# Ct1 X Assignment Solutions

3310 Words Apr 1st, 2015 14 Pages
CT1 – P XS – 13
Series X Solutions ActEd Study Materials: 2013 Examinations Subject CT1
Contents Series X Solutions If you think that any pages are missing from this pack, please contact ActEd’s admin team by email at ActEd@bpp.com. How to use the Series X Solutions Guidance on how and when to use the Series X Solutions is set out in the Study Guide for the 2013 exams.

Important: Copyright Agreement This study material is copyright and is sold for the exclusive use of the purchaser. You may not hire out, lend, give out, sell, store or transmit electronically or photocopy any part of it. You must take care of your material to ensure that it is not used or copied by anybody else. By opening this pack you agree to these conditions.
Using the simple discount formula of (1 - nd ) to get the discount factor:
 1  91 day discount  1   0.10   0.975  4  So:
1 v4

[1]

= 0.975 ﬁ (1 + i )

-

1 4

= 0.975

[1] [1]

ﬁ (1 + i ) = 0.975-4 ﬁ i = 10.66% Note that if you use

91 91 1 or rather than in both calculations, you will get 365 365.25 4 exactly the same answer.

The Actuarial Education Company

Page 2

CT1: Assignment X1 Solutions

Solution X1.4
The annual rate of interest, i , is given by:
(1 + i ) = 1.032 = 1.0609 ﬁ i = 6.09%

The present value of the annuity is given by:
 PV = 300a7 + 100v 7 a (4)
5

[1]

Now:
1  v 7 1  1.06097    5.90345 a7  d 1  1.06091

and: a (4) 
5

1  v5 i (4)

1  1.06095 1  1.06095   4.29685 0.059557 4 1.0609¼  1  

So:
PV  300  5.90345  100  1.06097  4.29685
= 1771.04 + 284.07 = \$2, 055.11 So the accumulated amount after 12 years is:
\$2, 055.11 ¥ 1.060912  \$4,178

[1]

[1]

[1]

Alternatively, this could be calculated as:
3007 (1 + i )5 + 100s (4) s
5

The Actuarial Education Company

CT1: Assignment X1 Solutions

Page 3

Solution X1.5
(i)
Level monthly annuity

Let X be the monthly amount paid, then the present value of the payments is:  1  1.0810  12 Xa (12)  12 X    83.4324 X  i (12)  10   where i (12)  12 1.08 [1]

1 12