Quantitative Aptitude PYQs

Concept:

Modulus function:

It is a function that gives the absolute value of a variable (x). It is defined as such that |x| = x, if x ≥  0 else |x| = - x

Calculation:

f(x) = |x - 2| + |3 - x| + |4 - x|, where x ∈ R.

For x < 2, f(x) = 2 - x + 3 - x + 4 - x = 9 - 3x

For 2 ≤ x < 3, f(x) = x - 2 + 3 - x + 4 - x = 5 - x

For 3 ≤ x < 4, f(x) = x - 2 + x - 3 + 4 - x = x - 1

For x ≥ 4, f(x) = x - 2 + 3 - x + x - 4 = 3x - 9

∴ Minimum value of f(x) is given by:

fmin =  \(\left{\begin{matrix} 3, x < 2 \3, 2 \leq x< 3 \2, 3 \leq x < 4 \3, 4 \leq x \end{matrix}\right.\)

∴ Minimum value of f(x) is at x = 3.

Concept:

Modulus function:

It is a function that gives the absolute value of a variable (x). It is defined as such that |x| = x, if x ≥  0 else |x| = - x

Calculation:

f(x) = |x - 2| + |3 - x| + |4 - x|, where x ∈ R.

For x < 2, f(x) = 2 - x + 3 - x + 4 - x = 9 - 3x

For 2 ≤ x < 3, f(x) = x - 2 + 3 - x + 4 - x = 5 - x

For 3 ≤ x < 4, f(x) = x - 2 + x - 3 + 4 - x = x - 1

For x ≥ 4, f(x) = x - 2 + 3 - x + x - 4 = 3x - 9

∴ Minimum value of f(x) is given by:

f_min =  \(\left{\begin{matrix} 3, x < 2 \3, 2 \leq x< 3 \2, 3 \leq x < 4 \3, 4 \leq x \end{matrix}\right.\)

∴ Minimum value of f(x) is 2.

Concept:

To find the remainder when the sum S is divided by 8, we need to calculate the value of S and then take the modulus of S with 8.

First, let's calculate the value of S:

S = 0! + 1! + 2! + 3! + 4! + ... + 100!

We can see that the factorial of any number greater than or equal to 5 will have at least one factor of 5, and therefore, it will have at least one factor of 2. This means that the factorial of any number greater than or equal to 5 will be divisible by 8.

4! is also divisible by 8

So, when we add up the factorials from 4! to 100!, they will all contribute a multiple of 8 to the sum. The remaining part of the sum, from 0! to 3!, can be calculated as follows:

0! + 1! + 2! + 3! = 1 + 1 + 2 + 6 = 10

Now, let's take the modulus of S with 8:

S = 10 (mod 8)

To find the remainder, we can divide 10 by 8:

10 = (1 × 8) + 2

So, the remainder when S is divided by 8 is 2.

∴ The correct answer is Option 03.

Calculation:

To find the remainder when S is divided by 60, we need to calculate the value of S modulo 60.

First, let's calculate the factorials from 0! to 9! to understand the pattern:

0! = 1

1! = 1

2! = 2

3! = 6

4! = 24

5! = 120

6! = 720

7! = 5040

8! = 40320

9! = 362880

Notice that for n ≥ 5, n! is divisible by 60 because it contains at least two factors of 2 and one factor of 5 (and 60 = 2 × 2 × 3 × 5). Therefore, for n ≥ 5, n! ≡ 0 (mod 60).

Now, let's calculate the sum S by considering the factorials from 0! to 4!:

S = 0! + 1! + 2! + 3! + 4!

= 1 + 1 + 2 + 6 + 24

= 34

Since S = 34, the remainder when S is divided by 60 is also 34.

Therefore, the correct answer is option 4) 34.

Concept:

Cosine Rule:

If a, b, c are three sides of a triangle opposite to angles A, B, C then:

cos A = \(\frac{b^2+c^2-a^2}{2bc}\)

cos B = \(\frac{a^2+c^2-b^2}{2ac}\)

cos C = \(\frac{a^2+b^2-c^2}{2ab}\)

Calculation:

Given, cos P = 1/3

Using, cosine rule, cos P = \(\frac{PQ^2+PR^2-QR^2}{2\times PQ \times PR}\)

⇒ \(\frac{1}{3}\) = \(\frac{(2n+2)^2+(2n+4)^2-(2n+6)^2}{2\times (2n+2) \times (2n+4)}\)

⇒ \(\frac{1}{3}\) = \(\frac{4n^2-16}{2(4n^2+12n+8)}\)

⇒ 8n² + 24n + 16 =12n² - 48

⇒ 4n² - 24n - 64 = 0

⇒ n² - 6n - 16 = 0

⇒ n2 - 8n + 2n - 16 = 0

⇒ n(n - 8) + 2(n - 8) = 0

⇒ (n - 8)(n - 2) = 0

⇒ n = 8

∴ The value of n is 8.

Concept:

Cosine Rule:

If a, b, c are three sides of a triangle opposite to angles A, B, C then:

cos A = \(\frac{b^2+c^2-a^2}{2bc}\)

cos B = \(\frac{a^2+c^2-b^2}{2ac}\)

cos C = \(\frac{a^2+b^2-c^2}{2ab}\)

Calculation:

Given, cos P = 1/3

Using, cosine rule, cos P = \(\frac{PQ^2+PR^2-QR^2}{2\times PQ \times PR}\)

⇒ \(\frac{1}{3}\) = \(\frac{(2n+2)^2+(2n+4)^2-(2n+6)^2}{2\times (2n+2) \times (2n+4)}\)

⇒ \(\frac{1}{3}\) = \(\frac{4n^2-16}{2(4n^2+12n+8)}\)

⇒ 8n2 + 24n + 16 =12n2 - 48

⇒ 4n2 - 24n - 64 = 0

⇒ n2 - 6n - 16 = 0

⇒ (n - 8)(n - 2) = 0

⇒ n = 8

Now, sides are:

PQ = 2n + 2 = 2⋅8 + 2 = 18

PR = 2n + 4 = 2⋅8 + 4 = 20 

QR = 2n + 6 = 2⋅8 + 6 = 22

∴ Smallest side is 18.

Concept:

sin 2x = 2sin x cos x

Calculation:

Given, sinx + cosx + tanx + cotx + secx + cosecx = 7 

⇒ \(\sin ^2 x \cos x+\cos ^2 x \sin x+\sin ^2 x+\sin x+\cos x+\cos ^2 x=7 \sin x \cos x\)

⇒ \( \sin x \cos x(\sin x+\cos x)+\sin x+\cos x+1=7 \sin x \cos x\)

⇒ \(\sin x+\cos x=\frac{7 \sin x \cos x-1}{1+\sin x \cos x}\)

⇒ \(\sin x+\cos x=\frac{7 \sin 2 x-2}{2+\sin 2 x}\)

Squaring both sides,

⇒ \(\sin ^2 x+\cos ^2 x+\sin 2 x=\left(\frac{7 \sin 2 x-2}{2+\sin 2 x}\right)^2\)

⇒ \(\sin 2 x=\frac{\left(49 \sin ^2 2 x+4-28 \sin 2 x\right)}{\left(4+\sin ^2 2 x+4 \sin 2 x\right)}-1\)

⇒ \(\sin 2 x=\frac{48 \sin ^2 2 x-32 \sin 2 x}{4+\sin ^2 2 x+4 \sin 2 x}\)

⇒ \(\sin 2 x\left(\sin ^2 2 x+4 \sin 2 x+4-48 \sin 2 x+32\right)=0\)

⇒ \(\sin 2 x\left(\sin ^2 2 x-44 \sin 2 x+36\right)=0\)

Since, sin 2x ≠ 0

⇒ sin2 2x - 44 sin2x + 36 = 0

∴On simplification, sin2 2x - 44 sin2x + 36 = 0

Concept:

sin 2x = 2sin x cos x

Calculation:

Given, sinx + cosx + tanx + cotx + secx + cosecx = 7 

⇒ \(\sin ^2 x \cos x+\cos ^2 x \sin x+\sin ^2 x+\sin x+\cos x+\cos ^2 x=7 \sin x \cos x\)

⇒ \( \sin x \cos x(\sin x+\cos x)+\sin x+\cos x+1=7 \sin x \cos x\)

⇒ \(\sin x+\cos x=\frac{7 \sin x \cos x-1}{1+\sin x \cos x}\)

⇒ \(\sin x+\cos x=\frac{7 \sin 2 x-2}{2+\sin 2 x}\)

Squaring both sides,

⇒ \(\sin ^2 x+\cos ^2 x+\sin 2 x=\left(\frac{7 \sin 2 x-2}{2+\sin 2 x}\right)^2\)

⇒ \(\sin 2 x=\frac{\left(49 \sin ^2 2 x+4-28 \sin 2 x\right)}{\left(4+\sin ^2 2 x+4 \sin 2 x\right)}-1\)

⇒ \(\sin 2 x=\frac{48 \sin ^2 2 x-32 \sin 2 x}{4+\sin ^2 2 x+4 \sin 2 x}\)

⇒ \(\sin 2 x\left(\sin ^2 2 x+4 \sin 2 x+4-48 \sin 2 x+32\right)=0\)

⇒ \(\sin 2 x\left(\sin ^2 2 x-44 \sin 2 x+36\right)=0\)

Since, sin 2x ≠ 0

⇒ sin2 2x - 44 sin2x + 36 = 0

⇒ sin 2x = \({-(-44) ± √{(-44)^2-4\times 1\times 36} \over 2}\)

⇒ sin 2x = \({44 ± √{1936-144} \over 2}\)

⇒ sin 2x = \({44 ± 16√{7} \over 2}\)

⇒ sin 2x = 22 ± 8√7

Now, sin θ ∈ [-1, 1]

⇒ sin 2x = 22 - 8√7 = a - b√c

∴ a = 22, b = 8 and c = 7

⇒ a - b + 2c = 22 - 8 + 2(7) = 28

∴ The value of a - b + 2c is 28.

Given:

The coefficients of the quadratic equation are:

a = 3 + 2√2

b = -(4 + 2√3)

c = 8 + 4√3

Concept:

The sum of roots and the product of roots of a quadratic equation

ax² + bx + c = 0 can be found as follows:

The sum of roots, S = -b/a

Product of roots, P = c/a

And

If x₁ and x₂ are roots, then the harmonic mean (HM) of the roots can be found

as follows:

HM = 2/(1/x₁ + 1/x₂) = 2/(S/P) = 2P/S

Calculation:

First, calculate the sum (S) and product (P) of roots.

⇒ S = -b/a = -(-[4 + 2√3])/(3 + 2√2) = (4 + 2√3)/(3 + 2√2)

⇒ P = c/a = (8 + 4√3)/(3 + 2√2)

Harmonic mean (HM) = 2P/S 

 \(⇒ HM = \frac{2 × [(8 + 4√3)/(3 + 2√2)] }{ [(4 + 2√3)/(3 + 2√2)]}\)

\(⇒ HM = \frac{2 × [(8 + 4√3)\times (3 + 2√2)] }{ [(4 + 2√3)\times (3 + 2√2)]}\)

\(⇒ HM = \frac{2 ×2 [(4 + 2√3)\times (3 + 2√2)] }{ [(4 + 2√3)\times (3 + 2√2)]}\)

∴  HM = 4

Given:

The coefficients of the quadratic equation are:

a = 3 + 2√2

b = -(4 + 2√3)

c = 8 + 4√3

Concept:

The sum of roots and the product of roots of a quadratic equation

ax² + bx + c = 0 can be found as follows:

The sum of roots, S = -b/a

Product of roots, P = c/a

And

If x₁ and x₂ are roots, then the geometric mean (GM) of the roots can be found

as follows:

GM = √(x₁.x₂)

Calculation:

First, calculate the sum (S) and product (P) of roots.

⇒ P = x₁.x₂ = c/a = (8 + 4√3)/(3 + 2√2)

⇒ GM = √{[(24 - 16√2 + 12√3 - 8√6)] / [(3² - (2√2)²)]}

⇒ GM = √{[(24 - 16√2 + 12√3 - 8√6)] / [9 - 8]}

⇒ GM = √{(24 - 16√2 + 12√3 - 8√6)}

We know that 

(a + b + c + d)² = a² + b² + c² + d² + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd

From option (1), i.e. √2(√6 - √3 + √2 - 1)

If we square this, we get

2×(6 - 2√18 + 2√12 - 2√6 + 3 - 2√6 + 2 - 2 + 1)

⇒ 24 - 16√2 + 12√3 - 8√6

Therefore, the correct option is 1) √2(√6 - √3 + √2 - 1).

Concept:

• If a, b, c are in AP, then 2b = a + c
• If a, b, c are in GP, then b2 = ac
• If a, b, c are in HP, then 2/b = 1/a + 1/c

Calculation:

Given, Δ(a, b, c, α) = \(\left|\begin{array}{ccc} a & b & a α+b \ b & c & b α+c \ a α+b & b α+c & 0 \end{array}\right|\)

Applying, C₃ → C₃ - (αC₁ + C₂) we get:

Δ(a, b, c, α) = \(​\left|\begin{array}{ccc} a & b & 0 \ b & c & 0 \ a α+b & b α+c & -(aα^2+2bα+c) \end{array}\right|\)

According to the question, Δ(a, b, c, α) = 0 for every α > 0

⇒ -(aα² + 2bα + c)(b² - ac) = 0

⇒ b2 - ac = 0, or aα2 + 2bα + c = 0

 

⇒ Δ(a, b, c, α) = 0 if a, b, c are in GP or, α is a root of ax2 + 2bx + c = 0 

∴ For Δ(a, b, c, α) = 0 for every α > 0, then a, b, c is are a GP.

Concept:

If x = a is root of the equation f(x) = 0, then the value of f(a) will also be equal to 0. 

Calculation:

Given, Δ(a, b, c, α) = \(\left|\begin{array}{ccc} a & b & a α+b \ b & c & b α+c \ a α+b & b α+c & 0 \end{array}\right|\)

Applying, C3 → C3 - (αC1 + C2) we get:

Δ(a, b, c, α) = \(​\left|\begin{array}{ccc} a & b & 0 \ b & c & 0 \ a α+b & b α+c & -(aα^2+2bα+c) \end{array}\right|\)

According to the question, Δ(a, b, c, α) = 0 for every α > 0

⇒ -(aα2 + 2bα + c)(b2 - ac) = 0

∴ Δ(7, 4, 2, α) = 0

⇒ -(7α2 + 2×4α + 2)(7² - 4×2) = 0

 

⇒ -(7α2 + 8α + 2)(49 - 8) = 0

⇒ 7α2 + 8α + 2 = 0

⇒ α is a root of the equation 7x2 + 8x + 2 = 0

∴ If Δ(7, 4, 2, α) = 0, then α is a root of 7x2 + 8x + 2 = 0.

Concept:

For finding maxima or minima of a function f(x),

• Put f'(x) = 0 to find the critical point.
• If f''(x) > 0, then function is minima and if f''(x) < 0, function has maxima.

Calculation:

Given that m(θ) = cot2θ + n2tan2θ + 2n

∴ m'(θ) = -2cotθcosec2θ + 2n2tanθsec2θ  = 0

⇒ n2tanθsec2θ = cotθcosec2θ

⇒ n2 ×  ×  =  × 

⇒ n2 =  = cot4θ 

⇒ n = cot2θ 

Now, m''(θ) = 2[2n2sec2θtan2θ + n2sec4θ + cosec4θ + 2cot2θcosec2θ]

= 2[2×cot4θ×sec2θ×tan2θ + cot4θsec4θ + cosec4θ + 2cot2θcosec2θ]

= 2[(cosec2θ)2 + (cotθcosecθ)2] > 0

⇒ m''(θ) > 0

⇒ At n = cot2θ, there is a minima.

Given that m(θ) = cot2θ + n2tan2θ + 2n

At n = cot2θ, there is a minima.

⇒ The minimum value of m(θ) = cot2θ + (cot2θ)2tan2θ + 2cot2θ

= cot2θ + cot⁴θtan2θ + 2cot2θ

= cot2θ + cot2θ + 2cot2θ [ ∵ cotθ = 1/ tanθ]

=  4cot2θ

= 4n

∴ The least value of m(θ) is 4n.

Concept:

For finding maxima or minima of a function f(x),

• Put f'(x) = 0 to find the critical point.
• If f''(x) > 0, then function is minima and if f''(x) < 0, function has maxima.

Calculation:

Given that m(θ) = cot2θ + n2tan2θ + 2n

∴ m'(θ) = -2cotθcosec²θ + 2n²tanθsec²θ  = 0

⇒ n2tanθsec2θ = cotθcosec2θ

⇒ n² × \(\frac{\sin θ}{\cos θ}\) × \(\frac{1}{\cos^2 θ}\) = \(\frac{\cos θ}{\sin θ}\) × \(\frac{1}{\sin^2 θ}\)

⇒ n² = \(\frac{\cos^4θ}{\sin^4 θ}\) = cot⁴θ 

⇒ n = cot²θ 

Now, m''(θ) = 2[2n²sec²θtan²θ + n²sec⁴θ + cosec⁴θ + 2cot²θcosec²θ]

= 2[2×cot⁴θ×sec²θ×tan²θ + cot⁴θsec⁴θ + cosec4θ + 2cot2θcosec2θ]

= 2[(cosec²θ)² + (cotθcosecθ)²] > 0

⇒ m''(θ) > 0

⇒ At n = cot2θ, there is a minima.

∴ At n = cot2θ, m(θ) attains the least value.

Formula used:

Equation of line passing through (x1, y1) & (x2, y2)

\(y-y_1 = \frac{y_2-y_1}{x_2-x_1}(x-x_1)\)

Calculation:

x = 0               ---(1)   

y = 0              ---(2)

x + y = 1 

⇒ y = 1 - x      ---(3)

At x = 0, y = 1 & at y = 0, x = 1

Therefore, E(1, 0) & F(0, 3) 

And

6x + y = 3

At x = 0, y = 3 & at y = 0, x = 3/6 = 1/2

6x + (1 - x) = 3       [From equation (3)]

5x = 2 ⇒ x = 2/5

From equation (3)

y = 1 - 2/5 = 3/5

Therefore, the co-ordinate of point C is (2/5, 3/5)

The remaining coordinate are O(0, 0), B(1/2, 0), D(0, 1)

By using the above formula, the equation of a line OC

\(y-0 = \frac{\frac{3}{5}-0}{\frac{2}{5}-0}(x-0)\)

y = (3/2)x

2y = 3x

∴ 3x - 2y = 0

Formula used:

Equation of line passing through (x₁, y₁) & (x₂, y₂)

\(y-y_1 = \frac{y_2-y_1}{x_2-x_1}(x-x_1)\)

Calculation:

x = 0               ---(1)   

y = 0              ---(2)

x + y = 1 

⇒ y = 1 - x      ---(3)

At x = 0, y = 1 & at y = 0, x = 1

Therefore, E(1, 0) & F(0, 3) 

And

6x + y = 3

At x = 0, y = 3 & at y = 0, x = 3/6 = 1/2

6x + (1 - x) = 3       [From equation (3)]

5x = 2 ⇒ x = 2/5

From equation (3)

y = 1 - 2/5 = 3/5

Therefore, the co-ordinate of point C is (2/5, 3/5)

The remaining coordinate are O(0, 0), B(1/2, 0), D(0, 1)

By using the above formula, the equation of a line BD

\(y-0 = \frac{1-0}{0-\frac{1}{2}}(x-\frac{1}{2})\)

y = -2(x - 1/2)

y = -2(2x - 1)/2

y = -2x + 1

∴ 2x + y - 1 = 0

Concept:

The general equation of ellipse is given by:

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) having foci at (± ae, 0) where e(eccentricity) is given by the relation:

a2 - b2 = a2e2 

Calculation:

Given, ellipse x2 + 4y2 = 1

⇒ \(\frac{x^2}{1}+\frac{y^2}{\frac{1}{4}}=1\)

⇒ a2 = 1, and b2 = 1/4

⇒ a = 1, and b = 1/2

Given, E and F are the foci

Now, for a point P(x, y) on the ellipse by definition of ellipse:

PE + PF = 2a

∴ PE + PF = 2

Concept:

The general equation of ellipse is given by:

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) having foci at (± ae, 0) where e(eccentricity) is given by the relation:

a² - b² = a²e² 

Calculation:

Given, ellipse x2 + 4y2 = 1

⇒ \(\frac{x^2}{1}+\frac{y^2}{\frac{1}{4}}=1\)

⇒ a² = 1, and b² = 1/4

⇒ a = 1, and b = 1/2

Now, a² - b² = a²e²  

⇒ 1 - 1/4 = e²/2

⇒ 3/4 = e²/2 

⇒ e = \(\frac{\sqrt{3}}{2}\)

∴ Foci = (± ae, 0) = \(\left(±\frac{\sqrt{3}}{2}, 0\right)\)

Also, end-points of latus rectum is \(\left(±\frac{\sqrt{3}}{2},± \frac{1}{4}\right)\)

Hence, \(\left(\frac{\sqrt{3}}{2}, 0\right)\), \(\left(\frac{\sqrt{3}}{2}, \frac{1}{4}\right)\) and \(\left(\frac{\sqrt{3}}{2},-\frac{1}{4}\right)\) lie on the latus rectum

∴ Points 1, 2 and 3 lie on latus rectum.

Formula used:

Area of circle = (Radius)²

Area of triangle = (1/2) base × height

Calculation:

Area of quarter circle = πa²/4

Area of Triangle = (1/2) × a × a = a²/2

Therefore

Area of the minor segment = πa2/4 - a²/2

Area of the minor segment = πa2/4 - a2/2

∴ Area of the minor segment = \(\frac{(\pi-2) a^2}{4}\)

Formula used:

Area of circle = (Radius)2

Area of triangle = (1/2) base × height

Calculation:

Area of quarter circle = πa2/4

Area of Triangle = (1/2) × a × a = a2/2

Therefore,

Area of the minor segment = πa2/4 - a2/2

Area of the minor segment = πa2/4 - a2/2 

Area of the major segment = πa2 - (πa2/4 - a2/2)

Area of the major segment = 3πa2/4 + a2/2

∴ Area of the major segment = \(\frac{(3 \pi+2) a^2}{4}\)