Grand Quiz Question #13
The 13th Question of the #LinkSnipped# is as follows 
[Worth 15 CE Coins]
Q. Fill numbers from 0 to 9 in the boxes below.
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!
[Worth 15 CE Coins]
Q. Fill numbers from 0 to 9 in the boxes below.
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!
Replies

Ambarish GaneshLargest 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.
Left the blue ones out!
The ages of X and W were not written in a down!(from question) ๐ 
Gandalf
 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 
Anil Jain  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 
silverscorpion16
204
97
The middle number is common between Cacross and Adown. 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 Dacross 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 Cacross in the previous box.
i.e., 29 + 97 + 16 + 64 = 206, whereas Cacross is 204.
This is the closest approx. i could come up with.
So, to answer the question, X's age was left out in Adown. 
Ankita KatdareThis question has been declared as invalid.
I apologize to all the participants.
This is one of my biggest goofups till date.
Coins of this question will not be credited to anyone.
I feel ashamed of my activity. ๐
PS: I again say sorry for behaving in a crude manner is preparing this question. ๐ญโ
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