Quantitative Aptitude PYQs

Formula used:

cos x + cos y = \(\rm 2 cos \frac{x + y}{2} cos \frac{x - y}{2}\)

1 + cos 2x = 2 cos²x

ABC is a trianlge, ∠A + ∠B + ∠C = 180° 

cos(A + B) + cos(A - B) = 2 cos A. cos B

Calculation:

cos 2A + cos 2B + cos 2C = -1

⇒ cos 2A + cos 2B + (1 + cos 2C) = 0

⇒ 2 cos(A + B) cos(A - B) + 2 cos² C = 0     ----(i)

From formula used ∠A + ∠B = 180° - ∠C  

cos (A + B) = cos (180° - C) = - cos C     ----(ii)

From equation (i) and (ii), we get 

⇒ 2(-cos C) cos(A - B)  + 2 cos² C = 0

⇒ -2 cos C [cos(A - B) -  cos C] = 0     ----(iii)

From (ii) and (iii), we get

⇒ -2 cos C [cos(A - B) + cos(A + B)] = 0

⇒ -2 cos C [cos(A + B) + cos(A - B)] = 0

⇒ -2 cos C  (2 cos A. cos B) = 0

∴ cos A cos B cos C = 0.

Formula used:

\(\rm A = \begin{vmatrix} a_{1} & a_{2} & a_{3}\ b_{1} & b_{2} & b_{3}\ c_{1} & c_{2} & c_{3}\ \end{vmatrix}\)

det(A) = \(\rm a_{1}\begin{vmatrix} b_{2} & b_{3}\ c_{2} & c_{3} \end{vmatrix} - a_{2} \begin{vmatrix} b_{1} & b_{3}\ c_{1} & c_{3} \end{vmatrix} + a_{3} \begin{vmatrix} b_{1} & b_{2}\ c_{1} & c_{2} \end{vmatrix}\)

= a₁(b₂ c₃ - b₃ c₂) - a₂(b₁c₃ - b₃c₁) + a₃ (b₁c₂ - b2c1)

Calculation:

\(\begin{vmatrix} cos C & tan A & 0\ sin B & 0 & -tanA\ 0 & sinB & cos C \end{vmatrix}\)

= cos C (0 + tan A sin B) - tan A (sin B cos C) 

= tan A sin B cos C - tan A sin B cos C

= 0

Concept:

The union of two or more sets is the set that contains all the elements

of each of the sets; an element is in the union if it belongs to at least

one of the sets.

n(A ∪ B) = n(A) + n(B) - n(A ⋂ B)

Calculation:

A = {first 250 natural number and multiple of 3}

⇒ A = {3, 6, 9, 12,.......750}

⇒ n(A) = 250

B = {first 200 even natural number}

B = {2, 4, 6, 8, 10.......400}

⇒ n(B) = 200

Therefore,

A ⋂ B = {6, 12,18 ....396)                   

⇒ n(A ⋂ B) = 396/6 = 66

⇒ n(A ⋂ B) =  66

We know that,

n(A ∪ B) = n(A) + n(B) - n(A ⋂ B)

⇒ n(A ∪ B) = 250 + 200 - 66

∴  n(A ∪ B) = 384

Formula used:

The sum of n terms of the AP is Sn = \(\rm \frac{n}{2}[2a + (n - 1)d]\)

Calculation:

S_k = \(\rm \frac{k}{2}(2a + (k-1)d)\)

⇒ S₃₀ = 15(2a + 29d)

⇒ S₂₀ = 10(2a + 19d)

⇒ S₁₀ = 5(2a + 9d)

We have to find the value of \(\rm \frac{S_{30}}{S_{20}-S_{10}}\)

⇒ \(\rm \frac{15(2a + 29d)}{10(2a + 19d)- 5(2a + 9d)}\)

⇒ \(\rm \frac{15(2a + 29d)}{10a + 145d}\)

⇒ \(\rm \frac{15(2a + 29d)}{5(2a + 29d)}\)

⇒ 3

∴ The value of \(\frac{S_{30}}{S_{20}-S_{10}}\) is 3.

Formula used:

Quadratic equation: ax² + bx + c = 0

let the two roots be r¹ and r²

Sum of the roots: r¹ + r² = \(\frac{-b}{a}\)

Calculation:

let the roots of the equation 4x² - (5k + 1)x + 5k = 0 be α and α + 1

α + α + 1 = \(\rm \frac{5k + 1}{4}\)

⇒ α = \(\rm \frac{5k - 3}{8}\)

⇒ α (α + 1) = \(\rm \frac{5k}{4}\)

⇒ \(\rm \frac{(5k - 3)(5k - 3 + 8)}{8\times 8} = \frac{5k}{4}\)

⇒ (5k − 3)(5k + 5) = 80k

⇒ (5k − 3)(k + 1) = 16k

⇒ 5k2 − 14k − 3 = 0

⇒ (k − 3)(5k + 1) = 0

⇒ k = 3, \(\rm \frac{-1}{5}\)

∴ One of the possible value is -1/5.

Formula used:

n! = n × (n - 1) × (n - 2) × ..... × 3 × 2 × 1

5! = 5 × 4 × 3 × 2 × 1 = 120

2! = 2 × 1 = 2

Calculation:

Four cases can possible: (3, 5, 7, 9, 3), (3, 5, 7, 9, 5), (3, 5, 7, 9, 7), (3, 5, 7, 9, 9)

For (3, 5, 7, 9, 3)

Number of arrangements = \(\rm \frac{5!}{2!}\)

⇒ \(\rm \frac{120}{2}\)

⇒ 60

Similarly for all 4 cases 60 ways 

⇒ 60 × 4

⇒ 240

Explanation:

Here we have to fill 4 entries it means we can form 2 × 2 matrices, 4 × 1 and 1 × 4 matrices

In 2 × 2 matrices, there are four places 

And each place has two ways to fill (1, 2) 

So, The total number of ways to fill the entries =  2 × 2 × 2 × 2 = 16.

In 4 × 1 matrices, there are four places

And each place has two ways to fill (1, 2) 

So, The total number of ways to fill the entries =  2 × 2 × 2 × 2 = 16.

Similarly in case of 1 × 4 matrices total number of ways = 16.

∴ Total number of matrices will be 48.

Concept:

The value i or the concept of i is used in explaining and expressing complex numbers. Complex numbers are numbers with a real and imaginary part. The imaginary part is defined with the help of i. Basically, “i” is the imaginary part which is also called iota.

Value of i is √-1 A negative value inside a square root signifies an imaginary value.

All the basic arithmetic operators are applicable to imaginary numbers. On squaring an imaginary number, we obtain a negative value.

i² = -1, i³ = -i, i⁴ = 1

Calculation:

Given

i = √-1

⇒ i² = -1

⇒ i-2n 

⇒ (i²)^-n

⇒ (-1)^-n

Since, n = 0, 1, 2, 3, ...

When n = even then i^-2n = 1

When n = odd then i-2n = -1

∴ Total possible values of the i-2n for n ϵ Z is two which are - 1 and 1.

Formula used:

\(\rm A = \begin{vmatrix} a_{1} & a_{2} & a_{1}\ b_{1} & b_{2} & b_{3}\ c_{1} & c_{2} & c_{3}\ \end{vmatrix}\)

det(A) = \(\rm a_{1}\begin{vmatrix} b_{2} & b_{3}\ c_{2} & c_{3} \end{vmatrix} - a_{2} \begin{vmatrix} b_{1} & b_{3}\ c_{1} & c_{3} \end{vmatrix} + a_{3} \begin{vmatrix} b_{1} & b_{2}\ c_{1} & c_{2} \end{vmatrix}\)

= a1(b2 c3 - b3 c2) - a2(b1c3 - b3c1) + a3 (b1c2 - b2c1)

Calculation:

\(\begin{vmatrix} 1 & -x & x\ 1 & 1 & -y\ 1 & z & 1 \end{vmatrix}\)

⇒  \(\begin{vmatrix} 1 & -\frac{a}{b - c} & \frac{a}{b - c}\ 1 & 1 & -\frac{b}{c - a}\ 1 & \frac{c}{a -b} & 1 \end{vmatrix}\) = Δ 

⇒ Δ = 1(1 + \(\rm \frac{bc}{(a - b)(c - a)}\)) + \(\rm \frac{a}{b - c}\)(\(\rm 1 + \frac{b}{c - a} \)) + \(\rm \frac{a}{b - c}\) \(\rm (\frac{c}{a - b} - 1)\)

⇒ Δ = 1 + \(\rm \frac{bc}{(a - b)(c - a)} + \frac{ab}{(b - c)(c - a)} + \frac{ac}{(b - c)(a - b)}\)

⇒ Δ = 1 + \( \frac{a}{(b - c)} \times \frac{(b - c)(a - b - c)}{(a - b)(c - a)} + \frac{bc}{(a - b)(c - a)}\)

⇒ Δ = \( \frac{ac - a^2 - bc + ab+ a^2 - ab - ac + bc}{(a - b)(c - a)}\)

∴ Δ =  0

Shortcut TrickPut x = 0, b = 1 and c = 2

⇒ \(x = \frac{a}{b-c}\) = 0, \(y = \frac{b}{c - a}\) = 1/2, \(z = \frac{c}{a - b}\) = -2

Let, \(\begin{vmatrix} 1 & -x & x\ 1 & 1 & -y\ 1 & z & 1 \end{vmatrix}\) = Δ 

⇒  Δ = \(\begin{vmatrix} 1 & 0 & 0\ 1 & 1 & -1/2\ 1 & -2 & 1 \end{vmatrix}\)

⇒  Δ = 1[1 - (-1/2)(-2)] 

∴   Δ = 0

Concept:

Matrix Multiplication:

• Two matrices can only be multiplied when the column number of the first matrix is equal to the row number of the second matrix
• If A is not an invertible matrix then |A| = 0
• If A is a singular matrix then |A| = 0

 

Calculation:

Given that,

\(A = \begin{bmatrix} 1 & 1 & 1\ 1 & 1 & 1\ 1 & 1 & 1 \end{bmatrix}\)

⇒ |A| = 1(1 - 1) - 1(1 - 1) + 1 (1 - 1) = 0

Hence, Inverse of A does not exist.

⇒ A² = \(\rm \begin{bmatrix} 1 & 1 & 1\ 1 & 1 & 1\ 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1\ 1 & 1 & 1\ 1 & 1 & 1 \end{bmatrix}\)

⇒ A² = \(\rm \begin{bmatrix} 3 & 3 & 3\ 3 & 3 & 3\ 3 & 3 & 3 \end{bmatrix}\)

⇒ A³ = A² . A

⇒ A³ = \(\begin{bmatrix} 3 & 3 & 3\ 3 & 3 & 3\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1\ 1 & 1 & 1\ 1 & 1 & 1 \end{bmatrix}\)

⇒ A³ = \(\rm \begin{bmatrix} 9 & 9 & 9\ 9 & 9 & 9\ 9 & 9 & 9 \end{bmatrix}\)

⇒ A³ = 9\(\rm \begin{bmatrix} 1 & 1 & 1\ 1 & 1 & 1\ 1 & 1 & 1 \end{bmatrix}\) ≠ A

Hence, statement (2) is incorrect.

Now, the value of 3A is

= 3 \(\rm \begin{bmatrix} 1 & 1 & 1\ 1 & 1 & 1\ 1 & 1 & 1 \end{bmatrix}\)

⇒ 3A = \(\rm \begin{bmatrix} 3 & 3 & 3\ 3 & 3 & 3\ 3 & 3 & 3 \end{bmatrix}\)

∴  3A = A²

Hence, statement (3) is correct.

Concept:

The distance of the center from any point present on the circumference of the circle is always same. (which is also called the radius of the circle)

The distance between two points (a, b) and (c, d) is defined as

√[(a - c)² + (b - d)²]

Calculation:

Let O(x, y) be the center of the circle.

Accordin to the concept used

OA = OB = OC 

⇒ OA2 = OB2  = OC2 

So, By taking OA² = OB² , we get 

⇒ (x − 5)² + (y + 8)² =  (x + 2)² + (y - 9)²

⇒ x² + 25 - 10x + y² + 64 + 16y = x² + 4 + 4x + y² + 81 - 18y

⇒ 89 - 10x + 16y = 85 + 4x - 18y

⇒ 89 - 85 - 10x + 16y - 4x + 18y = 0

⇒ 4 - 14x + 34y = 0 

⇒ 14x - 34y = 4      ----(1)        

And By taking OA2 = OC2 , we get 

⇒ (x − 5)2 + (y + 8)2 = (x - 2)² + (y - 1)²

⇒ x² + 25 - 10x + y² + 64 + 16y = x² + 4 - 4x + y² + 1 - 2y

⇒ 89 - 10x + 16y = 5 - 4x - 2y

⇒ 89 - 5 - 10x + 4x + 16y + 2y = 0

⇒ 84 - 6x + 18y = 0 

⇒ 6x - 18y = 84      ----(2)    

On solving (1) and (2), we get

y = -24 and x = -58

∴ The coordinates of the center of the circle is (-58, - 24).

Concept:

The general equation of the circle is x² + y² + 2gx + 2fy + c = 0, 

Whose, Radius = \(\rm \sqrt{g^{2} + f^{2} - c}\)

Calculation:

Let the required equation of circle be x2 + y2 + 2gx + 2fy + c = 0

Since, The given circle passes through (5, -8), (-2, 9) and (2, 1), we have

5² + (-8)² + 2g(5) + 2f(-8) + c = 0

⇒ 10g - 16f + c = -89      ----(1)

(-2)² + (9)² + 2g(-2) + 2f(9) + c = 0

⇒ - 4g + 18f + c = -85      ----(2)

2² + 1² + 2g(2) + 2f(1) + c = 0

⇒ 4g + 2f + c = -5       ----(3)

On applying (1) - (2), gives

14g - 34f = -4 

⇒ 7g - 17f = -2        ----(4)

On applying  (2) - (3), we get

• 8g + 16f = -80

⇒ g - 2f = 10     ----(5)

Solving equation (4) and (5), we get 

⇒ f = 24

From equation (5), we get 

⇒ g = 58

putting the value of f and g in (3), we get

⇒ 4(58) + 2(24) + c = -5

⇒ c = -285

Now the radius of circle = \(\rm \sqrt{g^{2} + f^{2} - c}\)

⇒ Radius = \(\rm \sqrt{3364 + 576 + 285}\)

⇒ Radius = \(\rm \sqrt{4225}\) = 65

∴ r > 60 is the correct relation.

Concept:

Equilateral triangle: A Triangle with all three sides equal.

Area = \(\rm \frac{√{3}}{4}a^{2}\)

Distance formula: Distance b/w two points A(x₁, y₁) and B(x₂, y₂) is,

\(d=\rm \sqrt{[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]}\)

Calculation:

Let ABC be the equilateral triangle such that,

A = (0, 0), B = (2, 2) and C = (x, y)

By distance formula we get,

AB = \( \sqrt{[(2 - 0)^{2} + (2 - 0)^2]}\) = \(\rm 2√{2}\)

We know that, Δ ABC is an equilateral triangle, then

⇒ AB = BC = AC = \(\rm 2√{2}\) units

Now, AC = BC

⇒ \(\rm \sqrt{[x^2+ y^{2}]}\) = \(\rm \sqrt{[(x - 2)^2+ (y - 2)^{2}]}\)

⇒  x2 + y2 = x2 + 4 - 4x + y2 + 4 - 4y     [∵ (a - b)² = a² - 2ab + b²]

⇒  8 - 4x - 4y = 0

⇒  y = 2 - x       -----(1)

AC = \(\rm 2√{2}\) units

\(\rm \sqrt{[x^2+ y^{2}]} = 2√{2}\) 

Squring both side

⇒ x2 + y2 = 8

⇒ x2 + (2 - x)2 = 8

⇒ x2 + 4 + x2 - 4x = 8

⇒  2x2 - 4x - 4 = 0

⇒ x2 - 2x - 2 = 0

⇒ x = 1 + \(\rm 2√{2}\)

Hence, from equation (1),

y = 2 - (1 + \(\rm 2√{2}\))

⇒ y = 1 - \(\rm 2√{2}\)

⇒ C(x, y) = (1 + \(\rm 2√{2}\), 1 - \(\rm 2√{2}\))

Which contain irrational coordinates. So statement 1 is correct.

Now, the area of an equilateral triangle,

A = \(\rm \frac{√{3}}{4}a^{2}\)

⇒ A = \(\rm \frac{√{3}}{4}(2√ 2)^{2}\) = 2√3 square units.

The area of the equilateral triangle contains root 3 which is always

irrational.

Hence, statement 2 is correct.

∴ Both 1 and 2 are correct.

Concept:

Equilateral triangle: A Triangle with all three sides equal.

Area = \(\rm \frac{√{3}}{4}a^{2}\)

Distance formula: Distance b/w two points A(x1, y1) and B(x2, y2) is,

\(d=\rm √{[(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}]}\)

Calculation:

Let ABC be the equilateral triangle such that,

A = (0, 0), B = (2, 2) and C = (x, y)

By distance formula we get,

AB = \( √{[(2 - 0)^{2} + (2 - 0)^2]}\) = \(\rm 2√{2}\)

We know that, Δ ABC is an equilateral triangle, then

⇒ AB = BC = AC = \(\rm 2√{2}\) units

Now, AC = BC

⇒ \(\rm √{[x^2+ y^{2}]}\) = \(\rm √{[(x - 2)^2+ (y - 2)^{2}]}\)

⇒  x2 + y2 = x2 + 4 - 4x + y2 + 4 - 4y     [∵ (a - b)2 = a2 - 2ab + b2]

⇒  8 - 4x - 4y = 0

⇒  y = 2 - x       -----(1)

AC = \(\rm 2√{2}\) units

\(\rm √{[x^2+ y^{2}]} = 2√{2}\) 

Squring both side

⇒ x2 + y2 = 8

⇒ x2 + (2 - x)2 = 8

⇒ x2 + 4 + x2 - 4x = 8

⇒  2x2 - 4x - 4 = 0

⇒ x2 - 2x - 2 = 0

⇒ x = 1 + √3

Hence, from equation (1),

y = 2 - (1 + √3)

⇒ y = 1 - √3

⇒ C(x, y) = (1 + √3, 1 - √3)

Hence, the required difference is

d = 1 + √3 - (1 - √3)

⇒ d = 1 + √3 - 1 + √3

⇒ d = 2√3

∴ The difference of coordinates of the third vertex is 2√3.

Calculation:

Mid point of BD is same as mid point of AC

Mid point of AC = \(\rm (\frac{1 + 3}{2} , \frac{3 + 5}{2})\) = (2, 4)

Equation of BD 

y - 2 = \(\rm \frac{2 - 4}{- 1 - 2}\) (x + 1)

⇒ y - 2 = \(\rm \frac{2}{3}\)(x + 1)

⇒ 3y - 6 = 2x + 2

⇒ 2x - 3y + 2 + 6 = 0

⇒ 2x - 3y + 8 = 0

∴ The equation of the diagonal BD is 2x - 3y + 8 = 0

Concept:

The diagonal of a parallelogram divides it into two parts of equal areas.

Area of triangle having three coordinates (x₁, y₁), (x₂, y₂), (x₃, y₃)

Area = (1/2) | [x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3(y1 – y2)] |

Calculation:

Here in the figure diagonal AC divides ABCD into two equal parts

Area(ABCD) = 2 × Area (ΔABC)    ----(i)

Area (ΔABC) = (1/2) | [1 (2 - 5) + (-1) (5 - 3) + 3 (3 - 2)] |

⇒ Area (ΔABC) = (1/2) | [ -3 - 2 + 3 ] |

⇒ Area (ΔABC) = (1/2) | - 2  | = 1

Now, From (i), we get 

Area(ABCD) = 2 × 1 = 2

∴ The area of parallelogram is 2 square unit.

Concept:

The equation of line passes through (x₁, y₁) and slope m is (y - y₁) = m(x - x₁)

If m1 and m₂ are the solpes of two perpendicular lines then m₁ × m₂ = -1

Slope of line ax + by + c = 0 is -a/b

Calculation:

Let the slope of Altitude BD is m

From figure, we can say that altitude BD ⊥ AC

⇒ Slope of AC × Slope of BD = -1

⇒ (-1/2) × m = -1

⇒ m = 2     ----(i)

Now, We have to find the coordinate of B 

The point of intersection of AB and BC will give B.

x - 2 = 0 

⇒ x = 2

and y + 1 = 0 

⇒ y = -1

So, The coordinate of B is (2, -1)

Now, The equation of altitude BD is 

(y + 1) = 2(x - 2)

⇒ y + 1 = 2x - 4

⇒ 2x - y - 4 - 1 = 0

⇒ 2x - y - 5 = 0

∴ The equation of the altitude through B on AC is 2x - y - 5 = 0

Concept:

Circumcentre: The point of intersection of the perpendicular bisector of sides of a triangle is called the circumcentre of a triangle.

Mid Point Formula: The mid point of (a, b) and (c, d) is [(a + c)/2, (b + d)/2]

The equation of line passes through (x1, y1) and slope m is (y - y1) = m(x - x1)

If m1 and m2 are the solpes of two perpendicular lines then m1 × m2 = -1

Slope of line ax + by + c = 0 is -a/b

Explanation:

Here in the figure,

MO and NO are the perpendicular bisectors of the given triangle 

So, The point of intersection of MO & NO gives the circumcentre

Now, Point of intersection of AB and BC gives B

⇒ B = (2, -1)

Point of intersection of AC and BC gives C 

⇒ C = (6, -1)

Point of intersection of AB and AC gives A

⇒ A = (2, 1)

Now, Point M = Midpoint of BC 

⇒ Point M = [(2 + 6)/2, (-1 - 1)/2] = (4, -1)

Now, Point N = Midpoint of AC 

⇒ Point N = [(2 + 6)/2, (-1 + 1)/2] = (4, 0)

Now, The slope of line BC = 0

⇒ Slope of line MO = ∞ 

⇒ Equation of line MO is (y + 1) = ∞ (x - 4)

⇒ x - 4 = 0     ----(i)

Now, The slope of line AC = -1/2

⇒ Slope of line NO = 2

⇒ Equation of line NO is (y - 0) = 2 (x - 4)

⇒ y = 2x - 8    ----(ii)

On solving the equations (i) and (ii), we get 

x = 4 and y = 0

∴ The coordinate of circumcente is (4, 0).

Explanation:

We have two fixed endpoints of the latus rectum.

As we know that the focus lies on the midpoint of the segment joining the endpoints of the latus rectum.

So, the focus of the parabola is fixed.

So, the distance between the focus and the directrix is fixed.

As we know the directrix is parallel to the latus rectum.

So, the slope of the directrix is fixed.

So, we can have two lines that are parallel to the latus rectum and are at a fixed distance from the focus.

So, we have one focus and two possible directrices.

∴ A maximum of two parabolas can be drawn.

Concept:

If the equation of parabola is in the form of (y – k)² = 4p(x – h) then 

Endpoints of lactus rectum = (h + p, k ± 2p), 

Focus of parabola = (h + p, k)

Calculation:

Here, Given end point of latus rectum are (-2, 4) and (-2, -4)

So, h + p = -2       ----(1)

and k + 2p = 4       ----(2)

and k - 2p = -4       ----(3)

On adding (2) and (3), we get

2k = 0

⇒ k = 0

⇒ p = 2

Now From (i),

⇒ h = -4

Now the focus is (h + p, k)

⇒ Focus is (-4 + 2, 0)

⇒ Focus is (-2, 0)(h+p,k

∴ Statement 2 is correct

Now the equation of parabole is (y – 0)2 = 4 × 2 (x + 4)

⇒ y2 = 8x + 32

Which do not passes through the origin (0, 0) on satisfying.

∴ Statement 1 is incorrect.

∴ Only statement 2 is correct.