IIT JEE Practice Question and Answer

Q: Three persons A, B and C are standing in a queue. There are five persons between A and B and eight persons between B and C. If there be three persons ahead of C and 21 persons behind A, what could be the minimum number of persons in the queue? 5797 0

  • 1
    41
    Correct
    Wrong
  • 2
    40
    Correct
    Wrong
  • 3
    28
    Correct
    Wrong
  • 4
    27
    Correct
    Wrong
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Answer : 3. "28"
Explanation :

Answer: C) 28 Explanation: Three persons A, B, C can be arranged in a queue in six different ways, ie ABC, CBA, BAC, CAB, BCA, ACB. But since there are only 3 persons ahead of C, so C should be in front of the queue. Thus, there are only two possible arrangements, ie CBA and CAB.   We may consider the two cases as under: Case I:  ←3C↔8B↔5A→21   Clearly, number of persons in the queue = (3+1+8+1+5+1+21=) 40   Case II:  ←3C A↔5B   Number of persons between A and C                     = (8 - 6) = 2    ∵[C↔8B A→21B]   Clearly number of persons in the queue = (3+1+2+1+21) = 28   Now, 28 < 40. So, 28 is the minimum number of persons in the queue.

Q: Boy, man, uncle, fireman, grandfather are... 5783 0

  • 1
    Feminine titles
    Correct
    Wrong
  • 2
    Masculine titles
    Correct
    Wrong
  • 3
    Both Feminine and Masculine titles
    Correct
    Wrong
  • 4
    None of the above
    Correct
    Wrong
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Answer : 2. "Masculine titles"
Explanation :

Answer: B) Masculine titles Explanation: As they are the titles of male gender, they are called as Masculine titles.

Q: In a certain coding language, iF GO = 32 & SHE = 49 then SOME will be equal to ? 5736 0

  • 1
    56
    Correct
    Wrong
  • 2
    58
    Correct
    Wrong
  • 3
    62
    Correct
    Wrong
  • 4
    64
    Correct
    Wrong
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Answer : 1. "56"
Explanation :

Answer: A) 56 Explanation: If we count letters till the end of the alphabet including that letter in reverse order then Here the values are nothing but the values corresponding to the alphabets when taken in a reverse order,i.e,A=26 ,B=25...,Z=1G = 20 and O = 12 totaling 32S = 8, H = 19 and E = 22 totaling 49 So SOME = 8 + 12 + 14 + 22 = 56.

Q: 2 is to 55 as 6 is to 5733 0

  • 1
    99
    Correct
    Wrong
  • 2
    44
    Correct
    Wrong
  • 3
    9
    Correct
    Wrong
  • 4
    33
    Correct
    Wrong
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Answer : 1. "99"
Explanation :

Answer: A) 99 Explanation: Given 2 is to 55 2 + 3 = 5  5 x 11 = 55 Similarly, 6 6 + 3 = 9 9 x 11 = 99   Hence, 2 is to 55 as 6 is to 99.

Q: What is the number of digits in 333? Given that log3 = 0.47712? 5704 0

  • 1
    12
    Correct
    Wrong
  • 2
    13
    Correct
    Wrong
  • 3
    14
    Correct
    Wrong
  • 4
    15
    Correct
    Wrong
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Answer : 2. "13"
Explanation :

Answer: B) 13 Explanation:  Let   Let x=333 = 333    Then, logx = 33 log3     = 27 x 0.47712 = 12.88224    Since the characteristic in the resultant value of log x is 12   ∴The number of digits in x is (12 + 1) = 13    Hence the required number of digits in 333is 13.

Q: Convent : Cloister :: Eyrie : ? 5703 0

  • 1
    Show
    Correct
    Wrong
  • 2
    Open
    Correct
    Wrong
  • 3
    Hide
    Correct
    Wrong
  • 4
    Nest
    Correct
    Wrong
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Answer : 4. "Nest"
Explanation :

Answer: D) Nest Explanation: Here the first two words are synonyms. Similarly, Eyrie means a large nest. So, the synonym of Eyrie is nest.   Hence, Convent : Cloister :: Eyrie : Nest.

Q: Amar can do a piece of work in 10 days. He works at it for 4 days and then Arun finishes it in 9 days. In how many days can Amar and Arun together finish the work ? 5696 0

  • 1
    4 days
    Correct
    Wrong
  • 2
    8 days
    Correct
    Wrong
  • 3
    3 days
    Correct
    Wrong
  • 4
    6 days
    Correct
    Wrong
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Answer : 4. "6 days"
Explanation :

Answer: D) 6 days Explanation: 4/10 + 9/x = 1     => x = 15    Then both can do in  1/10 + 1/15 = 1/6     => 6 days

Q: A group of 10 representatives is to be selected out of 12 seniors and 10 juniors. In how many different ways can the group be selected if it should have at least one senior ? 5674 0

  • 1
    ²²C₁₀ + 1
    Correct
    Wrong
  • 2
    ²²C₉ + ¹⁰C₁
    Correct
    Wrong
  • 3
    ²²C₁₀
    Correct
    Wrong
  • 4
    ²²C₁₀ - 1
    Correct
    Wrong
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Answer : 4. "²²C₁₀ - 1"
Explanation :

Answer: D) ²²C₁₀ - 1 Explanation: The total number of ways of forming the group of ten representatives is ²²C₁₀. The total number of ways of forming the group that consists of no seniors is ¹⁰C₁₀ = 1 way The required number of ways = ²²C₁₀ - 1

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