ } \right)}} = - 2\)

Concept:

\(\tan \left( {A + B} \right) = \frac{{\tan A; + \tan B}}{{1 - \tan A\tan B}}\)

Calculation:

tan 54° = tan (45° + 9°)

As we know that, \(\tan \left( {A + B} \right) = \frac{{\tan A; + \tan B}}{{1 - \tan A\tan B}}\)

\(\Rightarrow {\rm{;tan}}54^\circ = \tan \left( {45^\circ + 9^\circ } \right) = \frac{{\tan 45^\circ + \tan 9^\circ }}{{1 - \tan 45^\circ \tan 9^\circ }} = \frac{{1 + \tan 9^\circ }}{{1 - \tan 9^\circ }}\)

\(\Rightarrow {\rm{;tan}}54^\circ = \frac{{\cos 9^\circ + \sin 9^\circ }}{{\cos 9^\circ - \sin 9^\circ }}\)

Concept**:**

sin 2A = 2 sin A × cos A

cos 2A = cos² A – sin2 A

Calculation**:**

Given: p = X cos θ - Y sin θ and q = X sin θ + Y cos θ.

⇒ p2 = X² cos² θ + Y² sin² θ – 2 XY cos θ sin θ     ...1)

⇒ q² = X² sin² θ + Y² cos² θ + 2 XY cos θ sin θ     ...2)

⇒ 4pq = 4 [X² cos θ sin θ + XY cos² θ – XY sin² θ – Y² sin θ cos θ]     ...3)

By adding (1), (2) and (3), we get

⇒ p² + 4pq + q² = X² [1 + 2 sin 2θ] + Y² [1 - 2 sin 2θ] + 4 XY cos 2θ

As it is given that, p² + 4pq + q² = AX² + BY²

⇒ X² [1 + 2 sin 2θ] + Y² [1 - 2 sin 2θ] + 4 XY cos 2θ = AX² + BY²

⇒ A = 1 + 2 sin 2θ, B = 1 - 2 sin 2θ and cos 2θ = 0.

⇒ θ = π/4, A = 3 and B = - 1.

Concept**:**

sin 2A = 2 sin A × cos A

cos 2A = cos2 A – sin2 A

Calculation**:**

Given: p = X cos θ - Y sin θ and q = x sin θ + Y cos θ.

⇒ p2 = x2 cos2 θ + Y2 sin2 θ – 2 XY cos θ sin θ     ...1)

⇒ q2 = x2 cos2 θ + Y2 sin2 θ + 2 XY cos θ sin θ     ...2)

⇒ 4pq = 4 [X2 cos θ sin θ + XY cos2 θ – XY sin2 θ – Y2 sin θ cos θ]     ...3)

By adding (1), (2) and (3), we get

⇒ p2 + 4pq + q2 = X2 [1 + 2 sin 2θ] + Y2 [1 - 2 sin 2θ] + 4 XY cos 2θ

As it is given that, p2 + 4pq + q2 = AX2 + BY2

⇒ X2 [1 + 2 sin 2θ] + Y2 [1 - 2 sin 2θ] + 4 XY cos 2θ = AX2 + BY2

⇒ A = 1 + 2 sin 2θ, B = 1 - 2 sin 2θ and cos 2θ = 0.

⇒ θ = π/4, A = 3 and B = - 1.

Concept**:**

sin 2A = 2 sin A × cos A

cos 2A = cos2 A – sin2 A

Calculation**:**

Given: p = X cos θ - Y sin θ and q = x sin θ + Y cos θ.

⇒ p2 = x2 cos2 θ + Y2 sin2 θ – 2 XY cos θ sin θ     ...1)

⇒ q2 = x2 cos2 θ + Y2 sin2 θ + 2 XY cos θ sin θ     ...2)

⇒ 4pq = 4 [X2 cos θ sin θ + XY cos2 θ – XY sin2 θ – Y2 sin θ cos θ]     ...3)

By adding (1), (2) and (3), we get

⇒ p2 + 4pq + q2 = X2 [1 + 2 sin 2θ] + Y2 [1 - 2 sin 2θ] + 4 XY cos 2θ

As it is given that, p2 + 4pq + q2 = AX2 + BY2

⇒ X2 [1 + 2 sin 2θ] + Y2 [1 - 2 sin 2θ] + 4 XY cos 2θ = AX2 + BY2

⇒ A = 1 + 2 sin 2θ, B = 1 - 2 sin 2θ and cos 2θ = 0.

⇒ θ = π/4, A = 3 and B = - 1.

Concept:

sin2 x + cos2 x = 1

cos (A - B) = cos A × cos B + sin A × sin B

Calculation:

Given: cos (θ – α) = a and cos (θ – β) = b.

As we know that, sin2 x + cos2 x = 1

\(\Rightarrow ;\sin \left( {\theta - \alpha } \right) = \sqrt {1 - {a^2}} ;and\sin \left( {\theta - \beta } \right) = \sqrt {1 - {b^2}}\)

cos (α – β) = cos [(θ - β) – (θ - α)]

As we know that, cos (A - B) = cos A × cos B + sin A × sin B

⇒ cos (α – β) = cos (θ - β) × cos (θ - α) + sin (θ - β) × sin (θ - α)

\(\Rightarrow ;\cos ;\left( {\alpha - \beta } \right) = ;ab + \sqrt {1 - {b^2}} \times \sqrt {1 - {a^2}}\)

Concept:

sin2 x + cos2 x = 1

cos (A - B) = cos A × cos B + sin A × sin B

sin (A - B) = sin A × cos B – cos A × sin B

Calculation:

Given: cos (θ – α) = a and cos (θ – β) = b.

As we know that, sin2 x + cos2 x = 1

\(\Rightarrow ;\sin \left( {\theta - \alpha } \right) = \sqrt {1 - {a^2}} ;and;\sin \left( {\theta - \beta } \right) = \sqrt {1 - {b^2}}\)

From previous question, we have \(\cos ;\left( {\alpha - \beta } \right) = ;ab + \sqrt {1 - {b^2}} \times \sqrt {1 - {a^2}}\)

sin (α – β) = sin [(θ - β) – (θ - α)]

As we know that, sin (A - B) = sin A × cos B – cos A × sin B

⇒ sin (α – β) = sin (θ - β) × cos (θ - α) - cos (θ - β) × sin (θ - α)

\(\Rightarrow ;\sin ;\left( {\alpha - \beta } \right) = a\sqrt {1 - {b^2}} - b\sqrt {1 - {a^2}}\)

\(\Rightarrow ;{\sin ^2}\left( {\alpha - \beta } \right) + 2;ab\cos \left( {\alpha - \beta } \right) = {a^2} + {b^2}\)

Concept:

sin² x + cos² x = 1

sin 2x = 2 sin x cos x

Calculation:

Given: sin α + cos α = p

By squaring both the sides, we get

⇒ sin² α + cos² α + 2 sin α cos α = p²

As we know that, sin² x + cos² x = 1 and sin 2x = 2 sin x cos x

⇒ 1 + sin 2α = p²

⇒ sin 2α = p² – 1

As we can write cos² 2α = 1 – sin² 2α

⇒ cos2 2α = 1 – (p² – 1)²

∴ The value of cos² 2α is p² (2 – p²).

Concept:

sin^-1 (x) = cosec^-1 (1/x)

sec^-1 y + cosec^-1 y = π/2, y ≤ - 1 or y ≥ 1.

Calculation:

Given: \({\sin ^{ - 1}}\frac{4}{5} + {\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2}\)

As we know that, sin^-1 (x) = cosec^-1 (1/x)

\(\Rightarrow ;{\sin ^{ - 1}}\frac{4}{5} + {\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2} = cose{c^{ - 1}};\left( {\frac{5}{4}} \right) + ;{\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2}\)

As we know that, sec^-1 y + cosec^-1 y = π/2, y ≤ - 1 or y ≥ 1

\(\Rightarrow ;{\sin ^{ - 1}}\frac{4}{5} + {\sec ^{ - 1}}\frac{5}{4} - \frac{\pi }{2} = \frac{\pi }{2} - \frac{\pi }{2} = 0\)

Concept:

\(\sin 2\theta = \frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}\)

\(\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}\)

\(\tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\)

\({\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x; - ;y}}{{1 + xy}}} \right),;if;x > 0,;y > 0\)

Calculation:

Given: \({\sin ^{ - 1}}\frac{{2p}}{{1 + p2}} - {\cos ^{ - 1}}\frac{{1 - {q^2}}}{{1 + {q^2}}} = {\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}}\)

Let, p = tan A, q = tan B and x = tan C

\(\Rightarrow {\rm{;si}}{{\rm{n}}^{ - 1}}\frac{{2\tan A}}{{1 + {{\tan }^2}A}} - {\cos ^{ - 1}}\frac{{1 - {{\tan }^2}B}}{{1 + {{\tan }^2}B}} = {\tan ^{ - 1}}\frac{{2\tan C}}{{1 - {{\tan }^2}C}}\)

\(\Rightarrow ;2A - 2B = 2C\)

⇒ A – B = C

⇒ tan^-1 p – tan^-1 q = tan^-1 x

As we know that, \({\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\frac{{x; - ;y}}{{1 + xy}}} \right)\)

\(\Rightarrow ;{\tan ^{ - 1}}p - {\tan ^{ - 1}}q = {\tan ^{ - 1}}\left( {\frac{{p - q}}{{1 + pq}}} \right) = {\tan ^{ - 1}}x\)

\(\Rightarrow ;x = \frac{{p - q}}{{1 + pq}}\)