Quantitative Aptitude PYQs

Concept:

\(\sec \theta =\frac{1}{\cos \theta }~and~\tan \theta =~\frac{\sin \theta }{\cos \theta }\)

sin² x + cos² x = 1

Calculation:

Given: \(cosec~\theta =\frac{29}{21}\)

As we know that, sin² x + cos² x = 1

\(\Rightarrow \sin \theta =\frac{21}{29}~and\cos \theta =~\frac{20}{29}\)

\(\Rightarrow 4\sec \theta +4\tan \theta =~4\times \frac{29}{20}+4\times \frac{21}{20}~=10\)

Concept:

If A, G and H is the arithmetic, geometric and harmonic mean respectively between any two numbers a and b.

Then A ≥ G ≥ H

where,

A = (a + b)/2,

G = √(ab) and

H = 2ab/(a + b)

Calculation:

Let a = cos θ and b = sec θ

\(\Rightarrow A=\frac{\cos \theta +\sec \theta }{2}\)

and, \(G=\sqrt{\cos \theta \times \sec \theta }\)

According to the concept used

\(\Rightarrow A\ge G\)

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

\(\Rightarrow \cos \theta +~\sec \theta ~\ge 2\)

Hence, the statement I is true.

Similarly, Let a = tan θ and b = cot θ

\(\Rightarrow A=\frac{\tan \theta +\cot \theta }{2}\)

and, \(G=\sqrt{\tan \theta \times \cot \theta }\)

According to the concept used

\(\Rightarrow A\ge G\Rightarrow ~\frac{\tan \theta +\cot \theta }{2}\ge 1\Rightarrow \tan \theta +~\cot \theta ~\ge 2\)

Hence, statement II is also true.

Calculation:

Let CD be the vertical flagstaff and AB is the ladder as shown in the figure below



Given: AB(a) = 9 m, CD = h m, BC = (h - 9) and ∠ DAC = 60°.

In Δ ACD, ∠ ADC = 180° - (150°) = 30°

In Δ ABD

As, AB = BD

⇒ ∠ DAB = ∠ ADB = 30°

⇒ ∠ DAC = 60° = ∠ BAC + ∠ DAB = ∠ BAC + 30°

⇒ ∠ BAC = 30°

In Δ ABC,

\(\Rightarrow \sin 30{}^\circ =\frac{h-9}{9}=\frac{1}{2}~\Rightarrow h=13.5~m\)

Concept:

The length of a chord subtending angle θ at the centre of a circle with radius r is given by: \(2r× \sin \left( \frac{\theta }{2} \right)\)

Calculation:

Given: r = 1 unit

⇒ The length of a chord subtending angle θ at the centre of a circle with radius 1 unit = \(2× \sin \left( \frac{\theta }{2} \right)\)

Alternate Solution:

In Triangle AOB,

Apply cosine rule

\(\cos {\rm{\theta }} = {\rm{;}}\frac{{{\rm{A}}{{\rm{O}}^2} + {\rm{;B}}{{\rm{O}}^2}{\rm{;}} - {\rm{;A}}{{\rm{B}}^2}}}{{2{\rm{;}} × {\rm{AO;}} × {\rm{BO}}}}\)

Where, AO = BO = Radius = 1

\(\cos {\rm{\theta }} = {\rm{;}}\frac{{1 + {\rm{;}}1{\rm{;}} - {\rm{;A}}{{\rm{B}}^2}}}{{2{\rm{;}} × 1{\rm{;}} × 1}}\)

⇒ 2 cos θ = 2 – AB²

⇒ AB² = 2 - 2 cos θ = 2 (1 - cos θ) = 2 × 2 sin² (θ/2)

∴ AB = 2 × sin (θ/2) = Length of chord

Concept:

\(2\times {{\tan }^{-1}}x={{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)~,\~-1<x<1\)

Calculation:

Given: \(\tan \left{ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right}\) ,

here x = 1/3 and - 1 < x < 1

\(\Rightarrow 2\times {{\tan }^{-1}}\left( \frac{1}{3} \right)={{\tan }^{-1}}\left( \frac{2\times \frac{1}{3}}{1-{{\left( \frac{1}{3} \right)}^{2}}} \right)~\)

\(\Rightarrow {{\tan }^{-1}}\left( \frac{3}{4} \right)\)

\(\Rightarrow \tan \left{ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right}=\tan \left( {{\tan }^{-1}}\left( \frac{3}{4} \right) \right)=\frac{3}{4}\)

∴ The value of \(\tan \left{ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right}\)is 3/4.

Concept:

Let \(\vec{u}~and~\vec{v}\) be two vectors. Then the scalar projection of vector \(\vec{u}~on~\vec{v}\) is given by: \(\frac{\vec{u}\cdot ~\vec{v}}{\left| {\vec{v}} \right|}\)

Calculation:

Given: \(\vec{a}=\hat{i}-2\hat{j}+\hat{k}\) and \(\vec{b}=4\hat{i}-4\hat{j}+7\hat{k}\)

\(\Rightarrow ~\vec{a}\cdot ~\vec{b}=\left( \hat{i}-2\hat{j}+\hat{k} \right)\cdot \left( 4\hat{i}-4\hat{j}+7\hat{k} \right)=4+8+7=~19\)

\(\Rightarrow \left| {\vec{b}} \right|=\sqrt{{{4}^{2}}+{{\left( -~4 \right)}^{2}}+{{7}^{2}}}=9\)

∴ The scalar projection of vector \(\vec{a}~on~\vec{b}=~\frac{\vec{a}\cdot ~\vec{b}}{\left| {\vec{b}} \right|}=\frac{19}{9}\)

Concept:

Let \(\vec{u}\) be a vector then \({{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}\)

Calculation:

Let \(\vec{u}~and~\vec{v}\) be two non-zero vectors.

Given: The magnitude of the sum of two non-zero vectors is equal to the magnitude of their difference.

\(\Rightarrow \left| \vec{u}+~\vec{v} \right|=~\left| \vec{u}-~\vec{v} \right|\)

\(\Rightarrow {{\left| \vec{u}+~\vec{v} \right|}^{2}}=~{{\left| \vec{u}-~\vec{v} \right|}^{2}}\)

As we know that, if \(\vec{u}\) be a vector then \({{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}\)

\(\Rightarrow \left( \vec{u}+~\vec{v} \right)\cdot \left( \vec{u}+~\vec{v} \right)=\left( \vec{u}-~\vec{v} \right)\cdot \left( \vec{u}-~\vec{v} \right)\)

\(\Rightarrow {{\left| {\vec{u}} \right|}^{2}}+{{\left| {\vec{v}} \right|}^{2}}+2~\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\times \cos \theta =~{{\left| {\vec{u}} \right|}^{2}}+{{\left| {\vec{v}} \right|}^{2}}-2~\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\times \cos \theta \)

\(\Rightarrow 4\times ~\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\times \cos \theta =0\)

∵ \(\vec{u}~and~\vec{v}\) be two non-zero vectors ⇒ \(\left| {\vec{u}} \right|\times \left| {\vec{v}} \right|\ne 0\)

⇒ cos θ = 0 ⇒ θ = 90°

Hence, \(\vec{u}~and~\vec{v}\) are perpendicular to each other.

Concept:

Let \(\vec{u}\) be a vector then \({{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}\)

Calculation:

\(\left( \vec{a}+\vec{b} \right)\cdot \left( \vec{a}-\vec{b} \right)=\{{\left| {\vec{a}} \right|}^{2}}-\left| {\vec{a}} \right|\times \left| {\vec{b}} \right|\times \cos \theta +\left| {\vec{a}} \right|\times \left| {\vec{b}} \right|\times \cos \theta -{{\left| {\vec{b}} \right|}^{2}}=~{{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}\)

Hence, statement 1 is true.

\(\left( \left| \vec{a}+\vec{b} \right| \right)\left( \left| \vec{a}-\vec{b} \right| \right)=\sqrt{{{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}+2\times \left| {\vec{a}} \right|\times \left| {\vec{b}} \right|\times \cos \theta }~\times ~\sqrt{{{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}-2\times \left| {\vec{a}} \right|\times \left| {\vec{b}} \right|~\times \cos \theta }\)

If cos θ = 1

\(\Rightarrow \left( \left| \vec{a}+\vec{b} \right| \right)\left( \left| \vec{a}-\vec{b} \right| \right)={{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}\)

But if cos θ ≠ 1

\(\Rightarrow \left( \left| \vec{a}+\vec{b} \right| \right)\left( \left| \vec{a}-\vec{b} \right| \right)\ne {{\left| {\vec{a}} \right|}^{2}}-{{\left| {\vec{b}} \right|}^{2}}\)

Hence, statement 2 is not true .

\({{\left| \vec{a}\cdot \vec{b} \right|}^{2}}+{{\left| \vec{a}\times \vec{b} \right|}^{2}}=\{{\left| {\vec{a}} \right|}^{2}}\times {{\left| {\vec{b}} \right|}^{2}}\times {{\cos }^{2}}\theta +~{{\left| {\vec{a}} \right|}^{2}}\times {{\left| {\vec{b}} \right|}^{2}}\times {{\sin }^{2}}\theta =~{{\left| {\vec{a}} \right|}^{2}}{{\left| {\vec{b}} \right|}^{2}}\)

Hence, statement 3 is true.

Concept:

If  \(\vec{a}\) and \(\vec{b}\) be two vectors, then \(\left| \vec{a}\times ~\vec{b} \right|\) represents the area of a parallelogram with sides \(\vec{a}\) and \(\vec{b}\)

Hence, statement 1 is false.

If  \(\vec{a}\times \vec{b}=\vec{0}\) and it is given that \(\vec{a}\ne \vec{0},~\vec{b}\ne \vec{0},\) ⇒  \(\vec{a}\) and \(\vec{b}\) are parallel to each other i.e \(\vec{a}=\lambda \vec{b}\)

Hence, statement 2 is true.

Concept:

If \(\vec{a}\) is a unit vector then \(\left| {\vec{a}} \right|=1\)

If \(\vec{u}\) be a vector then \({{\left| {\vec{u}} \right|}^{2}}=~\vec{u}\cdot ~\vec{u}\)

Calculation:

Given: If \(\vec{a}:and:\vec{b}\) are unit vectors and θ is the angle between them

As we know that, if \(\vec{u}\) be a vector then \({{\left| {\vec{u}} \right|}^{2}}=\vec{u}\cdot ~\vec{u}\)

\(\Rightarrow {{\left| \vec{a}-~\vec{b} \right|}^{2}}=\left( \vec{a}-~\vec{b} \right)\cdot \left( \vec{a}-~\vec{b} \right)={{\left| {\vec{a}} \right|}^{2}}+{{\left| {\vec{b}} \right|}^{2}}-2\times \left| {\vec{a}} \right|\times ~\left| {\vec{b}} \right|\times \cos \theta \)

∵ \(\vec{a}:and:\vec{b}\) are unit vectors 

\(\Rightarrow {{\left| \vec{a}-~\vec{b} \right|}^{2}}=2-2\times \cos \theta \)

As we know that, \(2-2\times \cos \theta =4\times {{\sin }^{2}}\left( \frac{\theta }{2} \right)\)

\(\Rightarrow {{\left| \vec{a}-~\vec{b} \right|}^{2}}=~4\times {{\sin }^{2}}\left( \frac{\theta }{2} \right)\)

\(\Rightarrow \frac{{{\left| \bar{a}-\bar{b} \right|}^{2}}}{4}=~{{\sin }^{2}}\left( \frac{\theta }{2} \right)\)

Concept:

• We have seen that Ax + By + C = 0 represents a straight line which cuts off intercepts \(\frac{-C}{A}\) and \(\frac{-C}{B}\) from the axes of coordinates.
• If A vanishes, but not B or C, the intercept on the axes of x is infinitely great. The equation of the straight line then reduces to the form of y = constant and hence represents a straight line parallel to Ox.
• Similarly, if B vanishes, but not A or C, the straight line meets the axis of y at an infinite distance and is, therefore, parallel to it.
• Any equation of the form Ax + By + C = 0, will represent a straight line on the xy-plane, when both a and b are not zero simultaneously, i.e., at least one of a and b is non-zero.

Concept:

If the equation of a line is given by: xcos α + ysin α = p then the slope of the line is given by: \(-\frac{\cos \alpha }{\sin \alpha }\)

Let θ be the angle between formed between two lines L₁ and L₂ with slope m₁ and m₂ respectively, then \(\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}\times {{m}_{2}}} \right|\)

Calculation:

Given: L₁: x cos α + y sin α = a and L₂: x sin β - y cos β = a are two equations of line.

As we know that, if the equation of a line is given by: xcos α + ysin α = p then the slope of the line is given by: \(-\frac{\cos \alpha }{\sin \alpha }\)

\(\Rightarrow {{m}_{1}}=-\frac{\cos \alpha }{\sin \alpha }~and~{{m}_{2}}=\frac{\sin \beta }{\cos \beta }\)

As we know that, if θ be the angle between formed between two lines L₁ and L₂ with slope m₁ and m₂ respectively, then \(\tan \theta =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}\times {{m}_{2}}} \right|\)

\(\Rightarrow \tan \theta =\frac{\frac{\sin \beta }{\cos \beta }+~\frac{\cos \alpha }{\sin \alpha }}{1-\frac{\sin \beta \times \cos \alpha }{\cos \beta \times \sin \alpha }}=~\frac{\cos \left( \beta -\alpha \right)}{\sin \left( \beta -\alpha \right)}\)

\(\Rightarrow \tan \theta =\cot \left( \beta -\alpha \right)=\tan \left( \frac{\pi }{2}-\left( \beta -\alpha \right) \right)\)

\(\Rightarrow \theta =~\frac{\left( \pi -2\beta +2\alpha \right)}{2}\)

Concept:

If P (a, b) and Q (c, d) are two points, then the distance between these two points P and Q is given by: \(\sqrt{{{\left( c-a \right)}^{2}}+{{\left( d-b \right)}^{2}}}\)

Calculation:

Given: P(m cos 2α, m sin 2α) and Q(m cos 2β, m sin 2β) are two points.

As we know that, If P (a, b) and Q (c, d) are two points, then the distance between these two points P and Q is given by: \(\sqrt{{{\left( c-a \right)}^{2}}+{{\left( d-b \right)}^{2}}}\)

\(\Rightarrow d=\sqrt{{{m}^{2}}\left{ {{\left( \cos 2\beta -\cos 2\alpha \right)}^{2}}+{{\left( \sin 2\beta -\sin 2\alpha \right)}^{2}} \right}}\)

\(\Rightarrow {{d}^{2}}={{m}^{2}}\times \left( 2-2\cos \left( 2\alpha -2\beta \right) \right)\)

As we know that, \(2-2\times \cos \theta =4\times {{\sin }^{2}}\left( \frac{\theta }{2} \right)\)

\(\Rightarrow {{d}^{2}}=4{{m}^{2}}\times {{\sin }^{2}}\left( \alpha -\beta \right)\)

⇒ d = |2m × sin (α - β)|

Concept:

All the sides of an equilateral triangle are equal.

Calculation:

Let,

A = (-1, -1),

B = \(\left( -\sqrt{3},~\sqrt{3} \right)\) and

C = (x, y)

∵ They are vertices of an equilateral triangle,

⇒ AB = BC = AC.

\(A{{B}^{2}}={{\left( -\sqrt{3}+1 \right)}^{2}}+{{\left( \sqrt{3}+1 \right)}^{2}}=8~units\)

Similarly,

\(B{{C}^{2}}={{\left( x+\sqrt{3} \right)}^{2}}+{{\left( y-\sqrt{3} \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2\sqrt{3}~\left( x-y \right)+6=8\)   ----(1)

\(A{{C}^{2}}={{\left( x+1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2\left( x+y \right)+2=8\)     ----(2)

From (1) and (2), we get

\(\Rightarrow ~2\sqrt{3}~\left( x-y \right)+6=~2\left( x+y \right)+2\)

⇒ x = y = 1.

Concept:

For an ellipse Eccentricity of an ellipse is given by: \(e=\frac{c}{a}\) where \(c=\sqrt{{{a}^{2}}-{{b}^{2}}}\)

Calculation:

Let B and B' be the endpoints of the minor axis with coordinates (0, b) and (0, - b).

Foci at points R and S with co-ordinates (-ae, 0) and (+ae, 0) respectively.

By joining the line segment BS and B'S as shown in the diagram below ∠BSB' = π/2

Let m₁ is the slope of BS ⇒ \({{m}_{1}}=\frac{b-0}{0-ae}=~-\frac{b}{ae}\)

Let m₂ is the slope of B'S ⇒ \({{m}_{2}}=\frac{-b \ - \ 0}{0~-~ae}=\frac{b}{ae}\)

Since, The angle between BS and B'S is π/2.

⇒ m₁ × m₂ = - 1.

\(\Rightarrow {{m}_{1}}\times {{m}_{2}}=~-\frac{b}{ae}\times \frac{b}{ae}=-1\)

\(\Rightarrow {{e}^{2}}=\frac{{{b}^{2}}}{{{a}^{2}}}\)     ---(1)

As we know that, \(e=\frac{c}{a}\) where \(c=\sqrt{{{a}^{2}}-{{b}^{2}}}\)

\(\Rightarrow {{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}=1-{{e}^{2}}\)     ----(from equation (1))

\(\therefore {{e}^{2}}=\frac{1}{2}\Rightarrow e=\frac{1}{\sqrt{2}}\)

Concept:

If a, b and c are direction ratio’s of a line then direction cosines of the line are given by:

\(l=\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}},~m=~\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}~and~n=~\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\)

Calculation:

Given: A point on a line has coordinates (p + 1, p - 3, √2p)

⇒ x = p + 1 ⇒ x - 1 = p     ---(1)

⇒ y = p - 3 ⇒ y + 3 = p     ---(2)

⇒ z = √2 × p ⇒ \(\frac{z}{\sqrt{2}}\)  = p   ---(3)

From (1), (2) and (3) we can say that

\(\frac{x-1}{1}=\frac{y+3}{1}=\frac{z-0}{\sqrt{2}}=p\)

∴ Direction ratios are: 1, 1, √2

⇒ Direction cosines are: \(\frac{1}{2},\frac{1}{2},\frac{1}{\sqrt{2}}\)

Calculation:

Given: \(\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+2}{7}\)

Hence, the point on the line satisfies the condition: \(\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+2}{7}=k\)

⇒ x = k + 1, y = 2k + 3 and z = 7k - 2.

⇒ Points lying on the line \(\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+2}{7}\) is of the form (k + 1, 2k + 3, 7k - 2)

If k = 1 then, the point (2, 5, 5) lies on the given line.

Calculation:

Given: the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) lies on the plane 2x - 4y + z = 7

Any point lying on the line will lie on the plane too as line lies on the plane.

Let, A be a point with co-ordinates (4, 2, k) and it lies on the line \(\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}\) as it satisfies the given equation of line.

⇒ The point A also lies on the plane. Hence it will satisfy the equation of plane 2x - 4y + z = 7

⇒ 8 - 8 + k = 7

⇒ k = 7

Concept:

If a line L makes positive angle α, β and γ with x, y and z - axis respectively. Then direction cosines of the line L are : l = cos α , m = cos β and n = cos γ

Calculation:

∵ It is given that the straight line passing through the point (1, 1, 1) makes an angle 60° with the positive direction of z-axis, and the cosine of the angles made by it with the positive directions of the y-axis and the x-axis are in the ratio √3 :

1. ⇒ l = k, m = √3 k and n = cos 60° = 1 / 2.

As we know that, l² + m² + n² = 1.

\(\Rightarrow {{l}^{2}}+{{m}^{2}}+{{n}^{2}}={{k}^{2}}+3{{k}^{2}}+\frac{1}{4}=1\)

\(\Rightarrow k=~\pm ~\frac{\sqrt{3}}{4}\)

So, direction cosines of Line 1 are: \(\left( \frac{\sqrt{3}}{4},~\frac{3}{4},\frac{1}{2} \right)\)

So, direction cosines of Line 2 are: \(\left( -\frac{\sqrt{3}}{4},~-\frac{3}{4},\frac{1}{2} \right)\)

\(\Rightarrow \cos \theta =\left( \frac{\sqrt{3}}{4}\times -\frac{\sqrt{3}}{4} \right)+\left( \frac{3}{4}\times -\frac{3}{4} \right)+\left( \frac{1}{2}\times ~\frac{1}{2} \right)=~-\frac{1}{2}\)

⇒ θ = 60°

Concept:

Equation of line passes through the 3 points (x_1, y₁, z₁), (x_2, y₂, z_2), and (x₃, y₃, z₃)

\(\rm \frac{x_3\ - \ x_2}{x_3 \ - \ x_1}=\frac{y_3\ -\ y_2}{y_3\ -\ y_1}=\frac{z_3\ -\ z_2}{z_3\ -\ z_1}\)

Calculation:

Given: the points (x, y, -3), (2, 0, -1) and (4, 2, 3) lie on a straight line

\(\Rightarrow \frac{3+1}{3+3}=\frac{2-0}{2-y}=\frac{4-2}{4-x}\)

\(\Rightarrow \frac{4}{6}=\frac{2}{4-x}\Rightarrow x=1\)

\(\Rightarrow \frac{4}{6}=\frac{2}{2-y}\Rightarrow y=-1\)

Additional Information

 Equation of line passes through the 3 points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3)

\(\rm \frac{x_2\ - \ x_3}{x_2 \ - \ x_1}=\frac{y_2\ -\ y_3}{y_2\ -\ y_1}=\frac{z_2\ -\ z_3}{z_2\ -\ z_1}\)

\(\rm \frac{x_1\ - \ x_2}{x_1 \ - \ x_3}=\frac{y_1\ -\ y_2}{y_1\ -\ y_3}=\frac{z_1\ -\ z_2}{z_1\ -\ z_3}\)