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
41
Multiply: sqrt(6) * sqrt(6)
Answer:
6
Multiplying a square root by itself cancels out the radical, effectively squaring the square root. sqrt(6) * sqrt(6) is sqrt(36), which immediately simplifies to 6.
42
Calculate: sqrt(2) * sqrt(8)
Answer:
4
Use the multiplication property of radicals to combine them into one root: sqrt(2 * 8) = sqrt(16). The square root of 16 perfectly evaluates to 4.
43
Evaluate: sqrt(5) * sqrt(20)
Answer:
10
Multiply the numbers inside the roots to get sqrt(100). The principal square root of 100 is 10, completing the simplification.
44
Multiply the radicals: sqrt(3) * sqrt(12)
Answer:
6
When multiplying square roots, multiply the numbers inside the radicals together. This gives sqrt(36), and since 36 is a perfect square, the square root is 6.
45
Subtract: sqrt(18) - sqrt(8)
Answer:
sqrt(2)
Simplify both terms first: sqrt(18) = 3*sqrt(2) and sqrt(8) = 2*sqrt(2). Subtracting them gives 3*sqrt(2) - 2*sqrt(2) = 1*sqrt(2), which is just sqrt(2).
46
Add the radicals: sqrt(12) + sqrt(27)
Answer:
5*sqrt(3)
First simplify each radical: sqrt(12) is 2*sqrt(3) and sqrt(27) is 3*sqrt(3). Now that they have the same radicand, add them: 2*sqrt(3) + 3*sqrt(3) = 5*sqrt(3).
47
Evaluate: 7*sqrt(7) - sqrt(7)
Answer:
6*sqrt(7)
The second term has an implicit coefficient of 1. Subtracting 1 from 7 gives 6. The radical part remains the same, producing 6*sqrt(7).
48
Add: 2*sqrt(5) + 4*sqrt(5)
Answer:
6*sqrt(5)
Since the radicands (number under the root) are the same, treat sqrt(5) as a variable. Add the coefficients 2 and 4 to get 6, giving a result of 6*sqrt(5).
49
Subtract: 5*sqrt(3) - 2*sqrt(3)
Answer:
3*sqrt(3)
Because both terms have sqrt(3), they are like radicals. Simply subtract the coefficients: 5 - 2 = 3, maintaining the radical to get 3*sqrt(3).
50
Add the radicals: sqrt(2) + 3*sqrt(2)
Answer:
4*sqrt(2)
You can add radicals just like like terms when the numbers inside the roots are identical. Adding an implicit 1*sqrt(2) to 3*sqrt(2) yields 4*sqrt(2).