Quantitative Aptitude PYQs

Shortcut TrickSince We have to take any point on the given ellipse 

Take y = 0 

Form the equation of ellipse 25x2 + 16y2 = 400

25x2 = 400

x2 = 16

We get x = ±4

So we take P as (4, 0)

Now, Applying the distance formula we get 

PQ + PR = \(\sqrt{[(4 - 0)^2 + (0 - 3)^2] }+ \sqrt{[(4 - 0)^2 + (0 + 3)^2] }\)

⇒ PQ + PR = 5 + 5 = 10.

 

Alternate Method Concept:

Ellipse: The locus of a point that moves such that its distance from a fixed point and a fixed straight line is always less than one,

(Eccentricity = e < 1) ⇔ \(\frac{{{\rm{;}}{{\bf{x}}^2}}}{{{{\bf{a}}^2}}} + \frac{{{{\bf{y}}^2}}}{{{{\bf{b}}^2}}} = 1\) (b > a)

Sum of the focal distance: The sum of the focal distance of any point on an ellipse is constant and equal to the length of the major axis of the ellipse. 

 

Sum of the focal distance = PS + PS’ = 2a for a > b

Sum of the focal distance = PS + PS’ = 2b for b > a

Focus: (0, ±be) for b > a

Eccentricity: \({\rm{e}} = {\rm{;}}\sqrt{ {1 - {\rm{;}}\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}} }\) for b > a

Calculation:

Given: 25x2 + 16y2 = 400

 \(\frac{{{{\rm{x}}^2}}}{{16}} + \frac{{{{\rm{y}}^2}}}{25} = 1\)

Here,

a2 = 16,

b2 = 25

⇒ a = 4 and b = 5 (a < b)

Eccentricity:

\({\rm{e}} = {\rm{;}}\sqrt {{1 - {\rm{;}}\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}}} = {\rm{;}\sqrt{ {1 - {\rm{;}}\frac{16}{{25}}}}} = {\rm{;}}\sqrt{\frac{{9}}{{25}}} = {\rm{;}}\frac{3}{5}\) 

Focus = (0, ± be) = (0, ± 3)

So, focus are (0, 3) and (0, -3)

Hence given point Q(0, 3) and R(0, -3) are the focus of the ellipse

PQ + PR = Sum of the focal distance = 2b = 2 × 5 = 10

We have

Circumcentre of Δ is (-1, -1)

Consider,

Equation (1) ⇒ x + 2 = 0

Equation (2) ⇒ y + 2 = 0

Equation (3) ⇒ kx + y + 2 = 0

From equation (1) & (2),

(x, y) = (-2, -2) = C1

From equation (2) & (3),

y = - 2

And, kx + (-2) + 2 = 0

x = 0

(x, y) = (0, 2) = C2

From equation (1) & (3),

x + 2 = 0 ⇒ x = -2

kx + y + 2 = 0

or -2k + y + 2 = 0

or y = (2k - 2)

∴ (x, y) = (-2, 2k - 2) = C3

From the given coordinate, the graph can be drawn as,

Since the given diagram shows that the Δ acts as a right-angle triangle, and for a right-angle triangle, the circumcentre lies at the midpoint of the hypotenuse.

Hence,

(-1, -1) = \((\frac{0-2}{2},\ \frac{-2-2K+2}{2})\)

⇒ -1 = - K

Hence, K = 1

Alternate Method

it is clear that the Δ acts as a right-angle triangle and in this Δ the circumcentre lies at the midpoint of the hypotenuse.

We have

Kx + y + 2 = 0

at x = y = -1

⇒ K(-1) + (-1) + 2 = 0

⇒ K = 1

Concept:

The coordinates of the point where the chord cut the parabola satisfIes  the equation of a parabola.

Calculation:

 

 

Given:

The equation of a parabola is y² = x.

The angle made by Chord OA with x-axis is θ

Let the length of the chord OA of the parabola is L 

So, Length of AM = L sinθ 

and Length of OM = L cosθ 

So, The coordinate of A = (L cos θ, L sin θ)

And this point will satisfy the equation of parabola y2 = x.

⇒ (Lsin θ)² = L cos θ

⇒L² sin² θ = L cos θ

⇒ L = cos θ. cosec2 θ

∴ The required length of chord is cos θ. cosec2 θ.

Concept:

Let A = (x1 ,y1) , B = (x2, y2), C = (x3 ,y3) be the points of triangle.

Area of \(\rm\triangle ABC\) = \(\rm\dfrac{1}{2}\begin{vmatrix} x_1 &y_1 &1 \ x_2 &y_2 &1 \ x_3&y_3 &1 \end{vmatrix}\)

If A, B, C be the points of a triangle and colinear then Area of \(\rm\triangle ABC\) = 0

Calculations:

Let A = (x1 ,y1) , B = (x2, y2), C = (x3 ,y3) be the points of triangle.

Area of \(\rm\triangle ABC\) = \(\rm\dfrac{1}{2}\begin{vmatrix} x_1 &y_1 &1 \ x_2 &y_2 &1 \ x_3&y_3 &1 \end{vmatrix}\)

Let A = (a, b), B = (c, d) and C = (a - c, b - d) be the points of triangle. They are collinear.

Area of \(\rm\triangle ABC\) = 0 

= \(\rm\dfrac{1}{2}\begin{vmatrix} x_1 &y_1 &1 \ x_2 &y_2 &1 \ x_3&y_3 &1 \end{vmatrix}\) = 0

⇒\(\rm\begin{vmatrix} x_1 &y_1 &1 \ x_2 &y_2 &1 \ x_3&y_3 &1 \end{vmatrix}\) = 0

⇒\(\rm\begin{vmatrix} a &b &1 \ c &d &1 \ a - c& b - d &1 \end{vmatrix}\) = 0

⇒ a(d - b + d) - b(bc - dc - ad + cd) + 1(bc - dc - ad + cd) = 0

⇒ ad - bc = 0

⇒ ad = bc

Concept:

The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

Calculations:

Given, ABC be a triangle. If D(2, 5) and E(5, 9) are the mid-points of the sides AB and AC respectively.

Hence, DE = \(\rm\sqrt{(2-5)^2+(5 -9)^2}\)

⇒ DE = \(\rm\sqrt{9+16}\)

⇒ DE = 5

The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side.”

Since, D(2, 5) and E(5, 9) are the mid-points of the sides AB and AC then their length is half of the third side.

⇒ BC = 2(DE)

⇒ BC = 2(5)

⇒ BC = 10

Concept:

If two nonvertical lines are perpendicular, then the product of their slopes is −1.

The slope of a line passing through the distinct points (x₁, y₁) and (x₂, y₂) is \(\rm \frac{y_2-y_1}{x_2-x_1}\)

 

Calculation:

 

 

Slope of line passing through points (0, k) and (3, 1)

 \(=\rm \frac{1-k}{3-0} \=\frac{1-k}{3}\)

3x - 4y - 5 = 0 

⇒4y = 3x - 5

⇒ y = \(\frac{3}{4}\rm x-\frac{5}{4}\) 

So, the slope of line 3x - 4y - 5 = 0 is 3/4

Now since line OP and 3x - 4y - 5 = 0 are perepndicular 

\(\rm \frac{1-k}{3}\times \frac{3}{4}=-1.....(\text {Product of slopes of perpendicular lines} )\ \Rightarrow 1-k=-4\ \Rightarrow k =5 \)

Hence, option (3) is correct.

Concept:

The angle between the lines whose slopes are m1 and m2 be given by,

\(\rm\tan;\theta = |\dfrac{m_1-m_2}{1+m_1m_2}|\)

 

Calculations:

The angle between the lines whose slopes are m1 and m2 be given by,

\(\rm\tan;\theta = |\dfrac{m_1-m_2}{1+m_1m_2}|\)

Given, m1 = 2 - √3 and m2 = 2 + √3

\(\rm\tan;\theta = |\dfrac{(2-\sqrt3)-(2+\sqrt3)}{1+(2-\sqrt3)(2+\sqrt3)}|\)

⇒ \(\rm\tan;\theta = |\dfrac{-2\sqrt3}{2}|\)

⇒ \(\rm\tan;\theta = |-\sqrt3|\)

⇒ \(\rm\tan;\theta =\pm \sqrt3\)

⇒ \(\rm\theta = \dfrac{\pi}{3}, \dfrac {2\pi} {3}\)

Concept:

The distance between the lines ax + by + c₁ = 0 and ax + by + c₂ = 0  is \(\rm \frac{|c_1-c_2|}{\sqrt{a^2+b^2}}\)

Area of square = (side)² 

 

Calculation:

Here, 3x - 4y - 5 = 0 and 3x - 4y + 15 = 0 are the equations of a pair of opposite sides of a square.

Distance between them =

 \(=\frac{|-5-15|}{\sqrt {3^2+4^2}}\= \frac{20}{\sqrt 25}\ =\frac{20}{5} \=4\)

Now area of square = 4² sq. units 

= 16 sq. units 

Hence, option (3) is correct.

Concept:

A hermitian matrix is a square matrix, which is equal to its conjugate transpose matrix.

The non-diagonal elements of a hermitian matrix are all complex numbers.

The complex numbers in a hermitian matrix are such that the element of the i^th row and j^th column is the complex conjugate of the element of the j^th row and i^th column.

Example:

\(\left[ {\begin{array}{*{20}{c}} {3}&{\rm{3+2i}}\ { {\rm{3-2i}}}&{2} \end{array}} \right]\)

Calculations:

Given, matrix \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {1 - {\rm{i}}}&{\rm{i}}\ { - {\rm{i}}}&{1 - {\rm{i}}} \end{array}} \right]\) where \(\rm i = \sqrt {-1},\)

⇒      \({\rm{\bar A}} = \left[ {\begin{array}{*{20}{c}} {1 + {\rm{i}}}&{\rm{-i}}\ { {\rm{i}}}&{1 + {\rm{i}}} \end{array}} \right]\)

⇒  \({\rm({\bar A})^T} = \left[ {\begin{array}{*{20}{c}} {1 + {\rm{i}}}&{\rm{i}}\ { {\rm{-i}}}&{1 + {\rm{i}}} \end{array}} \right]\)

 \({\rm({\bar A})^T}+ A = \left[ {\begin{array}{*{20}{c}} {1 + {\rm{i}}}&{\rm{i}}\ { {\rm{-i}}}&{1 + {\rm{i}}} \end{array}} \right]+\left[ {\begin{array}{*{20}{c}} {1 - {\rm{i}}}&{\rm{i}}\ { {\rm{-i}}}&{1 - {\rm{i}}} \end{array}} \right]\)

⇒ \({\rm({\bar A})^T}+ A = \left[ {\begin{array}{*{20}{c}} {2}&{\rm{2i}}\ { {\rm{-2i}}}&{2} \end{array}} \right]\)

Since, the diagonal elements of (A̅)T + A is real.

(A̅)^T + A is hermitian is correct statement.

Concept:

(x + y)ⁿ = ⁿΣ_r=0 nC_r x^{n – r} · y^r 

In independent term: Power of variables will be zero.

 

Calculation:

 \(\rm \left( \frac {2}{x^2} - \sqrt x \right)^{10}=\sum 10C_r(\frac{2}{x^2})^{10-r}(-x^{1/2})^r\ ⇒\sum 10C_r({2}{x^{-2}})^{10-r}(-x^{1/2})^r\ ⇒\sum 10C_r({2}^{10-r}{x^{-20+2r}})(-x^{\frac{r}{2}})\ ⇒\sum 10C_r({2}^{10-r}{x^{-20+2r+\frac{r}{2}}})(-1^{\frac{r}{2}})\ \)

Now, for independent term, -20 + 2r + r/2 = 0

⇒(4r + r)/2 = 20

⇒ 5r = 40

⇒ r = 8

So, the independent term=

 \(\rm 10 C_r\times2^{10-r}\times (-1)^{r/2}\ ⇒ 10 C_8\times2^{10-8}\times (-1)^{8/2}\ ⇒ \frac{10!}{8!2!}\times2^{2}\times 1\ ⇒\frac{10\times 9}{2}\times 4\ ⇒180\)

Hence, option (1) is correct.

Calculation:

(1 + 2x - x²)⁶ = a₀ + a₁x + a₂x² + ... + a₁₂x¹²

On putting x = -1 in the above expression, we get

(1 - 2 - 1)⁶ = \(\rm a_0-a_1+a_2-a_3...+a_{12}\)

⇒  \(\rm a_0+a_1+....+a_{12}= 64\) 

Hence, option (2) is correct.

Concept:

C(n, r) = \(\rm\dfrac{n!}{r!(n - r)!}\)

n! = n. (n - 1)!

Calculations:

Given, C(20, n + 2) = C(20, n - 2)

\(\rm\frac{20!}{(n+2)!(18-n)!} = \rm\frac{20!}{(n-2)!(22-n)!}\)

\(\rm\frac{1}{(n+2)!(18-n)!} = \rm\frac{1}{(n-2)!(22-n)!}\)

we know that

(n + 2)! =(n - 2)! (n - 1)n(n + 1)(n+2)

(22 - n)! = (18 - n)! (19 - n)(20 - n)(21 - n)(22 - n)

\(\rm\frac{1}{((n - 2)! (n - 1)n(n + 1)(n+2)(18-n)!} = \rm\frac{1}{(n-2)!(18 - n)! (19 - n)(20 - n)(21 - n)(22 - n))}\)

\(\rm\frac{1}{(n - 1)n(n + 1)(n+2)} = \rm\frac{1}{ (19 - n)(20 - n)(21 - n)(22 - n))}\)

\(\rm {(n - 1)n(n + 1)(n+2)} = \rm (19 - n)(20 - n)(21 - n)(22 - n)\)

Now , put n = 10 

L.H.S = (n - 1)n(n + 1)(n+2) = (9)(10)(11)(12) 

R.H.S = (19 - n)(20 - n)(21 - n)(22 - n) = (9)(10)(11)(12) 

L.H.S. = R.H.S.

Hence, if C(20, n + 2) = C(20, n - 2), then n = 10

Alternate MethodWe have,

C(20, n + 2) = C(20, n - 2)

⇒ \(^{20}C_{(n+2)}=^{20}C_{(n−2)}\)

Consider,

n = 20

x = (n + 2)

y = (n - 2)

x ≠ y,

∵ n = 20 and, n ≠ 0

x + y = n (Possible)

In that case,

(n + 2) + (n - 2) = 20

or, 2n = 20

Hence, n = 10

Calculation:

1. If in a triangle ABC, A = 2B and b = c, then it must be an obtuse-angled triangle. 

We have, in ΔABC 

A = 2B and b = c

⇒ Angles opposite to equal sides are equal

∴ ∠C = ∠B

Also, A = 2B = 2C

In ΔABC, ∠A + ∠B + ∠C = 180° 

⇒ 2C + C + C = 180°

\(⇒ C=\frac{180°}{4}=45°\)

⇒ B = 45° and A = 90° 

Thus, it shows that ΔABC is not an obtuse angle traingle.                      

Hence, statement 1 is incorrect.

2. There exists no triangle ABC with A = 40°, B = 65° and a/c = sin 40° cosec 15° 

We have A = 40°, B = 65° 

In ΔABC,

∠A + ∠B + ∠C = 180° 

⇒ 40° + 65° + ∠C = 180°

⇒ ∠C = 75°

Use sine rule in ΔABC,

\(\frac{a}{sin40°}=\frac{c}{sin75°}\)

\(\frac{a}{c}=\) sin 40° cosec 75° 

Hence, Statement 2 is incorrect.

Concept:

The matrix is singular if |A| = 0

Calculations:

Given the matrix \(\left[ {\begin{array}{*{20}{c}} 0&{\rm{k}}&4\ { - {\rm{k}}}&0&{ - 5}\ { - {\rm{k}}}&{\rm{k}}&{ - 1} \end{array}} \right]\) singular

We know that the matrix is singular iff |A| = 0

⇒\(\rm\begin{vmatrix} 0 &k &4 \ -k &0 &-5 \ -k &k &-1 \end{vmatrix} = 0\)

⇒ 0 - k (k - 5k) + 4 (- k²) = 0

⇒ 4k² - 4k² = 0

Therefore, we can take any value of k it makes the matrix singular.

Hence the total value of k is infinite.

Concept:

Covert Binary to Decimal:

For binary number with n digits;

D_n-1 …. D₃D₂D₁D₀

The decimal number is equal to the sum of binary digits (D_n) times their power of 2 (2ⁿ)

Decimal = D₀ × 2⁰ + D₁ × 2¹ + D₂ × 2² + ….

 

Calculation:

1101101 =

 \(1\times 2^0+0\times 2^1+1\times2^2+1\times2^3+0\times2^4+1\times2^5+1\times2^6\ =1+4+8+32+64\ =109\) 

1011011 = 

\(1\times 2^0+1\times 2^1+0\times2^2+1\times2^3+1\times2^4+0\times2^5+1\times2^6\ =1+2+8+16+64\ =91\)

(1101101 + 1011011)₂ = (109 + 91 )₁₀ = (200)₁₀

Hence, option (3) is correct.

Concept:

The log properties:

\(\rm \log;m^n = n;\log m\) and \(\rm \log ;(mn) = log;m + log;n\)

Calculations:

Given,  \(\rm \frac 1 {10} \log_5 1024 - \log_5 10 + \frac 1 5 \log_5 3125\) 

=  \(\rm \frac 1 {10} \log_5 2^{10} - \log_5 (2\times5) + \frac 1 5 \log_5 5^5\)

Using the log properties, we have

\(\rm \log;m^n = n;\log m\) and \(\rm \log ;(mn) = log;m + log;n\)

= \(\rm \frac 1 {10}\times 10; \log_5 2 - \log_5 (2)-log_5 5+ \frac 1 5\times 5; \log_5 5\)

= \(\rm \log_5 2 - \log_5 2-log_5 5+ \log_5 5\)

= 0

Concept:

If x = log_a b then x = \(\rm\dfrac{log;b}{log; a}\)

log (mn) = log m + log n

Calculations:

Given, x = logc (ab), y = loga (bc), z = logb (ca),

Consider,x = logc (ab) 

By logarithm property, we have

\(\rm x = \rm\dfrac {log;(ab)}{log;c}\), \(\rm y = \rm\dfrac {log;(bc)}{log;a}\)and \(\rm z = \rm\dfrac {log;(ca)}{log;b}\)

⇒ from this \(\rm xyz\neq1\) and \(\rm x+y+z \neq 1\)

Consider,

⇒1 + x = 1 + \(\rm\dfrac {log;(ab)}{log;c}\)

By usiing logarithm property  log (mn) = log m + log n, we have

⇒1 + x =  \(\rm\dfrac {\log ;c+log;(ab)}{log;c}\)

⇒ 1+ x = \(\rm\dfrac {\log ;abc}{log;c}\)

Similarly, 1 + y = \(\rm\dfrac {\log ;abc}{log;a}\) and 1 + z = \(\rm\dfrac {\log ;abc}{log;b}\)

Consider, \(\rm \dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z} = \rm\dfrac{log;c}{log;abc}+\dfrac{log;b}{log;abc} +\dfrac{log;a}{log;abc}\)                                                     

= \(\rm\dfrac{log;c+log;a+log;b}{log;abc}\)

= \(\rm\dfrac{log;abc}{log;abc}\)

= 1

Concept:

The two matrices are equal. Hence, their corresponding elements are also equal.

Calculations:

Given, \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\rm x + y}& \rm y\ {\rm 2x}&{\rm x - y} \end{array}} \right],;\rm B = \left[ {\begin{array}{*{20}{c}} 2\ { - 1} \end{array}} \right]\) and \(\rm C = \left[ {\begin{array}{*{20}{c}} 3\ 2 \end{array}} \right]\)

To find the value of the determinant of matrix A, we need to find x and y first.

Given, AB = C

\(\rm \left[ {\begin{array}{*{20}{c}} {\rm x + y}& \rm y\ {\rm 2x}&{\rm x - y} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2\ { - 1} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 3\ { 2} \end{array}} \right] \)

⇒\(\rm \left[ {\begin{array}{*{20}{c}} {\rm 2(x + y)-y}\ {\rm 4x-x+y} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 3\ { 2} \end{array}} \right] \)

⇒\(\rm \left[ {\begin{array}{*{20}{c}} {\rm 2x + y}\ {\rm 3x+y} \end{array}} \right]=\left[ {\begin{array}{*{20}{c}} 3\ { 2} \end{array}} \right] \)

The two matrices are equal. Hence, their corresponding elements are also equal.

2x + y = 3....(1)

3x + y = 2....(2)

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

⇒ x = - 1  and y = 5

Matrix A becomes,\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\rm x + y}& \rm y\ {\rm 2x}&{\rm x - y} \end{array}} \right]\)= \(\left[ {\begin{array}{*{20}{c}} {\rm 4}& \rm 5\ {\rm -2}&{\rm -6} \end{array}} \right]\)

Determinant of A becomes, |A|  = \(\begin{vmatrix} 4 &5 \ - 2&-6 \end{vmatrix}\) = - 14

Calculations:

Given,1.5 ≤ x ≤ 4.5

So critical points are x = 1.5 = \(\frac 3 2\) or x = 4.5 = \(\frac 9 2\)

⇒ 2x - 3 = 0 or 2x - 9 = 0

Changing inequality into equality form

∴ (2x - 3)(2x - 9) = 0

As we can see in the wavy curve value of (2x - 3)(2x - 9) is negative.

(2x - 3)(2x - 9) ≤ 0

Hence option 4 is the correct answer.

Concept:

Relation Based on Logarithmic Comparison:

• A relation R is defined on the set S × S by xRy if log_ax > log_ay.
• The base of the logarithm is a = 1/2, which is a value between 0 and 1.
• For 0 < a < 1, the logarithmic function log_ax is a strictly decreasing function.
• Important property: If log_ax > log_ay, then it implies x < y when 0 < a < 1.
• Hence, the given relation reduces to xRy ⇔ x < y.
• Relation properties:
• Reflexive: A relation is reflexive if xRx for all x in S. But x < x is never true ⇒ Not reflexive.
• Symmetric: If x < y, then y < x is false ⇒ Not symmetric.
• Transitive: If x < y and y < z ⇒ x < z ⇒ Transitive.

 

Calculation:

Given,

Relation: xRy ⇔ log_ax > log_ay

Base of logarithm: a = 1/2

⇒ Since 0 < a < 1, log_ax is a decreasing function

⇒ log_ax > log_ay ⇔ x < y

⇒ So, xRy ⇔ x < y

⇒ Check reflexivity: x < x ⇒ false ⇒ Not reflexive

⇒ Check symmetry: If x < y, then y < x ⇒ false ⇒ Not symmetric

⇒ Check transitivity: If x < y and y < z ⇒ x < z ⇒ Transitive

∴ The relation is transitive only.