Mathematics/General Ability MCQs
Topic Notes: Mathematics/General Ability
MCQs and preparation resources for competitive exams, covering important concepts, past papers, and detailed explanations.
Plato
- Biography: Ancient Greek philosopher (427–347 BCE), student of Socrates and teacher of Aristotle, founder of the Academy in Athens.
- Important Ideas:
- Theory of Forms
- Philosopher-King
- Ideal State
61
If the square of a number is equal to 5 times the number minus 6, what is the number?
Answer:
2 or 3
The mathematical equation is x^2 = 5x - 6, which rearranges to x^2 - 5x + 6 = 0. Factoring gives (x-2)(x-3) = 0, so the possible numbers are 2 or 3.
62
The product of two consecutive odd integers is 35. What are the positive integers?
Answer:
5 and 7
Let the integers be x and x+2. x(x+2) = 35, or x^2 + 2x - 35 = 0. Factoring yields (x+7)(x-5) = 0. The positive root is x = 5. So the integers are 5 and 7.
63
A positive number and its square sum to 42. Find the number.
Answer:
6
Let the number be x. The equation is x^2 + x = 42, or x^2 + x - 42 = 0. Factoring gives (x+7)(x-6) = 0. Since the number is positive, x = 6.
64
The area of a rectangle is 24 sq units, and its length is 2 units more than its width. Find the width.
Answer:
4 units
Let the width be w. The length is w + 2. Area = w(w + 2) = 24. This gives w^2 + 2w - 24 = 0. Factoring yields (w+6)(w-4) = 0. Since width must be positive, w = 4 units.
65
The sum of the squares of two consecutive natural numbers is 25. The numbers are:
Answer:
3 and 4
Let the numbers be x and x+1. Then x^2 + (x+1)^2 = 25. This expands to 2x^2 + 2x + 1 = 25, or 2x^2 + 2x - 24 = 0. Dividing by 2 gives x^2 + x - 12 = 0. The factors are (x+4)(x-3) = 0. The natural number is 3, making the numbers 3 and 4.
66
The product of two consecutive positive integers is 30. What are the integers?
Answer:
5 and 6
Let the integers be x and x+1. The equation is x(x+1) = 30, or x^2 + x - 30 = 0. Factoring gives (x+6)(x-5) = 0. Since they are positive integers, x must be 5. The integers are 5 and 6.
67
Find the quadratic equation whose roots are -3 and -4.
Answer:
x^2 + 7x + 12 = 0
The sum of the roots is (-3) + (-4) = -7. The product is (-3) * (-4) = 12. Using x^2 - (sum)x + product = 0, we get x^2 - (-7)x + 12 = 0, which is x^2 + 7x + 12 = 0.
68
What is the quadratic equation with a sum of roots equal to 7 and a product equal to 12?
Answer:
x^2 - 7x + 12 = 0
Given Sum (S) = 7 and Product (P) = 12. Substitute these into the formula x^2 - Sx + P = 0 to get x^2 - 7x + 12 = 0.
69
If the roots of an equation are 2 + sqrt(3) and 2 - sqrt(3), what is the equation?
Answer:
x^2 - 4x + 1 = 0
Sum = (2 + sqrt(3)) + (2 - sqrt(3)) = 4. Product = (2 + sqrt(3))(2 - sqrt(3)) = 2^2 - (sqrt(3))^2 = 4 - 3 = 1. The equation is x^2 - Sx + P = 0, which becomes x^2 - 4x + 1 = 0.
70
Find the equation whose roots are 1/2 and 2.
Answer:
2x^2 - 5x + 2 = 0
Sum of roots = 1/2 + 2 = 5/2. Product of roots = 1/2 * 2 = 1. Equation is x^2 - (5/2)x + 1 = 0. Multiplying the entire equation by 2 to clear the fraction yields 2x^2 - 5x + 2 = 0.