1. If [latex]{sec}^{2}[/latex]θ + [latex]{tan}^{2}[/latex] θ = 7, then the value of θ when 0° ≤ θ ≤ 90°, is

A. 60° B. 30° C. 0° D. 90°

Answer: Option A
Explanation: a [latex]{sec}^{2}[/latex]θ + [latex]{tan}^{2}[/latex] θ = 7 1+ [latex]{tan}^{2}[/latex] θ + [latex]{tan}^{2}[/latex] θ = 7 2[latex]{tan}^{2}[/latex] θ = 6 [latex]{tan}^{2}[/latex] θ = 3 tan θ = [latex]\sqrt{3}[/latex]
2. The simplified value of [latex]{(sec x sec y + tan x tan y)}^{2}[/latex] – [latex]{(sec x tan y + tan x sec y)}^{2}[/latex](sec x tan y + tan x sec y):

A. –1 B. 0 C. [latex]{sec}^{2}[/latex]x D. 1

Answer: Option D
Explanation: = [latex]{sec}^{2}[/latex] x [latex]{sec}^{2}[/latex] y + [latex]{tan}^{2}[/latex] x [latex]{tan}^{2}[/latex] y + 2 (sec x tan y)(tan x sec y ) = [latex]{sec}^{2}[/latex] x [latex]{sec}^{2}[/latex] y + [latex]{tan}^{2}[/latex] x. [latex]{tan}^{2}[/latex] y - [latex]{sec}^{2}[/latex] x [latex]{tan}^{2}[/latex] y - [latex]{tan}^{2}[/latex] x [latex]{sec}^{2}[/latex] y = [latex]{sec}^{2}[/latex] x [latex]{sec}^{2}[/latex] y - [latex]{sec}^{2}[/latex] x. [latex]{tan}^{2}[/latex] y - [latex]{tan}^{2}[/latex] x [latex]{sec}^{2}[/latex] y + [latex]{tan}^{2}[/latex] x [latex]{tan}^{2}[/latex] y = [latex]{sec}^{2}[/latex] x ([latex]{sec}^{2}[/latex] y - [latex]{tan}^{2}[/latex] y) - [latex]{tan}^{2}[/latex] x([latex]{sec}^{2}[/latex] y - [latex]{tan}^{2}[/latex] y) = ([latex]{sec}^{2}[/latex] y - [latex]{tan}^{2}[/latex] y)[latex]{sec}^{2}[/latex] x - [latex]{tan}^{2}[/latex] x = 1 x 1 = 1
3. sin θ + Cosec θ = 2, then the value of [latex]{sin}^{100}[/latex]θ + [latex]{cosec}^{100}[/latex]θ is equal to:

A. 1 B. 2 C. 3 D. 100

Answer: Option B
Explanation: sin θ + [latex]\frac{1}{sin θ}[/latex] = 2 [latex]{sin}^{2}[/latex]θ - 2 sin θ + 1 = 0 (sin θ - 1) = 0 sin θ = 1 [latex]{sin}^{100}[/latex]θ + [latex]{cosec}^{100}[/latex]θ = 1 + 1 = 2
4. [latex]{cos}^{2}[/latex]α + [latex]{cos}^{2}[/latex]β , then the value of [latex]{tan}^{3}[/latex]α + [latex]{sin}^{5}[/latex]β is:

A. -1 B. 0 C. 1 D. [latex]\frac{1}{√3}[/latex]

Answer: Option B
Explanation: [latex]{cos}^{2}[/latex]α + [latex]{cos}^{2}[/latex]β = 2 1 - [latex]{sin}^{2}[/latex]α + 1 - [latex]{sin}^{2}[/latex]β = 2 [latex]{sin}^{2}[/latex]α + [latex]{sin}^{2}[/latex]β = 0 [latex]{sin}^{2}[/latex]α = 0 & [latex]{sin}^{2}[/latex]β = 0 α = 0 & β = 0 [latex]{tan}^{3}[/latex]α + [latex]{sin}^{5}[/latex]β = 0
5. If θ is a positive accute angle and tan 2θ tan 3θ = 1, then the value of (2 [latex]{cos}^{2 }[/latex] [latex]\frac{5θ}{2}[/latex]-1):

A. [latex]\frac{-1}{2}[/latex] B. 0 C. 1 D. [latex]\frac{1}{2}[/latex]

Answer: Option C
Explanation: tan20 tan30 = 1 tan 3θ = [latex]\frac{1}{tan2θ}[/latex] tan 3θ = cot 2θ tan 3θ = tan (9θ - 2θ) 3θ = 9θ - 2θ 5θ = 9θ θ = [latex]{18}^{°}[/latex] 2 [latex]{cos}^{2}[/latex] [latex]\frac{5θ}{2}[/latex] - 1 = 2 [latex]{cos}^{2}[/latex] 45 - 1 = 2 x [latex]\frac{1}{2}[/latex] - 1 = 1 - 1 = 0

1. If tan A = cot A = x , then the value of x is?

A. [latex]\frac{1 + 2 {cos}^{2} A}{sin A cos A}[/latex] B. [latex]\frac{{sin A cos A}}{1 - 2 {cos}^{2} A}[/latex] C. [latex]\frac{{sin A cos A}}{1 + 2 {cos}^{2} A}[/latex] D. [latex]\frac{1 - 2 {cos}^{2} A}{sin A cos A}[/latex]

Answer: Option D
Explanation: tan A - cot A = x [latex]\frac{sin A}{cos A}[/latex] - [latex]\frac{cos A}{sin A}[/latex] = x [latex]\frac{{sin}^{2} A - {cos}^{2} A}{sin A cos A}[/latex] = x [latex]\frac{1 - 2 {cos}^{2} A}{sin A cos A}[/latex] = x
2. What is equation of the line passing through the point (-1, 3) and having x- intercept of 4 units?

A. 3x - 5y = 12 B. 3x + 5y = 12 C. 3x - 5y = - 12 D. 3x - 5y = - 12

Answer: Option B
Explanation: Slope = [latex]\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}[/latex] = [latex]\frac{(0 - 3)}{(4 - (-1))}[/latex] = [latex]\frac{- 3}{5}[/latex] Equation of line ⇒ (y - 0) = [latex]\frac{- 3}{5}[/latex](x - 4) 5y = -3x + 12 ⇒ 3x + 5y = 123
3. 25% discount is offered on an item. By applying a promo code the customer wins 8% cash back. What is the effective discount?

A. 35.75% B. 35% C. 31% D. 12.5%

Answer: Option C
Explanation: Effective Discount = D1 + D2 - [latex]\frac{D1 × D2}{100}[/latex] = 25 + 8 - [latex]\frac{25 × 8}{100}[/latex] = 31%
4. Prabhat has done [latex]\frac{1}{2}[/latex] of a job in 12 days. Santosh completes the rest of the job in 6 days. In how many days can they together do the job?

A. 12 days B. 4 days C. 8 days D. 16 days

Answer: Option C
Explanation: Prabhat has done [latex]\frac{1}{2}[/latex] of a job in 12 days ∴Prabhat will do 1 work in 12 x 12 = 24 days. Similarly, Santosh will do 1 work in 12 days. No of days work required to complete the work together = [latex]\frac{1}{24}[/latex] + [latex]\frac{1}{12}[/latex] = [latex]\frac{1 + 2}{24}[/latex] = [latex]\frac{3}{24}[/latex] = [latex]\frac{1}{8}[/latex] ∴ Required Time = 8 days
5. x and y are two numbers such that their mean proportion is 16 and third proportion is 128. What is the value of x and y?

A. 8 and 16 B. 16 and 32 C. 8 and 32 D. 16 and 16

Answer: Option C
Explanation: [latex]\frac{x}{y}[/latex] + [latex]\frac{y}{128}[/latex] ⇒ [latex]{y}^{2}[/latex] = 128x ----------- (1) And, [latex]\sqrt{xy}[/latex] = 16 ----------- (2) ⇒ xy = 256 from (1) and(2) x = 8, y = 32

1. In the first 32 overs of a cricket match, the run rate was 7.2 runs/over. What is the required run rate in the remaining 18 overs to reach the target of 297 runs?

A. 4.3 B. 4.9 C. 3.1 D. 3.7

Answer: Option D
Explanation: Total runs made till 32 overs = 32 × 7.2 = 230.4 runs Remaining runs to be made = 297 - 230.4 = 66.6 runs ∴ Reauired run rate = [latex]\frac{66.6}{18}[/latex] = 3.7
2. Chord AB of a circle when extended meets the tangent to the circle at point P. PT is the tangent touching the circle at point T. If lengths of PT and PB are 6 cm and 4 cm respectively, what is the length of PA?

A. 12 cm B. 18 cm C. 27 cm D. 9 cm

Answer: Option D
Explanation: P[latex]{T}^{2}[/latex] = PA x PB 36 = 4 x PA PA = 9 cm
3. A student multiplied a number by 4/5 instead of 5/4. What is the percentage error in the calculation?

A. 56.25% B. 18% C. 28.13% D. 36%

Answer: Option D
Explanation: Let the no. be 20x[LCM of 4 x 5] Wrong Multiplication = 20x × [latex]\frac{4}{5}[/latex] = 16x Right Multiplication = 20x × [latex]\frac{5}{4}[/latex] = 25x % error = [latex]\frac{25x - 16x}{25x}[/latex]x 100 = 36%
4. Ruchir walks at 20 km/hr and Rukma cycles at 25 km/hr towards each other. What was the distance between them when they started if they meet after 48 minutes?

A. 54km B. 45km C. 36km D. 27km

Answer: Option C
Explanation: In 1 hour they travel together (20 + 25) kms towards each other. ⇒ 60 min → 45 km 48 min → [latex]\frac{45}{60}[/latex] x 48km = 36 km
5. A vendor buys chikoos at 15 for Rs 8 and then sells at 10 for Rs 6. What will be the result?

A. 12.5 percent loss B. 11.11 percent gain C. 12.5 percent gain D. 11.1% loss

Answer: Option C
Explanation: CP of 30 chikoo = Rs 16 SP of 30 chikoo = Rs 18 P = [latex]\frac{18 - 16}{16}[/latex] x 100 = 12.5%