Posted by intooder@yahoo.com (192.112.251.48) on October 08, 2003 at 16:11:34:
In Reply to: can you hellllllllllllllllllllllme Please posted by on October 08, 2003 at 02:03:47:
I believe this is the logical approach to follow. For brevity, for test 1 - P1 = Passed, and F1 = Failed; for test 2 - P2 = Passed, and F2 = Failed.
From the statement of the problem:
P1 - F1 = 39.................(1)
Also,
P2 - F2 = 31.................(2)
The problem additionally states:
P2 = (P1 - 7) + F1/3........ (since 7 of the original passees failed (P1 - 7) still pass, and additionally a third of the original failures pass)
Also,
F2 = 7 + 2*F1/3.........(what's left of the above equation for P2)
Now substitue these values for P2 and F2 in eq (2), and we get:
((P1 - 7) + F1/3) - (7 + 2*F1/3) = 31
Simplifying this gives:
3*P1 - F1 = 135...............(3)
Now subtract eq (1) from eq (3) which gives:
2*P1 = 96
Therefore P1 = 48,
F1 = 9 .......(from eq (1)),
P2 = 44 ......(substituting values for P1 and F1),
F2 = 13 ......(substituting value for F1).
To verify this is correct, we also know:
(P1 + F1) = (P2 + F2)....... (number of students remain the same between the 2 tests)
Substituting values for all 4 variables gives:
(48 + 9) = (44 + 13)
(57) = (57) ....... which proves our theory correct.