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
91
In how many ways can a team of 2 members be selected from 6 people?
Answer:
15
Step 1: Selecting a team does not involve order, so use combinations. Step 2: We need to choose 2 from 6, which is 6C2. Step 3: 6C2 = (6 × 5) / (2 × 1) = 15 ways.
92
If 10Cx = 10C3, what is a possible value of x other than 3?
Answer:
7
Step 1: By the property of combinations, nCx = nCy implies either x = y or x + y = n. Step 2: Here, n = 10 and y = 3. Step 3: Since x is not 3, x + 3 = 10, which means x = 7.
93
What is the value of nCn?
Answer:
1
Step 1: Use the combination formula nCr = n! / (r!(n - r)!). Step 2: Substitute r = n: n! / (n!(n - n)!) = n! / (n! × 0!). Step 3: Since 0! = 1, it simplifies to n! / n! = 1.
94
Calculate the value of 10C8.
Answer:
45
Step 1: Use the property nCr = nC(n - r). Step 2: 10C8 = 10C(10 - 8) = 10C2. Step 3: 10C2 = (10 × 9) / (2 × 1) = 90 / 2 = 45.
95
Evaluate 5C3 (Combinations of 5 objects taken 3 at a time).
Answer:
10
Step 1: The formula for nCr is n! / (r!(n - r)!). Step 2: Substitute n = 5 and r = 3: 5! / (3! × 2!). Step 3: (5 × 4 × 3!) / (3! × 2 × 1) = 20 / 2 = 10.
96
In how many ways can 6 people be seated on 3 chairs?
Answer:
120
Step 1: Seating people implies arrangement, so use permutations. Step 2: We need to arrange 3 people out of 6, which is 6P3. Step 3: 6P3 = 6 × 5 × 4 = 120 ways.
97
What is the relation between nPr and nCr?
Answer:
nPr = nCr × r!
Step 1: nCr = n! / (r!(n - r)!). Step 2: nPr = n! / (n - r)!. Step 3: Multiply nCr by r!: [n! / (r!(n - r)!)] × r! = n! / (n - r)! = nPr. So, nPr = nCr × r!.
98
Find the value of nP0.
Answer:
1
Step 1: Use the permutation formula: nPr = n! / (n - r)!. Step 2: Substitute r = 0: n! / (n - 0)! = n! / n!. Step 3: Any non-zero number divided by itself is 1. Thus, nP0 = 1.
99
How many different words can be formed using all the letters of the word 'CAT'?
Answer:
6
Step 1: The word 'CAT' has 3 distinct letters. Step 2: Number of arrangements of n distinct items is n!. Step 3: 3! = 3 × 2 × 1 = 6 words.
100
In how many ways can 3 books be selected and arranged on a shelf out of 8 different books?
Answer:
336
Step 1: Since order matters for arranging on a shelf, use permutations. Step 2: Find 8P3 = 8! / (8 - 3)! = 8! / 5!. Step 3: 8 × 7 × 6 = 336 ways.