Quantitative Aptitude PYQs

Concept:

\(\rm i = \sqrt {-1}\)

i² = -1 , i³ = - i, i⁴ = 1, i⁶ = - 1 , i⁸ = 1 , i⁹ = i, i ¹² = 1, and i¹⁵ = - i

 

Calculations: 

Given determinant is \(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{i}}^2}}&{{{\rm{i}}^3}}\ {{{\rm{i}}^4}}&{{{\rm{i}}^6}}&{{{\rm{i}}^8}}\ {{{\rm{i}}^9}}&{{{\rm{i}}^{12}}}&{{{\rm{i}}^{15}}} \end{array}} \right|\)

Since, we have, 

\(\rm i = \sqrt {-1}\)

i2 = -1 , i3 = - i, i4 = 1, i6 = - 1 , i8 = 1 , i9 = i, i 12 = 1, and i15 = - i

=\(\left| {\begin{array}{*{20}{c}} {\rm{i}}&{{{\rm{-1}}}}&{{{\rm{-i}}}}\ {{{\rm{1}}}}&{{{\rm{-1}}}}&{{{\rm{1}}}}\ {{{\rm{i}}}}&{{{\rm{1}}}}&{{{\rm{-i}}}} \end{array}} \right|\)

=i(i - 1) + 1(-i - i) - i (1 + i)

= i² - i - 2i - i - i²

= - 4i

Concept:

The product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix.

 

Calculations:

Given \(A = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\ {\rm{h}}&{\rm{b}}&{\rm{f}}\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]_{3\times 3}\) and \({\rm{B}} = \left[ {\begin{array}{*{20}{c}} {\rm{x}}\ {\rm{y}}\ {\rm{z}} \end{array}} \right]_{3\times1}\)

The product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix.

i.e AB is posible and its order is \(\rm 3\times 1\)

Now, To obtain the entry in row 1, column 1 of \displaystyle AB,$\text{}$AB, multiply the first row in \displaystyle AA by the first column in \displaystyle BB, and add.

To obtain the entry in row 2, column 1 of \displaystyle AB,$\text{}$AB, multiply the second row in \displaystyle AA by the first column in \displaystyle BB, and add.

To obtain the entry in row 3, column 1 of \displaystyle AB,$\text{}$AB, multiply the third row in \displaystyle AA by the first column in \displaystyle BB, and add.

\(AB = \left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\ {\rm{h}}&{\rm{b}}&{\rm{f}}\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]_{3\times 3}​​\left[ {\begin{array}{*{20}{c}} {\rm{x}}\ {\rm{y}}\ {\rm{z}} \end{array}} \right]_{3\times1}\)

 

\(AB = \left[ {\begin{array}{*{20}{c}} {{\rm{ax}} + {\rm{hy}} + {\rm{gz}}}\ {{\rm{hx}} + {\rm{by}} + {\rm{fz}}}\ {{\rm{gx}} + {\rm{fy}} + {\rm{cz}}} \end{array}} \right]_{3\times1}\)

Concept:

Number of ways to arrange n things in r places is given by, ⁿC_r 

 

Calculation:

In word ABLE, there are 2 vowels and 2 consonants.

Total number of letters = 4

Total number of even place = 2

There are 2 vowels to be filled in 2 places.

Hence, the number of ways = ²C₂ = 1

The vowels can arrange among themselves in 2! = 2 ways.

Now, the 2 consonants can fill the remaining 2 places in 2! = 2 ways.

Therefore, total number of ways = 1 × 2 × 2 = 4 ways.

Hence, option (2) is correct.

Concept:

The maximum number of intersection by n circles is given by 2 \(\times\) ⁿC₂.

2 \(\times\) ⁿC₂ =  \(\rm2\times \dfrac {n(n-1)}{2}\) = n (n-1)

Hence, the maximum number of intersection by n circles is given by n(n - 1)

 

Calculations:

To Find: The maximum number of points of intersection of 5 non- overlapping circles.

We know that, 

The maximum number of intersection by n circles is given by 2 \(\times\) ⁿC₂.

2 \(\times\) ⁿC₂ =  \(\rm2\times \dfrac {n(n-1)}{2}\) = n (n-1)

Hence, the maximum number of intersection by n circles is given by n(n - 1)

Here, n = 3

The maximum number of points of intersection of 3 non- overlapping circles = 3 (3 - 1) = 6

Hence, The maximum number of points of intersection of 5 non- overlapping circles = 6

Concept:

If n_c = 2, number of points of intesection = 2, if n_c = 3, number of points of intesection = 6, where n_c = number of circles 

The mathematical formula can be made for the maximum number of intersection by n circles is given by 2 × nC2.

\(\rm nC_r=\frac{n!}{(n-r)!r! }\)

 

Calculation:

The maximum number of points of intersection of 5 non- overlapping circles = 

 \(2 \times \rm 5C_2\⇒ 2\times \frac{5!}{3!2!}\ ⇒ 2\times \frac{5\times 4}{2}\⇒ 20\)

Hence, option (3) is correct.

Concept:

\(\rm 1+tan^2x=sec^2x\)

 

Calculation:

a sin2 x + b cos2 x = c

Divide by cos²x, we get 

\(\begin{array}{l} \rm a \tan ^{2} x+b=c \sec ^{2} x=c\left(1+\tan ^{2} x\right) \ \rm a \ tan^2 {x}-c\ tan^2 x=c-b\ \rm {\tan ^{2} x}=\frac{c-b}{a-c} \end{array}\)

Hence, option (1) is correct.

Concept:

\(\rm 1+tan^2x=sec^2x\)

 

Calculation:

b sin² y + a cos² y = d 

Divide by cos²y, we get 

\(\begin{array}{l} \rm b \tan ^{2} y+a=d \sec ^{2} y=d\left(1+\tan ^{2} y\right) \ \rm b \ tan^2 {y}-d\ tan^2 y=d-a\ \rm {\tan ^{2} y}=\frac{d-a}{b-d} \end{array}\)

Hence, option (3) is correct.

Calculation:

Here,  p tan x = q tan y.

⇒p/q = tan y / tan x

We have, \(\rm {\tan ^{2} y}=\frac{d-a}{b-d}\) and \(\rm {\tan ^{2} x}=\frac{c-b}{a-c}\)

\(\begin{array}{c} \rm \frac{p^{2}}{q^{2}}=\frac{\tan ^{2} y}{\tan ^{2} x}=\frac{\frac{d-a}{b-d}}{\frac{c-b}{a-c}} \ = \rm \frac{(a-c)(d-a)}{(c-b)(b-d)} \ \end{array}\)

\(\Rightarrow \rm​​\frac{p^2}{q^2}\rm = \frac {(a - d)(c - a)}{(b - c)(d - b)}\)

Hence, option (2) is correct.

Concept:

\(\sin^2\theta +\cos^2\theta=1\)

 

Calculation:

\(\rm \frac {t_3 - t_5}{t_5 - t_7}\)= 

\( \frac{\sin ^{3} \theta+\cos ^{3} \theta-\sin ^{5} \theta-\cos ^{5} \theta}{\sin ^{5} \theta+\cos ^{5} \theta-\sin ^{7} \theta-\cos ^{7} \theta} \ = \frac{\sin ^{3} \theta+\cos ^{3} \theta-\sin ^{3} \theta\left(\sin ^{2} \theta\right)-\cos ^{3} \theta(\cos ^{2} \theta)}{\sin ^{5} \theta+\cos ^{5} \theta-\sin ^{5}\theta(\sin^2\theta) -\cos ^{5} \theta(\sin^2\theta) }\ = \frac{\sin ^{3} \theta+\cos ^{3} \theta-\sin ^{3} \theta\left(\sin ^{2} \theta\right)-\cos ^{3} \theta(\cos ^{2} \theta)} {\sin ^{5} \theta+\cos ^{5} \theta-\sin ^{5}\theta(\sin^2\theta) -\cos ^{5} \theta(\sin^2\theta) }\ = \frac{\sin ^{3} \theta(1-\sin^2\theta)+\cos ^{3}(1-cos^2\theta)} {\sin ^{5} \theta(1-\sin^2\theta)+\cos ^{5} \theta(1-\cos^2\theta) }\ \)

\(\begin{array}{l} =\frac{\sin ^{3} \theta \cdot \cos ^{2} \theta+\cos ^{3} \theta\left(\sin ^{2} \theta\right)}{\sin ^{5} \theta \cdot \cos ^{2} \theta+\cos ^{5} \theta\left(\sin ^{2} \theta\right)} \ =\frac{\sin ^{3} \theta \cdot \cos ^{2} \theta+\cos ^{3} \theta \cdot \sin ^{2} \theta}{\sin ^{2} \theta \cdot \sin ^{3} \theta \cdot \cos ^{2} \theta+\sin ^{2} \theta \cdot \cos ^{3} \theta \cdot \cos ^{2} \theta} \end{array}\)

\(=\frac{\sin^{2} \theta \cdot \cos ^{2} \theta(\sin\theta+\cos\theta)}{\sin ^{2} \theta \cdot \cos ^{2} \theta\left(\sin ^{3} \theta+\cos ^{3} \theta\right)}\ =\frac{\sin\theta+\cos\theta}{\sin ^{3} \theta+\cos ^{3} \theta}\ =\rm \frac{t_1}{t_3}\)

Hence, option (1) is correct.

Concept:

\(\rm2\sin \theta cos \theta= sin 2\theta \)

 

Calculation:

Here, t_n = sinⁿ θ + cosⁿ θ

 t1² - t2 =

 \(\rm (\sin \theta +cos \theta)^2 -(sin^2\theta +cos^2\theta )\ =sin^2\theta +cos^2\theta+2\sin \theta cos \theta-sin^2\theta -cos^2\theta\ =2\sin \theta cos \theta\= sin 2\theta \)

Hence, option (2) is correct.

Concept:

sin 45°  = cos 45°  = 1/√2

 

Calculation: 

\(\begin{aligned} \rm t_{10} &=\sin ^{10} \theta+\cos ^{10} \theta \ &=\left(\frac{1}{\sqrt{2}}\right)^{10}+\left(\frac{1}{\sqrt{2}}\right)^{10} \ &=2\left(\frac{1}{\sqrt{2}}\right)^{10} \ &=2\left(\frac{1}{2}\right)^{5} \ &=\frac{1}{2^{4}} \ &=\frac{1}{16} \end{aligned}\)

Hence, option (3) is correct.

Concept:

sin A + sin B = \(\rm 2sin (\frac{A+B}{2})cos(\frac{A-B}{2})\)

cos(90 - A) = sin A

 

Calculation:

sin α + cos β  = sin 15 + cos 15

= sin 15 + cos (90 - 75)

= sin 15 + sin 75

 \(=\rm 2sin (\frac{75+15}{2})cos(\frac{75-15}{2})\ =2sin (45)cos30 \=2\times\frac{1}{\sqrt2}\times \frac{\sqrt 3}{2}\=\frac{\sqrt 3}{\sqrt2}\)

 

Hence, option (4) is correct.

Concept:

sin A + sin B = \(\rm 2sin(\frac{A+B}{2})cos (\frac{A-B}{2})\)

Sin (A + B) = Sin A Cos B + Cos A Sin B

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

Calculation:

sin 7α - cos 7β = sin 105 - cos 105

sin (90 + 15) - cos (90 + 15)

= sin 90 cos15 + cos 90 sin 15 - (cos 90 cos 15 - sin 90 sin 15)

= cos 15 + sin 15

= sin 15 + cos 15 

= sin 15 + sin 75

\(=\rm 2sin (\frac{75+15}{2})cos(\frac{75-15}{2})\ =2sin (45)cos30 \=2\times\frac{1}{\sqrt2}\times \frac{\sqrt 3}{2}\=\frac{\sqrt 3}{\sqrt2}\)

Hence, option (4) is correct.

Concept:

Sin (A + B) = Sin A Cos B + Cos A Sin B

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

 

Calculation:

 

sin (α + 1°) + cos (β + 1°) = sin (15 + 1) + cos (15 + 1)

= sin 15 cos 1 + cos 15 sin 1 + cos 15 cos 1 - sin 15 sin 1

= cos 1 (sin 15 + cos 15) + sin 1 (cos 15 - sin 15 )  ....(1)

 

sin 15 + cos 15

= sin 15 + sin 75

\(=\rm 2sin (\frac{75+15}{2})cos(\frac{75-15}{2})\ =2sin (45)cos30 \=2\times\frac{1}{√2}\times \frac{√ 3}{2}\=\frac{√ 3}{√2}\)

And 

cos 15 - sin 15

= cos (90 - 75) - sin 15

= sin 75 - sin 15

\(=\rm 2cos (\frac{75+15}{2})sin(\frac{75-15}{2})\ =2cos (45)sin30 \=2\times\frac{1}{√2}\times \frac{1}{2}\=\frac{1}{√2}\)

So, (1) will become

cos 1 (√3/√2) + sin 1 (1/√2) 

= 1 / √2 (√3 cos 1° + sin 1°)

Hence,  option (4) is correct.

Concept:

\(\begin{array}{l} \rm \sin x+\sin y=2 \sin \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right) \ \rm \cos x-\cos y=-2 \sin \left(\frac{x+y}{2}\right) \cdot \sin \left(\frac{x-y}{2}\right) \end{array}\)

 

Calculation:

Here, sin x + sin y = cos y - cos x

\(\begin{array}{l} \rm 2 \sin \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right) =\rm-2 \sin \left(\frac{x+y}{2}\right) \cdot \sin \left(\frac{y-x}{2}\right) \end{array}\ \rm=2 \sin \left(\frac{x+y}{2}\right) \cdot \sin \left(\frac{x-y}{2}\right)\ \Rightarrow \sin \left(\frac{x+y}{2}\right) \cdot \cos \left(\frac{x-y}{2}\right)=\sin \left(\frac{x+y}{2}\right) \cdot \sin \left(\frac{x-y}{2}\right)\ \text{divide by }\cos (\frac{x-y}{2})\ \Rightarrow \sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x+y}{2}\right)\tan(\frac{x-y}{2})\ \Rightarrow \tan(\frac{x-y}{2})=1\)

Hence, option (3) is correct.

Concept:

For multiplication of two matrices: 

• The number of columns of the 1^st matrix must equal the number of rows of the 2^nd matrix.
• The result will have the same number of rows as the 1^st matrix, and the same number of columns as the 2^nd matrix.
• Example: To multiply an "m × n" matrix by " l × p" matrix, the n must be the same as I, and the result will be "m × p" matrix.

 

Calculation:

Given: A is a matrix of order 3 × 5 and B is a matrix of order 5 × 3

Number of rows in A = 3

Number of column in A = 5

Number of rows in B = 5

Number of column in B = 3

 

The order of AB = number of row is A × number of columns in B

= 3 × 3

And, 

The order of BA = number of row is B × number of columns in A

= 5 × 5

Hence, option (3) is correct.

Concept:

If common ratio = r  = \(\rm \frac{a_2}{a_1}=\frac{a_3}{a_2}=...=\frac{a_n}{a_{n-1}}\), then sequence a₁, a₂, a₃ …. a_n is an G.P

If common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1 , then sequence a₁, a₂, a₃ …. a_n is an A.P

 

Calculation:

For statement 1:

p², q² and r² (where p, q, r > 0) are in GP

 \(⇒\rm\frac{q^2}{p^2}=\frac{r^2}{q^2}\ ⇒ \frac{q}{p}=\frac{r}{q}\)

So, p, q, and r are in also in GP

For statement 2:

For  ln p, ln q, and ln r to be in AP

ln q - ln p = ln r - ln q

⇒2ln q = ln p + ln r

⇒ln q² = ln p + ln r

⇒ q2 = p r

⇒ (q²)² = p² r²

So, p2, q2 and r2 (where p, q, r > 0) are in GP

So, statements 1 and 2 both are correct.

Hence, option (3) is correct.

Concept:

\(\rm cot(α +β )=\frac{\cotα \cot β-1}{cotα +\cot β}\)

For quadratic equation \(\rm ax^2+bx+c=0\), roots are given by \(\rm x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

Calculation:

cot α and cot β are the roots of the equation x2 - 3x + 2 = 0

Roots 

 \(\rm = {-(-3) \pm \sqrt{(-3)^2-4(2)} \over 2}\ = {(3) \pm \sqrt{9-8} \over 2}\ =\frac{3\pm1}{2}\ =\frac{3+1}{2} \ OR \ \frac{3-1}{2}\ =2 \ OR \ 1 \)

Let, cot α = 2, cot β = 1

\(\rm cot(α +β )=\frac{\cotα \cot β-1}{cotα +\cot β}\ =\frac{2\times 1-1}{2+1}\ =\frac{1}{3}\)

Hence, option (2) is correct.

Concept:

Consider a quadratic equation: ax² + bx + c = 0.

Let, α and β are the roots.

Quadratic equation in the form of roots: x² – (α+β)x + (αβ) = 0

 

Calculation:

Given, αβ = α²β²

⇒ αβ  = 1

Here, α + β = α² + β²   

⇒ α + β = (α² + β²) + 2α β  - 2α β 

⇒ α + β = (α + β)² - 2α β

⇒ α + β = (α + β)² - 2(1)      ....(∵ αβ  = 1)

⇒ (α + β)² - (α + β) - 2 = 0

The above equation is a quadratic equation of variable (α + β)

⇒ α + β will have two values 

So, the number of such quadratic equations will be two 

Hence, option (2) is correct.

Concept:

arg (z) = arg (x + iy) = \(\rm tan^{-1}(\frac{y}{x})\)

arg (z₁ / z₂) = arg z₁  - arg z₂ 

Calculation:

Here, complex number: \(\rm \frac {1 - i√ 3}{1 + i √ 3} \)

⇒ arg (\(\rm \frac {1 - i√ 3}{1 + i √ 3}\)) = arg(1 - i√3) - arg(1 + i√3) 

⇒ \(\rm tan^{-1}(\frac{-\sqrt3}{1})-\rm tan^{-1}(\frac{\sqrt3}{1})\ \)

⇒ (- 60°) - 60°

⇒ -120°

It is measured in a clockwise direction.

Now, the argument of the complex number \(\rm \frac {1 - i\sqrt 3}{1 + i \sqrt 3}\)= (360° - 120°) = 240° 

Hence, option (1) is correct.