November 2001 - Course 1 SOA Solutions

31. E
The general solution of the differential equation may be determined as follows:
1
∫ dy = ( k1 k2)
dt
y
∫ −
ln y = k − k t+
C
(C is a constant)
( ) ( 1 2)
c ( k1 k2)
t
() = e e −
y t
When t = 0 ,
c
y( 0)
= e , so
() ( ) ( k ) 1 k2
t
y t = y 0 e −
1
Now we are given that if k1
= 0, y( 8) = y( 0)
.
2
Therefore,
1
−8k2
y( 0) = y( 8) = y( 0)
e
2
1 −8k2
= e
2
8k2
= ln( 2)
1
k2
= ln ( 2 )
8
24( k1−k2)
[Note the problem also gives y( 24) = 2y( 0) = y( 0)
e , but that information is not
needed to determine k
2
].
32. B
Observe
[ N ]
[ N ]
Pr 1 ≤ ≤4 ⎡1 1 1 1 ⎤ ⎡1 1 1 1 1 ⎤
Pr ⎡⎣N
≥1 N ≤4⎤ ⎦ = = + + + + + + +
Pr ≤ 4 ⎢
⎣6 12 20 30⎥ ⎦
⎢
⎣2 6 12 20 30⎥
⎦
10 + 5 + 3 + 2 20 2
= = =
30 + 10 + 5 + 3 + 2 50 5
Course 1 Solutions 20 November 2001