Suppose I assume the people as a,b,c,d,e such that a<b<c<d Timmy)
6[a+b+c+d+e] = 840 a+b+c+d+e = 840/6 = 140
a+b+c = 60 (20*3)
a+b+d = 72 (24*3)
a+b+e = 75
a+b+2c+d+e = 165
a+b+c+d+e = 140
c = 25
a+b = 35
Now a+b+e = 75
e is 40 in that case.
Please point out what I’m missing here
Sir I have a doubt, please help!
Suppose I assume the people as a,b,c,d,e such that a<b<c<d Timmy)
6[a+b+c+d+e] = 840 a+b+c+d+e = 840/6 = 140
a+b+c = 60 (20*3)
a+b+d = 72 (24*3)
a+b+e = 75
CAT-holics 
Sir I have a doubt, please help!
Suppose I assume the five people as a,b,c,d,e such that a<b<c<da+b=35
Why is e (which is the highest, that is Timmy) not 43 but 40 in this method? What am I missing here? Someone please point out
Not sure where you got 40 from randomly, could you explain?
regards
J
Suppose I assume the people as a,b,c,d,e such that a<b<c<d Timmy)
6[a+b+c+d+e] = 840 a+b+c+d+e = 840/6 = 140
a+b+c = 60 (20*3)
a+b+d = 72 (24*3)
a+b+e = 75
a+b+2c+d+e = 165
a+b+c+d+e = 140
c = 25
a+b = 35
Now a+b+e = 75
e is 40 in that case.
Please point out what I’m missing here
Sir I have a doubt, please help!
Suppose I assume the people as a,b,c,d,e such that a<b<c<d Timmy)
6[a+b+c+d+e] = 840 a+b+c+d+e = 840/6 = 140
a+b+c = 60 (20*3)
a+b+d = 72 (24*3)
a+b+e = 75
a+b+2c+d+e = 165
a+b+c+d+e = 140
c = 25
a+b = 35
Now a+b+e = 75
e is 40 in that case.
Please point out what I’m missing here
What makes you assume a+b+e = 75? It could equally be a+c+d. Hence cannot take the third-highest or -lowest values; they would not be unique.
regards
J
Why have you assumed a + b + e = 75? That could equally be a + c + d?
regards
J