Quantitative Aptitude PYQs

Given:

\(\rm 6x^2 - 25x + \frac{6}{x^2} + \frac{25}{x} + 12 = 0 \)

Formula used:

If n - \(\frac {1}{n}\) = p then n² + \(\frac{1}{{n^2}}\) = p² + 2

Calculation:

\(\rm 6x^2 - 25x + \frac{6}{x^2} + \frac{25}{x} + 12 = 0 \)

⇒ 6(x² + \(\frac{1}{{x^2}}\)) − 25(x − \(\frac {1}{x}\)) + 12 = 0

Let, x − \(\frac {1}{x}\) be 'a'

From the above concept, (x² + \(\frac{1}{{x^2}}\)) = a² + 2

⇒ 6(a² + 2) − 25a + 12 = 0

⇒ 6a² + 12 − 25a + 12 = 0

⇒ 6a² − 25a + 24 = 0

⇒ 6a² − 16a − 9a + 24 = 0

⇒ 2a(3a − 8) − 3(3a − 8) = 0

⇒ (2a − 3) (3a − 8) = 0

⇒ a = \(\frac {3}{2}\) or \(\frac {8}{3}\)

Since, \(\frac {8}{3}\) is not given 

∴ The required value of x − \(\frac {1}{x}\) = \(\frac {3}{2}\).

Given:

\(\rm 6x^2 - 25x + \frac{6}{x^2} + \frac{25}{x} + 12 = 0 \)

Formula used:

If n - \(\frac {1}{n}\) = p then n2 + \(\frac{1}{{n^2}}\) = p2 + 2

Calculation:

\(\rm 6x^2 - 25x + \frac{6}{x^2} + \frac{25}{x} + 12 = 0 \)

⇒ 6(x2 + \(\frac{1}{{x^2}}\)) − 25(x − \(\frac {1}{x}\)) + 12 = 0

Let, x − \(\frac {1}{x}\) be 'a'

From the above concept, (x2 + \(\frac{1}{{x^2}}\)) = a2 + 2

⇒ 6(a2 + 2) − 25a + 12 = 0

⇒ 6a2 + 12 − 25a + 12 = 0

⇒ 6a2 − 25a + 24 = 0

⇒ 6a2 − 16a − 9a + 24 = 0

⇒ 2a(3a − 8) − 3(3a − 8) = 0

⇒ (2a − 3) (3a − 8) = 0

⇒ a = \(\frac {3}{2}\) or \(\frac {8}{3}\)

Case:1 a = \(\frac {3}{2}\)  

⇒ (x2 + \(\frac {1}{x^2}\)) = (a² + 2) = (3/2)² + 2

⇒ (x2 + \(\frac {1}{x^2}\)) = (9/4 + 2) = 17/4 (not in the option)

Case:2  a = \(\frac {8}{3}\) 

⇒ (x2 + \(\frac {1}{x^2}\)) = (8/3)² + 2

⇒ (x2 + \(\frac {1}{x^2}\))  = (64/9 + 2) = 82/9

∴ The require value of x² + \(\frac {1}{x^2}\) is \(\frac {82}{9}\).

Concept:

Pythagoras theorem: In a right-angled theorem, 

Hypotenuse² = Base² + Perpendicular²

Where R is the Radius of a circle of centre O

• cos (90° - θ) = sin θ
• sin²θ + cos²θ = 1
• cos 90° = 0

 

Calculation:

We have, AB = 15 cm, BC = 9 cm, AC = 12 cm

12² + 9² = 144 + 81 = 225 = 15²

Hence, Δ ABC is a right-angled triangle in which ∠C = 90° 

As we know, ∠A + ∠B + ∠C = 180° 

⇒ ∠A + ∠B = 90°         (∵ ∠C = 90°)

⇒ ∠A = 90° - ∠B           

Now, the required value

cos² A + cos² B + cos² C 

⇒ cos2 A + cos2 (90° - A) + cos2 90°  

⇒ cos2 A + sin2 A + 0           [∵ cos 90° = 0]

⇒ 1 + 0                                [∵ sin²θ + cos²θ = 1]

∴  cos2 A + cos2 B + cos2 C = 1

Concept:

Pythagoras theorem: In a right-angled theorem, 

Hypotenuse2 = Base2 + Perpendicular2

Where R is the Radius of a circle of centre O

• cos (90° - θ) = sin θ
• sin2θ + cos2θ = 1
• cos 90° = 0

 

Calculation:

We have, AB = 15 cm, BC = 9 cm, AC = 12 cm

122 + 92 = 144 + 81 = 225 = 152

Hence, Δ ABC is a right-angled triangle in which ∠C = 90° 

As we know, ∠A + ∠B + ∠C = 180° 

⇒ ∠A + ∠B = 90°         (∵ ∠C = 90°)

⇒ ∠A = 90° - ∠B           

Now, the required value

sin2 A + sin2 B + sin2 C 

⇒ sin2 A + sin2 (90° - A) + sin2 90°  

⇒ sin2 A + cos2 A + 1           [∵ sin 90° = 1]

⇒ 1 + 1 = 2                          [∵ sin2θ + cos2θ = 1]

∴  sin2 A + sin2 B + sin2 C = 2

Concept:

Pythagoras theorem: In a right-angled theorem, 

Hypotenuse2 = Base2 + Perpendicular2

Where R is the Radius of a circle of centre O

Calculation:

We have, AB = 12 cm, BC = 9 cm, AC = 15 cm

122 + 92 = 144 + 81 = 225 = 152

Hence, Δ ABC is a right-angled triangle in which ∠C = 90°           

Using the above concept, the radius of the circle

OA = OB = OC = 15/2 = 7.5 cm

∴ The radius of the circle is 7.5 cm

Concept used:

tan θ = Perpendicular/Base

sin θ = Perpendicular/hypotenuse

The height of the cloud in the lake from point P is also 2H metres  

Calculation:

In an Δ PAC

tan 30° = AC/AP = 1/√3

⇒ AP = H√3

In an Δ PAB

tan θ = AB/AP = (H + 2H)/H√3

⇒ tan θ = √3 = tan 60° 

∴ θ = 60°

Concept used:

tan θ = Perpendicular/Base

sin θ = Perpendicular/hypotenuse

The height of the cloud in the lake from point P is also 2H metres  

Calculation:

In an Δ PAC

sin 30° = AC/PC

⇒ 1/2 = H/PC

⇒ PC = 2H

∴ The value of PC is 2H metres.

Given:

Total freuesncy (N) = 200

Mean = 1.46

Formula used:

Mean = Sum/Numbers

Calculation:

Now,

⇒ 46 + p + q + 25 + 10 + 5 = 200

⇒ p + q + 86 = 200

⇒ p + q = 114                    -----(1)

Mean = [(p + 2q + 140)/200] 

⇒ [(p + 2q + 140)/200] = 1.46       (Mean = 1.46 given)

⇒ p + 2q = (292 − 140)

⇒ p + 2q = 152                  -----(2)

Subtracting equation (1) from equation (2),

⇒ p + 2q − p − q = (152 − 114)

⇒ q = 38

Now, on putting the value of q in equation (1), 

⇒ p + 38 = 114

⇒ p = 76

∴ The required value of p is 76.

Given:

Total freuesncy (N) = 200

Mean = 1.46

Formula used:

Mean = Sum/Numbers

Calculation:

Now,

⇒ 46 + p + q + 25 + 10 + 5 = 200

⇒ p + q + 86 = 200

⇒ p + q = 114                    -----(1)

Mean = [(p + 2q + 140)/200] 

⇒ [(p + 2q + 140)/200] = 1.46       (Mean = 1.46 given)

⇒ p + 2q = (292 − 140)

⇒ p + 2q = 152                  -----(2)

Subtracting equation (1) from equation (2),

⇒ p + 2q − p − q = (152 − 114)

⇒ q = 38

∴ The required value of q is 38.

Given:

Total freuesncy (N) = 200

Mean = 1.46

Formula used:

Median: 

The smallest value for which the cumulative frequency is higher than 50%.

Calculation:

 

Here, N = 200, N/2 = 200/2 = 100

So, the cumulative frequency of just more than N/2 is 122.

∴ The median is 1.

Shortcut Trick

Put p = q

⇒ tan θ = \(\rm \frac{\sqrt{p^2-p^2}}{p} = 0\)

⇒ θ = 0° 

⇒ sec θ + cos θ + 2 = 1 + 1 + 2 = 4

From option (2);

\(\rm \frac{(p+q)^2}{pq} =\frac{(p+p)^2}{p\times p} = \frac{4p^2}{p^2} = 4\)

Alternate Method

Concept:

1. Pythagoras Theorem: In a right-angled triangle, 

Base² + Perpendicular² = Hypotenuse²

2. tan θ = P/B, sec θ = H/B, cos θ = B/H

3. (a + b)² = a² + b² + 2ab

Calculation:

We have,

tan θ = \(\rm \frac{\sqrt{q^2-p^2}}{p}\) 

⇒ AB² + BC² = AC²

⇒ \(\left(\sqrt{q^2 \ - \ p^2} \right)^2 + p^2 = AC^2\)

⇒ q² - p² + p² = AC²

⇒ AC² = q²

⇒ AC = q

⇒ sec θ + cos θ + 2

⇒ \(\frac{q}{p} \ + \ \frac{p}{q} \ + \ 2\)

⇒ \(\frac{q^2 \ + \ p^2 \ + \ 2pq}{pq}\)

⇒ \(\frac{(p \ + \ q)^2}{pq}\)

∴ If tan θ = \(\rm \frac{\sqrt{q^2-p^2}}{p}\) ; 0 < θ < 90°, then what is sec θ + cos θ + 2 equal to \(\frac{(p \ + \ q)^2}{pq}\)

Given:

11 sin θ + 60 cos θ = 61; 0 < θ < 90°

Shortcut TrickConcept:

If a sin θ + b cos θ = √(a² + b²)

⇒ tan θ = a/b, cot θ = b/a

Where, a, b & c are Pythagorean triples

Calculation:

We have, 11 sin θ + 60 cos θ = 61

11, 60 & 61 are Pythagorean triples

⇒ \(√{660(\tanθ+\cotθ)} = √{[660( \frac{11}{60} +\frac{60}{11})]}\)

⇒ \(√{660\frac{(121+3600)}{660}}\)

⇒ √3721 = 61​​

Additional Informationsin θ = a/√(a²+b²), cos θ = b/√(a² + b²)

Given:

Number of sides in regular polygon = 15

Length of sides = 1 unit

Concept used:

The angle at the centre of the polygon = \(\frac{360^\circ}{n}\)

Where n is the number of sides

tanθ = \(\frac{Perpendicular}{{base}}\)

Calculation:

We know that, 

The angle at the centre of a polygon of n sides = \(\frac{360^\circ}{n}\)

The angle at the centre of the polygon of 15 sides = \(\frac{360^\circ}{15} = 24^\circ\)

As OC is the radius & AB is the tangent, hence 

∠ OCA = ∠ OCB = 90°  

Therefore, ΔOAC is the right-angled triangle  

tan 12° = \(\frac{AC}{OC}=\frac{1}{{2(OC)}}\)

⇒ 2 OC = \(\frac{1}{{tan12° }}\) = cot 12° 

⇒ Diameter = cot 12°     [∵ diameter = 2 radius]

∴ The diameter of a circle is cot 12°

Given:

x = \(\rm \frac{1+\sin θ}{\cosθ}\),  

Formula Used:

• sec²θ - tan²θ = 1
• sec θ = 1/cos θ, tan θ = sin θ / cos θ
• (a - b)(a + b) = a² - b²

 

Calculation:

We have,

⇒ \(\frac{\tanθ\ +\ \secθ\ -\ 1}{\tanθ\ -\ \secθ\ +\ 1}\)

⇒ \(\frac{\secθ\ +\ \tanθ\ -\ (sec^2θ \ -\ \tan^2θ)}{\tanθ\ -\ \secθ\ +\ 1}\)

⇒ \(\frac{\secθ\ +\ \tanθ\ -\ [(secθ \ -\ \tanθ)(secθ \ + \ \tanθ)]}{\tanθ\ -\ \secθ\ +\ 1}\)

⇒ \(\frac{(\secθ\ +\ \tanθ)\ [1 - (secθ \ + \ \tanθ)]}{\tanθ\ -\ \secθ\ +\ 1}\)

⇒ secθ + tanθ 

⇒ \(\frac{1}{cosθ} \ + \ \frac{sinθ}{cosθ}\)

⇒ \(\frac{1\ + \ sinθ}{cosθ}\)

⇒ x

∴ If x = \(\rm \frac{1+\sin θ}{\cosθ}\), then the value of  \(\frac{\tanθ+\secθ-1}{\tanθ-\secθ+1}\) is equal to x.

Given:

 \(\rm \frac{\tanθ+\sinθ}{\tanθ-\sinθ}= \frac{\secθ+1}{\secθ-1}\)

 \(\frac{\cos^2θ-\sin^2θ}{\cos^2θ+\sin^2θ}= \frac{2\tanθ}{\tan^2θ+1}\)

Calculation:

Statement I:

We have,

⇒ \(\rm \frac{\tanθ\ +\ \sinθ}{\tanθ\ -\ \sinθ}\)

⇒ \(\rm \frac{\frac{sinθ}{cos\theta}\ +\ \sinθ}{\frac{sinθ}{cos\theta}\ -\ \sinθ}\)

⇒ \(\rm \frac{sin\theta\left[\frac{1}{cos\theta}\ +\ \ 1\right]}{sin\theta\left[\frac{1}{cos\theta}\ -\ 1\right]}\)

⇒ \(\rm \frac{\sec\theta \ +\ \ 1}{sec\theta\ -\ 1}\)

Hence, a statement I is correct.

Statement II;

We have,

⇒ \(\frac{\cos^2θ\ -\ \sin^2θ}{\cos^2θ\ +\ \sin^2θ}\)

⇒ \(\frac{\cos^2θ\ \left[1 \ - \ \frac{sin^2θ}{cos^2\theta}\right]}{\cos^2θ\ \left[1 +\ \frac{sin^2θ}{cos^2\theta}\right] }\)

⇒ \(\frac{\ 1\ -\ \tan^2θ}{\ 1\ +\ \tan^2θ}\)

Hence, Statement II is incorrect.

∴ Statement I is correct but statement II is incorrect.

Given:

\(\frac{\sin θ + \cosθ }{\sin θ -\cos θ}\) = 5

Calculation:

\(\frac{\sin θ + \cosθ }{\sin θ -\cos θ}\) = 5

⇒ sinθ + cosθ = 5sinθ − 5cosθ 

⇒ 6cosθ = 4sinθ 

⇒ \(\frac{cosθ }{{sinθ }}\) = \(\frac{2 }{{3 }}\)

⇒ cos θ = 2k, sin θ = 3k 

Now, the required value 

⇒ \(\frac{2\sin θ + 3\cosθ }{3\sin θ -2\cos θ}\) = \(\frac{(2 \times 3k + 3 \times 2k)}{{(3 \times 3k - 2 \times 2k)}}\)

⇒ \(\frac{2\sin θ + 3\cosθ }{3\sin θ -2\cos θ}\) = \(\rm \frac{12}{5}\)

∴ The required value is \(\rm \frac{12}{5}\).

Formula used:

1. (a + b)(a - b) = a² - b²
2. cosec2θ - cot2θ = 1
3. 1/sin θ = cosec θ

 

Calculation:

\(\rm\frac{1}{cosecθ-\cotθ}-\frac{1}{\sinθ}=x\)

⇒ \(\rm\frac{(cosecθ+\cotθ)}{(cosecθ-\cotθ)(cosecθ+\cotθ)}-\frac{1}{\sinθ}=x\)

⇒ \(\rm\frac{(cosecθ+\cotθ)}{(cosec^2θ-\cot^2θ)}-\frac{1}{\sinθ}=x\)

Since, cosec²θ - cot²θ = 1

⇒ \(\rm\frac{(cosecθ+\cotθ)}{(cosec^2θ-\cot^2θ)}-\frac{1}{\sinθ}=x\)

⇒ cosec θ + cot θ − cosec θ = x

⇒ x = cot θ      ----(1)

The value of \(\rm\frac{1}{cosecθ+\cotθ}-\frac{1}{\sinθ}\)

⇒ \(\rm\frac{(cosecθ-\cotθ)}{(cosecθ+\cotθ)(cosecθ-\cotθ)}-\frac{1}{\sinθ}\)

⇒ \(\rm\frac{(cosecθ-\cotθ)}{(cosec^2θ-\cot^2θ)}-\frac{1}{\sinθ}\)

⇒ cosec θ − cot θ − cosec θ 

⇒ - cot θ  = -x      [From equation (1)]

∴ The required value is -x.

Shortcut Trick

Since the given trigonometric expression does not have variables in the options, its value is independent of θ.

Let us assume any convenient angle, say θ = 0°.

We know that sin 0° = 0 and cos 0° = 1.

Substitute θ = 0° in the given expression: 2(0 + 1) − 3(0 + 1)

⇒ 2(1) − 3(1) = 2 − 3 = −1

∴ The correct answer is -1.

Alternate Method

Given:

Expression: 2 sin⁶θ + 2 cos⁶θ − 3 sin⁴θ − 3 cos⁴θ

Formula Used:

a³ + b³ = (a + b)(a² + b² − ab)

a² + b² = (a + b)² − 2ab

sin²θ + cos²θ = 1

Calculation:

⇒ 2(sin⁶θ + cos⁶θ) − 3(sin⁴θ + cos⁴θ)

⇒ 2[(sin²θ)³ + (cos²θ)³] − 3[(sin²θ)² + (cos²θ)²]

⇒ 2[(sin²θ + cos²θ)(sin⁴θ + cos⁴θ − sin²θ cos²θ)] − 3[(sin²θ + cos²θ)² − 2 sin²θ cos²θ]

Substituting sin²θ + cos²θ = 1:

⇒ 2[1 × (sin⁴θ + cos⁴θ − sin²θ cos²θ)] − 3[1² − 2 sin²θ cos²θ]

⇒ 2[(sin²θ + cos²θ)² − 2 sin²θ cos²θ − sin²θ cos²θ] − 3[1 − 2 sin²θ cos²θ]

⇒ 2[1 − 3 sin²θ cos²θ] − 3[1 − 2 sin²θ cos²θ]

⇒ 2 − 6 sin²θ cos²θ − 3 + 6 sin²θ cos²θ

⇒ 2 − 3 = −1

∴ The correct answer is -1.

Additional Information

Standard Trigonometric Identities

sin²θ + cos²θ = 1  |  1 + tan²θ = sec²θ  |  1 + cot²θ = cosec²θ

Derived Trigonometric Results

sin⁴θ + cos⁴θ = 1 − 2 sin²θ cos²θ

sin⁶θ + cos⁶θ = 1 − 3 sin²θ cos²θ

Value Substitution Method

When a trigonometric expression evaluates to a constant in the options, substituting standard angles like 0°, 30°, 45°, 60°, or 90° helps bypass complex algebraic steps.

Given:

3600 sec2θ + 121 cosec2θ

Concept:

a sec²θ + b cosec²θ 

Then the maximum value is ∞ 

and minimum value is \((\sqrt a \ + \ \sqrt b)^2\)

Calculation:

We have,

⇒ 3600 sec2θ + 121 cosec2θ

⇒ \((\sqrt {3600} \ + \ \sqrt{121} )^2\)

⇒ (60 + 11)²

⇒ (71)²

⇒ 5041

∴ The minimum value of 3600 sec2θ + 121 cosec2θ is 5041.

Given:

\(\sqrt{\left(\frac{1+\sinθ}{\cosθ}\right)^2+\left(\frac{\cosθ}{1+\sinθ}\right)^2-2}\)

Formula used:

• (a - b)2  = a² + b² - 2ab
• (a + b)(a - b) = a² - b²
• sin θ/cos θ =  tanθ
• 1/cos θ = sec θ

 

Calculation:

\(\sqrt{\left(\frac{1+\sinθ}{\cosθ}\right)^2+\left(\frac{\cosθ}{1+\sinθ}\right)^2-2}\)

⇒ \(\sqrt{\left(\frac{1+\sinθ}{\cosθ}\right)^2+\left(\frac{\cosθ}{1+\sinθ}\right)^2-2\left(\frac{1+\sinθ}{\cosθ}\right)\left(\frac{cosθ}{1+\sinθ}\right)}\)

⇒  \(\sqrt{\left(\frac{1+\sinθ}{\cosθ}-\frac{\cosθ}{1+\sinθ}\right)^2}\)          [∵ (a - b)2  = a2 + b2 - 2ab]

⇒ \(\left(\frac{1+\sinθ}{\cosθ}-\frac{\cosθ}{1+\sinθ}\right)\)

Since, sin θ/cos θ =  tanθ, 1/cos θ = sec θ

⇒ \(\left({sec\ θ+tan\ θ}-\frac{1}{sec \ θ+\tan \ θ}\right)\)    

⇒  \(\left({sec\ θ+tan\ θ}-\frac{(sec \ θ-\tan \ θ)}{(sec \ θ+\tan \ θ)(sec \ θ-\tan \ θ)}\right)\)

⇒ \(\left({sec\ θ+tan\ θ}-\frac{(sec \ θ-\tan \ θ)}{(sec^2\ θ-\tan^2 \ θ)}\right)\)

⇒ sec θ + tan θ - sec θ + tan θ        [∵ sec2θ − tan2θ = 1] 

⇒ 2 tanθ 

**∴ The required value is 2 tanθ.**​