1. If the cost price of 50 oranges is equal to the selling price of 40 oranges, then the profit percent is?

A. 5 B. 10 C. 20 D. 25

Answer: Option D
Explanation: 50 CP = 40 SP [latex]\frac{CP}{SP}[/latex] = [latex]\frac{40}{50}[/latex] = 10 Profit Profit % = [latex]\frac{10}{40}[/latex]= × 100 = 25%
2. A number is as much greater than 36 as is less than 86. Find the number :

A. 38 B. 43 C. 61 D. 73

Answer: Option C
Explanation: Let the number be z. then, z – 36 = 86 – z ⇒ 2z = 86 + 36 ⇒ 2z = 122. ⇒ z = 61. Hence, the required number is 61
3. In the annual examination, Ankita got 10% less marks than Eakta in Mathematics. Ankita got 81 marks. The marks of Eakta are?

A. 90 B. 87 C. 88 D. 89

Answer: Option A
Explanation: 10% = [latex]\frac{1}{10}[/latex] Ankita Eakta 9 10 ↓×9 ↓×9 81 90 Hence, marks obtained by Eakta = 90
4. The ratio of the number of boys to that of girls in a school is 4 : 1. If 75% of boys and 70% of the girls are scholarship-holders, then the percentage of students who do not get scholarship is?

A. 50% B. 28% C. 75% D. 26%

Answer: Option D
Explanation: Let the number of boys = 400 Let the number of girls = 100 Total number of students who do not get scholarship = 400 × [latex]\frac{25}{100}[/latex] + 100 × [latex]\frac{30}{100}[/latex] = 100 + 30 = 130 Required percentage = [latex]\frac{130}{500}[/latex]= × 100 = 26%
5. If the price of a commodity is increased by 50%. By what fraction must its consumption be reduced so as to keep the same expenditure on its consumption?

A. [latex]\frac{1}{4}[/latex] B. [latex]\frac{1}{3}[/latex] C. [latex]\frac{1}{2}[/latex] D. [latex]\frac{2}{3}[/latex]

Answer: Option B
Explanation: % change =[latex]\frac{R}{(100 ± R)}[/latex] × 100% Required answer = [latex]\frac{50}{(100 + 50)}[/latex] = [latex]\frac{1}{3}[/latex]

1. If x = 2 then the value of x3 + 27x2 + 243x + 631 is:

A. 1211 B. 1231 C. 1233 D. 1321

Answer: Option C
Explanation: Given equation, f(x) = x[latex]^{3}[/latex]+ 27x[latex]^{2}[/latex] + 243x + 631 ⇒ x(x[latex]^{2}[/latex] + 27x+ 243) + 631 Now, put the value of x = 2 ⇒ 2(2[latex]^{2}[/latex] + 27 × 2 + 243) + 631 ⇒ 2 (4 + 54 + 243) + 631 ⇒ 2(301) + 631 = 602 + 631 = 1233.
2. When 2x + [latex]\frac{2}{x}[/latex]2 = 3, then value of x[latex]^{3}[/latex] + [latex]\frac{1}{x}[/latex] + [latex]\frac{2}{x^{3}}[/latex] is

A. [latex]\frac{2}{7}[/latex] B. [latex]\frac{7}{8}[/latex] C. [latex]\frac{7}{2}[/latex] D. [latex]\frac{8}{7}[/latex]

Answer: Option B
Explanation: Given, 2x + [latex]\frac{2}{x}[/latex] = 3 or [latex]\frac{(x + 1)}{2}[/latex] = [latex]\frac{3}{2}[/latex] Cubing both sides, we get x[latex]^{3}[/latex]+ [latex]\frac{1}{x^{3}}[/latex] + 3 [latex]\frac{(x + 1) }{x}[/latex]= [latex]\frac{27}{8}[/latex] or, x[latex]^{3}[/latex] + [latex]\frac{1}{x^{3}}[/latex]+ 3 × [latex]\frac{3}{2}[/latex] = [latex]\frac{27}{8}[/latex] or, x[latex]^{3}[/latex] + [latex]\frac{1}{x^{3}}[/latex] = [latex]\frac{27}{8}[/latex] – [latex]\frac{9}{2}[/latex] = [latex]\frac{– 9}{8}[/latex] or, x[latex]^{3}[/latex] + [latex]\frac{1}{x^{3}}[/latex] + 2 = 2 – [latex]\frac{9}{8}[/latex]9 ⇒ [latex]\frac{7}{8}[/latex]7
3. If x = 332, y = 333, z = 335 then the value of x[latex]^{3}[/latex] + y[latex]^{3}[/latex] + z[latex]^{3}[/latex] – 3xyz = ?

A. 1000 B. 9000 C. 7000 D. 65000

Answer: Option C
Explanation: ∵ ( x[latex]^{3}[/latex] + y[latex]^{3}[/latex] + z[latex]^{3}[/latex] – 3xyz) = (x + y + z) (x[latex]^{2}[/latex] + y[latex]^{2}[/latex]+ z[latex]^{2}[/latex] – xy – yz – zx) = (x + y + z) × [latex]\frac{1}{2}[/latex] [(x – y)[latex]^{2}[/latex] + (y – z)[latex]^{2}[/latex] + (z – x)[latex]^{2}[/latex]] On putting the values, we get = 1000 × [latex]\frac{1}{2}[/latex] [(332 – 333)[latex]^{2}[/latex] + (333 – 335)[latex]^{2}[/latex] + (335 – 332)[latex]^{2}[/latex]] = 1000 × [latex]\frac{1}{2}[/latex] [1 + 4 + 9] = 7000
4. If [latex]\frac{x + 1}{x}[/latex] = 5, then [latex]\frac{2x}{3x^{2} – 5x + 3}[/latex] is equal to

A. 5 B. [latex]\frac{1}{5}[/latex] C. 3 D. [latex]\frac{1}{3}[/latex]

Answer: Option B
Explanation: x + [latex]\frac{1}{x}[/latex] = 5 ⇒ x[latex]^{2}[/latex] – 5x + 1 = 0 Equation multiplied by 3, then ⇒ 3x[latex]^{2}[/latex] – 15x + 3 = 0 ⇒ 3x[latex]^{2}[/latex] + 3 = 15x ∴ [latex]\frac{2x}{3x^{2} – 5x + 3}[/latex] = [latex]\frac{2x}{15x – 5x}[/latex] = [latex]\frac{2x}{10x}[/latex] = [latex]\frac{1}{5}[/latex]
5. If ab + bc + ca = 0, then the value of [latex]\frac{1}{a^{2}– bc }[/latex] + [latex]\frac{1}{b^{2} – ac}[/latex] + [latex]\frac{1}{c^{2} – ab}[/latex] is

A. 2 B. –1 C. 0 D. 1

Answer: Option C
Explanation: ab + bc + ca = 0 ⇒ ab + ca = –bc ∴ a[latex]^{2}[/latex] – bc = a[latex]^{2}[/latex] + ab + ca = a(a + b + c) Similarly, b[latex]^{2}[/latex] – ac = b(a + b + c) c[latex]^{2}[/latex] – ac = c(a + b + c) ∴ [latex]\frac{1}{a2 – bc }[/latex] + [latex]\frac{1}{b2 – ac}[/latex] + [latex]\frac{1}{c2 – ab}[/latex] = [latex]\frac{1}{a(a + b + c)}[/latex] + [latex]\frac{1}{b(a + b + c)}[/latex] + [latex]\frac{1}{c(a + b + c)}[/latex] = [latex]\frac{bc + ac + ab}{abc(a + b + c)}[/latex] = 0

1. Evaluate: (sin [latex]\frac{π}{6}[/latex] + cos [latex]\frac{π}{3}[/latex] – tan [latex]\frac{3π}{4}[/latex] + cosec [latex]\frac{2π}{2}[/latex])

A. 1 B. 0 C. 3 D. 5

Answer: Option A
Explanation: We know that: sin [latex]\frac{π}{6}[/latex] = sin 30° = [latex]\frac{1}{2}[/latex] , cos [latex]\frac{π}{3}[/latex] = cos 60° = [latex]\frac{1}{2}[/latex] ; tan [latex]\frac{π}{4}[/latex] = tan 45° = 1 and cosec [latex]\frac{π}{2}[/latex] = 1 ∴ sin [latex]\frac{π}{6}[/latex] + cos [latex]\frac{π}{3}[/latex] – tan [latex]\frac{3π}{4}[/latex] + cosec [latex]\frac{2π}{2}[/latex] = [latex]\frac{(1 + 1 – 13 + 12)}{2}[/latex] = [latex]\frac{1}{2}[/latex]
2. The least value of 9 cosec[latex]^{2}[/latex] A + 16 sin[latex]^{2}[/latex] A is

A. 7 B. 24 C. 25 D. 14

Answer: Option B
Explanation: 9 cosec[latex]^{2}[/latex] A + 16 sin[latex]^{2}[/latex]A = [latex]\frac{9}{sin^{2} A}[/latex] + 16 sin[latex]^{2}[/latex]A = ([latex]\frac{3}{sinA}[/latex])[latex]^{2}[/latex] + (4 sin A)[latex]^{2}[/latex] [ ∵ a[latex]^{2}[/latex] + b[latex]^{2}[/latex] = (a –b)[latex]^{2}[/latex] + 2ab] Let a = [latex]\frac{3}{sin A}[/latex] , b = 4 sin A = ([latex]\frac{3 – 4 sin A}{sin A}[/latex])[latex]^{2}[/latex] + [latex]\frac{2 × 3 × 4 sinA}{sin A}[/latex] = ([latex]\frac{3 – 4 sin A}{sin A}[/latex])[latex]^{2}[/latex] + 24 For the least value of ([latex]\frac{3 – 4 sin A}{sin A}[/latex]) should be 0. ∴ The least value will be 24.
3. If 2sin 2Θ – 3 = 0, then the value of Θ lies between

A. 0° < Θ < [latex]\frac{π}{2}[/latex] B. 0° < Θ ≥ [latex]\frac{π}{2}[/latex] C. π ≥ Θ > [latex]\frac{π}{2}[/latex] D. π < Θ ≥ 2π

Answer: Option A
Explanation: ∵ 2sin 2Θ – 3 = 0 or, sin2Θ = 3 (= sin 60°) 2 or, 2Θ = 60° ∴ Θ = 30° Hence, Θ lies between 0° < Θ < [latex]\frac{π}{2}[/latex]
4. The value of [latex]\frac{sin A}{1 + cos A}[/latex] + [latex]\frac{sin A}{1 – cos A}[/latex] is (0° < A < 90°)

A. 2 cosec A B. 2 sec A C. 2 sin A D. 2 cos A

Answer: Option A
Explanation: [latex]\frac{sin A}{1 + cos A}[/latex] + [latex]\frac{sin A}{1 – cos A}[/latex] = [latex]\frac{sin A (1 – cos A)}{(1 + cos A)}[/latex] + [latex]\frac{sin A (1 + cos A)}{ (1 – cos A)}[/latex] = [latex]\frac{sin A – sin A cos A + sin A + sin A cos A}{1 – cos^{2}A}[/latex] = [latex]\frac{2 sin A}{sin^{2} A}[/latex] [∵ 1 – cos[latex]^{2}[/latex] A = sin[latex]^{2}[/latex] A] = [latex]\frac{2}{sin A}[/latex] = 2 cosec A [ ∵ [latex]\frac{1}{sinA}[/latex] = cosec A]
5. If r sin Θ = 1, r cos Θ = √3, then the value of ( √3 tan Θ + 1) is

A. √3 B. [latex]\frac{1}{√3}[/latex] C. 1 D. 2

Answer: Option D
Explanation: Given, r sin Θ = 1 ...(i) r cos Θ = √3 ...(ii) Eq. (i) ÷ (ii), [latex]\frac{r sin Θ}{r cos Θ }[/latex] = [latex]\frac{1}{√3}[/latex] ⇒ tan Θ = [latex]\frac{1}{√3}[/latex] Now, √3 tan Θ + 1 = √3 × [latex]\frac{1}{√3}[/latex] + 1 = 1 + 1 = 2