IIT JEE Practice Question and Answer

Q: In a plant cell, DNA may be found in 2015 0

  • 1
    only in the nucleus and mitochondria.
    Correct
    Wrong
  • 2
    only in the nucleus and chloroplasts
    Correct
    Wrong
  • 3
    in the nucleus, mitochondria, and chloroplasts.
    Correct
    Wrong
  • 4
    only in the nucleus.
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 3. "in the nucleus, mitochondria, and chloroplasts."
Explanation :

Answer: C) in the nucleus, mitochondria, and chloroplasts. Explanation: All of the genetic information in a cell was initially thought to be confined to the DNA in the chromosomes of the cell nucleus. It is now known that small circular chromosomes, called extranuclear, or cytoplasmic, DNA, are located in two types of organelles found in the cytoplasm of the cell. These organelles are the mitochondria in animal and plant cells and the chloroplasts in plant cells. Chloroplast DNA (cpDNA) contains genes that are involved with aspects of photosynthesis and with components of the special protein-synthesizing apparatus that is active within the organelle. Mitochondrial DNA (mtDNA) contains some of the genes that participate in the conversion of the energy of chemical bonds into the energy currency of the cell—a chemical called adenosine triphosphate (ATP)—as well as genes for mitochondrial protein synthesis.

Q: 52 C in a P? 2014 0

  • Show AnswerHide Answer
  • Workspace

Answer :
Explanation :

52 C in a P represents 52 Cards in a Pack. Similar to  12 S of the Z  5 C in an OF

Q: A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red. 2014 0

  • 1
    23/42
    Correct
    Wrong
  • 2
    19/42
    Correct
    Wrong
  • 3
    7/32
    Correct
    Wrong
  • 4
    16/39
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 2. "19/42"
Explanation :

Answer: B) 19/42 Explanation: A red ball can be drawn in two mutually exclusive ways  (i) Selecting bag I and then drawing a red ball from it.   (ii) Selecting bag II and then drawing a red ball from it.   Let E1, E2 and A denote the events defined as follows: E1 = selecting bag I, E2 = selecting bag II A = drawing a red ball Since one of the two bags is selected randomly, therefore  P(E1) = 1/2  and  P(E2) = 1/2 Now, PAE1 = Probability of drawing a red ball when the first bag has been selected = 4/7   PAE2  = Probability of drawing a red ball when the second bag has been selected = 2/6  Using the law of total probability, we have   P(red ball) = P(A) = PE1×PAE1+PE2×PAE2                              = 12×47+12×26=1942

Q: What are the four main functions of a computer? 2013 0

  • 1
    Data, information, bits and bytes
    Correct
    Wrong
  • 2
    Hardware, software, modeling and operations
    Correct
    Wrong
  • 3
    Input, processing, output and storage
    Correct
    Wrong
  • 4
    Learning, thinking, intelligence and virtuosity
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 3. "Input, processing, output and storage"
Explanation :

Answer: C) Input, processing, output and storage Explanation: The four main functions of a computer are ::   1. Input, 2. Processing, 3. Output and 4. Storage

Q: The oxidation number of Cr in K2Cr2O7 is 2013 0

  • 1
    +6
    Correct
    Wrong
  • 2
    -5
    Correct
    Wrong
  • 3
    +3
    Correct
    Wrong
  • 4
    -4
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 1. "+6"
Explanation :

Answer: A) +6 Explanation: We will rewrite the formula as an equation. 2K + 2Cr + 7O = 0 The zero comes from there being no overall charge on the compound. Each oxidation number is represented by the element's symbol. Since this compound is not a peroxide or superoxide, the oxidation value for O is -2 K being in group 1 has a +1 charge and therefore has an oxidation number of K is +1 Now, by putting known values in we have: 2(+1) + 2(Cr) + 7(-2) = 0 2 + 2Cr -14 = 0 2Cr = +12 Cr = +6.

Q: If it is possible to make a number which is the square of a two-digit odd number using the second, the fourth and the sixth digits of the number  93217648 and using each only once, which of the following is that two-digit odd number? 2013 0

  • 1
    17
    Correct
    Wrong
  • 2
    19
    Correct
    Wrong
  • 3
    15
    Correct
    Wrong
  • 4
    21
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 2. "19"
Explanation :

Answer: B) 19 Explanation: Second, fourth and the sixth digits of the number are 3, 1 and 6. Using these digits only once we can make 361, which is a perfect square number. Hence, our answer is 19 x 19 = 361

Q: A person travels from K to L a speed of 50 km/hr and returns by increasing his speed by 50%. What is his average speed for both the trips ? 2013 0

  • 1
    55 kmph
    Correct
    Wrong
  • 2
    58 kmph
    Correct
    Wrong
  • 3
    60 kmph
    Correct
    Wrong
  • 4
    66 kmph
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 3. "60 kmph"
Explanation :

Answer: C) 60 kmph Explanation: Speed on return trip = 150% of 50 = 75 km/hr. Average speed = (2 x 50 x 75)/(50 + 75) = 60 km/hr.

Q: Opening in a wall to let in air or light 2013 0

  • 1
    Ceiling
    Correct
    Wrong
  • 2
    Door
    Correct
    Wrong
  • 3
    Window
    Correct
    Wrong
  • 4
    All the above
    Correct
    Wrong
  • Show AnswerHide Answer
  • Workspace

Answer : 3. "Window"
Explanation :

Answer: C) Window Explanation: The Opening in a wall to let in air or light is nothing but a Window.

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully

      Report Error

    Please Enter Message
    Error Reported Successfully