1. For what value of ‘y’, [latex]{x}^{2}[/latex] + 112x +[latex]{y}^{2}[/latex] is a perfect square?

A. [latex]\frac{1}{24}[/latex] B. [latex]\frac{1}{12}[/latex] C. [latex]\frac{1}{6}[/latex] D. [latex]\frac{1}{3}[/latex]

Answer: Option A
Explanation: [latex]{x}^{2}[/latex] + 112x +[latex]{y}^{2}[/latex] = [latex]{x}^{2}[/latex] + 2(124)x + [latex]{y}^{2}[/latex] By, [latex]{(a + b)}^{2}[/latex] = [latex]{a}^{2}[/latex] + 2ab + [latex]{b}^{2}[/latex] Here, a = x and b = y = [latex]\frac{1}{24}[/latex] So, for [latex]{x}^{2}[/latex] + 112x +[latex]{y}^{2}[/latex] to be perfect square y = [latex]\frac{1}{24}[/latex]
2. If x and y are natural numbers such that x + y = 2017, then what is the value of [latex]{(-1)}^{x}[/latex] + [latex]{(-1)}^{y}[/latex]?

A. 2 B. -2 C. 0 D. 1

Answer: Option C
Explanation: Sum of two numbers can be odd only if one of them is odd and other one is even. Let x be odd and y be even ⇒ [latex]{(-1)}^{x}[/latex] + [latex]{(-1)}^{y}[/latex] = -1 + 1 = 0 ∴ The value is 0
3. A root of equation a[latex]{x}^{2}[/latex] + bx + c = 0 (where a, b and c are rational numbers) is 5 + 3√3. What is the value of [latex]\frac{({a}^{2}+ {b}^{2} + {c}^{2})}{(a + b + c)}[/latex]?

A. [latex]\frac{35}{3}[/latex] B. [latex]\frac{37}{3}[/latex] C. [latex]\frac{-105}{11}[/latex] D. [latex]\frac{-105}{13}[/latex]

Answer: Option D
Explanation: Since one root of the equation is irrational, so another root will be 5 – 3√3 According to the problem statement, ⇒ [x – (5 + 3√3)] [x – (5 – 3√3)] = 0 ⇒[latex]{x}^{2}[/latex] – 10x – 2 = 0 ⇒ a = 1, b = -10 and c = -2, Put this in the required equations ∴ [latex]\frac{({a}^{2}+ {b}^{2} + {c}^{2})}{(a + b + c)}[/latex] = [latex]\frac{(1 + 100 + 4)}{ (1 – 10 – 2)}[/latex] = [latex]\frac{-105}{11}[/latex]
4. If the arithmetic mean of two numbers is 7 and the geometric mean of the same two number is 2√10. Then find the numbers x and y respectively, such that x > y.

A. 4,10 B. 2,5 C. 5,2 D. 10,4

Answer: Option D
Explanation: Given, Arithmetic Mean = 7 ⇒ [latex]\frac{(x + y)}{2}[/latex] = 7 ⇒ x + y = 14 ----(1) And, Geometric Mean = 2√10 ⇒ √xy = 2√10 ⇒ xy = [latex]{(2√10)}^{2}[/latex] = 40 Now, ⇒ [latex]{(x - y)}^{2}[/latex] = [latex]{(x + y)}^{2}[/latex] - 4xy = 142 - (4 × 40) = 196 - 160 = 36 ⇒ x - y = 6 ----(2) Now, adding equation (1) and (2) we get ⇒ 2x = 20 ⇒ x = 10 Now, by equation (1) ⇒ y = 14 - x = 14 - 10 = 4
5. Cost of 8 pencils, 5 pens and 3 erasers is Rs. 111. Cost of 9 pencils, 6 pens and 5 erasers is Rs. 130. Cost of 16 pencils, 11 pens and 3 erasers is Rs. 221. What is the cost (in Rs) of 39 pencils, 26 pens and 13 erasers?

A. 316 B. 546 C. 624 D. 482

Answer: Option
Explanation: Let the price of single pencil, pen, and eraser be x, y, and z respectively According to question, 8x + 5y + 3z = Rs. 111 ----(1) 9x + 6y + 5z = Rs. 130 ----(2) 16x + 11y + 3z = Rs. 221 ----(3) Subtracting equation (1) from (3) ⇒ (16x + 11y + 3z) - (9x + 6y + 5z) = 221 - 111 ⇒ 8x + 6y = 110 ⇒ 4x + 3y = 55 ----(4) Multiply the equation (2) by 3 and (3) by 5 and then subtracting equation (2) from (3) ⇒ (16x + 11y + 3z) × 5 - (9x + 6y + 5z) × 3 = 221 × 5 - 130 × 3 ⇒ 80x + 55y + 15z - 27x - 18y - 15z = 1105 - 390 ⇒ 53x + 37y = 715 ----(5) Multiply the equation (4) by 53 and (5) by 4 and then subtracting equation (4) from (5) ⇒ 212x + 159y - 212x - 148y = 2915 - 2860 ⇒ 11y = 55 ⇒ y = 5 By putting the value of y = 5 in equation (4) ⇒ 4x + 3 × 5 = 55 ⇒ x = 10 By putting the value of y = 5 and x = 10 in equation (1) ⇒ 8 × 10 + 5 × 5 + 3z = 111 ⇒ 80 + 25 + 3z = 111 ⇒ z = 2 ∴ Cost of 39 pencils, 26 pens and 13 erasers is 39x + 26y + 13z =39 × 10 + 26 × 5 + 13 × 2 = Rs. 546

1. If a + b + c = 9, ab + bc + ca = 26, [latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] = 91, [latex]{b}^{3}[/latex] + [latex]{c}^{3}[/latex] = 72 and [latex]{c}^{3}[/latex] + [latex]{a}^{3}[/latex] = 35, then what is the value of abc?

A. 48 B. 24 C. 36 D. 42

Answer: Option B
Explanation: Using algebraic identities, [latex]{(a + b + c)}^{2}[/latex] = [latex]{a}^{2}[/latex] + [latex]{b}^{2}[/latex] + [latex]{c}^{2}[/latex] + 2ab + 2bc + 2ca By putting the respective values given in question, ⇒ [latex]{(9)}^{2}[/latex] = [latex]{a}^{2}[/latex] + [latex]{b}^{2}[/latex] + [latex]{c}^{2}[/latex] + 2(ab + bc + ca) [∵ ab + bc + ca = 26] ⇒ [latex]{(9)}^{2}[/latex] = [latex]{a}^{2}[/latex] + [latex]{b}^{2}[/latex] + [latex]{c}^{2}[/latex] + 2(26) ⇒ [latex]{a}^{2}[/latex] + [latex]{b}^{2}[/latex] + [latex]{c}^{2}[/latex] = 81 - 52 = 29 Given equations, [latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] = 91 ----(1) [latex]{b}^{3}[/latex] + [latex]{c}^{3}[/latex] = 72 ----(2) [latex]{c}^{3}[/latex] + [latex]{a}^{3}[/latex] = 35 ----(3) On adding (1), (2) and (3) [latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] + [latex]{b}^{3}[/latex] + [latex]{c}^{2}[/latex] + [latex]{c}^{2}[/latex] + [latex]{a}^{3}[/latex] = 91 + 72 + 35 ⇒ 2([latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] + [latex]{c}^{3}[/latex]) = 198 ⇒ [latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] + [latex]{c}^{3}[/latex] = 99 Using algebraic identities, [latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] + [latex]{c}^{3}[/latex] - 3abc = (a + b + c) ([latex]{a}^{2}[/latex] + [latex]{b}^{2}[/latex] + [latex]{c}^{2}[/latex] - ab - bc - ca) By putting the respective values, ⇒ 99 - 3abc = 9 (29 - 26) [∵ ab + bc + ca = 26 and a + b + c = 9] ⇒ 3abc = 99 - 27 ⇒ abc = [latex]\frac{72}{3}[/latex] ∴ abc = 24
2. If [latex]\frac{x}{a}[/latex] + [latex]\frac{y}{b}[/latex] = 3 and [latex]\frac{x}{b}[/latex] – [latex]\frac{y}{a}[/latex] = 9, then what is the value of [latex]\frac{x}{y}[/latex]?

A. [latex]\frac{(b + 3a)}{(a – 3b)}[/latex] B.[latex]\frac{(a + 3b)}{(b – 3a)}[/latex] C. [latex]\frac{(1 + 3a)}{ (a + 3b)}[/latex] D.[latex]\frac{(a + 3{{b}^{2}})}{(b – 3{{a}^{2}})}[/latex]

Answer: Option
Explanation: Here[latex]\frac{x}{a}[/latex] + [latex]\frac{y}{b}[/latex] = 3 ⇒ bx + ay – 3ab = 0 ---- (1) Here[latex]\frac{x}{b}[/latex] – [latex]\frac{y}{a}[/latex] = 9 ⇒ ax – by – 9ab = 0 ---- (2) Multiplying equation 1 by ‘a’ and equation 2 by ‘b’ ⇒ abx + [latex]{a}^{2}[/latex]y – 3[latex]{a}^{2}[/latex]b = 0 ---- (3) ⇒ abx – [latex]{b}^{2}[/latex]y – 9a[latex]{b}^{2}[/latex] = 0 ---- (4) Subtracting equation 4 from 3, ⇒ y =[latex]\frac{3ab(a – 3b)}{({a}^{2} + {b}^{2})}[/latex], substituting value of y in equation 3, ⇒ x =[latex]\frac{ 3ab(b + 3a)}{({a}^{2}+ {b}^{2})}[/latex] ⇒ [latex]\frac{x}{y}[/latex] = [latex]\frac{(b + 3a)}{(a – 3b)}[/latex] ∴ The value of [latex]\frac{x}{y}[/latex] = [latex]\frac{(b + 3a)}{(a – 3b)}[/latex]
3. The value of x for which the expressions 19 - 5x and 19x + 5 become equal is ______.

A. [latex]\frac{7}{12}[/latex] B. [latex]\frac{-7}{12}[/latex] C. [latex]\frac{12}{7}[/latex] D. [latex]\frac{-12}{7}[/latex]

Answer: Option A
Explanation: According to the given information, 19 – 5x = 19x + 5 ⇒ 19 – 5 = 19x + 5x ⇒ 14 = 24x ∴ x = [latex]\frac{14}{24}[/latex] = [latex]\frac{7}{12}[/latex]
4. What is the sum of all the natural numbers from 1 to 35?

A. 620 B. 630 C. 650 D. 680

Answer: Option B
Explanation: We know that Formula for Sum of natural no's beginning from 1 = [latex]{n}^{2}[/latex] × (1+n) where n = last no. i.e. 35 ⇒ Sum = 352 × (1 + 35) ⇒ Sum = 35 × 18 = 630
5. The sum of three consecutive odd natural numbers is 93. The smallest of three numbers is:

A. 29 B. 31 C. 23 D. 27

Answer: Option A
Explanation: Let the three consecutive odd numbers be a, a + 2, a + 4. Given, the sum of three consecutive odd natural numbers is 93. ∴ a + a + 2 + a + 4 = 93 ⇒ 3a = 87 ⇒ a = 29

1. The price of 10 chairs is equal to that of 4 tables. The price of 15 chairs and 3 tables together is Rs. 4500. The total price of 12 chairs and 5 tables is:

A. Rs. 4750 B. Rs. 4840 C. Rs. 4500 D. Rs. 4900

Answer: Option D
Explanation: Let the price of chair be ‘a’. Given, the price of 10 chairs is equal to that of 4 tables. ∴ 10 × a = 4 × price of the table ⇒ Price of table = 2.5a Given, the price of 15 chairs and 3 tables together is Rs. 4500 ∴ 15a + 7.5a = 4500 ⇒ 22.5a = 4500 ⇒ a = Rs. 200 Price of table = 2.5a = Rs. 500 Total price of 12 chairs and 5 tables = 12 × 200 + 5 × 500 = Rs. 4900
2. A number is greater than twice its reciprocal by [latex]\frac{31}{4}[/latex]. Find the number.

A. 7 B. 8 C. 9 D. 6

Answer: Option B
Explanation: Let the number be ‘x’. A number is greater than twice its reciprocal by [latex]\frac{31}{4}[/latex] ⇒ x = 2([latex]\frac{1}{x}[/latex]) + [latex]\frac{31}{4}[/latex] ⇒ x - [latex]\frac{2}{x}[/latex]= [latex]\frac{31}{4}[/latex] ⇒ 4[latex]{(x -2)}^{2}[/latex] = 31x ⇒ 4[latex]{x}^{2}[/latex] - 31x - 8 = 0 ⇒ 4[latex]{x}^{2}[/latex] - 32x + x - 8 = 0 ⇒ 4x(x - 8) + 1(x - 8) = 0 ⇒ (x - 8)(4x + 1) = 0 x = 8 or x = - [latex]\frac{1}{4}[/latex] ∴ x = 8
3. If x = 2 then the value of [latex]{x}^{3}[/latex] + 27[latex]{x}^{2}[/latex] + 243x + 631 is:

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

Answer: Option C
Explanation: Given equation, f(x) = [latex]{x}^{3}[/latex] + 27[latex]{x}^{2}[/latex] + 243x + 631 ⇒ x([latex]{x}^{2}[/latex] + 27x+ 243) + 631 Now, put the value of x = 2 ⇒ 2(22 + 27 × 2 + 243) + 631 ⇒ 2 (4 + 54 + 243) + 631 ⇒ 2(301) + 631 = 602 + 631 = 1233.
4. If a = 2.234, b = 3.121 and c = –5.355, then the value of [latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] + [latex]{c}^{3}[/latex] – 3abc is

A. -1 B. 0 C. 1 D. 2

Answer: Option B
Explanation: a + b + c = 2.234 + 3.121 – 5.355 = 0 If a + b + c = 0, then[latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] + [latex]{c}^{3}[/latex] – 3abc = 0, which can be proved as under a + b = – c Cubing both sides, we get ⇒ [latex]{(a + b)}^{3}[/latex]= [latex]{(-c)}^{3}[/latex] ⇒ [latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] + 3ab(a + b) = – [latex]{c}^{3}[/latex] ⇒ [latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] + 3ab(– c) = – [latex]{c}^{3}[/latex] ⇒ [latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] – 3abc = – [latex]{c}^{3}[/latex] ⇒ [latex]{a}^{3}[/latex] + [latex]{b}^{3}[/latex] + [latex]{c}^{3}[/latex] – 3abc = 0
5. If [latex]{x}^{2}[/latex] + [latex]{y}^{2}[/latex] + 1 = 2x, then the value of [latex]{x}^{3}[/latex] + [latex]{y}^{5}[/latex] is

A. 2 B. 0 C. -1 D. 1

Answer: Option D
Explanation: [latex]{x}^{2}[/latex] + [latex]{y}^{2}[/latex] + 1 = 2x ⇒ [latex]{x}^{2}[/latex] + [latex]{y}^{2}[/latex] + 1 – 2x = 0 ⇒ [latex]{x}^{2}[/latex] – 2x + 1 + [latex]{y}^{2}[/latex] = 0 ⇒ [latex]{(x - 1)}^{2}[/latex] + [latex]{y}^{2}[/latex] = 0 In the above eq. the L.H.S. can only become zero when the base of terms; (x – 1) and y becomes zero because for any other value the sum of their squares will always be a positive integer. Taking (x – 1) = 0 and y = 0 Therefore, x = 1 and y = 0 ∴ [latex]{x}^{3}[/latex] + [latex]{y}^{5}[/latex] = 1 + 0 = 1.