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In a mathematical game, one hundred people are standing in a line and they are required to count off in fives as ' one, two, three, four, five, one , two, three, four, five' and so on from the first  person in the line. The person who says ‘five’ is taken out of the line. Those remaining repeat this procedure until only 4 people remain in the line. What was the original position in the line of the last person to leave?

  • Option 1)

    93

  • Option 2)

    95

  • Option 3)

    97

  • Option 4)

    98

  • Option 5)

    96

In first round 20 people will be out and last one that got out will be number 100. In second round 16 people will be out from remaining 80 people and last one out will be 99. In third round 12 will be out and 98 will remain.Similarly in every coming round 98 will be last one and not coming in muntiple of 5 and hence the last person will be 98 to go out .

 The sum of first 20 odd counting numbers is:

  • Option 1)

    20

  • Option 2)

    100

  • Option 3)

    400

  • Option 4)

    225

  • Option 5)

    313

How many zeros are there in the product of N, if N is represented by 2 \times 4 \times 6 \times 8 \times 10 \times ......X 100 ?

  • Option 1)

    36

  • Option 2)

    12

  • Option 3)

    24

  • Option 4)

    20

  • Option 5)

    10

How many zeros are there at the end of the product 33 \times 175 \times 180\times 12 \times 44 \times 80 \times 66?

  • Option 1)

    6

  • Option 2)

    2

  • Option 3)

    8

  • Option 4)

    4

  • Option 5)

    5

How many zeros will be there in the expansion of the expression 1^{1} \times 2^{2} \times 3^{3} \times 4^{4} \times.... X (100)^{100}?

  • Option 1)

    1200

  • Option 2)

    1232

  • Option 3)

    1300

  • Option 4)

    900

  • Option 5)

    100

A number when divided by 5 leaves a remainder 3. What is the remainder when the square of this number is divided by 5?

  • Option 1)

    4

  • Option 2)

    8

  • Option 3)

    6

  • Option 4)

    1

  • Option 5)

    9

 

(4^{61} + 4^{62} + 4^{63} + 4^{64}) is divisible by :

 

  • Option 1)

     

    6

     

  • Option 2)

     

    11

     

  • Option 3)

     

    13

     

  • Option 4)

     

    17

     

  • Option 5)

     

    3

     

    The solution of the  equation        is:

The last digit of the number 13 + 23 + 33 + 43 …. + 993 is :

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    3

  • Option 4)

    5

  • Option 5)

    2

      

 

The number of zeros in the product 5 x 10 x 25 x 40 x 50 x 55 x 65 x 125 x 80 is :

 

  • Option 1)

     

    12

     

  • Option 2)

     

    10

     

  • Option 3)

     

    13

     

  • Option 4)

     

    9

     

  • Option 5)

     

    8

     

The number of zeros in the product 5 x 10 x 25 x 40 x 50 x 55 x 65 x 125 x 80 is :
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