Maths MCQs for AD (ASF) – Series 10 (Q131-140)
Q131. In how many ways can the letters of the word ‘LEADER’ be arranged?
- A) 360
- B) 180
- C) 720
- D) None of these
Answer: A
LEADER has 6 letters with E repeated twice; arrangements = 6!/2! = 360.
Q132. How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 without repetition?
- A) 50
- B) 60
- C) 120
- D) None of these
Answer: B
Number of ways = 5P3 = 5×4×3 = 60.
Q133. In how many ways can a committee of 3 men and 2 women be selected from 6 men and 4 women?
- A) 120
- B) 140
- C) 100
- D) None of these
Answer: A
Number of ways = 6C3 × 4C2 = 20×6 = 120.
Q134. A bag contains 5 red and 4 blue balls. Two balls are drawn at random. Find the probability that both are red.
- A) 5/36
- B) 5/18
- C) 5/9
- D) None of these
Answer: B
Total ways = 9C2 = 36; favourable = 5C2 = 10; probability = 10/36 = 5/18.
Q135. A coin is tossed 3 times. Find the probability of getting exactly 2 heads.
- A) 3/8
- B) 1/2
- C) 1/4
- D) None of these
Answer: A
Total outcomes = 8; favourable outcomes (HHT, HTH, THH) = 3; probability = 3/8.
Q136. From a pack of 52 cards, one card is drawn at random. Find the probability that it is either a king or a queen.
- A) 2/13
- B) 1/13
- C) 4/13
- D) None of these
Answer: A
Kings + Queens = 8 cards; probability = 8/52 = 2/13.
Q137. How many different words can be formed with the letters of the word ‘OFFICE’?
- A) 720
- B) 180
- C) 360
- D) None of these
Answer: C
OFFICE has 6 letters with F repeated twice; arrangements = 6!/2! = 360.
Q138. In how many ways can 5 different books be arranged on a shelf?
- A) 100
- B) 60
- C) 120
- D) None of these
Answer: C
Number of arrangements = 5! = 120.
Q139. Two dice are thrown together. Find the probability that the sum of the numbers on them is 7.
- A) 1/9
- B) 1/12
- C) 1/6
- D) None of these
Answer: C
Favourable outcomes = 6; total outcomes = 36; probability = 6/36 = 1/6.
Q140. A committee of 4 is to be selected from 5 boys and 6 girls such that it has at least one boy. Find the number of ways.
- A) 300
- B) 320
- C) 315
- D) None of these
Answer: C
Total ways = 11C4 = 330; ways with no boy = 6C4 = 15; at least one boy = 330−15 = 315.
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