Quantitative Aptitude PYQs

Explanation:

Given:

4 sin-1 x + cos-1 x = π 

⇒ \(4( \frac{\pi }{2} - Cos^{-1}x) + (\frac{\pi }{2} - sin^{-1}x) = \pi \)

⇒ \(2\pi - 4cos^{-1}x + \frac{\pi }{2} - sin^{-1}x = \pi \)

⇒ 4cos^-1 x + sin^-1x = \(\frac{3\pi}{2}\)

∴ Option (c) is correct

Explanation:

Let I = cot² (sec^-1 2) + tan²(cosec^-1 3)

⇒ I = cot² (60°) + tan ² (cot^-1 \(\sqrt 9 -1 \)

⇒ I = \(\frac{1}{3} + tan^2( tan^{-1}\frac{1}{\sqrt8})\)

⇒ I = \(\frac{1}{3} + \frac{1}{8} = \frac{11}{24}\)

∴ Option (b) is correct

Explanation:

Given,

\(\rm \frac{a}{\cos A}=\frac{b}{\cos B}=\frac{c}{\cos C}\)

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

⇒b²+ c ² – a ² = c ² + a ² – b ² = a ² + b ² – c ²

⇒ a ² = b ² = c ²  

⇒ a = b = c

Hence, ABC is an equilateral triangle

Side of the equilateral triangle, a = 6 cm

The area of the equilateral triangle is,

\( \text{Area} = \frac{\sqrt{3}}{4} a^2 \)

⇒ \( \text{Area} = \frac{\sqrt{3}}{4} \times 6^2 \)

⇒ \( \text{Area} = \frac{\sqrt{3}}{4} \times 36 \)

⇒ \( \text{Area} = 9\sqrt{3} , \text{cm}^2 \)

∴ The area of the triangle is 9 × √3 square cm.

Explanation:

Given:

tanα and tanβ are roots of 7x² – 6x + 1 = 0

Hence, 

⇒ tanα + tanβ = 6/7...(1)

⇒ tanα.tanβ = 1/7...(2)

⇒ tan(α+β) = \(\frac{tanα + tanβ}{1- tanα . tanβ} = \frac{\frac{6}{7}}{1- \frac{1}{7}} = 1\)

⇒ α +β = 45°

⇒ 2α +2β = 90°

It means the third angle of a triangle is 90°.

Hence, the triangle is right-angled.

Since

⇒tanα - tanβ = \(\sqrt{(tanα + tanβ)^2 - 4tanα .tanβ} = \frac{2\sqrt2}{7}\)

Also

⇒ tan(α -β) = \(\frac{tanα - tanβ}{1+ tanα .tanβ}\) = \(\frac{1}{2\sqrt2}\)

⇒ tan2(α -β) = \(\frac{2tan(α -β)}{1- tan^2(α-β)}\)

 = \(\frac{4\sqrt7}{7}\)

Thus 2α≠2β 

∴ Option (c) is correct.

Concept:

Triangle Angle Sum Property:

• In a triangle, the sum of all interior angles is always equal to 180º.
• Given angles can be used to find the unknown angle using the formula: \( \angle A + \angle B + \angle C = 180^\circ \)
• Once all angles are known, trigonometric ratios can be used to relate the sides of the triangle.
• Law of Sines:
• The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant: \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)
• Here, a, b, and c are the sides opposite to angles A, B, and C respectively.

Calculation:

Given,

Angle A = 75º

Angle B = 45º

Using the triangle angle sum property, find angle C:

⇒ ∠C = 180º - (∠A + ∠B)

⇒ ∠C = 180º - (75º + 45º)

⇒ ∠C = 180º - 120º

⇒ ∠C = 60º

Now, using the Law of Sines:

\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)

⇒ \(a = \frac{sin 75°}{sin 60°}.c = \frac{\sqrt3 +1}{\sqrt6}.c\)

⇒ \(b = \frac{sin 45°}{sin 60°}.c = \frac{2c}{\sqrt6}\)

Now

⇒  2a - b = \(\frac{2\sqrt3 c}{\sqrt6} = \sqrt2 c\)

∴ Option (b) is correct.

Explanation:

Given

cot2x. cot3x = 1

⇒ cos2x. cos3x – sin2x.sin3x= 0

⇒ cos(2x+3x) = cos5x = 0    0 < x < π 

Hence 5x = \(\frac{\pi }{2}, \frac{3\pi }{2}, \frac{5\pi }{2}, \frac{7\pi }{2}, \frac{9\pi }{2}\)

⇒ x = \(\frac{\pi }{10}, \frac{3\pi }{10}, \frac{5\pi }{10}, \frac{7\pi }{10}, \frac{9\pi }{10}\)

Thus, only five solutions are possible

∴ the correct answer is option C

Explanation:

cos¹⁰⁰x – sin¹⁰⁰x = 1 is only possible when cos² x = 1 or sin² x = 0

because cos² x ∈ [0, 1]

⇒ cos¹⁰⁰x ∈  [0, 1]

Also  sin² x ∈ [0, 1] 

⇒  sin¹⁰⁰x ∈ [0, 1] 

⇒ Cos²x = 1

⇒ x = nπ ± 0°

x = nπ 

∴ The correct answer is option 1.

Explanation:

In triangle ABC, A + B + C = π 

⇒ tan(A + B + C) = tanπ  = 0

⇒ \(\frac{tan A+tan B+ tan C - tan A tan Btan C}{1- tan A.tan B- tan B.tan C - tan C.tan A} = 0\)

⇒  tanA. tanB. tanC = tanA + tanB + tanC

⇒ cotA. cotB. cotC = \(\frac{1}{tan A+tan B+tan C }\)

= 1/k

∴ Option (b) is correct

**Concept:**

**Product-to-Sum Formulas:**

- The product-to-sum formulas are used to express products of sines and cosines as sums.
- For sines, the formula is:

\\(\sin A \sin B = \frac{1}{2} [ \cos (A-B) - \cos (A+B) ] \\)

 

**Calculation:**

⇒ sin12°.sin48° = \\(\frac{2(sin12°.sin48°)}{2}\\)

= \\(\frac{1}{2} [Cos 36° -Cos60°]\\)

= \\(\frac{1}{2}[\frac{\sqrt5 +1}{4} -\frac{1}{2}] =\frac{\sqrt5 -1}{8}\\)

∴ **Option (c) is correct**

Explanation:

⇒ \(\frac{cos17°- sin17°}{cos17°+ sin17°} = \frac{cos17°(1-\frac{sin 17°}{Cos17°})}{cos17°(1+\frac{sin 17°}{Cos17°})}\) 

= \(\frac{1-\tan17°}{1+\tan17°}\) 

= \(\frac{tan 45° - tan17°}{1 + tan 45°. tan17°}\)

= tan 28° = Cot 62°

∴ the Correct answer is Option 4.

Explanation:

(I) Let tan 22.5° = tan 45/2 = t

⇒ tan45° = \(tan( 2 \times\frac{45}{2})\)

= \(\frac{2tan\frac{45}{2}}{1 -tan^2\frac{45}{2}}\)

⇒ 1 = \(\frac{2t}{1-t^2}\) ⇒  t² + 2t –1 = 0 

⇒ t = \(\frac{-2 ± √4+4}{2} = -1±√2\)

⇒ tan \(\frac{45}{2} = -1 ± √2 = -1 +√2\)   (In first quadrant tan is positive}

This is an irrational number. 

(II) cot22.5° = \(\frac{1}{tan 22.5} = \frac{1}{-1 +√2} = 1+√2\)

This is also an irrational number

(III) tan22.5° – cot22.5° = -1+ √2 -(1 +√ 2)

= -2

This is not an irrational number

∴ Option (c) is correct.

Explanation:

Given:

⇒ \(\rm \frac{x}{\cos θ}=\frac{y}{\cos \left(\frac{2\pi}{3}-θ\right)}=\frac{z}{\cos\left(\frac{2\pi}{3}+θ\right)}\) = k (say)

⇒ x = kcosθ

⇒ y = k \(cos\frac{2\pi }{3} -θ \)

⇒ z = k \(cos\frac{2\pi }{3} +θ \)

Now, x + y + z = k [cosθ + cos \((\frac{2\pi }{3} -θ) + cos(\frac{2\pi }{3} +θ)\)]

= k [cosθ + 2cos\(\frac{2\pi }{3}cosθ \)]

= k[cosθ + 2 (\(-\frac{1}{2}) cosθ \)]

= k[cosθ- cosθ ] =0

⇒ x + y + z =0

∴ Option (b) is correct.

Explanation:

Given:

p tan (θ - 30°) = q tan (θ + 120°)

⇒ p\q = \(\frac{tan(\theta + 120°)}{tan(\theta -30°)}\)

= \(\frac{sin(\theta+120)cos(\theta -30)}{cos(\theta+120)sin(\theta-30)}\)

⇒ \(\frac{p+q}{p-q} = \)(\frac{sin(\theta+120)cos(\theta -30)+cos(\theta+120)sin(\theta-30)}{sin(\theta+120)cos(\theta -30)-cos(\theta+120)sin(\theta-30)}\)

= \(\frac{sin[(\theta+120)+(\theta-30)]}{sin[(\theta+120)-(\theta-30)]}\)

= \(\frac{sin(90+2\theta)}{sin 150}\)

⇒ \(\frac{p+q}{p-q} = \frac{sin(90+2\theta)}{\frac{1}{2}}\)

= 2 cos2θ

∴ Option (d) is correct

Explanation:

(I) Given that P and Q are reflexive relations of A.

⇒ For any a∈A, (a, a) ∈ P and (a, a) ∈ Q 

⇒ (a, a )∈ P∩Q

⇒ P∩Q is reflexive relation.

So statement (I)is true

(II) Given that P and Q are symmetric relation

Let (a, b) ∈ (P∪Q)

⇒ (a, b) ∈ P or (a, b) ∈ Q.

⇒ (b, a) ∈ P or (b, a) ∈ Q

⇒  (b, a) ∈ (P∪Q)

⇒P∪Q is symmetric relation.

**So statement

(II) is true.**

(III) Given that P and Q are transitive

Let (a, b)∈ (P∩Q) and (b, c)∈ (P∩Q)

⇒ {(a, b)∈ P, (b, c) ∈ P} and {(a, b)∈Q, (b, c) ∈Q}.....     {∵ P and Q are transitive}

⇒ (a, c) ∈ P∩Q

Thus P∩Q is a transitive relation.

**So statement

(III) is also true.**

∴ Option (d) is correct.

Explanation:

Given:

⇒A∩B =10

Now, (A × B) ∩  (B × A) = 10² =100

Thus, there are 100 elements Common in A x B and B x A 

∴ Option (d) is correct

Explanation:

Let p(n) = 7ⁿ – 6n

= (6+1)ⁿ – 6n

= 1+6n +ⁿC₂6² + ⁿC₃6³+.......... 6ⁿ -6n)

= 1+ 6²(nC2+ nC3 × 6 +....... 6^n-2)

= 1+ 36(¹⁰⁰C₂ + 100C3 × 6 +........6⁹⁸).....(n =100)

= 1 + 36 k

where k = (100C2 + 100C3 × 6 +........698)

Thus When 7ⁿ – 6n is divided by 36, the remainder will be 1

∴ Option (b) is correct

Explanation:

Since there are 4 lines, the maximum number of intersection points are ⁴C₂ = 6.

The maximum number of intersections of the circle with 4 lines is 4 × 2 = 8.

Required intersection points = 6 + 8 = 14

∴ the correct answer is Option 3 ie 14

Explanation:

Given:

\(\frac{S_p}{S_q} =\frac{p^2}{q^2}\)

⇒ \(\frac{\frac{p}{2}[2a +(p-1)d]}{\frac{q}{2} [2a+(q-1)d]}\) = p²/q²

⇒ \(\frac{2a+(p-1)d}{2a+(q-1)d} =\frac{p}{q}\)

⇒ 2aq +q(p – 1)d = 2ap + p(q – 1)d

⇒ 2a(q – p) = d(q – p) 

⇒ 2a = d or d = 2a

Thus, The common difference is equal to twice of the first term

∴ The Correct answer is Option C

Explanation:

Given:

(x - 1)2 + (x - 3)2 + (x - 5)2 = 0

⇒ x² – 2x + 1 + x² – 6x + 9 + x² – 10x + 25 = 0

⇒ 3x² – 18x + 35 = 0

Now, discriminant D =  (–18)² –4 × 3 × 35 

= – 96 < 0

Thus, The given equation does not have any real roots. 

∴ Option (a) is correct

Explanation:

⇒ Total student = 240

⇒10 students failed in every subject Hence, remaining total = 230

According to question

⇒ a + b + c + d + e + f + g = 230 .....

(1) ⇒ b + c + d = 110.... (2) ⇒ e+ f + g = 60 ..... (3) Eq.

(i) – Eqs. [(ii) + (iii)], we get

⇒ a = 230 –(110 + 60) = 60  

∴ Option (a) is correct.