**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.

**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) a^{3}

c) 1

d) a

**3) If (tan θ + cot θ) = 2, and 0 ^{o }< θ < 90^{o}, then find the value of (tan^{6}θ + Cot^{9}θ).**

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}/ {2Cos^{2}(θ/2) – 1)}.**

a) sec^{2}θ – 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) 60^{0}

b) 30^{0}

c) 75^{0}

d) 45^{0}

**7) If (tanA + cotA)/(tanA – cotA) = 2 where 0 ^{0} < A < 90^{0}, then the value of sinA is?**

a) 1/2

b) sin 75^{0}

c) √3/2

d) sin 25^{0}

**8) A man observes that angle of elevation of the top of tower of height 120√3 m is 30 ^{o}.He starts moving towards the tower and after 2 minutes of walk, he observes that angle of elevation of top of tower is 60^{o}. 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 = 85 ^{o }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 cm^{2}

b) 32√210 cm^{2}

c) 64√145 cm^{2}

d) 32√110 cm^{2}

**Answers :**

**1) Answer: B) **

**2) Answer: D**

**3) Answer: A) **

Tan θ + cot θ = 2

Tan θ + 1/tanθ = 2

Tan^{2}θ + 1 = 2 tan θ

Tan^{2}θ – 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, Tan^{6}θ + Cot^{9 }θ = 1^{6 }+ 1^{9 }= 1 + 1 = 2

**4) Answer: B) **

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 – cos^{2}π /8) * (1 – cos^{2}3π /8)

= sin^{2}π /8 * sin^{2}3π /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)

**5) Answer: C) **

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

= {sin^{2}(θ/2) + cos^{2}(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)}

= (sec^{2}θ – tan^{2}θ)/ (secθ – tanθ)

= 1/ (secθ – tanθ)

**6) Answer: D) **

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

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

= (2m^{2} + 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)

= (2m^{2} + 2m + 1)/[(m + 1)(2m + 1) – m]

= (2m^{2} + 2m + 1)/(m + 1)(2m + 1) – m]

= (2m^{2} + 2m + 1)/(2m^{2} + 2m + 1)

= 1

= tan45^{0}

(A + B) = 45^{0}

**7) Answer: C) **

(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

tan^{2}A = 3

tanA = √3

tan A = tan 60^{0}

A = 60^{0}

sinA = sin60^{0}

= √3/2

**8) Answer: A) **

**9) Answer: D**

As PA and PB are tangents to the circle, so angle OAP = angle OBP = 90^{o}

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

So, angle COD = 95^{o}

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 = 180^{o}

x + x + 95^{o }= 180^{o}

2x = 180^{o }– 95^{o }= 85^{o}

x = 85^{o }÷ 2 = 42.5^{o}

**10) Answer: D**

If a, b, c is the length of the medians of a triangle then the area of the triangle = 4/3 √(s (s-a)(s-b)(s-c))

Where s = (a+b+c)/2

In the given question, s = (18 +28 +34)/2 = 40

Hence the area = 4/3 √(40 (40-18)(40-28)(40-34)) cm^{2}

= 4/3 √(40 X 22 X 12 X 6) = 4/3 X √(4^{2} X 2^{2} X 3^{2} X 110) cm^{2}

= 32√110 cm^{2}