# Quantitative Aptitude Questions Daily Quiz Day – 60

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1) If sec θ= 13/12 and  is acute, then what is the value of (√(cotθ+tanθ))?

(a) 13/(2√15)

(b) 12/(2√13)

(c) 13/(2√5)

(d) 2/13

2) If (1+tan2 θ) = 625/49  and θ is acute, then what is the value of (√(sinθ+cosθ))?

(a) 1

(b) 5/4 √(31/42)

(c)

(d) 5/7

3) If sin θ + sin2 θ = 1, then what is the value of (cos12 θ + 3 cos10 θ + 3 cos8 θ + cos6 θ – 1)?

(a) -1

(b) 0

(c) 1

(d) 2

4)If cot θ = √11 and θ is acute, then what is the value of (cosec2 θ + sec2 θ) /(cosec2 θ – sec2 θ)

(a) 2/3

(b) -6/5

(c) 3/4

(d) 7/6

5) If (cos θ/(1+sin  θ)) + (cos θ/(1- sin  θ))=2 and  is acute, then what is the value (in degrees) of θ?

(a) 30

(b) 45

(c) 60

(d) 90

Directions (6-10): The given pie chart shows the distribution (in degrees) of cars sold of different models by a company in 2015-16. 6) If the number of cars sold of model D is 40500, then how many more cars of model E are sold than that of A?

(a) 8100

(b) 16200

(c) 24300

(d) 13500

7) If the number of cars sold of model D is 40500, what is the ratio between the number of cars sold of model D and E?

(a) 9 : 5

(b) 6 : 5

(c) 11 : 9

(d) 9 : 7

8) If the number of cars sold of model D is 72900, then what is the total number of cars sold of all the models together by the company?

(a) 291600

(b) 208100

(c) 162000

(d) 214160

9) If number of cars sold of model C are 22000, then what will be the difference in the number of cars sold of model A and B?

(a) 800

(b) 1200

(c) 1000

(d) 1500

10) If 5% of the total number of cars sold of model E is 750, then what is the average number of cars sold of all the models?

(a) 15000

(b) 14400

(c) 16800

(d) 14000

secθ=13/12 [Triplet, 5, 12, 13]

cotθ=(12/ 5) & tanθ= (5/12)

So, √(cotθ+tanθ)=√(12/5+5/12)=13/√60

=13/(2√15)

(1+tan θ) = secθ = 625/49

sec θ = 25/7

sin θ = 24/25

cos θ = 7/25

√sin θ  + cos θ

= √31/5

sin θ = (1 – sin2 θ)

sin2 θ = cos4 θ

1 – cos2 θ = cos4 θ

1 = cos4 θ + cos2 θ

1 = (cos4 θ + cos2 θ )3

1 = cos12 θ + 3 cos10 θ + 3 cos8 θ + cos6 θ

cos12 θ + 3 cos10 θ + 3 cos8 θ + cos6 θ – 1 = 0

cot θ = √11 ⇒ cot2 θ = 11

(cosec2 θ + sec2 θ) /(cosec2 θ – sec2 θ) = 1/(cos2 θ – sin2 θ) = 1/ (cos 2θ) = 1/(2 cos2θ – 1)

cos2 θ = 1/2 ⇒ (1/(2×(1/2)-1)) = -(6/5)

put θ = 45°(cos θ/(1+sin  θ)) + (cos θ/(1- sin  θ)) = 2 √2

No. of cars = (40600/90) 30=13500

Required ratio = 40500/33750 = 162/135 = 6/5

Required No. of models cars = 810 × 360 = 291600