Sarah is a computer engineer and manager and works for a software company. She receives a fixed annual increase in the number of

Question

Sarah is a computer engineer and manager and works for a software company. She receives a

fixed annual increase in the number of projects she handles each year. She worked on 129

projects in the fourth year of her employment and 207 projects in her tenth year.

(a) How many projects did Sarah work on in her first year?

(b) If a project pays her $500 on average,

I. How much did she earn in her twelfth year?

II. How much did she earn in total in her first 12 years with the company

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Thu Giang 3 years 2021-08-14T04:18:51+00:00 1 Answers 24 views 0

Answers ( )

  1. bonexptip
    0
    2021-08-14T04:20:35+00:00
    Reply

    Answer:

    a) Number of projects in the first year = 90

    b) Earnings in the twelfth year = $116500

    Total money earned in 12 years = $969000

    Step-by-step explanation:

    Given that:

    Number of projects done in fourth year = 129

    Number of projects done in tenth year = 207

    There is a fixed increase every year.

    a) To find:

    Number of projects done in the first year.

    This problem is nothing but a case of arithmetic progression.

    Let the first term i.e. number of projects done in first year = a

    Given that:

    a_4=129\\a_{10}=207

    Formula for n^{th} term of an Arithmetic Progression is given as:

    a_n=a+(n-1)d

    Where d will represent the number of projects increased every year.

    and n is the year number.

    a_4=129=a+(4-1)d \\\Rightarrow 129=a+3d .....(1)\\a_{10}=207=a+(10-1)d \\\Rightarrow 207=a+9d .....(2)

    Subtracting (2) from (1):

    78 = 6d\\\Rightarrow d =13

    By equation (1):

    129 =a+3\times 13\\\Rightarrow a =129-39\\\Rightarrow a =90

    Number of projects in the first year = 90

    b)

    Number of projects in the twelfth year =

    a_{12} = a+11d\\\Rightarrow a_{12} = 90+11\times 13 =233

    Each project pays $500

    Earnings in the twelfth year = 233 \times 500 = $116500

    Sum of an AP is given as:

    S_n=\dfrac{n}{2}(2a+(n-1)d)\\\Rightarrow S_{12}=\dfrac{12}{2}(2\times 90+(12-1)\times 13)\\\Rightarrow S_{12}=6\times 323\\\Rightarrow S_{12}=1938

    It gives us the total number of projects done in 12 years = 1938

    Total money earned in 12 years = 500 \times 1938 = $969000

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