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A vampire enters a town and bites 2 person every night. The bitten 2 persons grows in to two big vampires and bites 4 persons the next night. These 4 vampires bites 6 people the next night and the 6 bites 8 and so on.......how many vampires will be there on the 5th night?

  • maths -

    The "bite rate" seems to be decreasing in your example.
    Fist one vampire bites two people (2 each)
    Then two bite 4 (2 each)
    Then 4 bite 6 (1.5 each)
    Then 6 bite 8 (1.33 each)
    If this trend continues, 8 bite 10, (1.25 each) on the fifth night.
    What do the vampires do who are more than one day old? Retire?

    It doesn't make any sense to me. You do not have a geometric progression.

  • maths -

    his qusetion was about a journey between two towns. During the start of the journey the odometer reads 39593 ( the reverse of the no is also the same) and at the end of the journey again the odometer reads a number similar to prior one....If the max speed is 90km/hr during the journey. what will be the avg speed thruout the journey?

  • maths -

    41 Vampires at the end of the 5th night

  • maths -

    how????

  • maths -

    I think almost everyone forgetting the first line of the problem." A VAMPIRES A TOWN AND BITES TWO PERSONS EVERY NIGHT ".

    1st night= 2 (bitten by original vampire)
    2nd night= 2 +4(bitten by 2 persons turned vampires previous night)
    3rd night= 2 +4(bitten by 2 vampires turned previous night) +6(bitten by 4 vampires turned previous night)

    (proceding in this manner...)

    4th night= 2+4+6+8
    5th night= 2+4+6+8+10.

    Thus total vampires would be 1(original)+No.of.people bitten.

    NO.of.people bitten= 2(1st night)+6(2nd night)+12(3rd night)+20(4th night) +30(5th night) = 70.
    Totally, 71 vampires.

    correct me if i am wrong.

  • maths -

    this solution is formed by considering the old vampires.

    No.of vampires in first night = 1
    No.of persons bitten in first night = 2

    No.of vampires in second night = 1+2
    No.of persons bitten in second night = 2+4

    No.of vampires in third night = 1+2+2+4
    No.of persons bitten in third night = 2+4+4+6

    No.of vampires in fourth night = 1+2+4+4+6+2+2+4
    No.of persons bitten in fourth night = 2+4+6+6+8+4+4+6

    No.of vampires in fifth night = 1+2+4+6+6+8+4+4+6+2+4+4+6+2+2+4

    therefore total no.of vampires in the fifth night is 65

  • maths -

    111.Its like convolution

  • math -

    1st night(N) 1 vampire(v)(since evry night it bites 2 vampires)
    2nd N 2 v
    3rd N 2 v 4v
    4th N 2 v 4v 6v
    5th n 2 v 4v 6v 8v

    totally 41 cuz the it takes them a day to bcom a vampire.

  • maths -

    65 is absolutely right bcz
    day no.of vam no.of vam bitten
    1 1 2
    2 1+2=3 2+4=6
    3 3+6(bitten) 6+10(by the new 6)
    4 9+16 16+24(by 16)
    5 25+40 notnecessary
    so the total no.of vampires are 25+40=65
    i can say all the other alternative answers are wrong..

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