Quantitative Aptitude PYQs

Concept:

The interior angle of a regular polygon is given by:  

\(\rm (n - 2) \times \frac{180^{\circ}}{n}\)

Where, n = number of sides

Interior Angle = 180°- Exterior Angle

Calculation:

Regular Octagon has 8 sides

so, n = 8

Interior Angle

= \(\rm (n - 2) \times \frac{180^{\circ}}{n}\)

= \(\rm (8 - 2) \times \frac{180^{\circ}}{8}\)

= \(\rm 6 \times \frac{180^{\circ}}{8}\)

= \(\dfrac{3\pi}{4}\)

Concept:

\(\rm \sin \theta = \frac{Perpendicular}{Hypotenuse}\)

\(\rm \cos \theta = \frac{Base}{Hypotenuse} \)

sin2 θ + cos2 θ = 1 

Calculation:

7 sinθ + 24 cosθ = 25

Dividing by 25 on both the sides, we get

\(\rm \frac{7}{25}\)sinθ + \(\rm \frac{24}{25}\)cosθ = 1      ....(i)

We know that,

sin² θ + cos² θ = 1  

sin θ.sin θ + cos θ.cos θ = 1      ....(ii)

On comparing equ (i) and (ii)

sin θ = \(\rm \frac{7}{25}\) 

cos θ = \(\rm \frac{24}{25}\)

Now, (sinθ + cosθ) 

= \(\rm \frac{7}{25}\)+ \(\rm \frac{24}{25}\)

= \(\rm \frac{31}{25}\)

Formula Used:

For a right-angle triangle:

Sin θ = P/H

Cos θ = B/H

Tan θ = P/B

Here, P = perpendicular, B = base & H = hypotenuses

And, H² = B²+ P²

2. tan(A + B) = (tanA + tanB)/(1 - tanA.tanB)

 

Calculation:

Length of the ladder CD = 5 mFrom given data, the triangle diagram can be drawn as

DB = Height of flagstaff

AC = Ladder = 6 m = DA = 6 m.

∠DCB = 75°

∠ABC = 90°

Now, ∠BDC = 180° - (90° + 75°) = 15° .

Then, ∠ADC = ∠ACD { Since, ΔADC is isosceles triangle}

⇒ ∠ACD = 15°

Hence,

⇒ ∠ACB = ∠DCB - ∠ACD = 75° - 15° = 60°

Now Δ ABC can be separately drawn as,

Since, ∆ABC is right angle tringle

⇒ sin 60° = AB/AC

⇒ (√3/2) = AB/6

⇒ AB = 3√3 m .

Hence, Height of flagstaff = DB = BA + AD = 3√3 + 6 m

Calculation:

Let AB be the tower and let AC and AD be its shadows when the angles of elevation of the sum are 60^o and 45^o, respectively. Then, ∠ACB = 60^o, ∠ADB = 45^o, CD  = x m

Let AB = 5(3 + √3) m and AC = y meter

In \(\rm \Delta \) CAB, We have,

\(\rm tan 60^{\circ} = \frac{AB}{AC}\)

√3 = \(\rm \frac{5(3 + √3 )}{AC}\)

AC = \(\rm \frac{5(3 + √3 )}{√3 }\)

On rationalizing, we get

AC = \(\rm \frac{5(3 + √3 )}{√3 }\)(\times\)(\rm \frac{√3 }{√3 }\)

= \(\rm \frac{5(3 √3 + 3)}{3 }\)

In \(\rm \Delta \) DAB, We have, 

\(\rm tan45^{\circ} = \frac{AB}{AD}\)

1 = \(\rm \frac{AB}{AD}\)

AB = AD

AD = 5(3 + √3) 

x + y = 5(3 + √3) 

x + \(\rm \frac{5(3 √3 + 3)}{3 }\) =  5(3 + √3) 

x =  5(3 + √3) - \(\rm \frac{5(3 √3 + 3)}{3 }\)

x = 10 m

Concept:

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

\(\rm \frac{sin \theta }{cos \theta } = tan \theta \)

Calculation:

3 cos θ = 4 sin θ

\(\rm \frac{3}{4} = \frac{sin \theta }{cos \theta } \)

\(\rm \frac{3}{4} = tan \theta \)

\(\rm tan (45^{\circ} + \theta) = \frac{tan 45^{\circ} + tan \theta }{1 - tan 45^{\circ}\times tan \theta }\)

= \(\rm \frac{1 + \frac{3}{4} }{1 - 1 \times \frac{3}{4}} \)

= \(\rm \frac{1 + \frac{3}{4} }{1 - \frac{3}{4}}\)

= \(\rm \frac{\frac{7}{4}}{\frac{1}{4}}\)

= 7

Concept:

(i)² = 1

(-i)⁴ = 1

Calculation:

\(\rm \left(\frac{1-i}{1+i}\right)^{n^2}=1\)

On rationalizing,

\(\rm \left (\frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} \right )^{n^{2}}\)= 1

\(\rm \left (\frac{(1 - i)^{2}}{1 - i^{2}} \right )^{n^{2}}\) = 1

\(\rm \left (\frac{1 + i^{2} - 2i}{1 - (-1)} \right )^{n^{2}}\) = 1

\(\rm \left (\frac{1 - 1 - 2i}{1 + 1} \right )^{n^{2}}\) = 1

\(\rm \left (\frac{- 2i}{2} \right )^{n^{2}}\)= 1

\(\rm \left (- i \right )^{n^{2}} \) = 1

if we put n = 2 then, 

\(\rm \left (- i \right )^{n^{2}} \) = 1

Satisfy the equation.

Hence, Option 1 is correct.

Concept:

Change of base rule:

log_b (M) = \(\rm \frac{log_{a}(M)}{log_{a}(b)}\)

Calculation:

 logcos x sin x = 1

\(\rm \frac{log sin x}{log cos x}\) = 1

log sin x = log cos x

sin x = cos x 

The smallest positive value of x for which sin x = cos x is x = \(\dfrac{\pi}{4}\)

Concept:

Using Scalar Multiple Property:

If all the elements of a row (or column) of a determinant are multiplied by a non-zero constant, then the determinant gets multiplied by the same constant.

Calculation:

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

Column 1 of all elements is multiplied by p then given matrices is also multiplied by p

pΔ = \(​​\)(\left| {\begin{array}{*{20}{c}} {{pa_1}}&{{b_1}}&{{c_1}}\ {{pa_2}}&{{b_2}}&{{c_2}}\ {{pa_3}}&{{b_3}}&{{c_3}} \end{array}} \right|\)

Again, column 3 of all elements is multiplied by q then given matrices is also multiplied by q

pqΔ = \(\left| {\begin{array}{*{20}{c}} {{pa_1}}&{{b_1}}&{{qc_1}}\ {{pa_2}}&{{b_2}}&{{qc_2}}\ {{pa_3}}&{{b_3}}&{{qc_3}} \end{array}} \right|\)

Concept:

(1 + x)ⁿ = C₀ + x. C₁ + x². C₂ + ...... + xⁿC_n

Calculation:

Let S = C1 + C2 + C3 + _ _ _ _ _ + Cn

⇒ S = ⁿC1 + ⁿC2 + ⁿC3 + _ _ _ _ _ + ⁿCn

⇒ S = 1 + nC1 + nC2 + nC3 + _ _ _ _ _ + nCn - 1

⇒ S = ⁿC₀ + nC1 + nC2 + nC3 + _ _ _ _ _ + nCn - 1          ----(1)

As we know, (1 + x)n = C0 + x. C1 + x2. C2 + ...... + x nC_n

Put x = 1, we get

⇒ (1 + x)n =  nC0 + nC1 + nC2 + nC3 + .... + nC_n = 2ⁿ      -----(2)

From equation (1) and (2), we get

S = nC0 + nC1 + nC2 + nC3 + _ _ _ _ _ + nCn - 1

∴ The value of C1 + C2 + C3 + _ _ _ _ _ + Cn is 2ⁿ - 1.

Concept:

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) 

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

Calculation:

Let Δ  = \(\left| {\begin{array}{*{20}{c}} {{a}}&{{b}}&{{c}}\ {{b}}&{{c}}&{{a}}\ {{c}}&{{a}}&{{b}} \end{array}} \right|\) 

= a (cb - a²) - b (b² - ca) + c (ba - c²)

= abc - a³ - b³ + abc + abc - c³

= -[a³ + b³ + c³ - 3abc]

= -(a + b + c)(a² + b² + c² - ab - bc - ca)      ....(i)

As we know,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ 4² = a2 + b2 + c2 + 2 × 0 

⇒ 16 = a2 + b2 + c2

From equ (i)

Δ = -[4 × (16 - 0)]

= - 64

Concept:

The minimum and maximum value of a sin x + b cos x 

-\(\rm \sqrt{a^{2} + b^{2}}\) ≤ a sin x + b cos x ≤ \(\rm \sqrt{a^{2} + b^{2}}\)

Calculation:

As we know, -\(\rm \sqrt{a^{2} + b^{2}}\) ≤ a sin x + b cos x ≤ \(\rm \sqrt{a^{2} + b^{2}}\)

⇒ -\(\rm \sqrt{2^{2} + 0^{2}}\) ≤  2sin x + 0 cos x ≤ \(\rm \sqrt{2^{2} + 0^{2}}\)

⇒ -2 ≤ 2sin x ≤ 2

⇒ -2 ≤ 2k + 1 ≤ 2

⇒ -3 ≤ 2k ≤ 1

⇒ -1.5 ≤ k ≤ 0.5

k = 0, -1 

Hence, Option 3 is correct.

Concept:

Let us consider sequence a1, a2, a3 …. an is a G.P.

So, common ratio = \(\rm r = \frac{a_{p + 1}}{a_{p}} \), Where p ∈ Positive Integer

Logarithmic formula: log a - log b = \(\rm \frac{log a}{log b}\)

Properties of Determinants:

• Determinant evaluated across any row or column is same.
• If all the elements of a row or column are zeroes, then the value of the determinant is zero.
• The determinant of an Identity matrix is 1.
• If rows and columns are interchanged then the value of the determinant remains the same (value does not change).
• If any two-row or two-column of a determinant are interchanged the value of the determinant is multiplied by -1.
• If two rows or two columns of a determinant are identical the value of the determinant is zero.

 

Calculation:

Let r be the common ratio of the given GP

Now,

\(\left| {\begin{array}{*{20}{c}} {{ln:a_1}}&{{ln:a_2}}&{{ln:a_3}}\ {{ln:a_4}}&{{ln:a_5}}&{{ln:a_6}}\ {{ln:a_7}}&{{ln:a_8}}&{{ln:a_9}} \end{array}} \right|\)

Apply, C₂ →  C₂ - C₁ and C₃→ C₃ - C₂

= \(\left| {\begin{array}{*{20}{c}} {{\ln:a_1}}&{{\ln:a_2 - \ln a_1}}&{{\ln:a_3 - \ln a _2}}\ {{\ln:a_4}}&{{\ln:a_5 - \ln a_4}}&{{\rm ln:a_6 - \ln a_5}}\ {{\rm ln:a_7}}&{\rm{ln:a_8 - \ln a_7}}&{\rm{ln:a_9 -\ln a_8}} \end{array}} \right|\)

= \(\left| {\begin{array}{*{20}{c}} {{\ln:a_1}}&{{\ln r}}&{{\ln r }}\ {{\ln:a_4}}&{{\ln r }}&{{\ln r}}\ {{\ln:a_7}}&{{\ln r }}&{{\ln r}} \end{array}} \right|\)      [Since \(\rm \frac{a_{p + 1}}{a_{p}} = r\) and log a_{p + 1} - log a_p = \(\rm log \frac{a_{p + 1}}{a_{p}} = log r\)]

= 0                                  [Since C₃ = C₂]

Concept:

The general form of quadratic equation: ax² + bx + c = 0

The discriminant of the equation is D = b2 - 4ac 

• D > 0, roots are unequal and real
• D < 0, roots are imaginary
• D = 0, roots are real and equal
• D ≥ 0, roots are real

Calculation:

Given : x2 + 2x + k = 0

On comparing with general form

a = 1, b = 2, c = k

According to the question roots are real

D ≥  0

⇒ b² - 4ac ≥ 0

⇒ 2² - 4 × 1 × k ≥ 0

⇒ 4 - 4k ≥ 0

⇒ 4 ≥ 4k

⇒ 1 ≥ k

∴ k ≤  1

Concept:

Product rule: The log of a product equals the sum of two logs.

\({\log _a}\left( {mn} \right) = ;{\log _a}m + ;{\log _a}n\)

Change of base rule: \({\log _m}n = ;\frac{1}{{{{\log }_n}m}}\)

Calculation:

\(\rm \dfrac{1}{log_2n}+\dfrac{1}{log_3n}+\dfrac{1}{log_4n}+{.....}+\dfrac{1}{log_{100}n}\)

= log_n2 + log_n3 + log_n4 + ..... + log_n100

= logn [2 × 3 × 4 × .... × 100]

= log_n 100!

= log_100! 100! 

= 1

Concept:

i² = -1 

Calculation:

Given z = 1 + i

We have to find the modulus of z + \(\rm \frac{2}{Z}\)

⇒ (1 + i) + \(\rm \frac{2}{1 + i}\)

On rationalizing the second term, we get

⇒ (1 + i) + \(\rm \frac{2}{1 + i}\)(\times\)(\rm \frac{1 - i}{1 - i}\)

⇒ (1 + i) + \(\rm \frac{2\times (1 - i)}{1 - i^{2}}\)

⇒ (1 + i) + \(\rm \frac{2\times (1 - i)}{1 - (-1)}\)

⇒ (1 + i) + \(\rm \frac{2\times (1 - i)}{2}\)

⇒ 1 + i + 1 - i

⇒ 2

∴ The modulus of \(\rm z+\frac{2}{z} = 2\).

Dimension Property of Matrix:

One property that is unique to matrices is the dimension property.

This property has two parts:

Part 1: The product of two matrices will be defined if the number of columns in the first matrix is equal to the number of rows in the second matrix.

Part 2: If the product is defined, the resulting matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix.

Explanation:

Let A be a matrix of order [a × b]    ----(i)

And B be a matrix of order [b × d]   ----(ii)

Since we have to find the multiplication of A and B 

So we let the column of A = Row of B  [which is b] 

From the properties of multiplication 

The order of AB will be [a × d] 

But, According to the question, the order of AB is [n × n]

i.e The value of a and d should be n

i.e a = d = n     ----(iii)

Now, From (i), (ii), and (iii), we get

The order of A = [n × b]

The order of B = [b × n]

Here the order of A and B are not the same 

But it can be the same when b = n

Here, We can say that to satisfy the above question the order of two matrices need not be the same.

Concept:

Prime Numbers = A number that is divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11)

The order of the matrix is defined by m × n

For an matrix m × n we have mn elements

Calculation:

All prime numbers less than 30 = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

Total number of  elements = 10

Since matrices of different orders are possible with elements comprising all prime numbers less than 30.

Since we have 10 elements the order of matrix can be as follows:

Total 4 matrices will be formed which will,

A = [5 × 2]

B = [2 × 5]

C = [10 × 1]

D = [1 × 10]

Hence Option 1 is correct.

Concept:

Suppose A is a square matrix of 2 rows and 2 columns

\(\rm A = \begin{bmatrix} a & b\ c & d \end{bmatrix}\)

The determinant is; |A| = [ad - bc]

Calculation:

\(A = \left| {\begin{array}{*{20}{c}} p&q\ r&s \end{array}} \right|\)

p, q, r and s are any four different prime numbers less than 20.

Here we have to find the maximum value of the determinant so we have to take maximum value for p and s and minimum value for q and r

All prime numbers less than 20 = 2, 3, 5, 7, 11, 13, 17, 19

So, p = 17, q = 2, r = 3 and s = 19

\(\rm A = \left| {\begin{array}{*{20}{c}} 17&2\3&19 \end{array}} \right|\)

|A| = [17 \(\times\) 19 - 2 \(\times\) 3]

= 323 - 6

= 317

Explanation:

If A and B are square matrices of order 2 such that det(AB) = det(BA)

This property is true regardless of whether A and B are unit matrices or not. Therefore, the condition det(AB) = det(BA) does not imply that either A or B must be a unit matrix.

∴ Option 4 is correct 

Additional Information Properties of determinants:

• The determinant of a diagonal matrix is the product of the diagonal entries.
• If A and B are both n×n matrices, then det(AB) = det(A) det(B).
• For a n×n matrix A, det(kA) = kn det(A).
• The determinant of a square matrix is the same as the determinant of its transpose.

Concept:

\(\rm cot (A + B) = \dfrac{cot A cot B - 1}{cot A + cot B}\)

Calculation:

Given,

(cot 2x ∙ cot 4x) - (cot 4x ∙ cot 6x) - (cot 6x ∙ cot 2x)

⇒ (cot 2x ∙ cot 4x) - cot 6x [cot 4x + cot 2x]

⇒ (cot 2x ∙ cot 4x) - cot (2x + 4x) [cot 4x + cot 2x]

⇒ (cot 2x ∙ cot 4x) -  \((\rm \dfrac{cot 2x .cot 4x - 1}{cot 2x + cot 4x})\)× [cot 4x + cot 2x]

⇒ (cot 2x ∙ cot 4x) - (cot 2x ∙ cot 4 x - 1)

⇒ (cot 2x ∙ cot 4x) - (cot 2x ∙ cot 4x) + 1

⇒ 1