Question 1124814

Let  J, A, N, L = scores for each person

(1)  (J+A+N+L)/4 = 75  —>   J+A+N+L=300
(2) J=A+10 —>   A = J-10
(3) J=N+25 —>   N = J-25
(4)  L=86 <br>

Using rightmost equations,  substitute into (1) for A, N, and L:

J+(J-10)+(J-25)+86 = 300<br>

You can solve this for J and you are done.<br>


I got  J=83, A=73, N=58  (and with L=86,  the average does check at 75)