If x1,x2,x3,x4 &x5 are 5 numbers ;
If Any xi comes in center i=1,2,3,4,5;
Then sum of two is sum of remaining two excluding xi;
If xi=1 remainings are 2,3,4 &5 :- 2+1+5( let it be Horizontal)=3+1+4(let it be vertical);
3+1+4 ( let it be Horizontal)=2+1+5(let it be vertical);
If xi=2 remainings are 1,3,4 & 5:- Solution not possible;
If xi=3 remainings are 1,2,4 &5 :- 1+3+5(H)=2+3+4(V);
2+3+4(H)=1+3+5(V);
If xi=4 remainings are 1,2,3 & 5:- Solution not possible;
If xi=5 remainings are 1,2,3,4 – (1+5+4)(H)=(2+5+3)(V);
(2+5+3)(H)=(1+5+4)
There are two possibilities in adding the five numbers to equal eight: 1,2,5 and 1,3,4. The number common to both permutations is 1. Therefore, the number in middle can only be 1.
Math1089 
1 3 5 horizontal
2 3 4 vertically. And vice versa
3 is common and sum is rest two is same both ways.
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Thank you so much for your solution. However, a complete solution is already provided, which includes your one.
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2+3+4=9. Therefore, wrong answer. The total should be 8.
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If x1,x2,x3,x4 &x5 are 5 numbers ;
If Any xi comes in center i=1,2,3,4,5;
Then sum of two is sum of remaining two excluding xi;
If xi=1 remainings are 2,3,4 &5 :- 2+1+5( let it be Horizontal)=3+1+4(let it be vertical);
3+1+4 ( let it be Horizontal)=2+1+5(let it be vertical);
If xi=2 remainings are 1,3,4 & 5:- Solution not possible;
If xi=3 remainings are 1,2,4 &5 :- 1+3+5(H)=2+3+4(V);
2+3+4(H)=1+3+5(V);
If xi=4 remainings are 1,2,3 & 5:- Solution not possible;
If xi=5 remainings are 1,2,3,4 – (1+5+4)(H)=(2+5+3)(V);
(2+5+3)(H)=(1+5+4)
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There are two possibilities in adding the five numbers to equal eight: 1,2,5 and 1,3,4. The number common to both permutations is 1. Therefore, the number in middle can only be 1.
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Thank you
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How did you solve problem number 10?
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