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  • Grand Quiz Question #13

    Ankita Katdare

    Administrator

    Updated: Oct 22, 2024
    Views: 1.0K
    The 13th Question of the #-Link-Snipped-# is as follows -

    [Worth 15 CE Coins]

    Q. Fill numbers from 0 to 9 in the boxes below.

    13


    Rules are:

    1. None of the digits can be repeated in the boxes.

    2. The boxes that have letters a, b, c and d can't have the digit '0'

    Information Given -

    a (across) - W's age.
    a (down) - Sum of the three ages in c - across
    b (down) - Y's age
    c (across) - Sum of ages of W, X, Y, Z
    c (down) - Z's age
    d (across) - X's age

    Show the numbers in the boxes and tell us whose age among, W, X, Y and Z was not written in a down?
    And yeah, tell us your approach in finding the solution!

    ---------------------------------------------------------------
    Fastest to answer correctly wins!


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Replies

  • Ambarish Ganesh

    Member • Nov 13, 2011

    Largest possible age: 98
    2nd largest : 87
    then 76, 65, 54, 43, 32, 21, 10. 10 is the least possible age.

    Trying 87 and 76, we get the other ages as 23 and 30, with 1 in the middle box.

    13

    Left the blue ones out!
    The ages of X and W were not written in a down!(from question) 😕
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  • Gandalf

    Member • Nov 13, 2011

    -----------
    | 4 | 1 | 6 |
    | 2 | 0 | 3 |
    | 9 | 5 | 7 |

    a and c got to be 1 and 2 since some of all would not exceed 298 and given the sum conditions
    16 + 63 + 29 + 95 = 203
    29 + 63 + 13 = 105


    i have not used 8 in my solution
    x and w are not written in down
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  • Anil Jain

    Member • Nov 13, 2011

    - | a | b |
    c | - | - |
    d | - | - |


    Given that a, c , d, c, can not be 0;
    For the rules its aparent that a, b , c, d are two digits ages only,
    Now, C has to be less than 399+; again, C is sum of all four ages and a is sum of 3 ages, so a is less than c
    Hence C has to be 2 and a has to be 1


    Now as c = W+X+Y+Z which is a three digit figure, so ony following combination seems possible which will satisfy all the conditions****




    - | 1 | 6 |
    2 | 0 | 0 |
    9 | 5 | - |


    Hence:




    w=16
    x=95
    y=60
    z=29
    c (Across)=w+x+Y+z = 200
    a (Down)=W=Y+Z = 105


    Hence w and x are written across.


    -CB
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  • silverscorpion

    Member • Nov 13, 2011

    |-|1|6|
    |2|0|4|
    |9|7|-|

    The middle number is common between C-across and A-down. This means that the difference between the sum of 4 ages and that of 3 ages is in the nineties.
    The maximum it can be is 98, (since digits should not be repeated).

    But whatever the case, we can see that D-across is where that left out age will come.
    Nevertheless, how much ever I try, I can't seem to accurately fill the boxes. There is a difference of 2 between the actual sum of 4 ages and C-across in the previous box.

    i.e., 29 + 97 + 16 + 64 = 206, whereas C-across is 204.

    This is the closest approx. i could come up with.
    So, to answer the question, X's age was left out in A-down.
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  • Ankita Katdare

    Administrator • Nov 13, 2011

    This question has been declared as invalid.
    I apologize to all the participants.
    This is one of my biggest goof-ups till date.
    Coins of this question will not be credited to anyone.

    I feel ashamed of my activity. 😔

    PS: I again say sorry for behaving in a crude manner is preparing this question. 😭​
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