M1S1.pdf (page 2 of 3) 1 Q Search Module 1 Section 1 Problems: Note: The problem
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M1S1.pdf (page 2 of 3) 1 Q Search Module 1 Section 1 Problems: Note: The problems at the end of each section are "warm-up exercises. These are generally not the type of problem that you will see on an actuarial exam. Actuarial exam type problems are generally harder problems that cover more than what is covered in any one section. These types of problems are at the end of each module. As with all math problems, strive to use correct notation For Problems 1-8, you are given the following accumulation function information: a(1) = 12, a(2)s 15, a(3)-20, and a(4) 3.0 (Remember that a(o) 1 for all accumulation functions.) 100 is deposited at time t = 0, Determine the accumulated amount at time t = 3. 1. 2. Determine the present value at timet-O of 60 at time t-4 The value at time t-2 is 300. Determine the accumulated value at time t = 3. 4, Determine the discounted value at time t = 1 of a value of 600 at time t-4. 5. Given 480 at time t1,plus 300 at time 3 a. Determine the (total) present value at timet0 of the two payments. b. Determine the (total) accumulated value at time t4 of the payments C. Determine the (total) value at time t = 2 of the payments 6. Determine the amount of interest earned trom time t-2 to time t4 if 500 is invested at time 0. 7. Determine the amount of interest earned from time t 2 to timet3 if 300 is invested at time t = 1. Determine the amount of interest earned from time t = 2 to time t·411240 is invested at time 1 and an additional 300 is invested at time t-3.Explanation / Answer
Solved the first four problem as per Chegg guidelines. post multiple question to get the remaining answers
1) At time t=0, amount = 100
At time t=3, amount = amount (at time t=0) * alpha(3) = 100 * 2 .0 = 200
Hence the acculumated amount at time t=3 will be 200
2) Let the present value be x
After time t=4, the amount become 60
60 = amount at time t=0 * alpha(4)
60 = amount at time t=0 * 3.0
amount at time (t=0) = 20
3)
Let the value be x
amount at time (t=2) = 200 = x * alpha(2)
Amount at time (t=3) = x * alpha(3)
200 = x * 1.5
x = 200/1.5
substituting the value in above equation we get
amount at time t=3 = 200/1.5 * 2 = 266.66
4)
Discounted value = 600 * alpha(1)/alpha(4) = 600 * 1.2/3 = 240
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