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Thread: Manahattan Challenge question

  1. #1
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    Manahattan Challenge question

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    In the game Cako, a player is awarded one tick for every third Alb captured, and one click for every fourth Berk captured. The total score is equal to the product of clicks and ticks. If a player has a score of 77, how many Albs did he capture?

    (1) The difference between Albs captured and Berks captured is 7.

    (2) The number of Albs captured is divisible by 4.

  2. #2
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    the answer is B.
    as we can say a definite NO with that second point.
    the first sentence is yes/might be.
    hence B is the answer

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    The answer is A. Please let me know if it is correct then I can provide the explanation.

  4. #4
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    I must say i would have dodged this question if faced on test-day because it took me a while to get it , here it is:

    The question stem tells us A/3 +B/4 = 77 (assuming, Albs and Berks captured are multiples of 3 & 4, respectively)

    (1) tells us A - B = 7
    Solving two equations gives us A=135 and B=128. BUT, so would A=136 & B=129, AND A=137 & B=130 give us 77 as per the rules of the game.
    So (1) is insufficient.

    (2) That A/4 is an integer tells us nothing about Berks. so clearly insuff.

    Together: A can only be 136 for it to be divisible by 4. So C it is.

  5. #5
    Within my grasp! sandeep_chads's Avatar
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    Hey Lav - your approach is right but I guess there is a small mistake

    A/3 + B/7 != 77 it is
    A/3 * B/7 = 77

    rest all looks good.

    I would have marked C and gone ahead because I would see 2 variables and 2 equations, however it turns into a quadratic equation and you always run a chance to have 2 solutions to it.
    Hey Harvard, I am right here!!
    rep me if I made some sense

  6. #6
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    Clicks Berks Ticks Albs
    1 4-7 77 231-233
    7 28-31 11 33-35
    11 44-47 7 21-23
    77 308-311 1 3-5

    I think the answer is A. Here's the explanation:

    Since the score is 77, we can have the clicks and ticks taking values as shown above. Also, it isn't stated that the Berk count is a multiple of 4. Likewise, the Alb count needn't bee a multiple of 3. So, if click count = x, then berk count can be 4x, 4x+1, 4x+2 or 4x+3. Likewise, if tick count = y, then alb count can be 3y, 3y+1 or 3y+2. That's the basis for the data in the above table.

    Now consider statement 1:

    Alb - Berk = 7. So obviously it can't be row1 or row 4, because of the huge differences. Row 3 also has widely-spaced Alb and Berk values. Since Albs < Berks in this row, we consider the smallest possible value of Berks which is 44, the corresponding value of Albs is 37, which is still outside the range of Albs (21-23). So Row 3 is also out. In Row 4, for Berks = 28, Albs = 35 which fits in the range. For higher values of Berks (29, 30, 31) Albs (36, 37, 38) would be outside the range in the table (33-35). So effectively in the entire table only 1 value set [Albs = 35, Berks = 28] fits. So statement 1 alone is sufficient.

    Now consider statement 2:

    No of Albs captured is divisible by 4. Thr's no such value in row2 and row 3. Row 1 and Row 2 both satisfy the condition. But we still can't zero in on one of these rows. So statement 2 alone is not sufficient.

    So answer is A.

  7. #7
    An Urch Guru Pundit Swami Sage
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    Let A denote the number of Albs
    Let B denote the number of Berks
    Let t denote the number of ticks
    Let c denote the number of clicks

    From the given portion of the question, we know the following:

    t = floor(A/3)
    c = floor(B/4)
    total = t*c = 77

    Since t and c must be non-negative integers whose product is 77, there are four possibilities:

    t c
    1 77 --> albs can be 3,4, or 5 / berks can be 308,309,310, or 311
    77 1 --> albs can be 231,232, or 233 / berks can be 4,5,6, or 7
    7 11 --> albs can be 21,22, or 23 / berks can be 44,45,46, or 47
    11 7 --> albs can be 33,34, or 35 / berks can be 28,29,30, or 31

    Moving on...

    (1) |A - B| = 7

    This eliminates three of the four aforementioned possibilities for t and c, but let's take a closer look at A and B...

    A can be 33,34, or 35
    B can be 28,29,30, or 31

    Only A = 35 and B = 28 gives a difference of exactly 7. SUFFICIENT!

    (2) A mod 4 = 0

    Without going through every possible number of Albs, both 4 and 232 are divisible evenly by 4. Not sufficient alone.


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