# Quantitative Aptitude Questions Daily Quiz Day – 29

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109 Dear Aspirants, Our SSC Crackers team is providing a new series of Quantitative Aptitude Questions for Upcoming Exam so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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1) a) a = 1, b = 0

b) a = 0, b = 1

c) a = 1, b = 2

d) a = -1, b = 2

2) a) 0

b) a3

c) 1

d) a

3) If (tan θ + cot θ) = 2, and 0o < θ < 90o, then find the value of (tan6θ + Cot9θ).

a) 2

b) 1

c) 0

d) -1

4) What is the value of [(1 + cos(π /8)) * (1 + cos(3π /8)) * (1 + cos(5π /8)) * (1 + cos(7π /8))]?

a) (1/4)

b) (1/8)

c) (1/16)

d) (1/2)

5) Simplify: {sin (θ/2) + cos (θ/2)}2/ {2Cos2(θ/2) – 1)}.

a) sec2θ – tan2θ

b) secθ – tanθ

c) 1/(secθ – tanθ)

d) 1/(sec θ + tan θ)

6) If tan A = m/(m + 1) and tanB = 1/(2m + 1), then (A + B) is equal to?

a) 600

b) 300

c) 750

d) 450

7) If (tanA + cotA)/(tanA – cotA) = 2 where 00 < A < 900, then the value of sinA is?

a) 1/2

b) sin 750

c) √3/2

d) sin 250

8) A man observes that angle of elevation of the top of tower of height 120√3 m is 30o.He starts moving towards the tower and after 2 minutes of walk, he observes that angle of elevation of top of tower is 60o. Find the speed of man.

a) 2 m/s

b) 3 m/s

c) 4 m/s

d) 5 m/s

9) There is a circle whose diameters AD and BC intersect each other at point O. Two tangents are drawn from a point P from outside of the circle at point A and B such that angle APB = 85o as shown in the figure. Find the value of ‘x’. a) 45

b) 60

c) 38.5

d) None of these

10) What is the area of the triangle if the lengths of its medians are 18cm, 28cm and 34cm?

a) 16√110 cm2

b) 32√210 cm2

c) 64√145 cm2

d) 32√110 cm2  Tan θ + cot θ = 2

Tan θ + 1/tanθ = 2

Tan2θ + 1 = 2 tan θ

Tan2θ – tan θ – tan θ + 1 = 0

Tanθ (tanθ – 1) – 1(tanθ – 1) = 0

(tanθ – 1)(tanθ – 1) = 0

(Tan θ – 1)2 = 0

Tan θ = 1

So, Cot θ = 1/Tan θ = 1

Therefore, Tan6θ + Cot9 θ = 16 + 19 = 1 + 1 = 2

cos7π /8 = cos(π – π /8) = – cosπ /8

cos5π /8 = cos(π – 3π /8) = – cos3π /8

= [(1 + cosπ /8) * (1 + cos3π /8) * (1 – cos3π /8) * (1 – cosπ /8)]

= (1 – cos2π /8) * (1 – cos23π /8)

= sin2π /8 * sin23π /8

= [1 – cos(π /4)]/2 * [1 – cos(3π /4)]/2

= (1/4) * (1 – 1/√2) * (1 + 1/√2)

= (1/4) * (1 – 1/2)

= (1/4) * (1/2)

= (1/8)

{sin (θ/2) + cos (θ/2)}2/{2Cos2(θ/2) – 1)

= {sin2(θ/2) + cos2(θ/2) + 2 sin(θ/2) cos(θ/2)}/(cosθ)

= (1 + sinθ)/cosθ

= 1/cosθ + sinθ/cosθ

= secθ + tanθ

= (secθ + tanθ) × {(secθ – tanθ)/ (secθ– tanθ)}

= (sec2θ – tan2θ)/ (secθ – tanθ)

= 1/ (secθ – tanθ)

tanA = m/(m + 1) and tanB = 1/(2m + 1)

(tanA + tanB) = [m/(m + 1)] + [1/(2m + 1)]

= (2m2 + 2m + 1)/(m + 1)(2m + 1)

tanA.tanB = [m/(m + 1)] * [1/(2m + 1)]

= m/(m + 1)(2m + 1)

tan(A + B) = (tanA + tanB)/(1 – tanA.tanB)

= (2m2 + 2m + 1)/[(m + 1)(2m + 1) – m]

= (2m2 + 2m + 1)/(m + 1)(2m + 1) – m]

= (2m2 + 2m + 1)/(2m2 + 2m + 1)

= 1

= tan450

(A + B) = 450

(tanA + cotA)/(tanA – cotA) = 2

(tanA + cotA) = 2(tanA – cotA)

tanA + cotA = 2tanA – 2cotA

3cotA = tanA

tanA = 3cotA

tanA/cotA = 3

tanA/(1/tan) = 3

tan2A = 3

tanA = √3

tan A = tan 600

A = 600

sinA = sin600

= √3/2 As PA and PB are tangents to the circle, so angle OAP = angle OBP = 90o

So, angle AOB = 360o – (90o + 90o + 85o) = 95o

So, angle COD = 95o

As OC and OD will be the radius of circle, so OC = OD, therefore, angle OCD = angle ODC = x

In triangle COD, angle OCD + angle ODC + angle COD = 180o

x + x + 95o = 180o

2x = 180o – 95o = 85o

x = 85o ÷ 2 = 42.5o