Quantitative Aptitude PYQs

Concept:

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

Calculation:

Given: \(\tan \theta = \frac{1}{2}\) and \(\tan \varphi = \frac{1}{3}\)

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

\(\Rightarrow ;\tan \left( {\theta + \phi } \right) = \frac{{\tan \theta + \tan \phi }}{{1 - \tan \theta \tan \phi }} = \frac{{\frac{1}{2} + \frac{1}{3}}}{{1 - \frac{1}{6}}} = 1 = \tan \frac{\pi }{4}\)

⇒ θ + φ = π/4

Concept:

sin² x + cos² x = 1

2 sin A sin B = cos (A - B) – cos (A + B)

cos 2A = cos² A – sin² A

Calculation:

Given: \(cosA = \frac{3}{4}\)

As we know that, sin² x + cos² x = 1

\(\Rightarrow ;{\sin ^2}A = 1 - {\cos ^2}A = 1 - \frac{9}{{16}} = \frac{7}{{16}}\)

\(\Rightarrow ;\sin A = \frac{{\sqrt 7 }}{4}\)

\(\Rightarrow ;\sin \left( {\frac{A}{2}} \right)\sin ;\left( {\frac{{3A}}{2}} \right) = \frac{1}{2} \times \left[ {2 \times \sin \left( {\frac{A}{2}} \right)\sin ;\left( {\frac{{3A}}{2}} \right)} \right]\)

As we know that, 2 sin A sin B = cos (A - B) – cos (A + B)

\(\Rightarrow ;\sin \left( {\frac{A}{2}} \right)\sin ;\left( {\frac{{3A}}{2}} \right) = \frac{1}{2} \times (\cos A - \cos 2A)\)      ...1)

As we know that, cos 2A = cos² A – sin² A

\(\Rightarrow ;\cos 2A = \frac{9}{{16}} - \frac{7}{{16}} = \frac{1}{8}\)

By substituting the value of cos A and cos 2A in equation (1), we get

\(\Rightarrow ;\sin \left( {\frac{A}{2}} \right)\sin \left( {\frac{{3A}}{2}} \right) = \frac{1}{2} \times \left( {\frac{3}{4} - \frac{1}{8}} \right) = \frac{5}{{16}}\)

Concept:

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

Calculation:

Given: tan 75° + cot 75°

⇒ tan 75° = tan (30° + 45°)

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

\(\Rightarrow ;\tan 75^\circ = \tan \left( {45^\circ + 30^\circ } \right) = \frac{{\tan 45^\circ + \tan 30^\circ }}{{1 - \tan 45^\circ \tan 30^\circ }} = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\)

As we know that, cot x = 1/tan x

\(\Rightarrow ;\cot 75^\circ = \frac{1}{{\tan 75^\circ }} = \frac{{\sqrt 3 ; - ;1}}{{\sqrt 3 ; + ;1}}\)

⇒ tan 75° + cot 75° = 4

Calculation:

Given: cos 46° cos 47° cos 48° cos 49° cos 50° ….

As we know that, cos 90° = 0.

⇒ cos 46° cos 47° cos 48° cos 49° cos 50° …. = cos 46° cos 47° cos 48° cos 49° cos 50° … cos 90°… = 0.

Concept:

\(\sin 18^\circ = \frac{{\sqrt 5 - 1}}{4}\)

If sin x = sin y then x = n π + (- 1)ⁿ y

Calculation:

Given: sin 2θ = cos 3θ, where \(0 < \theta < \frac{\pi }{2}\)

As we know that, cos x = sin (90° - x)

\(\Rightarrow ;\sin 2\theta = \sin \left( {\frac{\pi }{2} - 3\theta } \right)\)

As we know that, if sin x = sin y then x = n π + (- 1)ⁿ y

\(\Rightarrow ;2\theta = n\pi + {\left( { - 1} \right)^n};\left( {\frac{\pi }{2} - 3\theta } \right)\)

\(0 < \theta < \frac{\pi }{2}\)

\(\Rightarrow ~2\theta =\frac{\pi }{2}-~3\theta \Rightarrow ~\theta =\frac{\pi }{10}\)

\(\Rightarrow ;\sin \left( {\frac{\pi }{{10}}} \right) = \sin 18^\circ = \frac{{\sqrt 5 - 1}}{4}\)

Concept:

If α and β are the roots of the quadratic equation ax² + bx + c = 0. Then

\(\alpha + \beta = - \frac{b}{a};and;\alpha \times \beta = \frac{c}{a}\)

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

Calculation:

Given: tan 19° and tan 26° are the roots of the equation x² + px + q = 0.

As we know that, if α and β are the roots of the quadratic equation ax² + bx + c = 0. Then

\(\alpha + \beta = - \frac{b}{a};and;\alpha \times \beta = \frac{c}{a}\)

Here, a = 1, b = p and c = q.

⇒ tan 19° + tan 26° = - p and tan 19° × tan 26° = q.

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

\(\Rightarrow ;\tan \left( {19^\circ + 26^\circ } \right) = \frac{{\tan 19^\circ + \tan 26^\circ }}{{1 - \tan 19^\circ \times \tan 26^\circ }} = \tan 45^\circ = 1\)

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

Concept:

If a_n is the nth term of an AP and S_n is the sum of n terms of the same AP then a_n = S_n – S_{n - 1}

Calculation:

Given: S_n = n (n + 1)

⇒ S_{n - 1} = (n - 1) × (n) = n² - n

⇒ S_n = n² + n

As we know that, if a_n is the nth term of an AP and S_n is the sum of n terms of the same AP then a_n = S_n – S_{n - 1}

⇒ a_n = S_n – S_{n – 1} = 2n

⇒ a₄ = 2 × 4 = 8

Concept:

1 + tan² x = sec² x

(a + b)² = a² + b² + 2ab

Calculation:

⇒ (1 + tan α tan β)² = 1 + tan² α tan² β + 2 tan α tan β     ...1)

⇒ (tan α – tan β)² = tan² α + tan² β – 2 tan α tan β           ...2)

Adding (1) and (2), we get

⇒ (1 + tan α tan β)² + (tan α – tan β)²

⇒ 1 + tan2 α tan2 β + 2 tan α tan β + tan2 α + tan2 β – 2 tan α tan β 

⇒ 1 + tan2 α tan2 β + tan2 α + tan2 β

⇒ 1 + tan2 α + tan2 α tan2 β + tan2 β

⇒ (1 + tan2 α) + tan2 β (1 + tan2 α)

⇒ (1 + tan2 α)(1 + tan2 β)

⇒ sec² α sec² β

∴ (1 + tan α tan β)² + (tan α – tan β)² – sec² α sec² β is equal to 0.

Concept:

cosec² x – cot² x = 1

Calculation:

Given: p = cosec θ – cot θ and q = (cosec θ + cot θ)^-1

⇒ cosec θ + cot θ = 1/q

As we know that, cosec² x – cot² x = 1

⇒ (cosec θ + cot θ) × (cosec θ – cot θ) = 1

\(\frac1q \times p=1\)

⇒ p = q

Concept:

Sine rule: \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}\)

Calculation:

Given: The angles of a triangle ABC are in the ratio 1 : 2 : 3

Let, ∠ A = x, ∠ B = 2x and ∠ C = 3x

⇒ ∠ A + ∠ B + ∠ C = 180°     ...(By angle sum property)

⇒ x = 30°

⇒ ∠ A = 30°, ∠ B = 60° and ∠ C = 90°

As we know that, \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}\)

\(\Rightarrow ;\frac{a}{{\sin 30^\circ }} = \frac{b}{{\sin 60^\circ }} = \frac{c}{{sin \ 90^\circ }} = k\)

⇒ a = k/2, b = √3 k/2 and c = k

Hence, a : b : c = 1 : √3 : 2

Concept:

The perpendicular distance d from the point P (x₁, y₁) to the line ax + by + c = 0 is given by:

\(d = \left| {\frac{{a{x_1} + b{y_1} + c}}{{\sqrt {{a^2} + {b^2};} }}} \right|\)

Calculation:

Statement 1:

Given: Equation of line in normal is x cos θ + y sin θ = p and point (α, β).

By comparing the given equation of line in normal form with the standard equation of line ax + by + c = 0, we get a = cos θ, b = sin θ and c = - p.

Thus, the perpendicular distance between point (α, β) and the line x cos θ + y sin θ = p

\(\Rightarrow ;d = \left| {\frac{{\alpha \cos \theta + \beta \sin \theta - ;p}}{{\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta ;} }}} \right| = \left| {\alpha \cos \theta + \beta \sin \theta - p} \right|\)

Hence, statement 1 is false.

Statement 2:

Given: Equation of line is \(\frac{x}{a} + \frac{y}{b} = 1\) and point (α, β)

The given equation of line can be expressed as: b × x + a × y – ab = 0.

By comparing the given equation with the standard equation of line Ax + By + C = 0.

We get, A = b, B = a and C = - ab. Thus, the perpendicular distance between point (α, β) and the given line is:

\(d = \left| {\frac{{b \times \alpha + a \times \beta - ab}}{{\sqrt {{a^2} + {b^2};} }}} \right|\)

Hence, statement 2 is also false.

Concept:

The equation of a circle with centre at (h, k) and radius r units is given by: (x - a)² + (y - b)² = r².

Calculation:

Given: Equation of circle x² + y² = a² and equation of chord x + y = a.

Let AB is the chord represented by equation x + y = a.

⇒ y = a – x     ...1)

By substituting equation (1) in the equation x² + y² = a², we get the co-ordinates of A and B

Let, A = (a, 0) and B = (0, a).

If AB is the diameter of a circle, then the midpoint of AB will be the centre of the circle.

Let C be the mid-point of AB, then co-ordinates of C are: \(\left( {\frac{a}{2},\frac{a}{2}} \right)\)

Length of AB \(= \sqrt {{a^2} + {a^2}} = \sqrt {2;{a^2}}\)

∵ AB is the diameter

⇒ r² = (AB)²/4 = a²/2

So, the equation of circle with centre C and radius r units is given by

\(\Rightarrow ;{\left( {x - \frac{a}{2}} \right)^2} + {\left( {y - \frac{a}{2}} \right)^2} = {r^2} = \frac{{{a^2}}}{2}\)

⇒ x² + y² – ax – ay = 0

Explanation:

From the figure and the property of ellipse

PM = (a/e) - x

And, SP = e × PM

⇒ SP = e × [(a/e) - x]

⇒ SP = a - ex      ----(i)

Similarly from the figure and 

PN = (a/e) + x

And, S'P = e × PN

⇒ S'P = e × [(a/e) + x]

⇒ S'P = a + ex      ----(ii)

From equation (i) and (ii) we get

SP + S'P = a - ex + a + ex 

⇒ SP + S'P = 2a (Length of major axis)

 

Important Points In the case of vertical ellipse it must be equal to the length of the minor axis.

Concept:

I. The equation of a hyperbola with the centre at origin and foci on x – axis is given by:

\(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\)

II. The equation of a hyperbola with the centre at origin and foci on y – axis is given by:

\(\frac{{{y^2}}}{{{b^2}}} - \frac{{{x^2}}}{{{a^2}}} = 1\)

Calculation:

Given: 2x² – 3y² – 6 = 0

Now by dividing the given equation by 6 on both the sides, we get

\( \Rightarrow ;\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1\)

As we know that, the equation of a hyperbola with the centre at origin and foci on x – axis is given by: \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\)

Hence, the given equation represents a hyperbola.

Concept:

I. Standard equation of parabola symmetric about x – axis and vertex at origin is:

A. y² = 4ax, if a > 0

B. y² = - 4ax, if a < 0.

II. Standard equation of parabola symmetric about y – axis and vertex at origin is:

C. x² = 4ay, if a > 0

D. x² = - 4ay, if a < 0

Calculation:

Given: The two parabolas y² = 4ax and x² = 4ay

By squaring both the sides of y² = 4ax, we get

⇒ y⁴ = 16 × a² × x²     ...1)

So, by substituting the value of x² in equation (1), we get

⇒ y⁴ = 16 × a² × 4 × a × y

⇒ y = 4a     ...2)

So, by substituting (2) in y² = 4ax, we get

⇒ 16a² = 4ax ⇒ x = 4a

Hence, y² = 4ax and x² = 4ay intersect at point (4a, 4a) and as well as at point (0, 0) ∵ it satisfies both the equation y² = 4ax and x² = 4ay.

So, the given two curves intersect at two points on the line y = x.

Concept:

I. The diagonals of a rectangle bisect each other.

II. If C is the mid-point of line joining A (x₁, y₁) and B (x₂, y₂) then the co-ordinates of point C are: \(\left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)\)

Calculation:

Given: The points (1,3) and (5, 1) are two opposite vertices of a rectangle and the other tow vertices lie on the line y = 2x + c.

Let, A = (1, 3) and B = (5, 1).

As we know that, the diagonals of a rectangle bisect each other.

Let, C be the mid-point of AB where the line y = 2x + c and the diagonal AB.

⇒ The co-ordinates of C is: \(\left( {\frac{{1 + 5}}{2},\frac{{3 + 1}}{2}} \right) = \left( {3,;2} \right)\)

∵ The point C lies on the line y = 2x + c ⇒ The point C will satisfy the equation y = 2x + c.

⇒ 2 = 2 × 3 + c ⇒ c = - 4

Concept:

If a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 are 3 lines then these lines are said to be concurrent if:

\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\ {{a_2}}&{{b_2}}&{{c_2}}\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)

Calculation:

Given: the lines 3y + 4x = 1, y = x + 5 and 5y + bx = 3 are concurrent.

Now, by comparing the three lines with the standard equation of line ax + by + c = 0 we get:

⇒ a₁ = 4, b₁ = 3, c₁ = -1, a₂ = 1, b₂ = -1, c₂ = 5, a₃ = b, b₃ = 5 and c₃ = -3.

As we know that, if a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 are 3 lines then these lines are said to be concurrent if:

\(\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}}\ {{a_2}}&{{b_2}}&{{c_2}}\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0\)

\( \left| {\begin{array}{*{20}{c}} 4&3&{ - ;1}\ 1&{ - ;1}&5\ b&5&{ - ;3} \end{array}} \right| = ;14b - 84 = 0\)

⇒ b = 6.

Concept:

If L₁ and L₂ are two lines with slope m₁ and m₂ respectively such that m₁ × m₂ = - 1. Then the given lines are perpendicular to each other.

The equation of a line passing through the point (x₁, y₁) and having the slope ‘m’ is y – y₁ = m (x – x₁).

Calculation:

Given: Point (3, 2) and perpendicular to y = x.

So, by comparing the equation y = x with y = m₁x + c, we get m₁ (say) = 1.

As we know that, if two lines L₁ and L₂ are perpendicular to each other with slope m₁ and m₂ respectively, then m₁ × m₂ = -1.

So, let the slope of line perpendicular to y = x be m₂

⇒ m₁ × m₂ = 1 × m₂ = -1 ⇒ m₂ = -1.

So, the equation of line with slope m₂ and passing through point (3, 2) is given by

⇒ y – 2 = - 1 × (x - 3) ⇒ x + y = 5.

Concept:

I. Any triangle in which all the sides are of different length is called a scalene triangle.

II. Any triangle in which length of any two sides of the triangle is same is said to be an isosceles triangle.

III. Any triangle in which all the sides of the triangle are of equal length is said to be an equilateral triangle.

Calculation:

Given: The straight lines x + y – 4 = 0, 3x + y – 4 = 0 and x + 3y – 4 = 0 form a triangle.

Let, A be the point of intersection of line x + y – 4 = 0 and 3x + y – 4 = 0. So, by solving the two equations we get the co-ordinate of point A which is (0, 4)

Similarly, let B be the point of intersection of line 3x + y – 4 = 0 and x + 3y – 4 = 0. So, by solving the two equations we get the co-ordinate of point B which is (1, 1).

Similarly, let C be the point of intersection of line x + y – 4 = 0 and x + 3y – 4 = 0. So, by solving the two equations we get the co-ordinate of point C which is (4, 0).

So, Length of side AB \( = \sqrt {{1^2} + {3^2}} = \sqrt {10}\)

Similarly, Length of side BC \(= \sqrt {{3^2} + {1^2}} = \sqrt {10}\)

Similarly, Length of side AC \(= \sqrt {{4^2} + {4^2}} = 4\sqrt 2\)

So, as AB = BC ≠ AC.

Hence the triangle formed by the three given lines is an isosceles triangle.

Concept:

The Intercept made by the circle x² + y² + 2gx + 2fy + c = 0 on:

I. X-axis is given by: \(2\sqrt {{g^2} - c}\)

II. Y-axis is given by: \(2\sqrt {{f^2} - c}\)

Note: Intercepts are always positive.

Calculation:

Given: Equation of circle

x² + y² + 4x – 7y + 12 = 0

Now, by comparing the given equation of circle with the standard equation of circle x² + y² + 2gx + 2fy + c = 0, we get:

g = 2, f = - 7/2 and c = 12.

As we know that, the intercept made by the circle x² + y² + 2gx + 2fy + c = 0 on Y – axis is given by: \(2\sqrt {{f^2} - c}\).

\(\Rightarrow ;2\sqrt {{f^2} - c} = 2\sqrt {{{\left( { - \frac{7}{2}} \right)}^2} - 12} = 1\)