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Covers life contingency calculations using AM92 mortality and life table quantities to compute probabilities of death, including applications of uniform distribution of deaths. Includes a finance calculus question deriving accumulated amount from a time-varying force of interest and calculating present value. Concludes with a 25-year decreasing term assurance premium problem using specified mortality and interest assumptions, showing key actuarial summation steps.","CM1 Specimen Questions and Solutions  \nJuly 2020  \nQ1 Calculate 104 q[27]+1 using AM92 mortality. [3]  \n(Note: You should show your working, but intermediate steps can be shown using numerical values-no additional notation is required)  \nAdapted from CM1 April 2019 Q1  \nSolution:  \nThe question is asking for the probability that a life currently aged [27]+1 will die between the ages of 38 and 42  \n10|4q([27]+1) = 10p([27]+1) x 4q38  \n= L38/(L([27]+1)) x (1-L42/L38)  \n= (L38-L42) / L([27]+1)  \n= (9872.8954-9837.0661) / 9936.3549 [2]= 0.003606 [1]  \nOnly the last two lines are required for full credit.  \nQ2 Calculate 2.75 q84.5 using the method of uniform distribution of deaths.  \nBasis:  \nMortality ELT15(Females) [4]  \n(Note: You should show your working, but intermediate steps can be shown using numerical values-no additional notation is required)  \nCM1 April 2019 Q2 (although an extra mark has been allocated to allow for typing)  \nSolution:  \n2.75q84.5 = 1-2.75p84.5  \n= 1- (0.5p84.5) x (2p85) x (0.25p87) [1/2]  \n0.5p84.5 = 1- (0.5 x q84) / (1- (0.5 x q84))  \n= 1- (0.5 x 0.08757) / (1-0.5 x 0.08757) = 0.95421 [1]  \n2p85 = L87/L85  \n= 30651 / 38081 = 0.80489 [1/2]  \n0.25p87 = 1- (0.25 x q87)  \n= 1- (0.25 x 0.11859) = 0.97035 [1]  \n2. 75q84.5 = 1- (0.95421 x 0.80489 x 0.97035)  \n= 1-0.74526 = 0.25474 [1]  \nFull marks should be awarded if no notation is shown, provided the method used is clear.  \nQ3 The force of interest δ(t) is a function of time, and at any time t, measured in years is given by the formula:  \nδ(t) = 􀂮􀂭􀂯00.2412− 0.02t 0 t6  \n(i) Find an expression for A(t), the accumulated amount at time 􀝐 of a unit  \ninvestment made at time 􀝐 = 0 for 0 \u003Ct ≤ 6 . [2]  \nFor 6 \u003C t , A(t) can be written in the form:  \n􀜣 (􀝐) = 􀝁 􀢇+􀢈􀯧  \n(ii) Derive the values of 􀜽 and 􀜾 . [4]  \n(iii) Calculate the present value of $100 due at time t = 7. [2][Total 8]  \nSolution:  \n(i) A(t) =exp[ INT(0,t): (0.24-0.02t)dt ] [1]= exp[0.24 x t-0.01 x (t^2)] [1]  \n(ii) For t\u003C6, A(t) = exp[0.24 x t-0.01 x (t^2)] [1] So A(6) = exp (1.080) [0.5] Then for t>6, A(t) = A(6) x exp[0.12 x (t-6)]  \n= exp (1.080) x exp[0.12 x (t-6)]  \n= exp(0.36 + 0.12t) [1.5]  \nSo a=0.36, b=0.12 [1]  \n(iii) Present value = 100/ A(7)  \n= 100 exp ( - [ 0.36 + 0.12 x 7]) = 100exp(-1.2) [1]  \n=$30.12 [1]  \nQ4 A life insurance company issues 25-year decreasing term assurance policies to lives aged 40 exact. The death benefit, payable at the end of the year of death, is $500,000 in the first policy year, $480,000 in the second policy year thereafter reducing by $20,000 each year until the benefit is $20,000 in the twenty-fifth and final policy year. Premiums are payable annually in advance for 25 years or until earlier death.  \nShow that the annual premium per policy is approximately $643 using the basis below.  \nBasis: Mortality Rate of Interest Expenses  \nAM92 Ultimate 4% per annum Ignore  \n[6]  \nAdapted from CT5 September 2018 Q13(a) (although extra marks have been allocated for typing)  \nSolution:  \nP = (520000 x TA:40:\u003C25> -20000 x I(TA):40:\u003C25>) / adue:40:\u003C25> [1.5]  \nWhere TA:40:\u003C25> = EA:40:\u003C25> -v^25 x 25p40 = 0.38907-0.37512 x (8821.2612/9856.2863)  \n= 0.38907-0.33573 = 0.05334 [1.5]  \nAnd I(TA):40:\u003C25> = IA:40-v^25 x 25p40 x [ 25 x WL:65 + IA:65 ]  \n=7.95699-0.33573 x [25 x 0.52786 + 7.89442] = 0.87612 [2]  \nP= (520000 x 0.05334-20000 x 0.87612) / 15.884 = $643.06 [1]","cbCaipdp2myOIR89","https://ap.wps.com/l/cbCaipdp2myOIR89","pdf",135664,11,"English","en",105,"# Q1 Calculate 104 q[27]+1 using AM92 mortality\n# Q2 Calculate 2.75 q84.5 using uniform distribution of deaths\n# Q3 Force of interest δ(t): derive A(t) and present value at t=7\n# Q4 25-year decreasing term assurance: show premium approximately $643","[{\"question\":\"What is required to answer Q1 in the specimen set?\",\"answer\":\"Show working for the calculation of the probability using AM92 mortality, and note that only the last two lines are required for full credit.\"},{\"question\":\"How does Q2 use the method of uniform distribution of deaths?\",\"answer\":\"It computes 2.75 q84.5 using survival/death probabilities expressed through p and q terms derived from the stated mortality basis, with intermediate steps shown using numerical values.\"},{\"question\":\"What steps are used in Q4 to approximate the annual premium?\",\"answer\":\"The solution forms the premium using a term-assurance present value expression based on the decreasing death benefit, then computes the required actuarial factors from the given mortality/interest basis to arrive at approximately $643.06.\"}]","CM1 Specimen Questions and Solutions - July 2020 | PDF",1790495987,{"code":4,"msg":81,"data":82},"ok",{"site_id":75,"language":74,"slug":83,"title":65,"keywords":84,"description":66,"schema_data":85,"social_meta":139,"head_meta":141,"extra_data":143,"updated_unix":144},"cm1-specimen-questions-and-solutions-july-2020","",{"@graph":86,"@context":138},[87,101,121],{"@type":88,"itemListElement":89},"BreadcrumbList",[90,94,96,99],{"item":91,"name":92,"@type":93,"position":8},"https://docshare.wps.com","Home","ListItem",{"item":95,"name":9,"@type":93,"position":14},"https://docshare.wps.com/document/",{"item":97,"name":20,"@type":93,"position":98},"https://docshare.wps.com/document/exam/",3,{"item":100,"name":65,"@type":93,"position":19},"https://docshare.wps.com/document/cm1-specimen-questions-and-solutions-july-2020/399303/",{"url":100,"name":65,"@type":102,"image":103,"author":108,"headline":65,"publisher":110,"fileFormat":113,"inLanguage":74,"description":66,"dateModified":114,"datePublished":115,"encodingFormat":113,"isAccessibleForFree":116,"interactionStatistic":117},"DigitalDocument",{"url":104,"@type":105,"width":106,"height":107},"https://docshare.wps.com/thumbnails/cm1-specimen-questions-and-solutions-july-2020/399303.png","ImageObject",300,407,{"name":63,"@type":109},"Person",{"url":91,"name":111,"@type":112},"DocShare","Organization","application/pdf","2026-10-29","2026-09-27",true,{"@type":118,"interactionType":119,"userInteractionCount":72},"InteractionCounter",{"@type":120},"ViewAction",{"@type":122,"mainEntity":123},"FAQPage",[124,130,134],{"name":125,"@type":126,"acceptedAnswer":127},"What is required to answer Q1 in the specimen set?","Question",{"text":128,"@type":129},"Show working for the calculation of the probability using AM92 mortality, and note that only the last two lines are required for full credit.","Answer",{"name":131,"@type":126,"acceptedAnswer":132},"How does Q2 use the method of uniform distribution of deaths?",{"text":133,"@type":129},"It computes 2.75 q84.5 using survival/death probabilities expressed through p and q terms derived from the stated mortality basis, with intermediate steps shown using numerical values.",{"name":135,"@type":126,"acceptedAnswer":136},"What steps are used in Q4 to approximate the annual premium?",{"text":137,"@type":129},"The solution forms the premium using a term-assurance present value expression based on the decreasing death benefit, then computes the required actuarial factors from the given mortality/interest basis to arrive at approximately $643.06.","https://schema.org",{"og:url":100,"og:type":140,"og:title":65,"og:site_name":111,"og:description":66},"article",{"robots":142,"canonical":100},"index,follow",{"doc_id":61,"site_id":75},1791133681]